Frame: horizontal displacement and rotation at roller

Given data

Frame with $A(0,0)$ hinge, $B(0,4)$ (15 kN to the right), $C(0,5)$, $D(5,5)$ and $E(5,2)$ roller. Left column $AC = 5$ m ($EI$), beam $CD = 5$ m ($2EI$) with a UDL of 25 kN/m, right column $DE = 3$ m ($EI$). Required: horizontal displacement and rotation at the roller E.

Real loading reactions

$$\sum M_A = 0:\; E_y\times5 = 15\times4 + 125\times2.5 \Rightarrow E_y = 74.5\ \text{kN},\quad A_y = 125 - 74.5 = 50.5\ \text{kN},\quad A_x = 15\ \text{kN (left)}$$

Virtual cases

• Case 1: unit horizontal load at E (to the right): $A_x = 1$, $E_y = 0.4$ (up), $A_y = 0.4$ (down).
• Case 2: unit couple (anticlockwise) at E: $E_y = 0.2$ (down), $A_y = 0.2$ (up), $A_x = 0$.

Bending moments ($s$ along each member, in the order A to B to C to D to E)

Integrals ($\int Mm\,ds/EI_{member}$, in units of $1/EI$)

Results

$$\Delta_{Eh} = \frac{1435.83}{EI}\ \text{(to the right)},\qquad \theta_E = \frac{90.10}{EI}\ \text{(anticlockwise)}$$

Answer: horizontal displacement of E $= 1435.8/EI$ (m, with $EI$ in kNm$^2$) to the right; rotation at E $= 90.10/EI$ rad anticlockwise. Assumed geometry (as dimensioned in the figure): AB = 4 m and BC = 1 m, a 3 m right column, so the roller E is 2 m above A.