Structures and Materials- Section 8 Statically Indeterminate Structures

Section 8

Statically Indeterminate StructuresSo far, all of the exercises presented in this module have been statically determinate, i.e. there have been enough equations of equilibrium available to solve for the unknowns. This final section will be concerned with statically indeterminate structures, and two methods used to solve these problems will be presented.

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Contents• Statically Determinate and Indeterminate Structures

• Statically Indeterminate Structures 1. Stiffness Method

• Stiffness Method – Summary

• Stiffness Method - Example

• Statically Indeterminate Structures 2. Compliance Method

• Statically Indeterminate Structures Compliance Method – structure 1

• Statically Indeterminate Structures Compliance Method – structure 2

• Compliance Method – Summary

• Compliance Method- Example

• Compliance Method- Solution

• Compliance Method- Solution Structure 1

• Compliance Method- Solution Structure 2

• Temperature Effects

• Temperature Effects – Stress

• Temperature Effects – Example

• Temperature Effects – Solution

• Credits & Notices

Statically Determinate and Indeterminate Structures

Statically determinate structures

No. of Equilibrium Equations = No. of unknowns

In this situation we can are determine the unknown forces by using theprinciples of statics to determine the unknown forces, e.g. drawing free bodydiagrams and solving equilibrium equations.

Statically indeterminate structures

We cannot determine all the unknown forces using the principles of statics.

No. of Equilibrium Equations < No. of unknowns

2D system: 2j < m + r3D system: 3j < m + r System is statically indeterminate because we either have too many membersor over stiff support conditions giving too many reaction forces. To solve we need extra information. This information comes from the geometric characteristics of deformation during loading which gives additional equations.

Statically Indeterminate Structures1. Stiffness Method

A

C

P

L

B D This 2D structure has: j=4, m=3 and r=6

(Each joint B, C and D has two reactions)

2j < m+r 8 < 9

Structure is statically indeterminate

(Each bar has area A, Young's modulus E

Length LAC = L)

Free body diagram of joint A:

2PAB cos+PAC =P (Eqn 1)

PAB sin = PAD sin (Eqn 2)

We need more information to solve problem

A

P

PAB PAC

B PAD

Statically Indeterminate StructuresStiffness Method

Under load P the truss has deformed.

Member AB has extension 1

Member AC has extension 2

Assuming deformation is small (≈’)

AB cos = AC A’B cos = A’C

Hence we can write

1 =2 cos (Eqn 3)

We can obtain further information using

A

C

P

B D

2

1

(Eqn 4 - Hookes Law)PL

EA

A’

’

Statically Indeterminate StructuresStiffness Method

A

C

P

B D

2

1

1

2

1 2

2

2

2

3

Member AB cos

Member AC

Using Eqn3

cos coscos

cosUsing Eqn1 and Eqn2

2 cos 2 coscos

cos

2cos

AB AB AB

AC AC AC

ACAB

ABAC

ABAB AC AB

AB AD AB AD

P L P L

EA EAP L P L

EA EA

P LP L

EA EAP

P

PP P P P P

PP P P P

3

1

2cos 1AC

PP

A

P

PAB PAC

B PAD

Stiffness Method - Summary

1. Establish equilibrium equations (Eqns 1 and 2)

2. Select a suitable displacement () as the unknown quantity

3. Consider geometric characteristics of deformation (Eqn 3)

4. Establish constitutive relationship (Eqn 4 – Hooke’s Law) in terms of

5. Relate constitutive relationship to equilibrium equations

This method is also known as the consistent deformation method or compatibility method as the constitutive relationship must be compatible with the equilibrium equation.

Stiffness Method - Example

A channel metal truss consists of two aluminium vertical bars A, 4m, and B, 5 m with a horizontal steel bar C (6 m), as shown.

Bar C is rigid (i.e. no bending considered) and the bars are hinge connected. All bars have rectangular cross sections 80 mm x 40 mm for bar A and 50 mm x 20 mm for bar B.

A load P of 20 kN is hung from the bar C at such a place that the bar remains horizontal. The Young’s moduli of bars Al: 70 GPa Steel: 204 GPa Determine:

(i) Axial stress in each vertical bar(ii) Vertical displacement of bar C(iii) Horizontal position of the external load P with respect to point O.

