Flange Lateral Bending Stress ( f) Under the Wind Pressure · PDF fileFlange Lateral Bending Stress ( fl) ... Three dimensional model of the bridge was developed using both SAP2000

Flange Lateral Bending Stress (fl) Under the Wind Pressure By

Atorod Azizinamini, Ph.D.,P.E.

1- Introduction and Objectives

The purpose of preparing this document is to evaluate the application of AASHTO LRFD

Bridge Design Specification (Third edition) to calculate flange lateral bending stress, fl,

for a specific design example and compare the results to detail finite element analysis.

Specific objectives are as follows:

a) calculate the flange lateral bending stress using two and three dimensional finite

element analyses

b) calculate the flange lateral bending stress using code recommendations

c) calculate the magnification factor using detail finite element analysis approach

and that recommended by the code

d) Incorporate flange lateral bending stress term in constructability limit state check

for a three span continuous bridge designed using High performance steel

e) Provide preliminary conclusions with respect to advantage and shortcomings of

the procedures suggested by AASHTO LRFD Bridge Design Specification to

calculate flange lateral bending stress

2- Brief Summary of the Bridge Configuration

Bridge Considered is a three span continuous steel plate girders. Following is brief summary of the specific design information: No. of Spans = 3 Length Span No. (ft) _________________________________________________________________ 1 135 2 175

PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com

http://www.docudesk.com

3 135 No. of Lanes = 3 No. of Girders = 4 Skew Angle = 0 Dist. Curbline To Exterior Girder = 3 DECK DATA AND MATERIAL PROPERTIES _________________________________ Slab Thickness = 8.5 in Haunch Thickness = 3 in Sacrificial Wearing Surface = 0.5 in Concrete Compressive Strength (f'c) = 3000 psi COMPOSITE TYPE FOR ANALYSIS: Deck is Considered Composite Throughout LOADING _______ The Live Load Vehicle is the AASHTO HL-93 Loading ADDITIONAL DEAD LOADS _____________________ Superimposed Dead Load = 175 plf Future Wearing Surface = 20 psf Additional Girder Dead Weight = 10 % of Girder Dead Load

3- Wind Load Analysis

a- Elastic Three Dimensional Finite Element Analysis- Complete Bridge Model

Three dimensional model of the bridge was developed using both SAP2000 and

ANSYS5.7. The purpose of this exercise was to ensure the accuracy of the three

dimensional modeling. The modeling techniques used in developing three

dimensional model of the bridge is based on the past experiences gained from full

scale testing and modeling of the steel bridges in both laboratory and field. Both

model were subjected to same loading and produced approximately same results.


ANSYS5.7 model was selected to carryout all three dimensional analysis as one of

the objective of the work was to conduct geometrical non-linear analysis.

The code specified lateral load (non-factored) due to wind is 0.3 kips per linear foot. The

depth of the girder used is 54 inches. In the three dimensional finite element analysis a

pressure loading of 65 lb/ft2 was applied to outside girder (perpendicular to the web face),

which is approximately equivalent to 0.3 kips per linear ft wind load.

The stress contours and deflected shape of the bridge under the applied wind pressure is

shown in Figure 1. The close up of the model for the first span is shown in figure 2.

Deflected shape (plan view) Stress in x-direction

Figure 1 the deflection and stress of the bridge under the wind pressure


Figure 2 Flange stress under the wind pressure, a closer view- First Span

The maximum lateral deflection of the girder is approximately 10 inches (due to

unfactored wind load of 65 lb/ft2).

Summary of the flange lateral bending stresses obtained from three dimensional finite

element analysis of the entire bridge is shown in Table 1 (see column no. 7) for three

different locations along outside girder. Section 1 in table 1 is located in the first span. As

indicated in table 1, location of maximum flange lateral bending stresses is different for

various analyses types. Section 2 in table 1 is located over the pier and section 3 is

located at mid-span of middle span.


Table 1 Bottom flange stress under wind pressure from a 3-span bridge

Sec. Sy

(in3 ) M (kips-in)

Maximum Flange Lateral Bending

Stresses, (ksi)

Sec.

Mod.

