Explanation Given information: The frame is given in the Figure. The value of E is 200 GPa and I is 400 × 10 6 mm 4 . 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: Let apply a fictitious load P at joint C in the horizontal direction to find the deflection. The value of fictitious load P is zero. Sketch the frame with fictitious load P as shown in Figure 1. Let the equation for bending moment at distance x in terms of load P be M , the derivative of M with respect to P is ∂ M ∂ P . Find the reactions at the supports A and D : Summation of moments about A is equal to 0. ∑ M A = 0 D y ( 10 ) − 3 ( 10 ) ( 10 2 ) − 20 ( 10 ) − P ( 10 ) = 0 D y = 35 + P Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − A y + D y − 3 ( 10 ) = 0 − A y + 35 + P − 3 ( 10 ) = 0 A y = 5 + P Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − A x + 20 + P = 0 A x = 20 + P Find the equations for M and ∂ M ∂ P for the 2 segments of the frame as shown in Table 1. Segment x -coordinate M ∂ M ∂ P Origin Limits (m) AB A 0 − 10 ( 20 + P ) x x CB C 0 − 10 ( 35 + P ) x − 1.5 x 2 x The expression for horizontal deflection at C using Castigliano’s second theorem ( Δ C ) is shown as follows: Δ C = ∫ 0 L ( ∂ M ∂ P ) M E I d x (1) Here, L is the length of the beam. Rearrange Equation (1) for the limits 0 − 10 and 0 − 10 as follows. Δ C = 1 E I [ ∫ 0 10 ( ∂ M ∂ P ) M d x + ∫ 0 10 ( ∂ M ∂ P ) M d x ] Substitute the value of fictitious load P as zero in the Column 4 of Table 1