FEM-Design - StruSoftdownload.strusoft.com/FEM-Design/inst160x/...Verification Examples FEM-Design 16.0 1. Linear static calculations 1.1 Beam with two point loading at one-third of

Verification Examples FEM-Design 160

FEM-Design

Verification Examples

version 112016

StruSoft ABVisit the StruSoft website for company and FEM-Design information at

wwwstrusoftcom

TrademarksFEM-Design is a registered trademark of StruSoft

Contents1 Linear static calculations 4

11 Beam with two point loading at one-third of its span412 Calculation of a circular plate with concentrated force at its center613 A simply supported square plate with uniform load10

2 Second order analysis1221 A column with vertical and horizontal loads1222 A plate with in-plane and out-of-plane loads15

3 Stability analysis1931 Flexural buckling analysis of a beam modell with different boundary conditions1932 Buckling analysis of a plate with shell modell2133 Lateral torsional buckling of an I section with shell modell2434 Lateral torsional buckling of a cantilever with elongated rectangle section26

4 Calculation of eigenfrequencies with linear dynamic theory2841 Continuous mass distribution on a cantilever column2842 Free vibration shapes of a clamped circular plate due to its self-weight31

5 Seismic calculation3351 Lateral force method with linear shape distribution on a cantilever3352 Lateral force method with fundamental mode shape distribution on a cantilever3653 Modal analysis of a concrete frame building38

6 Calculation considering diaphragms4761 A simple calculation with diaphragms4762 The calculation of the shear center47

7 Calculations considering nonlinear effects5271 Uplift calculation52

711 A trusses with limited compression members52712 A continuous beam with three supports 58

72 Cracked section analysis by reinforced concrete elements69721 Cracked deflection of a simply supported beam69722 Cracked deflection of a cantilever beam74

73 Nonlinear soil calculation808 Cross section editor81

81 Calculation of a compound cross section819 Design calculations82

91 Required reinforcement calculation for a slab82References103Notes104

In this verification handbook we highlighted the analytical results with green and the finiteelement results with blue background for better comparison The analytical closed formulasare highlighted with a black frame

If the finite element mesh is not mentioned during the example it means that theautomatically generated mesh was used

1 Linear static calculations

11 Beam with two point loading at one-third of its span

Fig 111 left side shows the simple supported problem The loads the geometric and materialproperties are as follows

Force F = 150 kN

Length L = 6 m

Cross section Steel I beam HEA 300

The second moment of inertia in the relevant direction I1 = 18264ˑ10-4 m4

The shear correction factor in the relevant direction ρ2 = 021597

The area of the cross section A = 11253 cm2

Youngs modulus E = 210 GPa

Shear modulus G = 80769 GPa

The deflection of the mid-span based on the hand calculation (based on virtaul force theorem[1] see Fig 111 right side also)

e=2 MEI [ L

12 ]+ 2 F

ρGA [05L3 ]= 23

648FL3

FL3 ρGA

648150sdot63

210000000sdot18264ˑ10minus4+150sdot6

3sdot021597sdot80769000sdot0011253=003151 m=3151 mm

Figure 111 ndash The beam theory and the application of a virtual force

L3 L3 L3

M=+FL3

The first part of this equation comes from the bending deformation and the second part comesfrom the consideration of the shear deformation as well because FEM-Design uses Timoshenkobeam theory (see the Scientific Manual)

The deflection and the bending moment at the mid-span based on the linear static calculationwith three 2-noded beam elements (Fig 112 and Fig 113)

eFEM=3151 mm and the bending moment M FEM=300 kNm

The theoretical solution in this case (three 2-noded beam elements) must be equal to the finiteelement solution because with three beam elements the shape functions order coincides with theorder of the theoretical function of the deflection (the solution of the differtial equation)

Therefore the difference between the results of the two calculations is zero

Figure 112 ndash The finite element model

Figure 113 ndash The mid-span deflection [mm]

12 Calculation of a circular plate with concentrated force at its center

In this chapter a circular steel plate with a concentrated force at its center will be analyzed Firstof all the maximum deflection (translation) of the plate will be calculated at its center and thenthe bending moments in the plate will be presented

Two different boundary conditions will be applied at the edge of the plate In the first case theedge is clamped (Case I) and in the second case is simply supported (Case II) see Fig 121

The input parameters are as follows

The concentrated force P = 10 kN

The thickness of the plate h = 005 m

The radius of the circular plate R = 5 m

The elastic modulus E = 210 GPa

The Poissons ratio ν = 03

The ratio between the diameter and the thickness is 2Rh = 200 It means that based on thegeometry the shear deformation only have negligible effects on the maximum deflections It isimportant because FEM-Design uses the Mindlin plate theory (considering the sheardeformation see Scientific Manual for more details) but in this case the solution of Kirchhoffsplate theory and the finite element result must be close to each other based on the mentionedratio

The analytical solution of Kirchhoffs plate theory is given in a closed form [2][3]

Figure 121 ndash Clamped (Case I) and simply supported (Case II) circular plate with concentrated force

Case I

For the clamped case the maximum deflection at the center is

w cl=P R2

16π ( E h3

12(1minusν 2))

The reaction force at the edge

And the bending moment in the radial direction at the edge

M cl=P

With the given input parameters the results based on the analytical and the finite elementsolutions (with the default finite element mesh size see Fig 122) are

w cl=10sdot52

16π (210000000sdot0053

12(1minus032) )

=0002069m=2069mm w clFEM=204 mm

2π 5=0318

QrFEM=0318kNm

M cl=P

=0796kNm

mM clFEM=0796

Figure 122 ndash The clamped (Case I) and the simply supported (Case II) plate with the default mesh

Case II

For the simply supported case the maximum deflection in the center is

w ss=P R2

16π ( E h3

12(1minusν 2))(3+ν

1+ν )

w ss=10sdot52

16π (210000000sdot0053

12(1minus032) )

(3+031+03)=0005252 m=5252 mm w ssFEM=500 mm

2π 5=0318 kN m QrFEM=0318

Fig 123 shows the two deflected shape in side view The different boundary conditions areobvious based on the two different displacement shape The differences between the analyticalsolutions and finite element solutions are less than 5 but the results could be more accurate ifthe applied mesh is more dense than the default size

Based on the analytical solution the bending moments in plates under concentrated loads areinfinite It means that if more and more dense mesh will be applied the bending moment underthe concentrated load will be greater and greater Thus the following diagram and table (Fig124 and Table 121) shows the convergence analysis of Case I respect to the deflection andbending moment The deflection converges to the analytical solution (wcl = 207 mm) and thebending moment converges to infinite

Figure 123 ndash The deflected shape of Case I (clamped) and Case II (simply supported) with the default mesh

Figure 124 ndash Convergence analysis regarding to deflection and bending moment

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Table 121 ndash The convergence analysis

Average element size [m]

341 2034 273 05533 2057 391 04957 2060 362 032035 2068 433 027994 2072 549 0131719 2073 640 00531772 2075 1070 Local refinement 131812 2076 1430 Local refinement 2

Number of elements [pcs]

Deflection [mm]

Bending moment [kNmm]

13 A simply supported square plate with uniform load

In this example a simply supported concrete square plate will be analyzed The external load is auniform distributed load (see Fig 131) We compare the maximum displacements andmaximum bending moments of the analytical solution of Kirchhoffs plate theory and finiteelement results

The input parameters are in this table

The intensity of the uniform load p = 40 kNm2

The edge of the square plate a = 5 m

Poissons ratio ν = 02

The ratio between the span and the thickness is ah = 20 It means that based on the geometrythe shear deformation may have effects on the maximum deflection It is important becauseFEM-Design uses the Mindlin plate theory (considering the shear deformation see ScientificManual for more details) therefore in this case the results of Kirchhoffs theory and the finiteelement result could be different from each other due to the effect of shear deformations

Based on Kirchhoffs plate theory [2][3] the maximum deflection is in the center of the simplysupported square plate and its intensity can be given with the following closed form

wmax=000416p a4

( E h3

12(1minusν 2))

Figure 131 ndash The square plate with simply supported edges uniform load and default mesh size

The maximum bending moment in the plate if the Poissons ratio ν = 02

M max=00469 p a2

According to the input parameters and the analytical solutions the results of this problem are the following

The deflection at the center of the plate

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

The bending moment at the center of the plate

M max=00469sdot40sdot52=469

M maxFEM=4597kNm

Next to the analytical solutions the results of the FE calculations are also indicated (see Fig132 and 133) The difference is less than 3 and it also comes from the fact that FEM-Designconsiders the shear deformation also (Mindlin plate theory)

Figure 133 ndash The internal forces Mx ndash My ndash Mxy [kNmm]

Figure 132 ndash The deflected shape [mm] and the reaction forces [kNm] with the default mesh

2 Second order analysis

21 A column with vertical and horizontal loads

We would like to analyze the following column (see Fig 211) with second order theory Firstof all we make a hand calculation with third order theory according to Ref [6] and [8] withstability functions After this step we compare the results with FEM-Design In this moment weneed to consider that in FEM-Design second order analysis is implemented and the handcalulation will be based on third order theory therefore the final results wont be exactly thesame By the FEM-Design calulation we split the column into 3 bar elements thus the finiteelement number of the bars was 3 for more precise results