2D† Sap,

Constant

properties

2D† Sap,

specified

Properties

2D† Sap,

Specified

Properties

3D† Ansys, One

Girder Model

3D†† Ansys,

Entire Bridge

(1) (2) (3) (4) (5) (6) (7)

1 53 1092

(x=49.5ft) 832.5

15.7

(x=66.3 ft)

12.1

(x=43.5ft)

9.4

(x=45.1ft)

2

232 2213(x=135ft) 2941 12.7

(x=135ft) 8.4 (x=135ft) 8.4 (x=135ft)

3

49 1232

(x=222.5 ft) 535

10.9

(x=222.5ft)

9.5

(x=222.5ft)

8.2

(x=222.5ft)

† 0.3 k/ft applied to one girder and resulting stress divided by number of girders

†† 65 lb/ft2 which is equivalent of 0.3 k/ft applied to outside girder only. Analysis takes care of number of

girders

The maximum and minimum flange lateral bending stresses shown in figure 1 and 2 are

for first span and are approximately 9.4 ksi. The maximum and minimum flange lateral

bending stresses over the pier is 8.4 ksi. The maximum and minimum flange lateral

bending stresses in middle of the second span is 8.2 ksi.

b- Elastic Three Dimensional Finite Element Analysis- One Girder Model

A three dimensional model of one outside girder was also constructed using Ansys 5.7. In

this model the shell elements were used to model both the web and flanges using the


actual web and flange dimensions. The applied load was in the form of pressure and was

specified at the 65 pounds per square ft, which is equivalent to 0.3 kips per linear ft. From

this three dimensional model the maximum flange lateral bending stresses at the same

locations as that reported for complete model of the bridge were extracted and are shown

in column 6 of table 1. It should be noted that the stresses obtained from this three

dimensional analysis were divided by four (number of girders in the cross section).

c- Two Dimensional Analysis

One of the objectives of the work reported here was to determine the best approach to

calculate the flange lateral bending stresses. As a result, series of two dimensional

analyses were carried out.

Two different two dimensional analyses were carried out using Sap2000. In each analysis

one outside girder was modeled using two dimensional beam elements. One of the

analysis used constant section properties (section sizes at the middle of the first span)

over the entire length of the bridge. The other analysis used actual section properties

along the girder. Results of the Sap2000 in terms of moment for uniform cross section are

shown in figure 3. Section 2 is over the pier and section 3 is at middle of the middle span.

Figure 3 the moment diagram from the 2-D analysis with uniform section properties along the girder


The moment obtained from two dimensional analyses were converted to flange lateral

bending stresses using equation (1) shown below.

yf S

M=σ (1)

Where, Sy is the section modulus about the minor axis of one flange and M is the wind

induced moment per each girder (the moment shown in Figure 3 is for one girder. It is

assumed that total wind load is resisted by all four girders, which is in accordance with

results obtained from three dimensional finite element analyses).

Table 1 shows summary of the results obtained from two dimensional analyses in terms

of moment and stresses at three locations along one girder.

d- Calculating the Flange Lateral Bending Stresses using AASHTO

Recommendation

Section 4.6.2.7 of AASHTO LRFD in its commentary provides a recommendation on

how to calculate the resulting moments in the girders due to wind pressure. The

recommendation are as follows:

b

bw N

WLWLM

810

22

+= (2)

Using equation (2), the maximum moment for first span could be calculated as follows:

fkftklfftklf

M w −=+= 3.187)4(8

)135)(3.0(

10

)4.23)(3.0( 22

(3)

Using equation (2) the maximum moment for middle span could be calculated as follows:

fkftklfftklf

M w −=+= 1.303)4(8

)175)(3.0(

10

)1.23)(3.0( 22

(4)


The flange lateral bending stresses could then be obtained by dividing the resulting

moments by section modules. Table 2 provides summary of flange lateral bending

stresses using the AASHTO LRFD recommendation.

Table 2 Lateral flange bending stress under the wind pressure based on the AASHTO recommendations

Sec. Sy (in3 ) M (kips-in) Flange Lateral Bending Stresses, (ksi)

Section

Modules

AASHTO LRFD

Recommendation AASHTO LRFD Recommendation

1 53 2247.6 42.4

2 232 N.A. N.A.

