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Consistent Deformations - Force Method
Twice Statically Indeterminate Frame with Horizontal and Vertical Reactions as Redundants


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.

Frame structure to analyze
Figure 1 - Frame structure to analyze


The structure is statically indeterminate to the second degree (r=5, e=3, n = r-e = 5-3 = 2).


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

In this example, the horizontal and vertical reactions at support D are selected as the redundants to remove in order to obtain a primary determinate structure.

primary determinate structure
Figure 2 - Primary determinate structure


Calculate the support reactions of the primary structure.

Support reactions
Figure 3 - Support reactions

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

Moment diagram of 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 each redundant at the released support.

deflected shape
Figure 5 - Deflected shape of primary structure


Using the virtual work method, calculate the vertical translation of support D that corresponds to the redundants X1 and X2. Remove all loads and apply a unit force in the direction of each redundant, draw the moment diagram, m, and sketch the deflected shape due to the unit load.

For the redundant X1 acting in the direction of the reaction XD:


Figure 6(a) - Primary structure with unit load applied at X1


Figure 6(b) - Moment diagram m1 with X1 = 1 k


Figure 6(c) - Deflected shape with X1 = 1 k

Calculate the horizontal translation corresponding to the redundant X1, D1, at support D using the virtual work 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 m1 diagram:

Area No. Area/EI (A) (k-ft2)/EI Height (h) on m1 diagram (k-ft) Ai*hi (k2-ft3)/EI
A01 15 x -300/EI = -4500/EI 1/2 x 15 = 7.5 -33750/EI
A02 1/3 x 15 x -112.5/EI = -562.5/EI 3/4 x 15 = 11.25 -6328.125/EI
A03 1/2 x 15 x -300/EI = -2250/EI 0 0/EI
Total = Q(D1) = -40078.125/EI

Therefore, with Q = 1 k;
Q(D1) = -40078.125 (k2-ft3)/EI
D1 = -40078.125 (k-ft3)/EI

For the redundant X2 acting in the direction of the reaction YD:


Figure 7(a) - Primary structure with unit load applied at X2


Figure 7(b) - Moment diagram m2 with X2 = 1 k


Figure 7(c) - Deflected shape with X2 = 1 k

Calculate the horizontal translation corresponding to the redundant X2, D2, at support D.

Again, 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 m2 diagram:

Area No. Area/EI (A) (k-ft2)/EI Height (h) on m2 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(D2) = -208125/EI

Therefore, with Q = 1 k;
Q(D2) = -208125 (k2-ft3)/EI
D2 = -208125 (k-ft3)/EI

Calculate the flexibility coefficients, fdd, by determining the deformations of the primary structure when subjected to the redundant loads.

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

For f11, the coefficient at X1 due to the load applied at X1:

Area No. Area/EI (A) (k-ft2)/EI Height (h) on m1 diagram (k-ft) Ai*hi (k2-ft3)/EI
A11 1/2 x 15 x 15/EI = 112.5/EI 2/3 x 15 = 10 1125/EI
Total = Q(f11) = 1125/EI

Therefore, with Q = 1 k;
Q(f11) = 1125 (k2-ft3)/EI
f11 = 1125 (k-ft3)/EI

For f12, the coefficient at X1 due to the load applied at X2:

Area No. Area/EI (A) (k-ft2)/EI Height (h) on m2 diagram (k-ft) Ai*hi (k2-ft3)/EI
A11 1/2 x 15 x 15/EI = 112.5/EI 30 3375/EI
Total = Q(f12) = 3375/EI

Therefore, with Q = 1 k;
Q(f12) = 3375 (k2-ft3)/EI
f12 = 3375 (k-ft3)/EI

For f22, the coefficient at X2 due to the load applied at X2:

Area No. Area/EI (A) (k-ft2)/EI Height (h) on m2 diagram (k-ft) Ai*hi (k2-ft3)/EI
A21 15 x 30/EI = 450/EI 30 13500/EI
A22 1/2 x 30 x 30/EI = 450/EI 2/3 x 30 = 20 9000/EI
Total = Q(f22) = 122500/EI

Therefore, with Q = 1 k;
Q(f22) = 122500 (k2-ft3)/EI
f22 = 122500 (k-ft3)/EI

For f21, the coefficient at X2 due to the load applied at X1:

Area No. Area/EI (A) (k-ft2)/EI Height (h) on m1 diagram (k-ft) Ai*hi (k2-ft3)/EI
A21 15 x 30/EI = 450/EI 1/2 x 15 = 7.5 3375/EI
A22 1/2 x 30 x 30/EI = 450/EI 0 0/EI
Total = Q(f21) = 3375/EI

Therefore, with Q = 1 k;
Q(f21) = 3375 (k2-ft3)/EI
f21 = 3375 (k-ft3)/EI

note: f12 and f21 will always have the same result


The consistent deformation equations are:

D1 + f11* X1 + f21* X2 = 0   (1)

D2 + f12* X1 + f22* X2 = 0   (2)

These equations are set equal to zero since the pinned support at D does not allow translation in the direction of the redundants.


Using equation (1) and (2), simultaneously solve for X1 and X2:

-40,078.125 (k-ft3)/EI + 1125(k-ft3)/EI * X1 + 3375(k-ft3)/EI* X2 = 0   (1)
-208,125 (k-ft3)/EI + 3375 (k-ft3)/EI* X1 + 22,500 (k-ft3)/EI* X2 = 0   (2)
X1 = 14.32
X2 = 7.10

The answers imply that the applied unit loads need to be multiplied, respectively, by the corresponding result in order to obtain the final reactions.

The positive answers indicate that the reactions are in the same direction as the applied unit loads.


Impose the new, correct values of XD and 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 8 - 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 9 - Final shear diagram

Moment:

Figure 10 - Final moment diagram

Axial Load:

Figure 11 - Final axial load diagram

Deflected Shape:

Figure 12 - Deflected shape


Contact Dr. Fouad Fanous for more information.