1

BYDR. MAHDI DAMGHANI

2016-2017

Structural Design and Inspection-Energy method

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Suggested Readings

Reference 1 Reference 2

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Objective(s)

Familiarisation with force method (flexibility method)

Ability to solve indeterminate structures

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Introduction

Refer to chapter 5 of Reference 1Refer to chapter 4 of Reference 2This method is for solving indeterminate

structuresFocuses on the solution for the system

internal forces In this method the governing equations

express compatibility requirements in terms of the redundant forces and component flexibilities

Forces are determined first and then strains and displacements are found

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Procedure

This method is also called force method because: Internal forces are found

first Then displacements and

strains are calculated

1 •Determine degree of redundancy of the structure (NR)

2 •Remove NR members of the structure

3 •Represent each removed member by its associated force

4 •Calculate displacement for each force using unit load method

5 •Apply compatibility of displacements

6 •Solve a set of linear equations to calculate redundant forces

7 •Calculate displacements and strains

Let’s see this in some examples

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Example

Determine internal forces in the truss structure shown below. The cross-sectional area A and Young’s modulus E are the same for all members.

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Solution

Is the structure determinate or indeterminate?

What is the degree of redundancy? 3 members (m) 6 reaction forces (r) 4 joints=8 equilibrium

equations (2j)

NR=m+r-2j=1 At each node we have: 0,0 yx FF

1 2 3

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Solution

The redundant member is removed

The redundant member is replaced by a force R

Now the structure is determinate and can be solved under two forces P and R

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Solution

A B C

O

P

P

O

O

R

R1

2 31 32

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Solution

FA FC fA fC FAfAL/EA FCfCL/EA

P

A C

O

FCFA θ

cos2P

cos2P

cos21

cos21

3cos4EAPL

3cos4EAPL

32

1

)2(

cos2EAPL

AELfF

i ii

iiiO

1

A C

O

fCfA θ

cos2coscos

sinsinPFPFF

FFFF

ACA

CACA

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Solution

FA FB FC fA fB fC FAfAL/EA FBfBL/EA FCfCL/EA

cos2R

cos2

1

3cos4EARL

3cos4EA

RL

EARL

EARL

AELfF

i ii

iiiO

33

1

)3(

cos2

cos2R

A C

O

R

RFCFA

A C

O

1

1fCfA

1cos2

1

EARLR

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Solution

0)3()2(OO

0

cos21

cos21

33 EAPL

EARL

A B C

O

P

Rθ

3cos21

1PR

PRFFF

FFFFF

CAyO

CACAxO

coscos0

sinsin0

cos2RPFA

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1

)2(

cos2EAPL

AELfF

i ii

iiiO

EARL

EARL

AELfF

i ii

iiiO

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1

)3(

cos2

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Example

Calculate the forces in the members of the truss. All members have the same cross-sectional area A and Young’s modulus E.

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Solution

How many degrees of redundancy are in the structure? m=7 r=5 j=5

NR=m+r-2j=2

6

1

2

34

5 7

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Solution

2

1

4

R2

3

X1

X1

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Solution

2 4

R2

3

X1

X1

0

0)4()3()2(

)4()3()2(

CCC

ADADAD

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Solution

2

F=10

00

0

-14.14

0

11f=-0.71

0-0.71

0-0.71

1

EAAELfF

i ii

iiAD

1.276

1

)2(

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Solution

11f=-0.71

0-0.71

0-0.71

1

1

7

1

)3( 32.4 XEAAE

LfFi ii

iiAD

3

X1

X1

X1

X1

F=-0.71X1

0-0.71X1

0-0.71X1

X1

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Solution

11f=-0.71

0-0.71

0-0.71

1

2

6

1

)4( 7.2 REAAE

LfFi ii

iiAD

4

R2

R2

F=-2R2

0

R2R2

1.41R2 -1.41R2

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Solution

Therefore our first linear equation becomes;

Now we need another equation to solve two unknowns X1 and R2

This time we make sure vertical displacement at node C becomes zero

0)4()3()2(ADADAD

07.232.41.27 21 RX

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Solution

2

F=10

00

0

-14.14

0

f=-2

-1.410

1 1

1.41

EAAELfF

i ii

iiC

11.486

1

)2(

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Solution

1

7

1

)3( 7.2 XEAAE

LfFi ii

iiC

3

X1

X1

X1

X1

F=-0.71X1

0-0.71X1

0-0.71X1

X1

f=-2

-1.410

1 1

1.41

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Solution

2

6

1

)4( 62.11 REAAE

LfFi ii

iiC

R2

F=-2R2

-1.41R2

0

R2R2

1.41R2

4

R2

f=-2

-1.410

1 1

1.41

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Solution

Therefore our second linear equation becomes;

Now by solving the two linear equations simultaneously;

062.117.211.48 21 RX

0)4()3()2(CCC

062.117.211.48 21 RX

07.232.41.27 21 RXkNRkNX 15.3,28.4 21

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Example

Determine the reactions of the continuous beam in the beam structure below, using force method.

L L

w

A B C

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Solution

Is the structure determinate or indeterminate?

w

What is the degree of indeterminacy?3 equations (knowns)

A B C

0

0

0

z

y

x

M

F

F

4 reactions (unknowns)

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Solution

I intentionally choose support at B as the redundant

The released structure is then loaded by redundant member

Vertical displacement at B must be zerow

A B C

RA RC

RB=XB

0B

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Solution

To determine redundant, we superimpose deflection due to external load and a unit value of the redundant multiplied by the magnitude of redundant XB w

A C

A BC

wLwL

0B

10.5 0.5

BB

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Solution

w

A C

w

A BC

wLwL

0B

10.5 0.5

BB

00 BBBB X

EIL

EILw

BBB 4821,

38425 34

0

wLXR BB 25.1

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Solution

w

A C

w

A BC

wLwL

0B

XB1.25wL 1.25wL

BBBX

wLwLwLRA 8325.15.0 wLwLwLRC 8

325.15.0

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Solution

w

A B C

(3/8)wL(5/4)wL (3/8)wL

(3/8)wL(5/8)wL

(-5/8)wL (-3/8)wL

Shear

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Solution

w

A B C

(3/8)wL(5/4)wL (3/8)wL

(9/128)wL2

(-1/8)wL2

Moment

(9/128)wL2

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Example

How do you solve the following?

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Solution

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Example

How about this one?

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Solution

37

Example

How about this one?

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Solution

01221111 RRC

02222112 RRC

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Q1

Determine member forces and reaction at supports of truss structure shown below. EA is constant for all members.

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Q2

Determine forces in the truss structure shown below. EA is constant for all members.

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Q3

A continuous beam ABC is carrying a uniformly distributed load of 1 kN/m in addition to a concentrated load of 10 kN. Draw bending moment and shear force diagram. Assume EI to be constant for all members.

A B C

1kN/m

10kN

10m5m5m

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Q4

The beam ABC shown in Figure is simply supported and stiffened by a truss whose members are capable of resisting axial forces only. The beam has a cross-sectional area of 6000mm2 and a second moment of area of 7.2×106 mm4. If the cross-sectional area of the members of the truss is 400mm2, calculate the forces in the members of the truss and the maximum value of the bending moment in the beam. Young’s modulus, E, is the same for all members.

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Q5

Draw the quantitative shear and moment diagram and the qualitative deflected curve for the Frame shown below. EI is constant.