3 49 3637.4 74.2

As noted from table 2 the flange lateral bending stresses, using AASHTO LRFD

recommendations, are not calculated for section near pier. It is believed that the

AASHTO LRFD recommendations are only applicable for middle of each span.

4- Magnification Factor

It is assumed that the compression flange, when subjected to lateral loads, acts as a beam

column. The lateral load causes the flange to displace in lateral direction. As a result the

maximum moment in the compression flange will increase due to secondary effects.

AASHTO LRFD treats this magnification in a similar way that AISC building code treats

design of beam columns. Appendix A provides derivation of the moment magnification

factor as used in AISC building code.

As indicated in appendix A the general from of the magnification factor is as follows:


ek

m

P

PC

−1

Where P is the applied axial load to the beam column and Pek is the critical column

buckling load. Cm is a factor that accounts for different loading cases. Appendix A

provides more detail discussion of this magnification factor.

As noted from Appendix A the derivation of the moment magnification factor is based on

assuming a beam column being supported at both ends and subjected to end axial load

and some type of lateral load.

The magnification factor used in Equation 6.10.1.6-4 of the AASHTO LRFD is similar to

that shown above and is as follows:

cr

bm

F

f−1

85.0

According to AASHTO LRFD code the lateral flange bending stress should be magnified

only if the unbraced length between the cross frames exceeds that shown below:

yc

bm

bbpb

F

fRC

LL 2.1≤ (5)

Where

yctP F

ErL = (6)


For the design example under consideration the Lp is as follows:

ftinLP 6.650

290003.3 == (7)

Equation (5) is function of the compressive stress in the flange due to gravity load as

indicated in fbm term.

The Strength Limit State III is where wind load on structure WS influences the loading

combination. The Strength Limit State III could be written as follows:

1.25 fDC1 + 1.4 fWS

During the construction phase the fbm in equation (5) could be due to dead weight of the

steel girders or weight of the steel girders plus the wet concrete.

For the sake of discussion, the calculation of magnification factor is shown for middle

span.

a) magnification factor using self weight of concrete

This is a very unlikely event where concrete is placed and high wind loads are applied to

the girders before concrete has had a chance to harden. Therefore this scenario is a safe

guard for approximately 24 hour time period where concrete is cast but is not hardened

yet. For this scenario, for the middle span the maximum factored compressive stress, fbm

due to gravity load only is 25.8 ksi. This includes the weight of the girder plus the

concrete. The limiting unbraced length between cross frames would then become:

ftftLb 0.11

50

8.25)0.1(0.1

)6.6(2.11.23 =>= (8)


Since actual unbraced length between the cross frames is 23.1 ft and the limiting value is

11.0 ft, we must then magnify the fl term.

Substituting the numerical values, magnification factor becomes:

29.2

41

8.251

85.0 =−

=MF (10)

b) Magnification factor using only the weight of the girder

If concrete is not cast during severe wind condition, the fbm can be computed based on the

self weight of the steel girders. The moment induced in mid-span of the middle span, due

to weight of the girder only is 2021 ft-kips. The resulting compressive stress in the

compression flange could then be obtained as follows:

ksiS

Mf

xbm 0.3

833

)2021(25.1 === (11)

The limiting value for the unbraced length is calculated using Equation (5).

ftftLb 3.32

50

0.3)0.1(0.1

)6.6(2.11.23 =<= (12)

Therefore, for this scenario, there is no need to magnify the flange lateral bending stress

due to wind loads.

c- Magnification factor using Nonlinear Geometric Analysis

Nonlinear finite element analyses were carried out to account for second order effect

directly. ANSYS5.7 was used. Complete three dimensional model of the bridge was used

in the analysis.


Several scenarios were simulated in the nonlinear geometrical finite element analysis

using the full three dimensional model of the bridge. A) scenario where the dead weight

consisted of the weight of the girders only, b) scenario where dead weight consisted of

weight of the girders plus the weight of the wet concrete before it is hardened and c)

scenario where dead weight consisted of weight of the girder plus weight of the concrete

after concrete is hardened.

