Structural Analysis TYS Exp 249

8/17/2019 Structural Analysis TYS Exp 249

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ILD and Rolling Loads

2

T1 : Solution

Figure (ii) shows in I .L.D. for the bending moment of C .

Height of the I .L.D. at C = ( )

− ×= =

l

l units

Max. +ve B.M. at C .For the maximum positive (sagging) moment to occur at the section C , the loads should be place as shown

in figure (iii).∴ Max. +ve B.M. at C .

= 200 × + 80 × 1.37 kNm

= +452.46 kNmMax. –ve B.M. at C .For this condition the loads should be placed as shown in figure (iv)∴ Max. –ve B.M. at C .

=

− × + ×

kNm

= –443.66 kNm


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3Workbook

D

D

200 80

200 80

D

127

127

67

4 m 2 m3 m3 m

D A C B E

Mc A

A

C B

B

E

E

E

B AC

C

0.8m

Position of loads for max positive BM at C

Position of loads for max negative BM at C

0.8m

(iii)

(ii)

(I)

(iv)

1.37+

– –

1.26

T2 : Solution

The reactions V a , V b and the reacting moment M a for the positions of the unit load on the RHS of C and LHSof C are shown in figure.When the unit load is on RHS of C distant x from C .For this position of the unit load.For B.M. at C to be zero,

V b × 4 – 1 x = 0

∴ V b = x

∴ V a = 1 – x

M a = x

× 10 – 1(6 + x )

∴ M a =

x – 6

When the unit load is on LHS of C distant x from C .

For this position of the unit load.V b = 0, V a = 1 and M a = –1(6 + x )

Influence line diagram for V b When the unit load is on RHS of C distant x from C ,

V b = l – x

When x = 0, V a = 1, When x = 4, V a = 0When the unit load is on LHS of C , V a = 1


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4 Civil Engineering • Structural Analysis


Influence line diagram of M a When the unit load is on RHS of C distant x from C ,

M a = x

× 10 – 1(6 + x )

∴ M a =

x – 6

When x = 0, M a = –6 and When x = 4 m, M a = 0When the unit load is on LHS of C distant x from C

M a = –1(6 – x )When x = 0, M a = –6 and When x = 6, M a = 0Influence line diagram for shear force at D When the unit load is on RHS of C distant x from C ,

S d = 1 – x

When x = 0, S d = +1 and When x = 4 m, S d = 0When the unit load is on LHS of C distant x from C and less than 2 m, S d = +1When the unit load is on LHS of D , S d = 0Influence line for bending moment M d at D When the unit load is on RHS of C distant x from C

M d = x

× 6 – 1(2 + x ) = x

– 2

When x = 0, M d = –2 and When x = 4 m, M d = 0When the unit load is between C and D and distant x from C

M d = –1(2 – x )When x = 0, M d = –2 and When x = 2 m, M d = 0When the unit load is on LHS of D , M d = 0Maximum reaction at B due to live loadSee I .L.D. for V b For maximum value of V b the live load should cover atleast the part CB

V b (max) = Rate of loading × area of I LD covered by the live load

= 20 × × 4 × 1 = 40 kN

Maximum reaction at A due to live load

See I .L.D. for V a For maximum value of V a , the live load should cover the whole beam.

V a (max) = Rate of loading × area of I LD covered by the live load

=

× × + × × = 40 kN

Maximum Bending Moment M a at A due to live loadSee I .L.D. for M a


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5Workbook

For maximum value of M a , the live load should cover the whole beam.M a (max) = Rate of loading × area of I LD covered by the live load

= 20 × × 10 × 6 = 600 kNm

Maximum Shear Force S d at D due to live loadSee I .L.D. for S d For maximum value of S d , the live load should cover at least the part DB

S d (max) = Rate of loading × area of I LD covered by the live load

=

× × + × × = +80 kN

Maximum Bending Moment M d and D due to live loadSee I .L.D. for M d For maximum value of M d the live load should cover at least the part DB

