Analysis of Indeterminate Truss using Strain Energy Method

 

Analysis of Indeterminate Truss using Strain Energy Method 

Given problem

Explanation of solution:

First step: Make the given indeterminate structure into Basic Determinate Structure (BDS) by removing extra (redundant) reactions. In this given problem, there is one external redundancy (r-3=4-3=1, where r is the total number of reactions), and one internal indeterminacy (m+3-2j=6+3-2x4=9-8=1). So, ONE support reaction is to be removed and ONE member (internal reaction) is to be removed. In the solution shown here, vertical reaction at A (external redundant) and member BD (internal redundant) are selected. By removing these two, we get the Basic Determinate Structure (BDS). 

Now, the given indeterminate structure can be expressed as a superimposition (or addition) of three structures:

 1) BDS with given loading, 

2) BDS with redundant loading R₁, 

3) BDS with redundant loading R₂. 

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Second Step: Analysis of BDS with given loading: Using the method of joints, we can determine all the FIVE member forces under given loading of P. 

Start with Joint D: Apply vertical force equilibrium. The pin-joint at D is subjected to P force downwards. To balance that, member CD has to pull the joint by P force upwards. This creates tension in CD.  Again by applying horizontal force equilibrium on Joint D, we can get the force in member AD as 0(zero). 

Now consider Joint C. On Joint C, the force in member CD will act opposite (i.e downwards). We can obtain the force in inclined member AC by applying vertical force equilibrium at C. To balance the downward vertical force (P) of CD, the force in AC should push joint C upwards. The force should be such that it's vertical component should be P. Thus the force in AC should be √2P. This inclined force of √2P will cause a right ward push on joint C. To balance that the force in BC should pull joint C to the left. This creates a tension in member BC

Now consider the Joint A. On joint A, the member AC applies an inclined force √2P  pushing joint A downward. To balance this downward push, the member AB has to pull the joint A upwards. So, the member AB should apply P force upwards on Joint A. This creates a tension in member AB. Thus all member forces are determined without calculating the support reactions at A and B!!. The member forces are recorded under column F in tabular form

Third Step: Analysis of BDS with Redundant loading R₁: In place of unknown force R₁ use UNIT load along  R_1. By analysing the structure in the same way as above, we can conclude that the forces in all the members are zero except in member AB. The member forces are shown below. These member forces are tabulated under column f₁, as shown below 

Fourth Step: Analysis of BDS with Redundant loading R₂: In place of unknown force R₂ use UNIT load along  R_2. Again starting from joint D, we can conclude that for horizontal force equilibrium, the member force in AD should push the joint D by 1/√2 towards the right (because BD is pulling joint D towards left). This creates a compressive force of 1/√2 in member AD. For vertical force equilibrium, the member CD should push the joint D downwards to balance the upward pull along BD. This causes compression force of 1/√2 in member CD. Similar analysis gives same compressive forces of 1/√2 in members AB, BC, CD and AD. The force in inclined member AC will be +1 of tension. These results are shown below

Fifth Step: Applying compatibility equation: The final deflections along R₁ should be zero because there is a rigid hinged support at A. Therefore, the sum of deflections along R₁ in BDS under given loading + deflection along R₁ in BDS under R₁  + deflection along R₁ in BDS due to  R₂ should be zero. This is expressed as follows

Similarily, the final deflections along R₂ should be zero because there is a member BD. Therefore, the sum of deflections along R₂ in BDS under given loading + deflection along R₂ in BDS under R₁  + deflection along R₂ in BDS due to  R₂ should be zero. This is expressed as follows

These deflections are obtained from tabular format as follows

On substituting these deflections in the compatibility equations, we get

On solving these simultaneous equations, we get

To get the final member forces, we need to add the member forces in the three BDS as shown below

By comparing these values with the STAAD Pro modelling results (Using P=1 kN), we can conclude that the results are correct.