For the three cases described above, Table 3 gives the magnification factors to be used in

conjunction with flange lateral bending stresses at mid-span of the middle span (section 3

in table 2). The magnification factors reported in Table 3 is simply the ratios between

flange lateral stresses obtained from nonlinear finite element analysis divided by the

corresponding value from linear finite element analysis.

Table 3 Maginification factor for mid-span of middle span

Dead Load Considered fbm, ksi Magnification Factor

2D Analysis AASHTO Ansys, 3D nonlinear

Self weight of Girders only 3.0 1.0 1.01

Wet concrete & Girder

weight

25.8 2.29 1.31

Composite girder 1.0 1.03

Figures 4 through 5, gives maximum flange lateral stresses any where along the bridge as

obtained from nonlinear and corresponding linear finite element analysis for the three

construction scenarios described above. The magnification factors reported in Table 4 is

the maximum flange lateral bending stresses anywhere along the flanges obtained from


nonlinear finite element analysis divided by maximum flange lateral bending stress

anywhere along the bridge obtained from linear finite element analysis.


Linear Analysis

Nonlinear Analysis

Figure 4 The longitudinal stress contours in x-direction under wind and steel girder weight


Linear Analysis

Nonlinear Analysis

Figure 5 The longitudinal stress contours in x-direction under wind, steel girder weight and slab weight


Linear Analysis

Nonlinear Analysis

Figure 6 The longitudinal stress contours in x-direction under wind, steel girder weight and slab weight in composite condition


Table 4 Maginification factor for various construction scenarios

Dead Load Considered Magnification Factor

AASHTO Ansys, 3D nonlinear

Self weight of Girders only 1.0 1.06


weight

2.29 1.24

Composite girder 1.0 1.03

5- Summary

Previous sections provided different approaches for calculating the flange lateral bending

stresses. Methods used included

a) Three dimensional model of the entire bridge

b) Three dimensional model of one girder

c) Two dimensional model of girder using uniform sectional properties

d) Two dimensional model of the girder using specified section properties

e) AASHTO LRFD code recommendations

Table 5 provides summary of the flange lateral bending stresses for mid-span of the

middle span


Table 5- Unfactored flange lateral bending stresses due to 0.3 k/ft. wind load

Method Used Flange Lateral Bending Stresses, ksi

3-D Entire bridge 8.2

3-D One girder 9.5

2-D Uniform Section Properties 25.1

2-D Specified Section Properties 10.9

AASHTO Recommendations 74.2

Further, the magnification factors to be used in conjunction with flange lateral bending

stresses due to wind loads were calculated using different approaches. Methods used

included:

a) magnification factors calculated using dead weight of the girder only

b) magnification factor calculated using dead weight of wet concrete and steel girder

c) magnification factor calculated using weight of hardened concrete and steel girder

Table 6 provides summary of the magnification factor for flange lateral bending stresses

for mid-span of the middle span

Table 6 Maginification factor for mid-span of middle span

Dead Load Considered fbm, ksi Magnification Factor

2D Analysis AASHTO

Requirement

Nonlinear Analysis

Self weight of Girders only 3.0 1.0 1.01


weight

25.8 2.29 1.31

1.0 1.03


6- Constructibility Flexural Checks (AASHTO 6.10.3.2)

The Strength Limit State III is where wind load on structure appear. This load

combination is as follows:

1.25 fDC1 + 1.4 fWS

For the sake of discussion the calculations are shown for mid-span of middle span.

For this example, during construction, the maximum compressive stress in the

compression flange, fbu , due to factored (1.25 factor) gravity load is given by table 7

Table 7 Factored Stress in Compression Flange Due to Gravity Load

Dead Load Considered fbu, ksi

2D Analysis

Self weight of Girders only 3.0

Wet concrete & Girder weight 25.8

According to AASHTO equation 6.10.1.6-1 the flange lateral bending stress should be

less than 0.6Fyc.