M d (max) = Rate of loading × area of I LD covered by the live load

= 20 × × 6 × 2 = +120 kNm

D

D

D

D

D

C

C

C

C

C

A

A

A

A

A

B

B

B

B

B

+1

–

+

–

+

V b

V a

M a

S d

M d

1

1 16

2

1


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T1 : Solution

5 m

C

8 m

B A

24 m

y d

10 t

D

xH

V A VB

H

Let H be the horizontal reaction. Taking moments about A, we getV B × 24 = 10 x

V B =

=

x x

∴ V A =

−− = x x

Taking moments about C , the crown, we get

× x

= H × 5

or H = x

Let 10 t is acting on the left side of point D .Taking moments about D , the point 8 m from the left support

− x× 8 – Hy – 10 (8 – x ) = 0 ...(1)

Arches

3


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7Workbook

But y d = ( )

×− = × ×× x l x

l

or y d =

m

Substituting these values in equation (1), we get

− x× 8 – x × – 10 (8 – x ) = 0

⇒ 6(120 – 5 x ) – 40 x – 90 (8 – x ) = 0⇒ 720 –30 x – 40 x – 720 + 90 x = 0⇒ x = 0Now, assuming 10 t is acting on the right side of point D .

− x× 8 – x × – 10 ( x – 8) = 0

⇒ 6(120 – 5 x ) – 40 x – 90 ( x – 8) = 0

⇒ 720 –30 x – 40 x + 720 – 90 x = 0⇒ x = 9i.e. load of 10 t must be located at a distance of 9 m from the support.B.M. under the load

M = Hy – V A × 9

∵ H = 9t, y =

× × × =×

V A =

− × =

∴ M = 9 × 4.69 – 6.25 × 9 = –14.04 t-m

T2 : Solution

A two hinged parabolic arch carrying a central concentrated load of 100 kN is as shown in figure.Now, from equation of equilibrium ΣFy =0⇒ VA + VB = 100 kN ...(i)

⇒ ΣFx = 0⇒ HA = H B ...(ii)

ΣMB = 0⇒ VA × 100 – 100 × 50 = 0

⇒ VA = 50 kNFrom equation (i)

VB=50 kN

As, we know that, the two hinged arch is a indeterminatestructure with one degree of indeterminacy.So to find the horizontal reaction, we have to usecompatability equation.

Now take H A = H as redundant.∴ By compatability equation

l


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∂∂

= ∆AB = 0

∴ Equation of parabolic arch is

y = ( ) −l l

⇒ y =( )

( ) × − = 0.008x(100–x)

= 0.8x – 0.008x 2

= ( )−l l

⇒ tan θ =( )

( )

× −

= 0.008 × 2(50–x) = 0.016(50 – x)Bending moment at section x-x

Mx = V Ax – Hy = 50x – Hy∂∂ = –y

∵ U = ∫

⇒ ∂

∂ =

∂ ∂∂∫

= ( )( ) − −∫

⇒ 0 =

− +∫ ∫

⇒ H =∫

∫

H =( )

( )

−

−

∫

∫

=( )

( )

−

−

+ × −

∫

∫


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9Workbook

=

−

× − + × −

= = 97.656 kN

T3 : Solution

Let V a and V b be the vertical reactions at the supports A and B. Let H be the horizontal thrust at eachsupport.

V a l 1 = Hh 1 + Wa

∴ V a =

+

l l

Taking moments about C of the forces on the right side of C ,V b l 2 = Hh 2

∴ V b =

l

Adding equations (1) and (2),

V a + V b =

+ + l l l

But V a + V b = W

∴

+ + l l l

= W ;

+ l l = ( )

− − = l

l l

H = ( ) ( )

− −= +

l l

l l

l l l

; =( )

−+

l l

l l

T4 : Solution


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Equation of parabolic arch is

y = ( )− ...(i)and horizontal reaction

H = −∫ ∫

...(ii)

∴ Numerator

= =∫ ∫

= ( )

−∫

=

−

=

− − −

=

Denominator

= ∫ = ( ) − ∫ = ( ) −∫

= ( ) + −∫ =

+ −

=

+ −

=

Putting values in eq. (ii), we get

H =

− × =

− and V = 0


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M FCB = – [80 × 0.5 – 35] = –5 kN-mBeam ABBeam ABBeam ABBeam ABBeam AB