The input parameters

Elastic modulus E = 30 GPa

Normal force P = 2468 kN

Horizontal load q = 10 kNm

Cross section 02 m x 04 m (rectangle)

Second moment of inertia in the relevant direction I = 00002667 m4

Column length L = 4 m

According to Ref [6] and [8] first of all we need to calculate the following assistant quantities

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

Figure 211 ndash The column with vertical and horizontal loads

The constants based on this value for the appropriate stability functions

s=3294 c=0666 f =1104

With these values the bending moments and the shear forces based on third order theory and FEM-Design calculation

M clamped= f (1+c)q L2

12=1104(1+0666)

10sdot42

12=2452kNm M 2ndFEMclamped=2555 kNm

M roller=00 kNm M 2ndFEMroller=00 kNm

V clamped=[1+ f (1+c )6 ]( q L

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

V 2ndFEMclamped=2638kN

V roller=[1minus f (1+c )6 ](q L

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

V 2ndFEMroller=1361kNm

The differences are less than 5

Figure 212 ndash The shear [kN] and bending moment [kNm] diagram with 1st and 2nd order theory

Figure 213 ndash The lateral translations with 1st and 2nd order theory

22 A plate with in-plane and out-of-plane loads

In this chapter we will analyze a rectangular plate with single supported four edges The load isa specific normal force at the shorter edge and a lateral distributed total load perpendicular to theplate (see Fig 221) The displacement and the bending moment are the question based on a 2 nd

order analysis First of all we calculate the results with analytical solution and then we comparethe results with FE calculations

In this case the material and the geometric properties are the following

The dimensions of the plate a = 8 m b = 6 m

The specific normal force Nx = 1000 kNm

The lateral distributed load qz = 10 kNm2

The maximum displacement and moments based on the 1st order linear calculation

wmax=3538mm m x max=1805kNm

m m y max=2562

m xy max=1368kNm

In Chapter 32 the critical specific normal force for this example is

N cr=2860kNm

If the applied specific normal force is not so close to the critical value (now it is lower than the

Figure 221 ndash The single supported edges the lateral distributed load and the specific normal force

half of the critical value) we can assume the second order displacements and internal forcesbased on the linear solutions with the following formulas (with blue highlight we indicated theresults of the FE calculation)

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

=5441mm wmax 2nd FEM=5469mm

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

m m x max 2nd FEM=2850

m y max 2 nd=m y max1

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

m m y max 2 nd FEM=4030

m xy max 2nd=mxy max1

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

mm xy max 2nd FEM=2053

Figure 222 shows the problem in FEM-Design with the default mesh

The following figures show the moment distribution in the plate and the displacements withFEM-Design according to 1st and 2nd order theory

Figure 222 ndash The single supported slab with in-plane and out-of-plane loads

Figure 223 ndash The mx [kNmm] moment with 1st and 2nd order analysis

Figure 224 ndash The my [kNmm] moment with 1st and 2nd order analysis

Figure 225 ndash The mxy [kNmm] moment with 1st and 2nd order analysis

Figure 226 ndash The vertical translation [mm] with 1st and 2nd order analysis

3 Stability analysis

31 Flexural buckling analysis of a beam modell with different boundary conditions

The cross section is a rectangular section see Fig 311

The material C2025 concrete

The second moment of inertia about the weak axis I2 = 2667ˑ10-4 m4

The length of the column L = 4 m

The boundary conditions see Fig 311

The critical load parameters according to the Eulers theory are as follows and next to theanalytical solution [1] the relevant results of the FEM-Design calculation can be seen

Pinned-pinned boundary condition

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

Fixed-pinned boundary condition

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

Figure 311 ndash The buckling problem with the different boundary conditions and the cross section

a=02 m

Fixed-fixed boundary condition

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

Fixed-free boundary condition

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

The difference between the two calculations are less than 3 but keep in mind that FEM-Designconsiders the shear deformation therefore we can be sure that the Eulers results give highercritical values Fig 312 shows the first mode shapes of the problem with the differentboundary conditions

Figure 312 ndash The buckling mode shapes for different boundary conditionspinned-pinned fixed-pinned fixed-fixed fixed-free

32 Buckling analysis of a plate with shell modell

In this chapter we will analyze a rectangular plate with single supported four edges The load isa specific normal force at the shorter edge (see Fig 321) The critical force parameters are thequestions due to this edge load therefore it is a stability problem of a plate

The solutions of the differential equation of the plate buckling problem are as follows [6]

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

Figure 321 ndash The single supported edges and the specific normal force

According to the analytical solution the first five critical load parameters are

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

Next to these values we indicated the critical load parameters what were calculated with theFEM-Design

Figure 322 ndash The stability problem of a plate with single supported edges

The difference between the calculations less than 5

Figure 323 shows the first five stability mode shapes of rectangular single supported plate

Figure 323 ndash The first five stability mode shape of the described problem

33 Lateral torsional buckling of an I section with shell modell

The purpose of this example is calculate the lateral torsional critical moment of the followingsimple supported beam (see Fig 331)

The length of the beam is L = 10 m

The cross section see Fig 332

The warping constant of the section Iw = 125841 cm6

The St Venant torsional inertia It = 1534 cm4

The minor axis second moment of area Iz = 6027 cm4

The shear modulus G = 8077 GPa

In this case the critical moment can be calculated with the following formula based on theanalytical solution [6]

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

M cr=π 2sdot21000sdot6027

10002 radic 1258416027

+10002

sdot8077sdot1534

π 2sdot21000sdot6027

=4328kNcm

Figure 331 ndash The static frame of a simple supported beam loaded with bending moments of both ends

L=10 m

Figure 332 ndash The dimensions of the double symmetric cross section

In FEM-Design a shell modell was built to analyze this problem The bending moments in theshell model were considered with line loads at the end of the flanges (see Fig 333)

The supports provides the simple supported beam effects with a fork support for the shell model(see Fig 333)

From the FEM-Design stability calculation the critical moment value for this lateral torsionalbuckling problem is

M crFEM=4363kNcm

The critical shape is in Fig 334 The finite element mesh size was provided based on theautomatic mesh generator of FEM-Design

The difference between the two calculated critical moments is less than 1

Figure 333 ndash The FEM model with the supports and the loads (moments) at the ends

Figure 334 ndash The critical mode shape of the problem

34 Lateral torsional buckling of a cantilever with elongated rectangle section

The purpose of this example is calculate the critical force at the end of a cantilever beam (seeFig 341) If the load is increasing the state of the cantilever will be unstable due to lateraltorsional buckling

The cross section t = 40 mm h = 438 mm see Fig 341

The St Venant torsional inertia It = 8806246 mm4

The minor axis second moment of area Iz = 2336000 mm4

In this case (elongated rectangle cross section with cantilever boundary condition) the criticalconcentrated force at the end can be calculated with the following formula based on analyticalsolution

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

Figure 341 ndash The cantilever beam with concentrated load

In FEM-Design a shell modell was built to analyze this problem The concentrated load at theend of the cantilever was considered at the top of the beam (see Fig 342)

With the FEM-Design stability calculation the critical concentrated force value for this lateraltorsional buckling problem is

P crFEM=2400kN

The difference between the two calculated critical load parameters is less than 2

Figure 342 ndash The FE model of the cantilever beam with the default mesh

4 Calculation of eigenfrequencies with linear dynamic theory

41 Continuous mass distribution on a cantilever column

Column height H = 4 m

The cross section square with 04 m edge

The second moment of inertia I = 0002133 m4

The area of the cross section A = 016 m2

The shear correction factor ρ = 56 = 08333

The specific self-weight of the column γ = 25 kNm3

The mass of the column m = 1631 t

Based on the analytical solution [4] the angular frequencies for this case is

ω B=μ Biradic EIm H 3 μ B1=352μ B2=2203μ B3=617

if only the bending deformations are considered

The angular frequencies are [4]

Figure 411 ndash The cantilever with continuous mass distribution

ω S=μ Siradic ρGAm H

μ S1=05π μ S2=15π μS3=25π

if only the shear deformations are considered

Based on these two equations (considering bending and shear deformation) using the Foumlppltheorem the angular frequency for a continuous mass distribution column is

ω n2=

ω B2+

Based on the given equations the first three angular frequencies separately for bending and sheardeformations are

ω B1=352radic 30000000sdot000213316sdot43 =8716

ω S1=05π radic 08333sdot12500000sdot0161631sdot4

=79391s

=238181s

=396961s

According to the Foumlppl theorem the resultant first three angular frequencies of the problem are

ω n1=866391s

ω n2=531641s

ω n3=142581s

And based on these results the first three eigenfrequencies are (f = ω(2π))

f n1=137891s

f n2=846131s

f n3=2269231s

In FEM-Design to consider the continuous mass distribution 200 beam elements were used forthe cantilever column The first three planar mode shapes are as follows according to the FEcalculation

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

The first three mode shapes can be seen in Fig 412

The differences between the analytical and FE solutions are less than 2

Figure 412 ndash The first three mode shapes for the cantilever with continuous mass distribution

42 Free vibration shapes of a clamped circular plate due to its self-weight

In the next example we will analyze a circular clamped plate The eigenfrequencies are thequestion due to the self-weight of the slab