In addition the girder section must satisfy Equations 1 to 3 in Article 6.10.3.2.1 for the

compression flange and Equation 6.10.3.2.2 for the tension flange. In the current example

the web is non-compact according to the AASHTO Article 6.10.6.2.2.3, thus the

following equations should be checked.

ychflbu FRff φ≤+ (13)


ncflbu Fff φ≤+3

1 (14)

a) Calculations based on values obtained from code provisions

The flange lateral bending stress is fl is 74.2 ksi which exceeds the Fy and therefore

code requirement is violated regardless of which dead loads are considered in the

calculations

fl = 74.6 ksi x 1.4 = 104.4 ksi

fbu = 3 ksi or 25.8 ksi

b) Calculations based on values obtained from 2-D Analysis and magnification

factor from AASHTO – Dead Weight of Wet Concrete and Steel Girder only

fl = 10.9 ksi x 1.4 = 15.3 ksi ( Using specified section properties)

fbu = 25.8 ksi

Magnification factor = 2.29

Fnc = 36.2 ksi Compressive capacity of compression flange

ychflbu FRff φ≤+

25.8+ 15.3 (2.29) = 60.9 > 50 ksi N.G.

ncflbu Fff φ≤+3

1

25.8+ (1/3) 15.3 x 2.29 = 37.4 > 36.2 ksi N.G.


c) Calculations based on values obtained from 2-D Analysis and magnification

factor from AASHTO – Dead Weight of Steel Girder only

fl = 10.9 ksi x 1.4 = 15.3 ksi ( Using specified section properties)

fbu = 3 ksi

Magnification factor = 1.0

Fnc = 36.2 ksi Compressive capacity of compression flange

ychflbu FRff φ≤+

3 + 15.3 = 18.3 < 50 ksi O.K.

ncflbu Fff φ≤+3

1

3+ (1/3) 15.3 = 8.1 <36.2 ksi O.K.


Appendix A

Derivation of the Moment Magnification factor Used in

AISC Building Code


BEAM-COLUMNS

A beam-column is a member that is subjected to both axial force and bending moment. To begin our discussion, let us consider a simply supported beam-column subjected to an axial force P and uniformly distributed lateral load.

00 =∑M

022

2

=−−+ xwl

Pywx

M

w (intensity)

P x

y

EI = constant

P

P

w

V 2

wl

x

y M

o


xwlwx

PyM22

2

+−=

But,

''EIyM −=

→ xwlwx

PyEIy22

2'' +−=−

xwlwx

PyEIy22

2" −=+

Letting EI

Pk =2

xEI

wlx

EI

wyky

22" 22 −=+→

ph yyy +=

xKkxAy xxh cossin β+=

To find yp assume the following formula: CxCxCy p ++= 2

21

This results in:

42

22 22 Elk

wx

EIk

wlx

Elk

wy p −−=

Thus, the general solution is:

y=A sin Kx +Bcos Kx +K42

22 22 EIk

wx

EIk

wlx

EIk

w −−

The constant A and B are obtained using the following B.C’s.


40)0(

EIk

wBy =⇒=

Al

y ⇒= 0)2

('2

tan4

hl

EIk

w=

Introducing the notation 2

klu =

the general solution could be written as follows:

)(8

12

cos2

sintan16 2

2

4

4

xlxEIu

wl

l

ux

l

uxu

EIu

wly −−

−+=

from which the moment distribution along the length of the member is:

−+=−= 11

2cos

1

2sintan

4 2

22 uxux

uu

wlEIyM

For this problem, the maximum deflection occurs at the midspan.

ymax =

2

ly

ymax = 2

4

4

4

32cos

cos1

16 EIu

wl

u

u

EIu

wl −

−

Ymax can be written as:

Ymax =

−−4

24

5

)2sec2(12

384

5

u

uu

EI

wl

=

−−4

2

5

)2sec2(12

u

uuyo

where EI

wlyo 384

5 4

= is the deflection if only uniform lateral load had existed.

Several observations could be made from this equation.

1. The term in the brackets could be viewed as amplification factor due to axial load.


2. As axial load increases, the amplification factor increases as shown in figure below.


When P=0 A.F. (amplification factor), reduced to 1. When u = 2/π

).(2

2

l

EIP

π= , the value of A.F. approaches infinity. In other words, as

P approaches the Euler load, the lateral deflection of the member increases without bound.