M AB =

+ θ + θ − φ

EI EI EI

= θ− + + −

EI

= θ− +

EI

M BA =

+ θ + θ − φ

EI EI EI

= + 20 + 0 + EI θB – 0= 20 + EI θB

Beam BCBeam BCBeam BCBeam BCBeam BC (Ignore deflection(Ignore deflection(Ignore deflection(Ignore deflection(Ignore deflection ∆∆∆∆∆c c c c c ,,,,, θθθθθ

c c c c c = 10)= 10)= 10)= 10)= 10)

M BC = + θ EI

= − + θ EI

M CB = − θ EI

= − − θ EI

Equilibrium equation:Equilibrium equation:Equilibrium equation:Equilibrium equation:Equilibrium equation:

M BA + M BC = 0

⇒ (20+ EI θB )+( − + θ EI

) = 0

⇒

θ EI = 15

⇒ EI θB = 10 kN-m

∴ M AB = − + = − M BA = 20 + 10 = 30 kN-m

M BC = − + = − M CB = –5 – 10/2 = –10 kN-m

15 kN/m

A B

R A R B

4 m

30 kN/m15 kN/m


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13Workbook

R B 1 (4) + 15 = 15 (4) (2) + 30 ⇒ R B 1 = 33.75 kN∴ R A= 15 × 4 – R B 1 = 26.25 kN

80 kN

B B

R B 2

2 m

10 kN/m

30 kN/m

1.5 m

R B 2 = 80 kN ( ∵ Support C cannot given any reaction)∴ R B = R B 1 + R B 2 = 33.75 + 80 =113.75 kN

Assuming and

BC

30 kNm

15 kNm

A B C

SFD

BMD 10 kNm

33.75 kN

–

+

Alternate solution by moment distribution methodAlternate solution by moment distribution methodAlternate solution by moment distribution methodAlternate solution by moment distribution methodAlternate solution by moment distribution method

K AB : K BC =

= EI EI

Beam AB BA BC CB

D.F.

C.O.

F.E.M.

Bal.

– 1

– 5

– 5 – 10 kNm

2/3

– 20

+ 10

+ 30 kNm

1/3

– 35

+ 5

– 30 kNm

– 1/2

– 20

– 5 – 15 kNm

T2 : Solution

θ

θ


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From slope deflection equation

Assuming and

MAB = ( ) θ + θ

MBA = ( ) θ + θ

MBC = ( ) θ +

MCB = ( )θAt joint B,

MBA + M BC = 0

or ( ) θ + θ + θ = 0

or 4 θB + θA = 0

or θB =

θ− = − radians

= –0.001 radians⇒ θB = 0.001 radian (anticlockwise)

MAB = ( )

× × −

= 70 kNm

MBA = ( ) −

= 20 kNmMBC = 10000(–2 × 0.001) = –20 kNmMCB = 10000 × –0.001 = –10 kNm

Taking bottom face as reference face.


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15Workbook

T3 : Solution

P R –Deflected

shape

A

P

L/2 L/2

C B

Let force generated in the spring be R. Complementary strain energy ( U *) is given by,

U * =

+ ∫

x

EI

From the theorem of least work,

= 0

⇒

+∫

iii

x EI

= 0

Here the term ( i) denotes elongation in the elastic prop and term ( ii ) denotes deflection of beam end Ccorresponding to direction of reaction

Now, M =

⋅ ∀ < < ←

⋅ − − ∀ < <

x x

x

x x x

∴

=

∀ < <= ∀ < <

∀ < <

x x

x x

x x

∴ +∫ x EI = 0

⇒

− −+ ⋅ + ⋅∫ ∫ x x x

x dx x dx EI EI = 0

⇒

+ + ⋅ − ⋅ + ⋅ I I I I = 0


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⇒

+ + − +

EI EI EI EI = 0

⇒

+ − EI EI = 0

Given k =λ EI

⇒

+λ EI EI =

EI

⇒ + λ

=

⇒ R = λ + λ 5P

16

R

O=

kLE

3

I λ

when R =

=

λ λ +⇒ 4λ + 12 = 5 λ ⇒ λ = 12

T4 : Solution

The structure is statically indeterminate to one. Taking H as redundant and assuming end B to be ON rollersupport, H is given by,

H = δ

x

y

...( i)

where δL = ∆

∫ x ...( ii )