The density ρ = 785 tm3

The solution of the dynamic differential equation for the first two angular frequencies of aclamped circular plate are [5]

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

Figure 421 shows the problem in FEM-Design with the clamped edges and with the defaultmesh

According to the analytical solution the first two angular frequencies are

Figure 421 ndash The clamped circular plate and the default finite element mesh

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Based on the angular frequencies we can calculate the eigenfrequencies in a very easy way Nextto these values we indicated the eigenfrequencies what were calculated with the FEM-Design

Figure 422 shows the first two vibration mode shapes of the circular clamped plate

Figure 422 ndash The first two vibration shape mode of a clamped circular plate

5 Seismic calculation

51 Lateral force method with linear shape distribution on a cantilever

Inputs

The concentrated mass points 10 pieces of 10 t (see Fig 511)

The total mass m = 100 t

First of all based on a hand calculation we determine the first fundamental period

The first fundamental period of a cantilever column (length H) with a concentrated mass at theend (m mass) and EI bending stiffness [4]

T i=2π

radic 3 EImi H i

Figure 511 ndash The cantilever column with the concentrated mass points the first vibration shape [T=0765 s]the equivalent forces [kN] the shear force diagram [kN] and the bending moment diagram [kNm] with FEM-

Design

The fundamental period separately for the mass points from bottom to top

T 1=2π

radic 3sdot31000000sdot000213331sdot13

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

The approximated period based on these values according to the Dunkerley summary and theresult of FE calculation

T HC=radicsumi=1

T i2=07758s T FEM=0765s

The difference between the hand calculation and FEM-Design calculation is less than 2 forfurther information on the period calculation see Chapter 4

The base shear force according to the fundamental period of vibration (see Fig 511) and theresponse spectrum (see Fig 512)

F b=Sd (T 1)mλ=06588sdot10sdot10=6588kN

We considered the response acceleration based on the period from FE calculation to get a morecomparable results at the end Thus the equivalent forces on the different point masses are

F i=F b

z i mi

sum z j m j

=6588zi mi

1sdot1+2sdot1+3sdot1+4sdot1+5sdot1+6sdot1+7sdot1+8sdot1+9sdot1+10sdot1=6588

The equivalent forces from the bottom to the top on each point mass

F1=0120kN F 2=0240 kN F 3=0359kN F 4=0479 kN F 5=0599 kN

F6=0719kN F7=0838kN F 8=0958 kN F 9=1078kN F10=1198kN

These forces are identical with the FEM-Design calculation and the shear force and bendingmoment diagrams are also identical with the hand calculation

Figure 512 ndash The response spectrum [T = 0765 s Sd = 06588 ms2]

52 Lateral force method with fundamental mode shape distribution on a cantilever

Inputs

Figure 521 ndash The cantilever column with the concentrated mass points the first vibration shape withthe value of the eigenvector [T=0765 s] the equivalent forces [kN] the shear force diagram [kN] and

the bending moment diagram [kNm] with FEM-Design

F i=F b

sum s j m j

=6588si mi

F6=0683kN F7=0878kN F 8=1081kN F 9=1289 kN F10=1498kN

53 Modal analysis of a concrete frame building

In this chapter we show a worked example for modal analysis on a concrete frame buildingaccording to EN 1998-12008 with hand calculation and compare the results with FEM-DesignThis example is partly based on [4] The geometry the dimensions the material and the bracingsystem are in Fig 531-3 and in the following table

Inputs

Column heightTotal height h = 32 m H = 232=64 m

The cross sections Columns 3030 cm Beams 3050 cm

The second moment of inertia Ic = 0000675 m4 Ib = 0003125 m4

The concentrated mass points 12 pieces of 13358 t on 1st storey and12 pieces of 11268 t on 2nd storey (see Fig 532)

The total mass 1st storey m1 = 1603 t2nd storey m2 = 1352 t total mass M = 2955 t

Reduction factor for elastic modulus considering the cracking according to EN 1998-12008

α = 05

Behaviour factors q = 15 qd = 15

Accidental torsional effect do not considered ξ = 005 (damping factor)

Figure 531 ndash The concrete frame building with the columns and beams

The first exercise is the determination of the fundamental periods and mode shapes There areseveral hand calculation modes to get these values but in this chapter the details of the modalanalysis are important therefore we considered the first two fundamental periods based on FEM-Design calculation (see Fig 535) See the details and example on the eigenfrequencycalculation in Chapter 4

The dead loads and the live loads are considered in the mass points (see Fig 532)

Figure 532 ndash The frame building with the masses and bracings

Figure 533 ndash The side view of the building

According to the fundamental periods in Fig 535 the response accelerations from Fig 534are

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

The second step is to calculate the effective modal masses based on this formula

(ΦiTm ι )

ΦiT mΦ i

Figure 535 ndash The first two fundamental mode shapes [-] T1 = 0704 s T2 = 0252 s

Figure 534 ndash The considered design response specra according to EN 1998-12008

During the hand calculation we assume that the structure is a two degrees of freedom system inthe x direction with the two storeys because the first two modal shapes are in the same planesee Fig 535 Thus we only consider the seismic loads in one direction because in this way thehand calculation is more comprehensible

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

According to the assumption of a two degrees of freedom system the sum of the effective modalmasses is equal to the total mass

27232955

+23232955

Calculation of the base shear forces

F b1=S d1 m1=1115sdot2723=3036 kN F b2=S d2 m1

=1570sdot2323=365 kN

The equivalent forces come from this formula

p i=mΦi

Φ iT m ι

Φ iT mΦ i

The equivalent forces at the storeys respect to the mode shapes considering the mentioned twodegrees of freedom model

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

The equivalent forces on one frame from the six (see Fig 531)

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

The shear forces between the storeys respect to the two different mode shapes

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

The shear forces in the columns respect to the two different mode shapes

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

The bending moments in the columns respect to the two different mode shapes from the relevantshear forces (by the hand calculation we assumed zero bending moment points in the middle ofthe columns)

Mc1=[2530sdot3221525sdot32 2]=[4048

2440]kNm Mc2=[ 313sdot322minus371sdot322]=[ 5008

minus5936]kNm

The bending moments in the beams respect to the two different mode shapes

Mb1=[4048+24402440 ]=[6488

2440]kNm Mb2=[5008minus5936minus5936 ]=[minus0928

minus5936]kNm

The SRSS summation on the internal forces

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

Mb=[radic64882+(minus0928)2

radic24402+(minus5936)2]=[6489

2511]kNm

The CQC summation on the internal forces

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

And based on these values the results of the CQC summation

V c=[ radic25302+3132

+2sdot253sdot313sdot0007588

radic15252+(minus371)2+2sdot1525sdot(minus371)sdot0007588]=[2552

1567]kN

Mc=[radic40482+50082+2sdot4048sdot5008sdot0007588

radic24402+59362

+2sdot2440sdot5936sdot0007588]=[40832516]kNm

Mb=[radic64882+(minus0928)2+2sdot6488sdot(minus0928)sdot0007588

radic24402+(minus5936)2+2sdot2440sdot(minus5936sdot0007588)]=[6488

2507]kNm

The following displacements come from the FEM-Design calculation on the complete framestructure to ensure the comprehensible final results on the P-Δ effect

The displacements at the storeys respect to the two different mode shapes considering thedisplacement behaviour factor

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Based on these values the storey drifting respect to the two different mode shapes

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

SRSS summation on the story drifting

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P-Δ effect checking on the total building

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

V tot=[ 6sdotradic5062+6242

6sdotradic3052+(minus742)2]=[3059

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

After the hand calculation lets see the results from the FEM-Design calculation and comparethem to each other Fig 536 shows the effective modal masses from the FE calculationPractically these values coincide with the hand calculation

Fig 537 and the following table shows the equivalent resultant shear forces and the base shearforces respect to the first two mode shapes The differences between the two calculations areless than 2

Storey 1 equivalentresultant [kN]

Base shear force[kN]

Hand FEM Hand FEM Hand FEM

Mode shape 1 1206 1219 8198 8180 3036 3069

Mode shape 2 1830 1850 ndash 4452 ndash 4547 3650 3633

Figure 536 ndash The first two fundamental periods and the effectivemodal masses from FEM-Design

Fig 538-9 and the following table shows the internal forces after the differenr summationmodes (SRSS and CQC) The differences between the two calculations are less than 2

Column shear force[kN]

Column bending moment[kNm]

Beam bending moment[kNm]

Storey 1 Storey 2 Storey 1 Storey 2 Storey 1 Storey 2

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Figure 537 ndash The equivalent forces respect to the storeys and the base shear forces for the first two modeshapes [kN]

Figure 538 ndash The shear force [kN] and bending moment diagram [kNm] after the SRSS summation rule

Fig 5310 shows the Θ values from FEM-Design The differences between the hand calculationand FEM-Design are less than 3

Figure 539 ndash The shear force [kN] and bending moment diagram [kNm] after the CQC summation rule

Figure 5310 ndash The θ values at the different storeys from FEM-Design

6 Calculation considering diaphragms

61 A simple calculation with diaphragms

If we apply two diaphragms on the two storeys of the building from Chapter 53 then theeigenfrequencies and the periods will be the same what we indicated in Chapter 53

