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Consistent Deformations - Force Method
Indeterminate Beam with Vertical Reaction as Redundant


problem statement

Using the method of consistent deformations, determine the reactions, moments and shears under the loading conditions shown. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire beam.

Beam with applied loads
Figure 1 - Beam structure to analyze


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


To solve for this 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 vertical reaction at support B is selected as a redundant to remove in order to obtain a primary determinate structure.

beam with redundant support removed
Figure 2 - Primary structure


Calculate the support reactions of the released structure.

The resulting system, (B0 indicates the resulting deflection or deformation at the location of the removed redundant for the primary structure).

Support reactions
Figure 3 - Support reactions

Determine the moment diagram M0 due to the applied loads on the primary structure.

In this example, the cantilever method is used to develop the moment diagram. (See a Virtual Work Cantilver Example for a complete description of this step)

Moment diagram due to applied loads
Figure 4 - M0 - Moment diagram due to applied loads

note: The moment diagram due to the 6k load is segmented as shown due to the value of the moment diagram mb, which is found later in the problem.


Using the virtual work method, calculate the deflection at support B that correponds to the redundant YB. Remove all loads and apply a unit load in the direction of the redundant, draw the moment diagram, mb, and sketch the deflected shape due to the unit moment.

The resulting system, (fbb is the deformation caused by the unit load).

member forces due to unit load
Figure 5 - Primary structure with unit load applied and resulting deflected shape

Moment diagram due to unit load
Figure 6 - Moment diagram mb with YB = 1 k

Calculate the vertical translation corresponding to the redundant YB at support B using the following equation:

Calculate the deformation at the redundant, B0. Use the method of virtual work, calculate the areas on the M0 diagram and multiply each area by the corresponding heights, hi, measured at the centroid of this area on the mb diagram:

Area No. Area/EI (A) (k-ft2)/EI Height (h) on mb diagram (ft-k) Ai*hi (k2-ft3)/EI
A01 1/3 x 20 x -400/EI = -2666.67/EI 15 -40000/EI
A02 1/2 x 20 x -120/EI = -1200/EI 13.33 -16000/EI
A03 20 x -36/EI = -720/EI 10 -7200/EI
A04 1/2 x 6 x -36/EI = -108/EI 0 0
Total = Q(B0) = -63200/EI

Therefore, with Q = 1 k ;
Q(B0) = -63200 (k2-ft3)/EI
B0 = -63200 (k-ft3)/EI

Calculate the flexibility coefficient, fbb, by determining the deformation of the primary structure when subjected to the redundant load, YB = 1 k .

Again, using the method of virtual work, calculate the areas on the mb diagram and multiply by the corresponding heights, hi, measured at the centroid of each area:

Area No. Area/EI (A) (k-ft2)/EI Height (h) on mb diagram (k-ft) Ai*hi (k2-ft3)/EI
A11 1/2 x 20 x 20/EI = 200/EI 2/3 x 20 = 13.33 2666.67/EI
Total = Q(fbb) = 2666.67/EI

Therefore, with Q = 1 k ;
Q(fbb) = 2666.67 (k2-ft3)/EI
fbb = 2666.67 (k-ft3)/EI


The consistent deformation equation that corresponds to the redundant YB (the vertical reaction at support B) is:

B0 + fbb * YB = 0   (1)

This equation is set equal to zero since the pinned support at B does not allow any vertical translation.


Using equation (1), solve for YB:

-63200 (k-ft3)/EI + 2666.67 (k-ft3)/EI * YB = 0
YB = 23.7

Multiply the unit load, Q, at YB by 23.7 to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit force.


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

Resulting beam reactions
Figure 7 - Beam with support reactions


Shear:


Figure 8 - Final shear diagram

Moment:


Figure 9 - Final moment diagram

Deflected Shape:


Figure 10 - Deflected shape


Contact Dr. Fouad Fanous for more information.