Now,

∆ x

= x

EI

⇒

∆ x

=

+ x

EI

⇒ ∆ =

+ + x

x EI

At x = 0, ∆= 0⇒ C 2 = 0At x = L, ∆= 0

⇒ C 1 = −

EI


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(ii)(ii)(ii)(ii)(ii) Fixed end momentsFixed end momentsFixed end momentsFixed end momentsFixed end moments

=

× ×− = −

=

× ×=

=

×− = −

=

× =

=

×− = −

=

× =

(iii)(iii)(iii)(iii)(iii) Bending moment diagramBending moment diagramBending moment diagramBending moment diagramBending moment diagramTaking top face of the beam as reference face and outer face of the columns as reference face.Taking simply supported span AB

Maximum moment under the load = ( ) × × =

Fixed moment under the load =

− + × = 42.76925 kNm (hogging)

∴ Resultant moment under the load = 60 – 42.7692 = 17.2308 kNm (sagging)Taking simply supported span BC

Maximum moment = ( ) × =

Fixed moment at centre =

− + × = 73.816 (hogging)

∴ Resultant moment at the centre of span BC = 120 – 73.816 = 46.184 kNm (sagging)Taking simply supported span CD Maximum moment = 67.5 kNm (sagging)


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19Workbook

Fixed moment at the centre = ( ) − =

∴ Resultant moment at the centre of span CD = 67.5 – 36.79 = 30.71 kNm (sagging)

T6 : Solution

There are no external loads acting on the portal frame except the sway force of 100 kN from left to right. Soonly the sway analysis will be carried out.Distribution Factors:Distribution Factors:Distribution Factors:Distribution Factors:Distribution Factors:

=

× =

× =

× =

I I

I

I I

I I

I

I I

A

BC

4 m

100 kN

D4 m 3 m

θ

θ

∆

Let the frame ABCD deflect to the position AB 1 C 1 D due to the sway force.Let, CC 1 = ∆


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21Workbook

= –43.75( →) = 43.75 kN ( ←)Let the sway force S be acting from left to right and then

HA + H D + S = 0–25 – 43.75 + S = 0

⇒ S = 68.75 kN ( →)For a sway force of 68.75 kN, the sway moment are as per Col. (a)For the actual sway force of 100 kN, the corresponding sway moments will be

×

Taking tension as positive and compression as negative.


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θ

θ

FBD for Axial forceFBD for Axial forceFBD for Axial forceFBD for Axial forceFBD for Axial force

∵ cos θ = and sin θ =

∴ Axial force in CD = 36.36 cos θ + 63.64 sin θ = 67.272 kN

Axial force diagramAxial force diagramAxial force diagramAxial force diagramAxial force diagram

T7 : Solution

Fixed end moments

M fAB = × =

M fBA = –41.67 kNmM fBC = 41.67 kNmM fCB = –41.67 kNmM fCD = 41.67 kNmM fDC = –41.67 kNm


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T8 : Solution

Analysing the given beam by slope-deflection method.Fixed end MomentsFixed end MomentsFixed end MomentsFixed end MomentsFixed end Moments

= 0; = 0

= – ×

= –

= +

Let the deflection at the hinge be δ,Slope deflection equationsSlope deflection equationsSlope deflection equationsSlope deflection equationsSlope deflection equations

MAB = + δ + θ

= 0 +

δ θ −

MBA = + δ θ + = 0 +

δ θ

MBD = + δ θ + + =

δ θ + −

MDB = + δ + θ + =

δ θ + +

Joint equilibrium conditionsJoint equilibrium conditionsJoint equilibrium conditionsJoint equilibrium conditionsJoint equilibrium conditionsFrom the above figure it is clear that both M BA and M BD are equal to zero

MBA = 0 ...(i)MBD = 0 ...(ii)

Shear equation:Shear equation:Shear equation:Shear equation:Shear equation:

∑Fy = 0VA + VD – W = 0 ...(iii)

VA = – + [∵ MBA = 0]

⇒ VA = –


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Solving (v) and (vi), we get

θBA = and θBD =

∴ MAB = δ θ

= θ × θ

=

× θBA

=

× ×

MAB = WL = WL

and M DB = δ θ + +

=

θ + × θ +

= ( ) θ + θ +

=

+ +

= × + = WL

⇒ MBD = WL

Also, VA = = –

=

∴ VD = − =


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27Workbook

Maximum simply supported BM in BD =( )