62 The calculation of the shear center

In this example we show that how can we calculate the shear center of a storey based on theFEM-Design calculation We analyzed a bottom fixed cantilever structure made of threeconcrete shear walls which are connected to each other at the edges (see Fig 621) Thediaphragm is applied at the top plane of the structure (see also Fig 621 right side) If the heightof the structure is high enough then the shear center will be on the same geometry point where itshould be when we consider the complete cross section of the shear walls as a ldquothin-walledrdquo ldquoCrdquocross section (see Fig 621 left side) Therefore we calculate by hand the shear center of theldquoCrdquo profile assumed to be a thin-walled cross section then compare the solution what we can getfrom FEM-Design calculation with diaphragms

Secondly we calculate the idealized bending stiffnesses in the principal rigidity directions byhand and compare the results what we can calculate with FEM-Design results

Inputs

Height of the walls H = 63 m

The thickness of the walls t = 20 cm

The width of wall number 1 and 3 w1 = w3 = 40 m

The width of wall number 2 w2 = 60 m

The applied Youngs modulus of concrete E = 9396 GPa

Figure 621 ndash The geometry of the bracing core and the height of the bottom fixed structure (the diaphragm is lying on the top plane see the red line and hatch)

First of all lets see Fig 622 The applied cross section is a symmetric cross section In the webthe shear stress distribution comes from the shear formula regarding bending (see Fig 622)Therefore it is a second order polynom In the flanges the shear stress distribution is linearaccording to the thin walled theory With the resultant of these shear stress ditribution (see Fig622 V1 V2 and V3) the position of the shear center can be calculated based on the statical(equilibrium) equations

The shear stress values (see Fig 622)

τ =V SI t

=1sdot(02sdot4sdot3)

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

1sdot(02sdot4sdot3+02sdot3sdot15)

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

Based on these stresses the resultant in the flanges and in the web

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

V 2=23(τ maxminusτ )w2t+τ w2 t=

23(09164minus06665)6sdot02+06665sdot6sdot02=09997kN

Respect to the equilibrium (sum of the forces)

V=1kNasympV 2=09997kN

Figure 622 ndash The shear stress distribution in a thin-walled cross section if theshear force acting on the shear center

τ ma x

And also respect to the equilibrium (if the external load is acting on the shear center see Fig622) the sum of the moments

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

Thus the shear center is lying on the symmetry axis and it is xS=1601 m from the web (see Fig622) In FEM-Design the global coordinate system does not coincide with the symmetry axisof the structure (see Fig 621) Therefore we need no transform the results

Lets be a selected key node at the diaphragm in the global coordinate system (see Fig 621)

xm=0m ym=0m

Based on the unit forces (1 kN) and moment (1 kNm) on the key node the displacements of thekey node are as follows based on the FEM-Design calculation (see the theory manualCalculation considering diaphragm chapter also)

According to unit force on key node in X direction

uxx=15852mm u yx=072166 mm φ zx=029744sdot10minus4 rad

According to unit force on key node in Y direction

uxy=072166mm u yy=73314 mm φ zy=010328sdot10minus2 rad

According to unit moment on key node around Z direction

φ zz=016283sdot10minus3 rad

Based on these finite element results the global coordinates of the shear center of the diaphragmare

xS= xmminusφ zyφ zz

=0minus010328sdot10minus2

016283sdot10minus3 =minus6343m

yS= ym+φ zxφ zz

=0+029744sdot10minus4

016283sdot10minus3 =+01827 m

In FEM-Design the coordinates of the middle point of the web are (see Fig 621)

xmid=minus4919 m ymid=+0894m

With the distance between these two points we get a comparable solution with the handcalculation

xSFEM=radic( xSminusx mid )2+( ySminus ymid )

2=radic(minus6343minus(minus4919))

2+(01827minus0894)

2=1592 m

The difference between FEM and hand calculation is less than 1

The gravity center of the cross section (Fig 622) can be calculated based on the staticalmoments And of course the gravity center lying on the symmetry axis The distance of thegravity center from the web is

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

With the input Youngs modulus and with the second moments of inertia the idealized bendingstiffnesses in the principal directions can be calculated by hand

E I 1=9396sdot103sdot(02sdot63

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

E I 2=9396sdot103sdot(2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

With the finite element results we can calculate the translations of the shear center according tothe unit forces and moment on the key node (see the former calculation method)

The distances between the shear center and the selected key node are

Δ x=x Sminus xm=minus6343minus0=minus6343m=minus6343mm

Δ y= ySminus ym=+01827minus0=+01827 m=1827 mm

The translations of the shear center are as follows

uSxx=u xxminusφ zx Δ y=15852minus029744sdot10minus4sdot1827=15798mm

uSyx=u yx+φ zxΔ x=072166+029744sdot10minus4sdot(minus6343)=05330 mm

uSxy=u xyminusφ zy Δ y=072166minus010328sdot10minus2sdot1827=05330 mm

uSyy=u yy+φ zyΔ x=73314+010328sdot10minus2sdot(minus6343)=07803mm

Based on these values the translations of the shear center in the principal directions

u1=uSxx+uSyy

2+radic(uSxxminusuSyy

+uSxy2=

15798+078032

+radic(15798minus078032 )

+053302=

=18463mm

u2=uSxx+uSyy

2minusradic(uSxxminusuSyy

+uSxy2=

15798+078032

minusradic(15798minus078032 )

+053302=

=05138mm

According to these values the angles of the principal rigidity directions

α 1FEM=arctanu1minusuSxx

=arctan18463minus15798

0533=2657o

0533=minus6343o

The directions coincide with the axes of symmetries (see Fig 621-2) which is one of theprincipal rigidity direction in this case

Then with FEM-Design results we can calculate the idealized bending stiffnesses of thestructure

EI 1FEM=H 3

3sdot(051381000)E I 1FEM=1622sdot108 kNm2

EI 2FEM=H 3

7 Calculations considering nonlinear effects

71 Uplift calculation

711 A trusses with limited compression members

In this example a truss will be analyzed First of all we calculate the normal forces in the trussmembers and the maximum deflection for the gived concentrated loads After this step wecalculate the load multiplier when the vertical truss members reaches its limit compressionbearing capacity what we set See the inputs in the following table After the hand calulation wecompare the results with the FEM-Design nonlinear calculation results

Inputs

Column heightSpan H = 20 m L = 80 m

The cross sections KKR 80x80x6

The area of the cross sections A = 1652 mm2

The elastic modulus E = 210 GPa steel

The concentrated loads F = 40 kN

Limited compression of the vertical truss members Pcr = 700 kN

Figure 7111 ndash The truss with the concentrated loads and with thevirtual loads for the translation claculation

F F F F

δFδF=1

The normal forces in the truss members based on the hand calculation (without further details)are

N 1=N 17=minus100kN N 2=N 14=0 kN N 3=N15=+8485 kN

N 4=N 5=N 13=N 16=minus6000 kN N 6=N 10=+6000 kN

N 7=N 11=+2828kN N 8=N 12=minus8000 kN N 9=minus4000 kN

The normal forces in the truss members according to the vertical virtual force (see Fig 7111)

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 14=N 15=N 113=N 116=minus05kN N 16=N 110=+05kN

N 18=N 112=minus10kN

The normal forces in the truss members according to the horizontal virtual force (see Fig7111)

N 22=N 26=N 210=N 214=+10kN

N 21=N 23=N 24=N 25=N 27=N 28=N 29=N 211=N 212=N 213=N 215=N 216=N 217=0 kN

The hand calculation of the vertical translation at the mid-span with the virtual force method

EAsumi=1

N iδ N 1i li=0003841m=3841mm

The hand calculation of the horizontal translation at right roller with the virtual force method

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Figure 7112 ndash The truss with the concentrated loads in FEM-Design

Figure 7113 ndash The reaction forces

Figure 7114 ndash The normal forces in the truss members

Figure 7115 ndash The vertical translation at the mid-span and the horizontal translation at the right roller [mm]

The translations and the normal forces in the truss members based on the hand calculation areidentical with the FEM-Design calculation see Fig 7112-5

After this step we would like to know the maximum load multiplier when the vertical trussmembers reaches its limit compression bearing capacity what we set Pcr = 700 kN Themaximum compression force arises in the side columns see the hand calculation N1 = (ndash)100kN Therefore the load multiplier based on the hand calcualtion is λ = 70

Lets see the FEM-Design uplift calculation considering the limit compression in the verticalmembers

With λFEM = 700 multiplier the FEM-Design analysis gives the accurate result but with λFEM =701 (see Fig 7117) large nodal displacements occurred see Fig 7116 Thus by thisstructure if we neglect the effect of the side members the complete truss became a staticallyover-determinated structure FEM-Design solve this problem with iterative solver due to the factthat these kind of problems are nonlinear

Figure 7116 ndash Large nodal displacements when the side trussmembers reached the limit compression value [mm]

Figure 7117 ndash The two different analyzed load multiplier in FEM-Design

712 A continuous beam with three supports

In this example we analyse non-linear supports of a beam Lets consider a continuous beamwith three supports with the following parameters

