chapter 4: analysis of statistically indeterminate...

Structural Analysis

Chapter 4:

Analysis of statistically indeterminate structure

(Beam and Frame)

Statistical indeterminate beams/frame

• Beam are statistically indeterminate generally because of their support system

– which have number untraceable (reaction to support) more than 3

– static equilibrium is not the solution

• Method to solve the problem:

– Slope deflection method

– Moment distribution methodDetermine reaction force, further shear strength and bending moment

Slope deflection method

• The shear force and moment at the ends beam element are related to the end displacements rotations.

• Procedure

– Determine the fixed moment, moment result deflection and moment result deposit/support shift

– Determine slope at support

– Determine moment at support

• Slope – deflection equation

θ1

θ2

M1 M2

δ1 δ2

L

• Therefore bending moment diagram can be draw

• At any x cut, M=M1(1-x/L)-M2x/L

+veM1(1-x/L)

M1

M2

x

-ve

M2x/L

• Relationship between curve and moment for beam

)3(6/)6/2/(

)2(2/)2/(

)1(/)/1(,

3

2

32

1

3

2

2

1

212

2

2

2

BAxLxMLxxMEIyor

ALxMLxxMdx

EIdyor

LxMLxMMdx

yEIdTherefore

Mdx

yEId

• A and B can be defined using boundary method

When x= 0, y=0 placed in equation 3, B=0Also when x=0, dy.dx=θ1 in equation 2, A= -EIθ1

When x=L, y=δ, placed in equation (3)-EIδ = M1L

2/3 - M2L2/6 - EIθ1L equation (4)

Also when x = L ; dy/dx =θ2 place in equation (2)-EIθ2 = M1L/3 - M2L/2 - EIθ1 equation (5)

Rearrange equation (4) and (5)

M1=4EIθ1/L + 2EIθ2/L - 6EIδ/L2 equation (6)M2=4EIθ2/L + 2EIθ1/L - 6EIδ/L2 equation (7)

Or it can be write (simplified equation)

L

L

EIM

L

EIM

)32(2

)32(2

122

211

• Effect of Fixed end moment (FEM)(Page 4 of your lecturer notes)

– Defined as moment resultant at end to end outside tax incidence member that imposed when both its member is fixed

– When load is applied to the beam will produced FEM at both end

• Include effect of the FEM

)32(2

)32(2

1212

2121

L

EIMM

L

EIMM

F

F

Step of solution

• Write the slope for each end moment member

• Equilibrium equation obtainable to each joint when the total of joint member is zero (ΣM=0)

• Boundary condition placed at fixed end slope, where the displacement equal to zero

• After slope equation for value determined, all end moment defined with include slope value stated into equation

• Further member can be sorted for calculation reaction, shear strength and bending moment

• From here, shear force diagram and bending moment could be sketched

Example 4.1

• Determine the moment value and shear force to each support and draw shear force diagram (SFD) and bending moment (BMD) or structure beam below. EI value is constant

1. Finding the fixed end moment (FEM)

2. Check boundary condition

A = 0, Δ = 0 = 0, B = ?

3. Slope deflection equation

4. Equilibrium equation

5. End moment (From slope equation)

6. Shear body diagram and bending moment Diagram

i. Reaction

Exercise 4.1

• Determine the reaction at each support using slope deflection method and draw shear force diagram (SFD) and bending moment (BMD) for structure beam below. EI value is constant.

20 kN 20 kN

12.5 m 12.5 m 15 m 10 m

A B C

Exercise 4.2

• Determine moment for beam below using slope deflection method. Beam also have a 40mm displacement at support B and 30mm at support C. (EI = 400kN.m2)

6 kN/m30 kN

B C

4 m 4 m 3 m 3 m

A D

Moment distribution method

• Involve distribution moments to joint repetitively

• Accuracy depend on the number of repetition

• Value is dependant to

– fixed end moment (FEM)

– Factor bring side (Carry over factor-COF)

– Strength (Distribution factor-DF)

• Concept: determining element strength each structure which is depends on:

– Modulus flexibility/elasticity (E)

– Moment of inertia (I)

– Length of element/span (L)

MBA

A B

θAMAB

AAB

BAAB

L

EIM

L

EIFEMM

4

)32(2

L

EIM AB

4 4EI/L I stiffness where EI/L is stiffness factor

θA can be define using slope deflection method

ABBA MM2

1 Where ½ is carry over factor (COF)

K

M

K

KKK

MMMMM

KL

EM

KL

EM

KL

EM

AA

A

AADACAB

ADACABAA

AADA

AD

AACA

AC

AABA

AB

)(

)(

4

4

4

3

3

2

2

1

1

B C

D

A

MA

E3,I

3,L

3

• Replace θA = MA/ΣK into moment at A (MA) equation:

