So 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.

1. Section 8 Statically Indeterminate Structures So 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. © Loughborough University 2010. This work is licensed under a Creative Commons Attribution 2.0 Licence .

4. Statically Indeterminate Structures 1. Stiffness Method 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 L AC = L) Free body diagram of joint A: 2P AB cos  +P AC =P (Eqn 1) P AB sin  = P AD sin  (Eqn 2) We need more information to solve problem

5. Statically Indeterminate Structures Stiffness 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’  ’

6. Statically Indeterminate Structures Stiffness Method

9. Stiffness Method - Example Equilibrium of Bar C: Forces P A P B and P Bar A elongated by P A by amount  A Bar B elongated by P B by amount  B Bar C is remains horizontal hence  A =  B

10. Stiffness Method - Example

11. Statically Indeterminate Structures 2. Compliance Method 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

12. Statically Indeterminate Structures 2. Compliance Method = + Statically indeterminate Statically determinate Statically determinate structure structure 1 structure 2

13. Statically Indeterminate Structures Compliance Method – structure 1

14. Statically Indeterminate Structures Compliance Method – structure 1  V1 is vertical displacement at joint A So vertical displacement at joint C =  V1 because there is no load in member AC

15. Statically Indeterminate Structures Compliance Method – structure 2

16. Statically Indeterminate Structures Compliance Method – structure 2

17. Statically Indeterminate Structures Compliance Method

18. Statically Indeterminate Structures Compliance Method

20. 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 Cross sectional areas: Al = A AL St = A ST

21. Compliance Method- Solution This problem is statically indeterminate – The equilibrium equation is P AL + P ST = -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 Cross sectional areas: Al = A AL St = A ST

22. Compliance Method- Solution = + Statically Structure 1 Structure 2 Indeterminate Structure has been divided into 2 structures. These 2 are solved using statics.

23. Compliance Method- Solution Structure 1

24. Compliance Method- Solution Structure 2

25. Compliance Method- Solution Superpose 2 solutions

26. Temperature Effects So far only mechanical loads have been considered at room temperature (T 0 ). 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 (>T 0 ). A uniform expansion by an amount of  in which  is known as the coefficient of linear thermal expansion (material constant). Unit: 1/ 0 C (the reciprocal of degrees Celsius). R ecalling definition of the strain gives thermal strain as:

27. 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

28. 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.

29. 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 0 C from the room temperature of 20 0 C. E = 220 GPa and  =12e-6 o C -1 Determine the maximum thermal stress developed in the bar.  T L R R B A

30. 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 P AB . 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

31. Temperature Effects - Solution ST1 ST2 = +  T  ST 1  R R  T L R R B A  ST 2 

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