Frame horizontal deflection at roller support

Given data

Portal frame: A(0,0) hinge, B(0,3), C(3,3), D(3,0) roller. Beam BC: $I$, 3 m, with 100 kN at 1 m from B. Columns AB and CD: $2I$, 3 m. UDL 30 kN/m on AB acting to the right, and 40 kN horizontal at B to the right. Required: horizontal deflection of the roller D by virtual work.

Real loading: reactions

Horizontal: $A_x = 40 + 30\times3 = 130$ kN (to the left).

Moments about A (clockwise positive): the horizontal loads give $40\times3 + 90\times1.5 = 255$ kNm clockwise; the 100 kN load gives $100\times1 = 100$ kNm clockwise; the roller reaction $D_y\times3$ acts anticlockwise.

$$D_y = \frac{255 + 100}{3} = 118.33\ \text{kN (up)},\qquad A_y = 100 - 118.33 = -18.33\ \text{kN (i.e. 18.33 kN down)}$$

Virtual load

Unit horizontal load at D (to the right). The hinge A resists it with $A_x = 1$; the vertical reactions are zero. Hence $m = s$ in AB, $m = 3$ in BC and $m = 3 - s$ in CD (s measured from the start of each member).

Moments and integrals

Check: at C, $M = 355 - 118.33\times3 = 0$, as it must be at the roller column (no moment in CD).

$$\Delta_{Dh} = \frac{433.13 + 737.50 + 710.00}{EI} = \frac{1880.6}{EI}$$

Answer: horizontal deflection of the roller D $= 1880.6/EI$ (kNm$^3$/EI), to the right, where $EI$ is the stiffness of the beam BC (columns $2EI$).