AAD

AD

AAC

AC

AAB

AB

MK

KM

MK

KM

MK

KM

Expression Ki/ΣK known as distribution factor (DF), and ΣK<1, where:

Ki = Stiffness of member

ΣK = Total of stiffness for all member

• Distribution factor (DF) at the end fixed

• Distribution factor (DF) at the end pinned

A

KAB

∞

A

KAB

0

0)(

)(

AB

ABAB

K

KDF

1)0(

)(

AB

ABAB

K

KDF

Step of solution(Moment distribution method)

• Calculate distribution factor (DF)

• Fixed end Moment (FEM @MF)

• Moment distribution

– Table of distribution

• Finding the reaction

• Draw Shear Force Diagram & Bending Moment diagram

Joint A B C

Member AB BA BC CB

CF

DF

FEM

End moment

L

EIK

K

KDF

4,

Example 4.3

• Draw BMD and SFD for the beam below

1. Distribution factor (DF)

Joint Member K(4EI/L) ΣK DF(K/ΣK)

A AB 4EI/4.5 4EI/4.5+0 1

B

BA4EI/4.5

(4EI/4.5)+(8EI/3.5) =3.17EI

0.28

BC4E(2I)/3.5

0.72

C CB4E(2I)/3.5 8EI/3.5+∞

0

2. Fixed end moment (FEM)

3. DistributionJoint A B C

Member AB BA BC CB

CF 0.5 0.5 0.5 0

DF 1 0.28 0.72 0

FEM -45 +45 -66.86 +66.86

Distr. +45 +6.12 +15.74 0

COF +3.06 +22.5 0 +7.87

Distr. -3.06 -6.3 -16.2 0

COF -3.15 -1.53 0 -8.1

Distr. +3.15 +0.43 +1.1 0

COF +0.22 +1.58 0 +0.55

Distr. -0.22 -0.44 -1.14 0

COF -0.22 -0.11 0 -0.57

Distr. +0.22 +0.03 +0.08 0

COF +0.02 +0.11 0 +0.04

Distr. -0.02 -0.03 -0.08 0

End moment 0 67.36 -67.36 +66.65

+45 – 66.86 = -21.8621.86 X 0.28= +6.12

Exercise 4.3

• Draw SFD and BMD using moment distribution method

A

12.5m 12.5m 15m 10m

20kN 20kN

B C

1. Distribution factor (DF)

Joint Member K(4EI/L) ΣK DF(K/ΣK)

A AB 4EI/25 4EI/25+∞ 0

B

BA4EI/25

8EI/25

0.5

BC4EI/25

0.5

C CB4EI/25 4EI/25+∞

0

2. Fixed end moment (FEM)

kNmFEM

kNmFEM

kNmFEM

kNmFEM

CB

BC

BA

AB

7225

)10()15(20

4825

)10)(15(20

5.6225

)5.12()5.12(20

5.6225

)5.12)(5.12(20

2

2

2

2

2

2

2

2

3. Distribution

Joint A B C

Member AB BA BC CB

CF 0 0.5 0.5 0

DF 0 0.5 0.5 0

FEM -62.5 +62.5 -48 +72

Distr. 0 -7.25 -7.25 0

COF -3.63 0 0 -3.63

Distr. 0 0 0 0

End moment -66.13 +55.25 -55.25 +68.37

+62.5 – 48 = +14.5-14.5 X 0.5

4. Reaction

20kN

RA RB1

66.12 55.25

20kN

RB2 Rc

55.25 68.38

)(57.9;0

)(43.10

025.55)5.12(2012.66)(25

:0

1

kNRF

kNR

R

M

BY

A

A

B

)(47.7;0

)(53.12

0)(2538.68)15(2025.55

:0

2

kNRF

kNR

R

M

BY

C

C

B

Span A-B

Span B-C

4. SFD and BMD20kN 20kN

+

-

+

-

10.44

9.56

7.47

12.53

AB

C

C

+ +- - -

66.12

=66.12+10.44(12.5)=64.31

55.26

68.38

AB

C

=55.25+7.47(15)=64.31

AB

RA=10.43 RB1=9.57 RB2=7.47 RC2=12.53

Moment distribution method(Modify stiffness)

• To facilitate moment distribution process to be more precise

• Continues beam with either or both of them pin or roller

• Moment to pin or roller which is located at the end beam must null/empty

– No need to process bring side to support

– Modify should be done upper factor distribution

• Total moment at the end span , MAB = 0

AB

M

K1CK2

• At B support

For span beam have end fixed and one more end are joint with pin, stiffness for span have extension pin stated is ¾ from original stiffness while calculation factor distribution.