Inputs

Span length L = 2 m total length = 2x2 = 4 m

The cross sections Rectangle 120x150 mm

The elastic modulus E = 30 GPa concrete C2025

Intensity of distributed load (total partial) p = 10 kNm

In Case I the distribution of the external load and the nonlinearity of the supports differ fromCase II See the further details below (Fig 7121 and Fig 7128)

a) Case I

In this case the distributed load is a total load (Fig 7121) In the first part all of the threesupports behave the same way for compression and tension In the second part of this case themiddle support only bears tension We calculate in both cases the deflections shear forces andbending moments by hand and compared the results with FEM-Design uplift (nonlinear)calculations

In first part of this case the maximum deflection comes from the following formula consideringonly the bending deformations in the beam

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

The relevant results with FEM-Design

Figure 7121 ndash The beam with three supports and uniform distributed load

Figure 7122 ndash The deflection of the beam with three supports (total load)

The extremums of the shear force without signs

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

The extremums of the bending moment without signs

M midspan=9

128p L2

12810sdot22

=2812kNm M middle=18

10sdot22=50kNm

When the middle support only bear tension (second part of this case) basically under the totalvertical load (Fig 7121) the middle support is not active (support nonlinearity) Therefore itworks as a simply supported beam with two supports The deflection the shear forces and thebending moments are the following

The maximum deflection comes from the following formula considering only the bendingdeformations in the beam

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

Figure 7123 ndash The shear diagram of the beam with three supports (total load)

Figure 7124 ndash The bending moment diagram of the beam with three supports (total load)

The maximum of the shear force without sign

p (L+L)=12

10(2+2)=20kN

The extremum of the bending moment without sign

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

The differences between the calculated results by hand and by FEM-Design are less than 2

Figure 7125 ndash The deflection of the beam when the middle support only bear tension (total load)

Figure 7126 ndash The shear diagram of the beam when the middle support only bear tension (total load)

Figure 7127 ndash The bending moment diagram of the beam when the middle support only bear tension (total load)

b) Case II

In this case the distributed load is a partial load (Fig 7128) In the first part all of the threesupports behave the same way for compression and tension In the second part of this case theright side support only bears compression We calculate in both cases the deflections shearforces and bending moments by hand and compared the results with FEM-Design calculations

The extremums of the deflection come from the following formulas considering only thebending deformations in the beam (without signs)

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

Figure 7128 ndash The beam with three supports and uniform partial load

Figure 7129 ndash The deflection of the beam with three supports (partial load)

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

2sdot10=3828kNm M middle=

10sdot22=25 kNm

When the right side support only bear compression (second part of this case) basically under thepartial vertical load (Fig 7128) the right side support is not active (support nonlinearity)Therefore it works as a simply supported beam with two supports The deflection the shearforces and the bending moments are the following

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

Figure 71210 ndash The shear diagram of the beam with three supports (partial load)

Figure 71211 ndash The bending moment diagram of the beam with three supports (partial load)

The extremum of the shear force without sign

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Figure 71212 ndash The deflection of the beam when the right support only bear compression (partial load)

Figure 71213 ndash The shear diagram of the beam when the right support only bear compression (partial load)

Figure 71214 ndash The bending moment diagram of the beam when the right support only bear compression(partial load)

72 Cracked section analysis by reinforced concrete elements

721 Cracked deflection of a simply supported beam

Inputs

Span length Leff = 72 m

The cross section Rectangle b = 300 mm h = 450 mm

The elastic modulus of concrete Ecm = 31476 GPa C2530

The creep factor φ28 = 235

Effective elastic modulus of concrete Eceff = Ecm(1+φ28) = 9396 GPa

Mean tensile strength fctm = 2565 MPa

Elastic modulus of steel bars Es = 200 GPa

Characteristic value of dead load gk = 85 kNm

Characteristic value of live load qk = 120 kNm

Live load combination factor ψ2 = 06

Diameter of the longitudinal reinforcement ϕl = 18 mm

Diameter of the stirrup reinforcement ϕs = 8 mm

Area of longitudinal reinforcement Al = 5x182π4 = 12723 m2

Nominal concrete cover cnom = 20 mm

The cross sectional properties without calculation details

I stress stadium second moment of inertia II = 3075x109 mm4

II stress stadium second moment of inertia III = 2028x109 mm4

I stress stadium position of neutral axis xI = 2564 mm

II stress stadium position of neutral axis xII = 1973 mm

Figure 7211 ndash The simply supported RC beam

The calculation of deflection according to EN 1992-1-1

The load value for the quasi-permanent load combination

pqp=g k+ψ 2 qk=85+06sdot12=157kNm

The maximum deflection with cross sectional properties in Stadium I (uncracked)

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

The maximum deflection with cross sectional properties in Stadium II (cracked)

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

The maximum bending moment under the quasi-permanent load

M max=18

pqp Leff2=

157sdot722=10174 kNm

The cracking moment with the mean tensile strength

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

The interpolation factor considering the mixture of cracked and uncracked behaviour

ζ=max[1minus05( M cr

M max)2

0]=max[1minus05( 407410174)

0]=09198

This value is almost 10 it means that the final deflection will be closer to the cracked deflectionthan to the uncracked one

The final deflection with the aim of interpolation factor

w k=(1minusζ )w kI+ζ w kII=(1minus09198)1901+09198sdot2883=2804mm

First we modelled the beam with beam elements In FEM-Design we increased the division number of the beam finite elements to five to get the more accurate results

Fig 7212 shows the applied cross section and reinforcement with the defined inputparameters

Fig 7213 shows the deflection after the cracked section analysis The deflection of the beammodel in FEM-Design

w kFEM=3059 mm

The difference is less than 9

Figure 7212 ndash The cross section of the RC beam in FEM-Design

Figure 7213 ndash The deflection of the RC beam in FEM-Design with cracked section analysis

Secondly we modelled the beam with shell finite elements Fig 7214 shows the appliedspecific reinforcement with the defined input parameters with slab

Fig 7215 shows the deflection and the finite element mesh after the cracked section analysisThe deflection of the shell model in FEM-Design

w kFEM=270mm

Figure 7214 ndash The specific reinforcement with the shell model in FEM-Design

Figure 7215 ndash The deflection of the RC shell model in FEM-Design with cracked section analysis

722 Cracked deflection of a cantilever beam

Inputs

Figure 7221 ndash The cantilever RC beam

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w k=(1minusζ )w kI+ζ w kII=(1minus09474)1739+09474sdot2637=2590 mm

We modelled the beam with beam finite elements In FEM-Design we increased the divisionnumber of the beam finite elements to five to get the more accurate results

w kFEM=2754mm

Figure 7222 ndash The cross section of the RC cantilever in FEM-Design

Figure 7223 ndash The deflection of the RC cantilever in FEM-Design with cracked section analysis

73 Nonlinear soil calculation

This chapter goes beyond the scope of this document therefore additional informations arelocated in

FEM-Design ndash Geotechnical modul in 3D Theoretical background and verification andvalidation handbook

httpdownloadstrusoftcomFEM-Designinst150xdocuments3dsoilmanualpdf

8 Cross section editor

81 Calculation of a compound cross section

An example for compound cross section is taken from [7] where the authors calculated the crosssectional properties with the assumption of thin-walled simplifications The welded cross sectionis consisting of U300 and L160x80x12 (DIN) profiles In the Section Editor the exact coldrolled geometry was analyzed as it is seen in Figure 811

The following table contains the results of the two independent calculations with several crosssectional properties

Notation Ref [1] Section Editor

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Table 811 ndash The results of the example

Figure 811 ndash The analyzed cross section

9 Design calculations

91 Required reinforcement calculation for a slab

In this example we calculate the required reinforcements of a slab due to elliptic and hyperbolicbending conditions First of all the applied reinforcement is orthogonal and then the appliedreinforcement is non-orthogonal We calculate the required reinforcement with hand calculationand then compare the results with FEM-Design values

Inputs

The thickness h = 200 mm

The Poissons ratio of concrete ν = 02

The design value of compressive strength fcd = 20 MPa

The design value of yield stress of steel bars fyd = 4348 MPa

Nominal concrete cover cx = 20 mm cy = 30 mm

Effective heights dx = 175 mm dy = 165 mm

I) Elliptic bending

In the first case the bending condition is an elliptic bending In FEM-Design the model is a slabwith statically determinant support system and specific moment loads at its edges for the pureinternal force state (see Fig 9221)

Figure 9221 ndash The slab with the edge loads for pure stress state

Fig 9222 shows the constant internal forces in the slab due to the loads Fig 9223 shows theprincipal moments and their directions based on the FEM-Design calculation According to thepure stress state the principal moments and the directions are the same in each elements

First of all the reinforcement is orthogonal and the hand calculation and the comparison are the following

1 Orthogonal reinforcement ( φ =90 o )

The reinforcement is orthogonal and their directions concide with the local system (x=ξy=ϑ=η)

The moments in the slab (tensor of the applied moments)

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=mξη=+6 kNm m

Figure 9222 ndash The mx my and mxy internal forces in the slab [kNm]

Figure 9223 ndash The m1 and m2 principal moments and their directions in the slab [kNm]