= =

Net positive BM at C = – ×

= =


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T1 : Solution

1 . 5

m

1 . 5

m

40 kN

30 kN

2

1

30 kN

H A = 30 kN

R A

3 mX

X

3 m 3 m 3 m

J I H G F

A B C DE

K L M N

Take a section x – x as shown. The resulting free body diagram is as shown below

40 kN

P 1

30 kN

30 kN

P 2 A B

K

J I

Trusses

5


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29Workbook

For reaction R A, take moments about E ,R A (12) + 30 (3) = 40 (9) + 30 (3)

⇒ R A = 30 kNAlso, H A = 30 kN

Take moments about B ,P 1 (3) × 30 (3) = 0

⇒ P 1 = –30 kN = 30 kN (comp)Take moments about I , P 2 (3) = 30 (3) + 30 (3)⇒ P 2 = 60 kN (Tension)

T2 : Solution

w = 5 kN/m

1 0 m

H A = Hh

R A= wL 1 A

B

R B = wL 2

h 2

H B = H

h c = 7.5 m

O

L1 L2

L = 75 m

C

The position of lowest point 0 is given by,

L1 =

∆ − = − × = 25 m

L2 =

∆ − = + × = 50 m

(L – L1 = 75 – 25 = 5 0 m)

H =

ω = = 468.75kN

Equation of cable profile is given by,

y = ω = ×

x x = x

At x = L1 = 25 m, h 1 = 3.33 m

At x

= L2 = 50 m, h 2 = 13.33 mLength of the cable is given by,

S =

+ +

=

+ +

= 77.66 mCable tension at the two ends,

T A = ω + = × + =

T B = ω + = × + =


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T3 : Solution

Using Unit Load MethodUsing Unit Load MethodUsing Unit Load MethodUsing Unit Load MethodUsing Unit Load MethodLet R a and R f be the vertical reactions at A and F .

Taking moment about A,R f × 4 = ( ) × +

∴ R f = ( ) + kN ↑

∴ R a = ↓Considering the joint F we conclude P fb = 0

∴ P fe = ( ) + kN (compressive)Considering the joint A,

P ab = kN (tensile)P fa = 0

Joint B , resolving vertically,

P bc sin 45° =

∴ P bc = kN (tensile)

Resolving horizontally,

P be = cos 45°

= kN (compressive)

Joint C , Resolving horizontally,P cd = cos 45°

= kN (tensile)Resolving vertically,

P ce = cos 45°

= kN (compressive)Joint D , Resolving vertically,

P de sin 30° = 200P de = 400 kN (compressive)

To find the vertical deflection of the joint D , remove the given loading and apply a 1 kN load vertically at D .

The forces in the various members due to the application of a vertical load of 1 kN at D are shown in thetable.


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31Workbook

Taking tension as positive and compression as negative.

Member P PK l K l (m)

AB AFBCCDDEEFBFBEEC

200 2400

–400 6400

–200 2400

200 4800

–200(1+ 3200

12800

–200 2400

200 2400

0 0 8 0

3 3 3 3

3 – 3 4

6 6 2 6

3) 3) 3 3 + 4800

3 + 16000

3 – 3 4

Total

3 3 3 –2 8

3

4

44

–(1+ 4

0 0 4 0

Vertical deflection of D

δ = ∑ l = ( ) + × = 19.085 mmT4 : Solution

Step 1 :Step 1 :Step 1 :Step 1 :Step 1 : Apply external load after removing unit load and find member force.∑F H = 0

⇒ R A + R A + 60 = 0∑F V = 0

⇒ H D = 80 kN∑F D = 0

⇒ –R A + 8 + 80 × 6 = 0

⇒ R A = 60 kN⇒ R D = –120 kNAnalysis of Member forcesAnalysis of Member forcesAnalysis of Member forcesAnalysis of Member forcesAnalysis of Member forces

