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Consistent Deformations - Force Method
Once Statically Indeterminate Frame with Horizontal 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.

Frame structure to analyze
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 horizontal reaction at support D is selected as a redundant 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

note: colors represent one method of dividing the area in the moment diagram to be used in the visual integration.


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.

deflected shape
Figure 5 - Deflected shape


Using the virtual work method, calculate the horizontal translation of support D that corresponds to the redundant XD. 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 XD = 1 k


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

Calculate the translation, D0, 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 1/2 x 15 x 206.25/EI = 1546.875/EI 2/3 x -7.5 = -5 -7734.375/EI
A02 1/2 x 15 x 206.25/EI = 1546.875/EI 2/3 x -15 = -10 -15468.75/EI
A03 1/2 x 15 x 112.5/EI = 843.75/EI 5/6 x -15 = -12.5 -10546.875/EI
A04 1/2 x 15 x 112.5/EI = 843.75/EI 2/3 x -15 = -10 -8437.5/EI
A05 2/3 x 15 x 28.125/EI = 281.25/EI 1/2 x -15 = -7.5 -2109.375/EI
Total = Q(D0) = -44296.875/EI

Therefore, with Q = 1 k;
Q(D0) = -44296.875 (k2-ft3)/EI
D0 = -44296.875 (k-ft3)/EI

Calculate the flexibility coefficient, fdd, by determining the deformations of the primary structure when subjected to the redundant load, XD = 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 1/2 x 30 x -15/EI = -225/EI 2/3 x -15 = -10 2250/EI
A12 1/2 x 15 x -15/EI = -112.5/EI 2/3 x -15 = -10 1125/EI
fdd = 3375/EI

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


The consistent deformation equation that corresponds to the redundant X1 (the horizontal reaction at support D) is:

D0 + fdd * X1 = 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 horizontal direction.


Using equation (1), solve for X1:

-44,296.875 (k-ft3)/EI + 3375/EI (k-ft3)/EI* XD = 0
X1 = 13.125

Multiply the unit load, Q, at XD by 13.125 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 support reaction at X1 corresponding to the redundant XD, 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.