The first invariant of the tensor m x+m y=+24 kNm m

The calculation of the principal moments and their directions

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

2minusradic(m xminusm y

+m xy2=

+62=479 kNm m

α 0=arctanm1minusmx

=arctan1921minus16

6=2815o

Compare these results with Fig 9223 The difference is 0

The design moments (according to [9][10]) if the reinforcement (ξη) is orthogonal and theirdirections concide with the local co-ordinate system (xy)

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

1+cos90o+61minus2cos90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

This is a valid solution Because mud ξ+mudη=+36kNm mgtm x+m y=+24 kNm m

mud ξ=+22kNm m mudη=+14kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

1minuscos90ominus61+2cos90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Invalid solution Because mud ξ+mudη=+12 kNm mltm x+m y=+24 kNm m

Case ξ)

mud ξ=mξminusmξ ϑ

=16minus62

8=+115 kNmm

mudη=0

Invalid solution Because mud ξ+mudη=+115 kNm mltmx+m y=+24 kNm m

Case η)

mud ξ=0

mudη=mξ mϑminusmξ ϑ

mξ sin2φ+mϑ cos2φminusmξ ϑ sin 2φ=

16sdot8minus62

16sdotsin2 90o+8sdotcos290o

minus6sdotsin (2sdot90o)=+575

The results of the design moments based on FEM-Design are in Fig 9224 and 9225 Thedifference between the hand and FE calculation is 0

Figure 9224 ndash The mud ξ design moment for elliptic bending with orthogonal reinforcement [kNm]

Figure 9225 ndash The mud η design moment for elliptic bending with orthogonal reinforcement [kNm]

Calculation of the required reinforcement based on the valid design moments

In x ( ξ ) direction

Sum of the moments

mud ξ= f cd xc(d xminusxc

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

xc f cd=a sξ f yd 6403sdot20=asξ 4348 asξ=02945 mm2mm=2945mm2

Fig 9226 shows the required reinforcement in the relevant direction based on FEM-Designcalculation The difference is less than 1

In y ( η ) direction

Sum of the moments

mudη= f cd xc(d yminusxc

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

xc f cd=a sη f yd 4298sdot20=asη 4348 asη=01977mm2mm=1977 mm2

Figure 9226 ndash The asξ required reinforcement at the bottom for elliptic bending with orthogonalreinforcement [mm2m]

Secondly the reinforcement is non-orthogonal and the hand calculation and the comparison are the following

2 Non-orthogonal reinforcement ( φ =75 o between ξ and η )

The reinforcement is non-orthogonal and the ξ direction concides with the local x directionThus y=ϑ The angle between the ξ directional reinforcement and η directional reinforcement isφ=75o

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

The design moments (according to [9][10]) if the reinforcement (ξη) is non-orthogonal

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

1+cos 75o+61minus2cos75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

This is a valid solution Because mud ξ+mudη=+2992kNm mgtmx+m y=+24kNm m

mud ξ=+1735 kNm m mudη=+1257kNm m

Figure 9227 ndash The asη required reinforcement at the bottom for elliptic bending with orthogonalreinforcement [mm2m]

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

1minuscos 75ominus61+2cos75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

16sdotsin2 75o+8sdotcos2 75o

minus6sdotsin(2sdot75o)=+738

Invalid solution Because mud ξ+mudη=+738kNm mltm x+m y=+24 kNm m

Figure 9228 ndash The mud ξ design moment for elliptic bending with non-orthogonal reinforcement [kNm]

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

xc f cd=a sξ f yd 5029sdot20=asξ 4348 asξ=02313 mm2mm=2313 mm2

Figure 92210 ndash The asξ required reinforcement at the bottom for elliptic bending with non-orthogonalreinforcement [mm2m]

Figure 9229 ndash The mud η design moment for elliptic bending with non-orthogonal reinforcement [kNm]

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

xc f cd=a sη f yd 3854sdot20=asη 4348 asη=01773mm2mm=1773mm2

Figure 92211 ndash The asη required reinforcement at the bottom for elliptic bending with non-orthogonalreinforcement [mm2m]

II) Hyperbolic bending

In the second case the bending condition is a hyperbolic bending In FEM-Design the model is aslab with statically determinant support system and specific moment loads at its edges for thepure internal force state (see Fig 92212)

Fig 92213 shows the constant internal forces in the slab due to the loads Fig 92214 showsthe principal moments and their directions based on the FEM-Design calculation According tothe pure stress state the principal moments and the directions are the same in each elements

Firstly the reinforcement is orthogonal and the hand calculation and the comparison are the following

1 Orthogonal reinforcement

m x=mξ=+16 kNm m

m y=mϑ=mη=minus8kNm m

The first invariant of the tensor m x+m y=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+radic(16minus(minus8)2 )

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

minusradic(16minus(minus8)2 )

+62=minus942 kNm m

α 0=arctan1742minus16

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

sin 90o=minus2 kNm m

Invalid solution Because their have different signs

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

This is a valid solution at the bottom

mud ξ=+205 kNm m mudη=0 kNm m

Case η)

mud ξ=0

16sdot(minus8)minus62

16sdotsin2 90o+(minus8)sdotcos2 90o

minus6sdotsin(2sdot90o)=minus1025

This is a valid solution at the top

mud ξ=0 kNm m mudη=minus1025kNm m

In x ( ξ ) direction at the bottom

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Figure 92215 ndash The mud ξ design moment for hyperbolic bending with orthogonal reinforcement [kNm]

Figure 92216 ndash The mud η design moment for hyperbolic bending with orthogonal reinforcement [kNm]

In y ( η ) direction at the top

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

xc f cd=a sη f yd 3136sdot20=asη 4348 asη=01443 mm2mm=1443 mm2

Figure 92217 ndash The asξ required reinforcement at the bottom for hyperbolic bending with orthogonalreinforcement [mm2m]

Figure 92218 ndash The asη required reinforcement at the top for hyperbolic bending with orthogonalreinforcement [mm2m]

The reinforcement is non-orthogonal and the ξ direction concides with the local x directionThus y=ϑ

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

The design moments (according to the theory book) if the reinforcement (ξη) is orthogonal andtheir directions concide with the local co-ordinate system (xy)

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

1+cos 75o+61minus2 cos75o

sin 75o =+2064 kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

mud ξ=0 kNm m mudη=minus1440 kNm m

Figure 92220 ndash The mud η design moment for hyperbolic bending with non-orthogonal reinforcement [kNm]

Figure 92219 ndash The mud ξ design moment for hyperbolic bending with non-orthogonal reinforcement [kNm]

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

In η direction at the top

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces

Figure 92221 ndash The asξ required reinforcement at the bottom for hyperbolic bending with non-orthogonalreinforcement [mm2m]

Figure 92222 ndash The asη required reinforcement at the top for hyperbolic bending with non-orthogonalreinforcement [mm2m]

References[1] Beer FP Johnston ER DeWolf JT Mazurek DF Mechanics of materials McGraw-Hill 2012

[2] Timoshenko S Woinowsky-Krieger S Theory of plates and shells McGraw-Hill New York 1959

[3] Ventsel E Krauthammer T Thin plates and shells Marcel Dekker New York 2001

[4] Dulaacutecska E Jooacute A Kollaacuter L Tartoacuteszerkezetek tervezeacutese foumlldrengeacutesi hataacutesokra Akadeacutemiai Kiadoacute Budapest 2008

[5] Sadd MH Wave motion and vibration in continuous media Kingston Rhode Island 2009

[6] Ivaacutenyi M Halaacutesz O Stabilitaacutestan Műegyetemi Kiadoacute Budapest 1995

[7] Wagner W Gruttman F A displacement method for the analysis of flexural shear stresses inthin walled isotropic composite beams Computers and Structures Vol 80 pp 1843-1851 2002

[8] Majid KI Non-linear structures matrix methods of analysis and design by computers London Butterworths 1972

[9] Neacutemeth F Optimum design of steel bars in reinforced concrete slabs and the criticism of theEurocode 2 Vasbetoneacutepiacuteteacutes 3 pp 107-114 2001

[10] Wood RH The Reinforcement of Slabs in Accordance with a Pre-Determined Field of Moments Concrete Vol 2 No 2 pp 69-76 1968


































References

Notes

StruSoft ABVisit the StruSoft website for company and FEM-Design information at

wwwstrusoftcom

TrademarksFEM-Design is a registered trademark of StruSoft

Force F = 150 kN

Length L = 6 m

e=2 MEI [ L

12 ]+ 2 F

ρGA [05L3 ]= 23

648FL3

FL3 ρGA

648150sdot63

210000000sdot18264ˑ10minus4+150sdot6

L3 L3 L3

M=+FL3

Case I

w cl=P R2

16π ( E h3

12(1minusν 2))

M cl=P

w cl=10sdot52

16π (210000000sdot0053

12(1minus032) )

=0002069m=2069mm w clFEM=204 mm

2π 5=0318

QrFEM=0318kNm

M cl=P

=0796kNm

mM clFEM=0796

Case II

w ss=P R2

16π ( E h3

12(1minusν 2))(3+ν

1+ν )

w ss=10sdot52

16π (210000000sdot0053

12(1minus032) )

(3+031+03)=0005252 m=5252 mm w ssFEM=500 mm

2π 5=0318 kN m QrFEM=0318

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Deflection [mm]