∑ = 0

P BC – 80 = 0⇒ P BC = 80 kN

∑ = 0

⇒ P BA = 0

∑= 0

⇒ P AC cos θ + 60 = 0

⇒ P AC = −

= –100 kN

⇒ P AC = –100 kN

∑ = 0⇒ P AC sin θ + P AD = 0


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⇒ P AD = – P AC sin θ

= –(–100) × = +80 kN

⇒ P AD = 80 kN

∑ = 0⇒ P DC – 120 = 0⇒ P DC = 120 kNStep 2 :Step 2 :Step 2 :Step 2 :Step 2 : Apply unit load at C in vertical direction after removing external load and find out member forces

∑M D = 0, R ′A + R ′C = 0

∑F V = 0, H ′D = 1∑M D = 0

⇒ –R ′A × 8 + 1 × 6 = 0

⇒ R ′A =

= 0.75

⇒ R ′D = –0.75Analysis of Member ForcesAnalysis of Member ForcesAnalysis of Member ForcesAnalysis of Member ForcesAnalysis of Member ForcesLet θ be the angle between AC and CD .Member AB and BC will be a zero force member

k BC = 0k AB = 0

From

∑ = 0k CA sin θ + 1 = 0⇒ k CA = = –1.25

From ∑ = 0k CD = – k CA cos θ

⇒ k CD = ×

= 0.75

From ∑ = 0k DA = 1

Increase in length of AD= L × ∆T × α= 8 × 0.6 × 10 –5 × 120 m= 5.76 × 10 –3 m


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33Workbook

Hence

Member k (kN) P (kN) L (m) PkL AE (L T)kΔα

AB

BCCDDACA

0

00.75

1 –1.25

0

8012080

–100

6

868

10

200

200200200150

00

13.5 X 10 m16 X 10 m

41.67 X 10

–3

–3

–3

Σ = 71.17 X 10 –3

0

00

5.76 X 100

–3

5.76 X 10 –3

A mm2

( (

1 × ∆ = + × ∆ × α∑ ∑

⇒ ∆ = 76.93 × 10 –3 m = 76.93 mm+ve value of ∆means joint C will have deflection in the direction of unit load i.e. downwards.

T5 : SolutionNo. of redundant = 1

External redundancy = 1Choose R B as redundant

Steps for analysis are :Steps for analysis are :Steps for analysis are :Steps for analysis are :Steps for analysis are :1.1.1.1.1. Remove the redundant and analyze the truss due to external loads.

The member forces due to external loads are as shown in the figure below

D G C

B A

5 kN

10 kN

5 kN

–5 2

5 2 –10

F

10 kN

5

2.2.2.2.2. Remove the external load and apply unit load at B . [i.e. at support location below.]Analyse the truss due to this load. The member forces are as shown below.

D G C

B A

1

1

F

1

–1

–1 –1

2

2

1

–1


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Thus

Member p k pk k 2

AB

BCDG

GD

DA

GF

DF

AF

FC

0

00

0

5

–10

5 2

–5 2

0

0

–1 –1

–1

–1

0

0

2

2

0

00

0

–5

0

0

–10

0

0

11

1

1

0

0

2

2

∑ =

− =∑l l l

⇒ R B =

=∑

=1.875 kN

Member forces are tabulated as shown belowMember forces are tabulated as shown belowMember forces are tabulated as shown belowMember forces are tabulated as shown belowMember forces are tabulated as shown below

AB

BC

DG

GDDA

GF

DF

AF

FC

Member forces = p + kR BMember

0

–1.875

–1.875

–1.875+3.125

–10

+7.07

–4.42

2.65


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Matrix Methods of

Structural Analysis6

T1 : Solution

To generate the first column of the flexibility matrix, apply a unit force at coordinate 1.

δ11 =

=

I I

δ21 = 0

δ31 =

− = −

I I

δ41 = 0To generate the second column of the flexibility matrix, apply a unit force at coordinate 2.

δ12 = 0

δ22 =

=

I I

δ32 =

−− =

I I

δ42 = 0To generate the third column of the flexibility matrix, apply a unit force at coordinate 3.

δ13 =

− = I I

δ23 =

−− =

I I

δ33 =

=

I I

δ43 = 0All the elements of the fourth column of the flexibility matrix are zero, since the beam remains undeflectedwith a unit force is applied at coordinate 4. Hence,


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δ14 = δ24 = δ34 = δ44 = 0The required flexibility matrix [ δ] is given by the equation

[δ] =

− − − −

I

T2 : Solution

A B

2

1

3

L

Stiffness:Stiffness:Stiffness:Stiffness:Stiffness: Amount of force moment required to produce unit deflection rotaton.

