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.

Figure 1 - Frame structure to analyze

• determine the degree of indeterminacy

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

• select redundants and remove restraints

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.

Figure 2 - Primary determinate structure

• determine reactions and draw moment diagram for the primary 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 deflected shape

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.

Figure 5 - Deflected shape of primary structure

• calculate deformations at redundants

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 X_D:

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, h_i, 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 Y_D:

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, h_i, 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, f_dd, 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, h_i, measured at the centroid of each area:

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

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:

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:

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:

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

• write consistent deformation equations

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.

• solve consistent deformation equations

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.

• determine support reactions

Impose the new, correct values of X_D and Y_D, along with the other applied loads on the original structure. Calculate the remaining reactions using the three static equilibrium equations, (F_x = 0, F_y = 0 and M = 0).

Figure 8 - Frame with support reactions

• draw moment, shear, and axial load diagrams

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.