Example 4.5

• Determine end moment after moment distribution using the stiffness modifies method.

1. Distribution factor (DF)

Joint Member K(4EI/L) ΣK DF(K/ΣK)

A AB(3/4(4EI/4.5))

=0.667EI0.667+0 1

B

BA(3/4)4EI/4.5 =0.0667EI

0.0667EI+2.286EI= 2.953

0.23

BC4E(2I)/3.5=2.286EI 0.77

C CB4E(2I)/3.5=2.286EI

2.286EI+∞0

• Fixed end moment

• Moment distribution

Joint A B C

Member AB BA BC CB

CF 0.5 0 0.5 0

DF 1 0.23 0.77 0

FEM -45 +45 -66.86 +66.86

Distr. +45 +5.03 +16.83 0

COF 0 +22.5 0 +8.42

Distr. 0 -5.18 -17.32 0

COF 0 0 0 -8.6

Distr. 0 0 0 0

End moment 0 67.35 -67.35 +66.68

Moment distribution method (rigid non sway Frame)

• In general, approach steps involved is the same as to beam. More practical by using modified stiffness.

• Rigid non-sway frame is the state where deformation / rigid framework movement will not cause joint or extension or join framework shift (Δ= 0).

• Case-study to rigid frames this make equivalent as in the case beam.

• Shear force diagram

• Bending moment diagram

(+)

(+) (+)

(+)

(+)

(+)

(+)

(+)

(+)(+)

(+) (+)

(+)

(+)

(+)

(+) (+)

(+)

(+) (+)

(+)

(+)

Example

Using moment distribution method, determine the end moment, reaction and draw shear force diagram and bending moment diagram for the frame below (EI constant)

100kN

20kN/m 4m

4m2m

A B

C

1. Distribution factor (DF)

Joint Member K=(4EI/L) ΣK DF(K/ΣK)

A AB 4EI/6 4EI/6 +∞ 0

B

BA 4EI/6 4EI/6 + 4EI/4= 10EI/6

0.4

BC 4EI/4 0.6

C CB4EI/4 4EI/4+∞

0

2) Fixed End Moment (FEM)

3) Moment distribution

kNmwl

FEM

kNmwl

FEM

kNmL

bPaFEM

kNmL

PabFEM

CB

BC

BA

AB

7.2612

420

12

7.2612

420

12

4.446

42100

9.886

42100

22

22

2

2

2

2

2

2

2

2

Joint A B C

Member AB BA BC CB

COF 0 0.5 0.5 0

DF 0 0.4 0.6 0

FEM -88.9 +44.4 -26.7 +26.7

Distr. 0 -7.1 -10.6 0

CO -3.6 0 0 -5.3

Distr. 0 0 0 0

End moment -92.5 +37.3 -37.3 +21.4

Example 4.8

• Draw shear force and bending moment diagram to rigid frame structure as below.

2

3

44

4

34

4

3

4

3EI

L

EIKKBC

stiffness equation for span with pinned end

At joint B, the total of moment is 28.33-5.33+0 = +23

Therefore, Distribution value is

BA-23 X 0.25 = -5.75BC-23 x 0.5 = -11.5BD-23 x 0.25 = -5.75

Moment distribution method (rigid frame with sway)

• Sway – framework experience some movement

• Cause moment lurch to which shift happen

• Moment due to sway (MS) with shift could be phrased as

Δ P2

P1

Δ

A

B C

D

2

.6

L

EI

2

.6

L

EI

Δ

• If support end is pin or roller, then Moment due to sway (MS) becomes

• Sway to rigid frame will occur because of:

A

B C

D

0

32

DCS

CDS

M

L

EIM

Horizontal load Unsymmetrical vertical load

Unsymmetrical frame system (geometry or material (I))

• Solution: 2 stage and combined (overlap principle)

Unequal support

• First stage : No sway cause by constraint

• Second stage : sway framework due to Q counter direction to constraint discharge

Moment distribution can do as usual (FEM) Final moment consider as M1 (moment no sway)

horizontal to extension only (MS) Final moment consider as M2 (moment with sway)

• Stage 1 + Stage 2 = Actual moment

Actual moment is , M = M1 + X.M2 where X is correction value If F case from no sway and Q is from the sway case:

F – X.Q = 0X = F/Q

Example 4.9

• Draw shear force and bending moment diagram to rigid frame structure as below. The value of E is constant and assume EIΔ = 160

• Calculation for F value

*F = 40 + 10.40 – 6.75 = 43.65 kN

• Calculation for Q value

*Q = -(-22.24) – (-8.51) = 30.75 kN