3. If P remains constant, ymax is linear function of w. Thus for several

beam with constant axial load and varying lateral load, the law of superposition is held. However, if axial load is varied superposition does not hold. For the design purpose, the term in the brackets (A.F.) could be simplified. Expanding sec u in a power series:

....8064

277

720

61

24

5

2

11sec 8642 +++++= uuuuu

Substituting this in the expression for ymax yields: [ ]....1649.4067.01 42 +++= uouyy omzx Noting that:

eP

P

EI

Plklu

222

π===

= 2

2

l

EIPe

π

ymax =

+++ ...)(004.1)(003.11 2

eeo P

P

P

Py

Or approximately

ymax

+

++≈ ...1

2

eeo P

P

P

Py

Also from a power series (Maclaurin Series):


...11

1 2

0

+++==− ∑

=xxx

x

oo

m

m

Therefore:

−≈

eP

Pyy

1

10max

The maximum moment also occurs at midspan.

)1(sec4

)2

(2

2

max −== uu

wlLMM

(from expression on page __)

or

−=2

2

max

)1(sec2

8 u

uwlM

−=20max

)1(sec2

u

uMM

M0 is the moment, that would exist if only lateral load had been applied. Therefore, the term in brackets represents amplification factor due to axial load. Again using the power series, expansion for sec u, ,the expression for Mmax could be simplified as follows:

[ ]...0687.01694.04167.01 6420max ++++= uuuMM

Using the expression for u

22

l

EI

Pklu ==

eP

Pu

2

π=

= 2

2

l

EIPe

π

= 2

2

πlP

EI e


Mmax= M0

+

+

+

+ ...032.1031.1028.11

32

eee P

P

P

P

P

P

or

+

+

+

+= ...004.1003.11028.11

2

0maxeee P

P

P

P

P

PMM

or approximately,

Mmax =Mo

+

+

+

+ ...1028.11

2

Pe

P

P

P

P

P

ee

−

+=

e

e

P

PP

PMM

1

1028.110max

−

+=

e

e

P

PP

P

M1

028.01

0

or

−≈

eP

PMM

1

10max

Where the term in brackets is the design moment amplification factor.


BEAM COLUMN WITH A CONCENTRATED LATERAL LOAD

For this case the governing D.E.’s are:

Pyxl

alQEIy +−=− )(" for ax ≤≤0

Pyl

xlQaEIy +

−=− " for lxa ≤≤

For the case of 2

la = it can be shown that:

P

P

( )L

aLQ −

x

y M

P x

y

EI = constant

P

Q

L

a


=u

uMM o

tanmax (H.W.)

Where 4

QlM o = (Moment if only concentrated load had

existed)

and 2

klu =

To simplify the expression for maximum moment, we will use the power series expansion for tan u.

....315

17

15

2

3

1tan 753 ++++= uuuuu

++++= ...315

17

15

2

3

11 642

max uuuMM o

eP

Pklu

22

π==

+

+

+

+= ...811.0812.0823.01

32

maxeee

o P

P

P

P

P

PMM

+

+

+

+= ...985.0987.01823.01

2

0eee P

P

P

P

P

PM

+

+

+

+≈ ...1823.01

2

eeeo P

P

P

P

P

PM

−

+≈

e

eo

P

PP

PM

1

1823.01


−

+−≈

e

eeo

P

PP

P

P

P

MM1

823.01

max

[ ]

−

−≈

−

−≈

e

eo

e

eo

P

PP

P

M

P

PP

P

MM1

2.01

1

18.01

max

For convenience, let’s write this expression in the following form:

−

+

=

e

e

P

P

P

P

MM1

1

0max

ψ

Where andQL

M40 = ψ =0.2

Notice that for the previous case, maxM has the same form except that ψ =0.


So in summary:

Note that in each case, the amplification factor =

ek

m

P

PC

−1

Table below gives the summary of mC for different loading cases.