5 m Bar B

Bar A

Bar C

P

6m

4m

s O


Equilibirum equation Displacement of each bar

20

Displacement

4000 5000

70000 80 40 70000 50 20 8 2

44

Substituting in equilibrium equation

4

A A B BA B A B

A B

A B

A B A B

AA B B

B B B

P L P LP P P kN

EA EA

P P P P

PP P P

P P P P

4

164A

A A

kN

PP P P kN

Equilibrium of Bar C: Forces PA PB and P

Bar A elongated by PA by amount A

Bar B elongated by PB by amount B

Bar C is remains horizontal hence A=B


-2

-2

B C

B C

16000Stress in each bar 5 Nmm 5 MPa

80 404000

4 Nmm 4 MPa50 20

16000 4000 2Displacement of bar 0.29 mm

70000 80 40 7

Moments about O (LH end of C) P .L =P s

P .L 4000 6000 2 600s=

P 20000 1

A

B

A A

A

P L

EA

1200 mm 1.2 m

Load P is applied 1.2 m along C from LH end

Statically Indeterminate Structures2. Compliance Method

A

C

P

L

B D This 2D structure has: j=4, m=3 and r=6

2j < m+r 8 < 9

Structure is statically indeterminate.

To make structure statically determinate we need to remove a redundant reaction. Each joint B, C and D has two reactions (vertical and horizontal force) so we can remove one reaction force and problem is now statically determinate. We can remove vertical reaction at joint C.

2j = m+r 8 = 8

A

C

P

L

B D

Statically Indeterminate Structures2. Compliance Method

A

C

P

B D

A

C

P

B D

C B D

R

= +

Statically indeterminate Statically determinate Statically determinate

structure structure 1 structure 2

Statically Indeterminate StructuresCompliance Method – structure 1

A

C

P

L

B D

A

P

PAB PAC

B PAD

AC

AB AD AB AD

AB AC AB

AB AD AC

Vertical equilibrium at joint C

P =0


P sin = P sin P = P

P2P cos + P =P P =

2cos

We have determined the unknowns P , P and P


V1 is vertical displacement at joint A

So vertical displacement at joint C = V1

because there is no load in member AC

1

1 2

11 1 1

11 3

Extension of member AB =

cos 2 cos

coscos

cos 2 cos

U

AB AB ABU

UU V V

UV

P L P L PL

EA EA EA

PL

EA

A

C

P

B D

v1

u1


A

C B D

V2

U2

R

A

PAB R PAD

AC

AB AD AB AD

AB AB

AB AD AC

A

Vertical equilibrium at joint C

P =R


P sin = P sin P = P

-R2P cos + R =0 P =

2cos

We have determined the unknowns P , P and P

(Note sign for P

B )


A

C B D

V2

U2

R

2 2

2 2 2

22 3

Extension of member AC =

-RExtension of member AB =

2 cos

is vertical displacement at joint A ( cos )

cos 2 cos

Vertical displacement at joint C

AC ACAC

AB ABU

V U V

UV

P L RL

EA EA

P L L

EA EA

RL

EA

2V AC

A

PAB R PAD

Statically Indeterminate StructuresCompliance Method

1

Displacement of joint C is zero i.e. 0

There are 2 contributions to the displacement at C.

In structure 1 joint C displaced vertically downwards

In structure 2 joint C displaced vertically upward

c

v

2

1 2

3 3

3

3

s

0 ( )

2 cos 2 cos

2 cos

!!1 2cos

AC v

V AC vc

AC

PL RL RL

EA EA EA

P R R

PR P R

A

C

P

L

B D

Statically Indeterminate StructuresCompliance Method

A

C

P

L

B D

AB

AB

3

3

3

2

AB 3

P RP

2cos 2cos

2cos P

1 2cos

1 2cos 1

1 2cos

cosP

1 2cos

P R

PP

P

P

Compliance Method - Summary

The above procedure of the compliance method is summarised in the following four steps:

1. Establish equilibrium equation2. Remove enough reactions and designate them as redundant, to reduce a

statically indeterminate structure to a statically determinate one3. Calculate the displacements at the redundant due to actual loads4. Apply redundant load only to find the displacements5. Use superposition to add the displacements due both to the actual loads

and the redundant loads. The total displacements must be zero at the redundant supports.

This method is also known as the superposition method.

Compliance Method- Example

Two steel and aluminium tubular components of a length L are fitted concentrically. They are loaded with a compressive force P through rigid end plates as shown. By using compliance method determine:

1. The shortening of the assembly 2. Compressive forces in steel cylinder and aluminium tube

P

Al St Al Al St L

Cross sectional areas:

Al = AAL

St = AST

Compliance Method- Solution

This problem is statically indeterminate – The equilibrium equation is

PAL + PST = -P

There are no other equations from force equilibrium. We can solve this problem using the ‘compliance method’ by removing the reaction above the steel OR the aluminium. In my solution I remove the reaction force above the steel portion

P

Al St Al Al St L

Cross sectional areas:

Al = AAL

St = AST

Compliance Method- Solution

P

Al St Al L

P

Al St Al

Al St Al

R

= +

Statically Structure 1 Structure 2 Indeterminate

Structure has been divided into 2 structures. These 2 are solved using statics.