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

Force F = 150 kN

Length L = 6 m

e=2 MEI [ L

12 ]+ 2 F

ρGA [05L3 ]= 23

648FL3

FL3 ρGA

648150sdot63

210000000sdot18264ˑ10minus4+150sdot6

L3 L3 L3

M=+FL3

Case I

w cl=P R2

16π ( E h3

12(1minusν 2))

M cl=P

w cl=10sdot52

16π (210000000sdot0053

12(1minus032) )

=0002069m=2069mm w clFEM=204 mm

2π 5=0318

QrFEM=0318kNm

M cl=P

=0796kNm

mM clFEM=0796

Case II

w ss=P R2

16π ( E h3

12(1minusν 2))(3+ν

1+ν )

w ss=10sdot52

16π (210000000sdot0053

12(1minus032) )

(3+031+03)=0005252 m=5252 mm w ssFEM=500 mm

2π 5=0318 kN m QrFEM=0318

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Deflection [mm]

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

Force F = 150 kN

Length L = 6 m

e=2 MEI [ L

12 ]+ 2 F

ρGA [05L3 ]= 23

648FL3

FL3 ρGA

648150sdot63

210000000sdot18264ˑ10minus4+150sdot6

L3 L3 L3

M=+FL3

Case I

w cl=P R2

16π ( E h3

12(1minusν 2))

M cl=P

w cl=10sdot52

16π (210000000sdot0053

12(1minus032) )

=0002069m=2069mm w clFEM=204 mm

2π 5=0318

QrFEM=0318kNm

M cl=P

=0796kNm

mM clFEM=0796

Case II

w ss=P R2

16π ( E h3

12(1minusν 2))(3+ν

1+ν )

w ss=10sdot52

16π (210000000sdot0053

12(1minus032) )

(3+031+03)=0005252 m=5252 mm w ssFEM=500 mm

2π 5=0318 kN m QrFEM=0318

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Deflection [mm]

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

Case I

w cl=P R2

16π ( E h3

12(1minusν 2))

M cl=P

w cl=10sdot52

16π (210000000sdot0053

12(1minus032) )

=0002069m=2069mm w clFEM=204 mm

2π 5=0318

QrFEM=0318kNm

M cl=P

=0796kNm

mM clFEM=0796

Case II

w ss=P R2

16π ( E h3

12(1minusν 2))(3+ν

1+ν )

w ss=10sdot52

16π (210000000sdot0053

12(1minus032) )

(3+031+03)=0005252 m=5252 mm w ssFEM=500 mm

2π 5=0318 kN m QrFEM=0318

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Deflection [mm]

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

Case I

w cl=P R2

16π ( E h3

12(1minusν 2))

M cl=P

w cl=10sdot52

16π (210000000sdot0053

12(1minus032) )

=0002069m=2069mm w clFEM=204 mm

2π 5=0318

QrFEM=0318kNm

M cl=P

=0796kNm

mM clFEM=0796

Case II

w ss=P R2

16π ( E h3

12(1minusν 2))(3+ν

1+ν )

w ss=10sdot52

16π (210000000sdot0053

12(1minus032) )

(3+031+03)=0005252 m=5252 mm w ssFEM=500 mm

2π 5=0318 kN m QrFEM=0318

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Deflection [mm]

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

Case I

w cl=P R2

16π ( E h3

12(1minusν 2))

M cl=P

w cl=10sdot52

16π (210000000sdot0053

12(1minus032) )

=0002069m=2069mm w clFEM=204 mm

2π 5=0318

QrFEM=0318kNm

M cl=P

=0796kNm

mM clFEM=0796

Case II

w ss=P R2

16π ( E h3

12(1minusν 2))(3+ν

1+ν )

w ss=10sdot52

16π (210000000sdot0053

12(1minus032) )

(3+031+03)=0005252 m=5252 mm w ssFEM=500 mm

2π 5=0318 kN m QrFEM=0318

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Deflection [mm]

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

Case II

w ss=P R2

16π ( E h3

12(1minusν 2))(3+ν

1+ν )

w ss=10sdot52

16π (210000000sdot0053

12(1minus032) )

(3+031+03)=0005252 m=5252 mm w ssFEM=500 mm

2π 5=0318 kN m QrFEM=0318

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Deflection [mm]

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

0 5000 10000 15000 20000 25000 30000 3500000

Deflection

Bending moment

Number of elements

Deflection [mm]

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

wmax=000416p a4

( E h3

12(1minusν 2))

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

M max=00469 p a2

wmax=00041640sdot54

(30000000sdot0253

12(1minus022) )

=0002556 m=2556 mm wmaxFEM=2632 mm

M maxFEM=4597kNm

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

ρ=PP E

(π2 EIL2 )

(π230000000sdot00002667

42 )=0500

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

s=3294 c=0666 f =1104

12=1104(1+0666)

10sdot42

2 )=[1+1104(1+0666)

6 ](10sdot42 )=2613kN

2 )=[1minus1104 (1+0666)

6 ](10sdot42 )=1387kN

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

m m y max=2562

m xy max=1368kNm

N cr=2860kNm

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

wmax 2nd=wmax1

(1minus N x

N cr)=3538

(1minus10002860)

m x max 2nd=mx max1

(1minus N x

N cr)=1805

(1minus10002860)

=2776kNm

(1minus N x

N cr)=2562

(1minus10002860)

=3940kNm

(1minus N x

N cr)=1368

(1minus10002860)

=2104kNm

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

F cr1=π 2 E I 2

L2 =49348 kN F crFEM1=49109kN

F cr2=π 2 E I 2

(06992 L)2=100941kN F crFEM2=99746kN

a=02 m

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

F cr3=π 2 E I 2

(05 L)2=197392kN F crFEM3=193187 kN

F cr4=π 2 E I 2

(2 L)2=12337kN F crFEM4=12331kN

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

N cr=(m ba+

n2 am b )

2 π2( E h3

12(1minusν 2))

b2 m=12 3 n=123

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

m=1 n=1

N cr1=(1sdot68+

1281sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =286036kNm

N crFEM1=286258kNm

m=2 n=1

N cr2=(2sdot68+

1282sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =309377kNm

N crFEM2=310996kNm

m=3 n=1

N cr3=(3sdot68+

12 83sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =478456kNm

N crFEM3=488490kNm

m=4 n=1

N cr4=(4sdot68+

12 84sdot6)

2 π2(210000000sdot0053

12(1minus032) )

62 =732253kNm

N crFEM4=765558kNm

m=3 n=2

N cr5=(3sdot68+

2283sdot6 )

2 π2(210000000sdot0053

12(1minus032) )

62 =1069141kNm

N crFEM5=1080462kNm

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

M cr=π 2 E I z

L2 radic I w

+L2 G I t

π 2 E I z

10002 radic 1258416027

+10002

sdot8077sdot1534

=4328kNcm

L=10 m

M crFEM=4363kNcm

P cr=401 E I z

L2 radic G I t

P cr=401sdot210000sdot2336000

100002 radic 80770sdot8806246210000sdot2336000

=23687 N=2369 kN

P crFEM=2400kN

μ S1=05π μ S2=15π μS3=25π

ω n2=

ω B2+

=79391s

=238181s

=396961s

ω n1=866391s

ω n2=531641s

ω n3=142581s

f n1=137891s

f n2=846131s

f n3=2269231s

f FEM1=137801s

f FEM2=836361s

f FEM3=2233261s

ω nm=π 2

R2 β nm2radic(

12 (1minusν 2))

ρ h β 10=1015 β 11=1468

ω 10=π 2

52 10152radic(210000000sdot0053

12(1minus032) )

785sdot005=3183

f 10=50661s

f 10FEM=51291s

ω 11=π 2

52 14682radic(210000000sdot0053

12(1minus032) )

785sdot005=6658

f 11=10601s

f 11FEM=107311s

Inputs

T i=2π

radic 3 EImi H i

Design

T 1=2π

=001411s T 2=2π

=003990s

T 3=2π

=007330 s T 4=2π

=01129s

T 5=2π

=01577s T 6=2π

=02073s

T 7=2π

=02613s T 8=2π

=03192s

T 9=2π

=03809s T 10=2π

=04461s

T HC=radicsumi=1

T i2=07758s T FEM=0765s

F i=F b

z i mi

sum z j m j

=6588zi mi

Inputs

F i=F b

sum s j m j

=6588si mi

Inputs

α = 05

T 1=0704 s S d1=1115m

T 2=0252s S d2=157m

(ΦiTm ι )

ΦiT mΦ i

m1=([409 736 ][1603 0

0 1352][11])2

[409 736][1603 00 1352][409

736]=2723 t

27232955

m2=([676 minus445][1603 0

0 1352][11])2

[676 minus445][1603 00 1352][ 676

minus445]=2323 t

23232955

27232955

+23232955

=1570sdot2323=365 kN

p i=mΦi

Φ iT m ι

Φ iT mΦ i

p1=[1603 00 1352][409

736][409 736 ][1603 0

0 1352][11][409 736][1603 0

0 1352][409736]

1115=[12061830]kN

p2=[1603 00 1352][ 676

minus445][676 minus445][1603 0

0 1352][11][676 minus445][1603 0

0 1352][ 676minus445]