P

P

P

P

P

P

MB MA

2l

2l

1=mC

2.0 where

1

−=

+=

ψ

ψek

m P

PC

Form) d(Simplifie

4.04.06.0 ≥

−=

B

Am M

MC



BEAM-COLUMN WITH UNEQUAL END MOMENTS

AB MM >

Assume AB MM ≥ The D.E. for this case is as follows:

ABA Mx

l

MMPyEIy −+=+"

it can be shown that the solution to this D.E. is as follows:

( )2222

cossin

cos

EIk

M

lEIk

MMkx

EIk

M

klEIk

MklMy ABAABA −++++−=

=EI

Pk 2

P

P

l

MM BA +

x

y

M

MA

MA

MB

P P x

y

l

EI = Constant


In the previous cases, we knew the location of the maximum moment. However, in this case, the location is not too obvious. Remember that

"EIyM −= and maxM is found by evaluating "y at maximum location and substituting it in "EIyM −= .

The moment is the maximum where shear is zero, therefore to calculate the location where the moment is maximum, we could equate:

'"EIydx

dmV −== to zero.

For this problem, it can be shown that:

( )

++

−=kl

klM

MMM

MM B

ABA

B 2

2

max sin

1cos)(2/

If the member had been bent in a single curvature, maxM could be simply found by BB MM −= or

( ) ( )

+−=

kl

klMMMMMM BABA

B 2

2

maxsin

1cos/2/

for the case: MMM BA =−=

( )kl

klMM

2maxsin

cos12 −=

And its location is at midspan as shown below:


When a beam is subjected to axial load and unequal end moment, calculation of maxM could be simplified using the concept of equivalent moment. The figure below shows schematically the concept of equivalent moment.

M M

P P

l

M M

+

=

2l

2l

Mmax

Primary Moment

Secondary Moment

Total Moment


eqM could be found by equating the expressions for maxM for two cases

derived:

( ) ( ) ( )kl

klM

kl

klMMMMM eq

BABAB 22

2

sin

cos12

sin

1cos/2/ −=

+−

or

( ) ( )

( )kl

klMMMMM BABA

eq cos12

1cos/2/ 2

−+−

= ( BM )

or ( )Bmeq MCM =

where ( ) ( )

( )Coskl

klMMMMAC BAB

m −+−

=12

1cos/2/ 2α

Austin has shown that an expression approximating mC is as follows:

MA MB Meq Meq

P P P P

MB>MA

MA

MB

Meq Meq

Mmax Mmax


( ) 4.0/4.06.0 ≥−= BAm MMC This expression has been adopted by both AISC/ASD and LRFD. Thus the maxM for a beam subjected to axial load and unequal end moment could be calculated as follows:

( )kl

klMM eq 2max

sin

cos12 −=

( )kl

klMCM Bm 2max

sin

cos12 −= (Assuming AB MM ≥ )

The above expression could be simplified noting that

2sin2cos1 2 kl

kl =−

and

=2

cos2

sin4sin 222 klklkl

=2

secmax

klMCM Bm

Letting ukl =2

...720

61

24

5

2

11sec 642 ++++= uuuu

eP

Pklu

22

π==

...27.1268.123.11sec32

+

+

++=→

eee P

P

P

P

P

Pu


+

+++≈ ...123.11

2

eee P

P

P

P

P

P

−+≈

e

e

P

PP

P

1

123.11

e

ee

P

PP

P

P

P

−

+−≈

1

23.11

e

e

P

PP

P

−

+≈

1

23.01

eP

P−≈

1

1

Therefore:

e

Bm

P

PMC

M−

=1

max

Note that

e

m

P

PC

−1 is an amplification factor.

In all the cases considered, beams were simply supported. In the event that end conditions are different than simply supported it could be shown that

2

2

l

EIPe

π= would be replaced by: ( )2

2

kl

EIP ke

π= where k is the effective length

factor if only axial load had existed.


Therefore:

B

ke

m M

P

PC

M−

=1

max .


Documents

Flange Lateral Bending Stress ( f) Under the Wind Pressure · PDF fileFlange Lateral Bending Stress ( fl) ... Three dimensional model of the bridge was developed using both SAP2000