Compliance Method- SolutionStructure 1

P

Al St Al

1 1

11

1

Force equilibirum

(compression) 0

i.e. all of P is transmitted thorough the AL

Extension of Aluminium member =

Displacement at top of steel cylinder is t

AL ST

AL

AL ALAL

AL AL AL AL

RST

P P P

P L PL

E A E A

1

he

same as the extension of the AL member.

(upwards)RST AL

AL AL

PL

E A

Compliance Method- SolutionStructure 2

2

2

Force equilibirum - (compression)

The system is in equilibrium and so there is a reaction

(shown red) at the top of the Aluminium. The (total)

reaction = to maintain equilibrium.

Exten

ST

AL

P R

P R

R

22

22

2

2 2 2

sion Steel member =

Extension AL member =

Displacement at top of steel cylinder

(downwards)

ST STST

ST ST ST ST

AL ALAL

AL AL AL AL

RST

RST AL ST

AL AL ST ST

P L RL

E A E A

P L RL

E A E A

RL RL

E A E A

Al St Al

R R/2 R/2

Compliance Method- SolutionSuperpose 2 solutions

2

1

1 2

Displacement at top of steel cylinder = 0 i.e.

0 0

0

RSTi

i AL AL ST ST AL AL

AL AL ST ST AL AL

ST ST AL AL ST ST

ST ST

ST ST AL AL

ST STST ST ST

ST ST

RL RL PL

E A E A E A

RL RL PL

E A E A E A

R E A R E A P E A

P E AR

E A E A

P E AP P P R

E A E

1 2

AL AL

AL ALAL AL AL

ST ST AL AL

A

P E AP P P P R

E A E A

P

Al St Al L

Temperature Effects

So far only mechanical loads have been considered at room temperature (T0). If a bar is heated up, even without the involvement of mechanical loads, the bar will deform or expand. For the bars made of isotropic and homogeneous material, such expansion will take place in all three dimensions. Imagine that a simple rectangular bar of length L is heated to an arbitrary temperature T (>T0). A uniform expansion by an amount of

in which is known as the coefficient of linear thermal expansion (material constant). Unit: 1/0C (the reciprocal of degrees Celsius). Recalling definition of the strain gives thermal strain as:

0T T Tx y z T T

L

0L T T

Temperature Effects

Thermal strain is a dimensionless and is positive in expansion and negative in contraction. There is no shear thermal strain or distortion. Thermal strain of a moderate amount is reversible and disappears when temperature source is removed (elastic behaviour).

The mechanical properties of a material do not change when temperature fluctuates moderately. When both thermal and mechanical loads are present, its overall strain is calculated by

0

If loading is tensile

T

T TE

Temperature Effects - Stress

Thermal strain does not produce stress if a structure is not constrained as in the case of statically determinate structures. If a structure is constrained like statically indeterminate structures, thermal stress will be developed and is calculated by

A Young’s modulus decreases when the increase of temperature becomes very significant. It is noticeable that thermal stress doesn’t depend on the cross-sectional area unlike the mechanical stress.

0T TE E T T

Temperature Effects - Example

Consider a mild steel bar AB completely fixed at both ends as shown in the figure. The length of the bar is L and the cross-sectional area is A. The bar is uniformly heated up to 60 0C from the room temperature of 20 0C.

E = 220 GPa and =12e-6 oC-1

Determine the maximum thermal stress developed in the bar.

T

L

R

R

B

A

Temperature Effects - Solution

We are unable to evaluate the value of the reaction force using statics – this is a statically indeterminate problem.

The internal force is PAB.

We can solve the problem by using the compliance method. To do this we remove one of the reactions and allow free expansion.

We can then apply a load R to give a displacement equal to the expansion. This force will be the required force allowing us to calculate the thermal stress.

T

L

R

R

B

A

Temperature Effects - Solution

ST1 ST2

1 0

2

1 2

3 6

ST1: 40

ST2:

Displacement at A is zero

0 40 0

40

forceStress in bar = = 40

areaNotice stress is not function of area!

40 220 10 12 10

106 MPa

ST

ST

ST ST

L T T L

RL

EA

RLL

EAR EA

RE

A

= +

T

ST1 R

R

T

L

R

R

B

A

ST2

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