Structures and Materials- Section 8 Statically Indeterminate Structures

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 .

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
Compliance Method – Summary
Compliance Method- Example
Compliance Method- Solution
Compliance Method- Solution Structure 1
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 the
principles of statics to determine the unknown forces, e.g. drawing free body
diagrams 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 + r
3D system: 3j < m + r
System is statically indeterminate because we either have too many members
or 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 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

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’  ’

Statically Indeterminate Structures Stiffness Method

Stiffness Method - Summary
Establish equilibrium equations (Eqns 1 and 2)
Select a suitable displacement (  ) as the unknown quantity
Consider geometric characteristics of deformation (Eqn 3)
Establish constitutive relationship (Eqn 4 – Hooke’s Law) in terms of 
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:
Axial stress in each vertical bar
Vertical displacement of bar C
Horizontal position of the external load P with respect to point O.

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

Stiffness Method - Example

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

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

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

Statically Indeterminate Structures Compliance Method

Compliance Method - Summary
The above procedure of the compliance method is summarised in the following four steps:
Establish equilibrium equation
Remove enough reactions and designate them as redundant, to reduce a statically indeterminate structure to a statically determinate one
Calculate the displacements at the redundant due to actual loads
Apply redundant load only to find the displacements
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 Cross sectional areas: Al = A AL St = A ST

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

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

Compliance Method- Solution Superpose 2 solutions

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:

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

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.

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

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

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

This resource was created by Loughborough University and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. © 2010 Loughborough University. Except where otherwise noted this work is licensed under a Creative Commons Attribution 2.0 Licence . The name of Loughborough University, and the Loughborough University logo are the name and registered marks of Loughborough University. To the fullest extent permitted by law Loughborough University reserves all its rights in its name and marks, which may not be used except with its written permission. The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. Credits

Structures and Materials- Section 8 Statically Indeterminate Structures

by The Engineering Centre for Excellence in Teaching and Learning on Jun 17, 2010

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