Consistent Deformations - Force Method | Index of Examples | CCE Homepage

Consistent Deformations - Force Method
Indeterminate Truss


problem statement

Using the method of consistent deformations, determine the vertical and horizontal reactions at A and E and the resulting member loads for the truss in the accompanying figure.

The member properties are A = 2 in2 and E = 29x103 ksi.

The truss with applied loads
Figure 1 - Truss structure to analyze


The structure is indeterminate to the first degree (r=4, e=3, n=r-n=4-3=1).


To solve for the single degree of indeterminacy, the structure has to be reduced to a statically determinate and stable structure. This is done by removing a redundant support.

In this example, the horizontal reaction at support E is selected as a redundant to remove in order to obtain a primary determinate structure.

truss diagram with redundant support removed
Figure 2 - Primary determinate structure


Calculate the support reactions of the primary structure, then determine the individual member forces by using either the method of sections, or the method of joints.

Resulting member forces
Figure 3 - Support reactions and member forces


Use the virtual work method, calculate the horizontal translation of support E that corresponds to the removed redundant XE. Remove all loads and apply a unit force in the direction of the redundant.

Unit load applied to truss
Figure 4 - Primary structure with unit load applied

Calculate the member forces due to the unit load.

member forces due to unit load
Figure 5 - Member forces due to unit load

Since the horizontal translation at E is equal to zero in the original structure, the horizontal translation of the released structure at support E must be countered by a force at E which causes an equal translation in the opposite direction.

Since the member forces are known in the released structure, and the virtual structure with a unit load applied at E, the deflection at support E (E0) caused by the applied loads on the released structure is determined by the following equation.

where m is equal to the number of members, n is the force in the member due to the virtual load, N is the force in the member due to the applied loads, L is the length, A is the area, and E represents Young's Modulus of Elasticity.

The force required to counter this deflection is found by determining the deflection caused by a unit load in the positive direction at the removed support, and multiplying this answer by the unknown support reaction.


The consistent deformation euqation that correspondes to the redundant XE (the horizontal reaction at support E) is:

E0 + fee*XE = 0   (1)

where E0 is the deflection at E of the primary structure, fee is the deflection of the released structure at E caused by the unit load, and XE is the value of the unknown redundant at E.


The deflection of the released structure under applied loads, E0.
note: for trusses, only the members which are subjected to the unit load need to be included

Member n(k) N(k) L(ft) nNL/AE (k2-ft)/AE
AB 1 -25.83 4 -103.33/AE
BC 1 -25.83 4 -103.33/AE
CD 1 -44.167 4 -176.67/AE
DE 1 -44.167 4 -176.67/AE
Total = E0 = -560/AE

The deflection of the released structure at E caused by the unit load, fee.

Member n (k) N (k) L(ft) nNL/AE (k2-ft)/AE
AB 1 1 4 4/AE
BC 1 1 4 4/AE
CD 1 1 4 4/AE
DE 1 1 4 4/AE
Total = fee = 16/AE

Using equation (1), solve for XE:

-560 (ft2-k)/AE + 16 (ft2-k)/AE*XE = 0
XE = 35

Multiply the unit load by this value to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit load.


Impose the value of the calculated XE along with the other applied loads on the original truss. Calculate the remaining reactions using the trhee static equilibrium equations, (Fx = 0, Fy = 0 and M = 0).

Resulting truss reactions
Figure 6 - Truss with resultant support reactions

The resulting member forces are now determined by using the method of sections, or method of joints. However, a much easier method is to use superposition and add the effects caused by the redundant load on the released structure.

In this example, the 35 k load will cause only the top chord of the truss to experience a 35 k tensile load in each member. This result can be added to those found in figure 3 above. The final results,

final member forces
Figure 8 - Truss with final member forces


Contact Dr. Fouad Fanous for more information.