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Consistent Deformations - Force Method
Once Statically Indeterminate Frame with Vertical Reaction as Redundant
problem statement
Using the method of consistent deformations, determine the reactions, draw moment, shear, and axial load diagrams for the frame in the accompanying figure. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire structure.

Figure 1 - Frame 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 D is selected as a redundant to remove in order to obtain a primary determinate structure.
Figure 2 - Primary determinate structure
Calculate the support reactions of the primary structure.
Figure 3 - Support reactions
Determine the moment diagram M0 due to the applied loads on the primary structure.
Figure 4 - Mo - Moment diagram of primary structure
Sketch an approximate deflected shape for the primary structure, (see Fig. 4). Label the deflection that occurs in the direction of the redundant at the released support.
Figure 5 - Deflected shape of primary structure
Using the virtual work method, calculate the vertical translation of support D that corresponds to the redundant YD. Remove all loads and apply a unit force in the direction of the redundant, draw the moment diagram, md, and sketch the deflected shape due to this unit load.

Figure 6(a) - Primary structure with unit load applied

Figure 6(b) - Moment diagram md with YD = 1 k

Figure 6(c) - Deflected shape with YD = 1 k
Calculate the translation,
D0 (see Fig. 5),
at Support D using the following equation:
Using the method of virtual work, calculate areas on the M0 diagram and multiply each area by the corresponding heights, hi, measured at the centroid of this area on the md diagram:
| Area No. | Area/EI (A) (k-ft2)/EI | Height (h) on md diagram (k-ft) | Ai*hi (k2-ft3)/EI |
| A01 | 15 x -300/EI = -4500/EI | 30 | -135000/EI |
| A02 | 1/3 x 15 x -112.5/EI = -562.5/EI | 30 | -16875/EI |
| A03 | 1/2 x 15 x -300/EI = -2250/EI | 5/6 x 30 = 25 | -56250/EI |
| Total = Q( |
-208125/EI | ||
Therefore, with Q = 1 k
;
Q(
D0) = -208125 (k2-ft3)/EI
D0 = -208125 (k-ft3)/EI
Calculate the flexibility coefficient, fdd, by determining the deformations of the primary
structure when subjected to the redundant load, YD = 1 k
.
Again, using the method of virtual work, calculate areas on the md diagram and multiply by the corresponding heights, hi, measured at the centroid of the area:
| Area No. | Area/EI (A) (k-ft2)/EI | Height (h) on md diagram (k-ft) | Ai*hi (k2-ft3)/EI |
| A11 | 15 x 30/EI = 450/EI | 30 | 13500/EI |
| A12 | 1/2 x 30 x 30/EI = 450/EI | 2/3 x 30 = 20 | 9000/EI |
| Total = Q(fdd) = | 22500/EI | ||
Therefore, with Q = 1 k
;
Q(fdd) = 22500 (k2-ft3)/EI
fdd = 22500 (k-ft3)/EI
The consistent deformation equation that corresponds to the redundant YD (the vertical reaction at support D) is:
D0 + fdd* YD = 0 (1)
This equation is set equal to zero since the pinned support at D does not allow any translation in the direction of the redundant, i.e., in the vertical direction.
Using equation (1), solve for YD:
-208,125 (k-ft3)/EI + 22,500 (k-ft3)/EI* YD = 0
YD = 9.25
Multiply the unit load, Q, at YD by 9.25 to get the final reaction. The positive answer indicates that this reaction is in the direction of the applied unit load.
Impose the value of the calculated support reaction at YD 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).
Figure 7 - Frame with support reactions
All reactions are now known and are used to draw the complete shear and moment diagrams for the structure.
Shear:

Figure 8 - Final shear diagram
Moment:

Figure 9 - Final moment diagram
Axial Load:

Figure 10 - Final axial load diagram
Deflected Shape:

Figure 11 - Deflected shape
Contact Dr. Fouad Fanous for more information.