Explanation Given information: The frame is given in the Figure. The horizontal deflection at C is Δ C = 1 in . The value of E is 29,000 ksi. Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below. For summation of forces along x -direction is equal to zero ( ∑ F x = 0 ) , consider the forces acting towards right side as positive ( → + ) and the forces acting towards left side as negative ( ← − ) . For summation of forces along y -direction is equal to zero ( ∑ F y = 0 ) , consider the upward force as positive ( ↑ + ) and the downward force as negative ( ↓ − ) . For summation of moment about a point is equal to zero ( ∑ M at a point = 0 ) , consider the clockwise moment as negative and the counter clockwise moment as positive. Calculation: Consider the real system. Draw a diagram showing all the given real loads acting on it. Let the bending moment due to real load be M . Sketch the real system of the frame and the member end forces as shown in Figure 1. Find the reactions at the supports A and D : Summation of moments about A is equal to 0. ∑ M A = 0 D y ( 20 ) − 7 ( 20 ) = 0 D y = 7 k ↑ Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − A y + D y = 0 − A y + 7 = 0 A y = 7 k ↓ Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − A x − D x + 7 = 0 A x = D x = 3.5 k ← Consider the virtual system. Draw a diagram of frame without the given real loads. For horizontal deflection apply unit load at the joint in the horizontal direction. Let the bending moment due to virtual load be M v . Sketch the virtual system of the frame with unit load at joint C and the member end forces as shown in Figure 2. Find the reactions at the supports A and D : Summation of moments about A is equal to 0. ∑ M A = 0 D y ( 20 ) − 1 ( 20 ) = 0 D y = 1 k ↑ Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − A y + D y = 0 − A y + 1 = 0 A y = 1 k ↓ Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − A x − D x + 1 = 0 A x = D x = 0.5 k ← Find the equations for M and M v for the 3 segments of the frame as shown in Table 1. Segment x -coordinate M ( k-ft ) M v ( k-ft ) Origin Limits (ft) AB A 0 − 20 3.5 x 0.5 x BC B 0 − 20 70 − 7 x 10 − 1 x DC D 0 − 20 3.5 x 0.5 x Find the horizontal deflection at C using the virtual work expression: 1 ( Δ C ) = ∫ 0 L M v M E I d x (1) Rearrange Equation (1) for the limits 0 − 20 , 0 − 20 , and 0 − 20 as follows