1570=[ 8198minus4452]kN

pf1=[1206 61830 6]=[2010

3050]kN

pf2=[ 81986minus4452 6]=[ 1366

minus7420]kN

V1=[201+305305 ]=[506

305]kN V 2=[1366minus742minus742 ]=[ 624

minus742]kN

V c1=[506 23052]=[2530

1525]kN V c2=[ 624 2minus742 2]=[ 313

minus371]kN

Mc1=[2530sdot3221525sdot32 2]=[4048

minus5936]kNm

Mb1=[4048+24402440 ]=[6488

minus5936]kNm

V c=[ radic25302+3132

radic15252+(minus371)2]=[2549

1569]kN Mc=[radic40482+50082

radic24402+59362]=[4079

2511]kNm

radic24402+(minus5936)2]=[6489

2511]kNm

α 12=T 2

=02520704

r 12=8ξ 2

(1+α 12)α 123 2

(1minusα 122)

2+4ξ 2α 12(1+α 12)

8sdot0052(1+0358)035832

(1minus03582)2+4sdot0052

sdot0358 (1+0358)2=0007588

r=[ 1 00075880007588 1 ]

V c=[ radic25302+3132

1567]kN

radic24402+59362

2507]kNm

u1=qd[ 9541715]=15[ 954

1715]=[14312573]mm u2=qd[ 0818

minus0540]=15[ 0818minus0540]=[ 1227

minus0810]mm

Δ1=[ 14312573minus1431]=[1431

1142]mm Δ2=[ 1227minus0810minus1227]=[ 1227

minus2037]mm

Δ=[ radic14312+12272

radic11422+(minus2027)2]=[1436

1160]mm

P tot=[(m1+m2)gm2 g ]=[(1603+1352)981

1352sdot981 ]=[28991326]kN

1883]kN

θ 1=P tot1Δ1

V tot1 h=

2899sdot14363059sdot3200

=00425 θ 2=P tot2Δ2

V tot2 h=

1326sdot11601883sdot3200

=00255

Mode shape 1 1206 1219 8198 8180 3036 3069

SRSSHand

2549 1569 4079 2511 6489 2511

SRSSFEM

2578 1589(3718+4532)2=

41252751 5933 2751

CQCHand

2550 1567 4083 2516 6488 2507

CQCFEM

2580 1586(3721+4536)2=

41292746 5932 2746

Inputs

τ =V SI t

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=06665kNm2

τ max=V S max

(02sdot63

4sdot623

12minus

4sdot583

12 )sdot02

=09164kNm 2

V 1=V 3=τ t w1

06665sdot02sdot42

=02667 kN

τ ma x

V 1 w2=V 2 xS

xS=V 1 w2

=02667sdot609997

=1601 m

xm=0m ym=0m

yS= ym+φ zxφ zz

016283sdot10minus3 =+01827 m

2+(01827minus0894)

2=1592 m

x G=S y

2(02sdot4sdot2)02(4+6+4)

=1143m

4sdot623

12minus

4sdot583

12 )=1692sdot108 kNm2

02sdot43

12+2(02sdot4(2minus1143)2)+

6sdot023

12+02sdot6(1143)2)=4585sdot107 kNm2

u1=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=18463mm

u2=uSxx+uSyy

+uSxy2=

15798+078032

+053302=

=05138mm

0533=2657o

0533=minus6343o

EI 1FEM=H 3

EI 2FEM=H 3

Inputs

F F F F

δFδF=1

N 11=N 117=minus05kN N 12=N 114=N19=0 kN

N 13=N 115=N 17=N 111=+07071 kN

N 22=N 26=N 210=N 214=+10kN

EAsumi=1

N iδ N 1i li=0003841m=3841mm

EAsumi=1

N iδ N 2i li=00006918m=06918mm

Inputs

a) Case I

emax=21384

10sdot24

30000000sdot012sdot015312

=00008642m=08642mm

V 1=38

p L=38

10sdot2=75kN V 2=58

p L=58

10sdot2=125kN

M midspan=9

128p L2

12810sdot22

10sdot22=50kNm

emax=5

384p (L+L)4

10sdot(2+2)4

30000000sdot012sdot015312

=003292m=3292 mm

p (L+L)=12

10(2+2)=20kN

M max=18

p (L+L)2=18

10sdot(2+2)2=20kNm

b) Case II

emaxasymp21384

( p 2)L4

102sdot24

10 2sdot24

EI=0001461 m=1461mm

eminasymp5

384( p 2)L4

EIminus

( p 2)L4

102sdot24

EIminus

10 2sdot24

EI=00006173m=06173mm

16p L=

10sdot2=875 kN V 2=9

16p L=

10sdot2=1125 kN

V 3=116

1610sdot2=125kN

M midspan=( 7

16p L)

2 p=( 7

1610sdot2)

10sdot22=25 kNm

emidspan=5

384p L4

10sdot24

EI=0002058m=2058 mm

eright=1

24p L4

10sdot24

EI=0006584 m=6584 mm

p (L)=12

10sdot2=10 kN

M max=18

10sdot22=50 kNm

Inputs

w kI=5

384pqp Leff

Eceff I I

384157sdot724

9396000sdot0003075=001901m=1901mm

w kII=5

384pqp Leff

E ceff I II

384157sdot724

9396000sdot0002028=002883 m=2883 mm

M max=18

pqp Leff2=

157sdot722=10174 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05( 407410174)

0]=09198

w kFEM=3059 mm

w kFEM=270mm

Inputs

k 5ϕ18

w kI=18

pqp Leff4

E ceff I I

157sdot44

9396000sdot0003075=001739 m=1739 mm

w kII=18

pqp Leff4

E ceff I II

157sdot44

9396000sdot0002028=002637m=2637mm

M max=12

pqp Leff2=

157sdot42=1256 kNm

M cr= f ctm

hminus xI

=25650003075

045minus02564=4074kNm

M max)2

0]=max[1minus05(40741256)

0]=09474

w kFEM=2754mm

A [cm2] 8676 8630

yG [cm] 1210 1442

zG [cm] 1920 1922

yS [cm] 139 07230

zS [cm] 1006 1036

Iy [cm4] 113799 114312

Iz [cm4] 45133 43729

Iyz [cm4] 30132 30535

It [cm4] 4883 5211

Iw [cm6] - 2030820

Inputs

I) Elliptic bending

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

+radic( 16minus82 )

+62=1921 kNm m

m2=mx+m y

+m xy2=

+62=479 kNm m

=arctan1921minus16

6=2815o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos90o+6

sin 90o=+14 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos90ominus6

sin 90o=+2 kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 22000=20 xc(175minusxc

2 ) xc=6403 mm

Sum of the forces

Sum of the moments

2 ) 14000=20 xc(165minusxc

2 ) xc=4298 mm

Sum of the forces

m x=mξ=+16 kNm m

m y=mϑ=mη=+8kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16minus8cos 75o

sin 75o =+1735kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

1+cos75o+6

sin 75o=+1257 kNm m

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16+8cos 75o

sin 75o =+937kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

1minuscos75ominus6

sin 75o=+458kNm m

Case ξ)

=16minus62

8=+115 kNmm

mudη=0

Case η)

mud ξ=0

16sdot8minus62

Sum of the moments

2 ) 17350=20 xc(175minusxc

2 ) xc=5029mm

Sum of the forces

In η direction

Sum of the moments

2 ) 12570=20 xc(165minusxc

2 ) xc=3854 mm

Sum of the forces

m x=mξ=+16 kNm m

m1=m x+m y

2+radic(mxminusm y

+mxy2=

16+(minus8)2

+62=1742 kNm m

m2=mx+m y

+m xy2=

16+(minus8)2

+62=minus942 kNm m

6=arctan

1921minus166

=1332o

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2cosφsinφ

=16+8cos 90o

sin 90o =+22kNm m

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 90o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 90o

sin 90o =+10 kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 90ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 10250=20 xc(165minusxc

2 ) xc=3136 mm

Sum of the forces

m x=mξ=+16 kNm m

m xy=mξ ϑ=+6kNm m

Case a)

mud ξ=mξminusmϑ

cosφ1+cosφ

+mξ ϑ

1minus2 cosφsinφ

=16+8cos 75o

mudη=mϑ

11+cosφ

+mξ ϑ

1sinφ

=minus81

1+cos 75o+6

Case b)

mud ξ=mξ+mϑ

cosφ1minuscosφ

minusmξ ϑ

1+2cosφsinφ

=16minus8cos 75o

sin 75o =+378kNm m

mudη=mϑ

11minuscosφ

minusmξ ϑ

1sinφ

=minus81

1minuscos 75ominus6

Case ξ)

=16minus62

minus8=+205kNm m

mudη=0

Case η)

mud ξ=0

Sum of the moments

2 ) 20500=20 xc(175minusxc

2 ) xc=5959 mm

Sum of the forces

Sum of the moments

2 ) 14400=20 xc(165minusxc

2 ) xc=4423 mm

Sum of the forces


































References

Notes

Documents

FEM-Design - StruSoftdownload.strusoft.com/FEM-Design/inst160x/...Verification Examples FEM-Design 16.0 1. Linear static calculations 1.1 Beam with two point loading at one-third of