Truss horizontal displacement at C by virtual work

Given data and assumptions

Square panel truss: A(0,0), D(5,0), B(0,5), C(5,5); members AB, BC, CD, AD and diagonal BD. A is a hinge and D a roller. Loads at C: 50 kN downward and 100 kN horizontal to the right (away from the frame). $AE$ constant. Required: horizontal displacement of C.

Step 1: Real forces (tension +)

Reactions: $\sum M_A = 0$: $D_y\times5 = 50\times5 + 100\times5 \Rightarrow D_y = 150$ kN (up); $A_y = -100$ kN (down); $A_x = -100$ kN.

Joint C: CD carries the 50 kN vertical load: $F_{CD} = -50$ kN. The horizontal 100 kN is taken by BC: $F_{BC} = +100$ kN. Joint B: $F_{BD}\cos45^\circ = -F_{BC}\Rightarrow F_{BD} = -141.42$ kN; $F_{AB} = -F_{BD}\sin45^\circ = +100$ kN. Joint A / D: $F_{AD} = +100$ kN.

Step 2: Virtual forces (unit horizontal load at C to the right)

$k_{BC} = 1$, $k_{BD} = -\sqrt2 = -1.414$, $k_{AB} = 1$, $k_{AD} = 1$, $k_{CD} = 0$.

Step 3: Table

$$\Delta_{Ch} = \sum\frac{kFL}{AE} = \frac{2914.2}{AE}$$

Answer: horizontal displacement of C $= \dfrac{2914.2\ \text{kNm}}{AE}$, towards the right (direction of the unit load).