Dk = degree of kinematic indeterminacy

[No. of independent displacements (translational and rotational) possible in a structure]

Size of stiffness matrix = k [D k , D k ]

To determine [k] matrix for the beam

[k ] =

k 11 = amount of moment required when far end is fixed =

I

k 12 = moment due to unit settlement = ( ) I

k 12 = k 21 (As matrix is symmetric and validity of Maxwell’s reciprocal theorem)k 13 = 0 (as axial deformation won’t produce)

k 22 = ( ) ↓ I

k 23 = 0 = k 32 (same reason as for k 21 = k 12 )k 13 = k 31

Stiffness Matrix :Stiffness Matrix :Stiffness Matrix :Stiffness Matrix :Stiffness Matrix :

[k ] =

− −

I I

I I


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37Workbook

Flexibility Matrix :Flexibility Matrix :Flexibility Matrix :Flexibility Matrix :Flexibility Matrix :

[δ] =

I I

I I

T3 : Solution

The two independent displacement components are rotation at B and C. Hence degree of freedom is two.Let us assign Co-ordinate to the independent displacement components

1 2

Let us restrain the complete structure and find out the forces developed in the restrained structure in theCoordinate directions.

M FAB = –30M FBA = +30M FBC = –15M FCB = +15

⇒ Net moment in direction (1) = M 1 = 30 – 15 = 15 tmNet moment in direction (2) = M 2 = 15 tm

External moment at joint B and C are zero⇒ [∆] = –[ k ]–1 [P ]Let us give unit displacement in Co-ordinate direction (1) without giving any displacement in direction (2)

⇒ k 11 =

= I

I l

= 0.8 E I

k 21 = I

l = 0.2 E I

Let us give unit displacement along (2) without giving displacement along (1)⇒ k 12 = 0.2 E I

⇒ k 22 = 0.4 E I

Hence, stiffness matrix is

[k ] =

I I

I I

⇒ [∆] =

− I I

I I

⇒ [∆] = − − I


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Once the displacement are known, the end moments can be calculated using slope deflection equations

M AB =

θ + θ + θ −

I

l l

M AB = − + − − I

I = –32.1 tm

M BA =

− × + + −

I

I = 25.7 tm

M BC =

× − + − −

I

I I = –25.7 tm

M CB = 0The free body diagram will be as shown below

24 t 12 t 32.1 25.725.7

= 11.36 t = 3.43 t

12.64 t 8.57 t 12 – (32.1 – 25.7) 6 –25.7

10 10

The loading on the beam is as shown below:

24 t 12 t

BC

5 m 5 m 5 m 5 m

A

12.64 t 19.93 t 3.43 t

T4 : Solution

As the frame cannot sway, rotation at B and C are the unknown displacements.Let as assign Co-ordinates at (B) and (C) as (1) and (2).

B

C

A

1 2

All the joints are locked and moments at (1) and (2) are calculated due to external

M 1 =

× × ×− = –5.6 tm

M 2 = 8 tmas no external moments are acting at (1) and (2) hence,

[∆] = –[ k ]–1 [P ]


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39Workbook

Deriving Stiffness matrixDeriving Stiffness matrixDeriving Stiffness matrixDeriving Stiffness matrixDeriving Stiffness matrixWhen unit rotation is given in direction (1) without giving rotation in direction (2).

k 11 =

+ I I

= 2.8 E Ι

k 21 = E Ι

When unit displacement is given in direction (2) without giving displacement in direction (1) we have.⇒ k 12 = E Ι

⇒ k 22 = 2 E Ι

Hence, the stiffness matrix is

[k ] =

I I

I 2 I

⇒ [∆] = –[ k ]–1[P ]

⇒

θ θ

=

−

I

I

M AB =

− + +

I

I = –1.93 tm

Note :Note :Note :Note :Note : M FAB = − × ×

= –3.6 tm

M BA =

× + +

I

I = 5.74 tm

M BC = ( )

× − + −

I

I I = –5.74 tm

M CB =

− × + +

I

I I = 0

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Structural Analysis TYS Exp 249