Frame: roller displacement and hinge rotation

Given data

Portal frame: node 0 (hinge) at (0,0), node 1 at (0,9), node 2 at (6,9), node 3 (roller) at (6,0). Members 0-1 and 1-2: $2EI$; member 2-3: $EI$. UDL 20 kN/m acting horizontally (to the left) on the right column 2-3 (9 m). Required: horizontal displacement of the roller 3 and rotation at the hinge 0.

Real loading: reactions

Total horizontal load $= 20\times9 = 180$ kN (left), so $H_0 = 180$ kN (to the right). Moments about node 0: the UDL resultant acts at 4.5 m height, giving $180\times4.5 = 810$ kNm; the roller reaction $V_3\times6$ balances it, so $V_3 = 135$ kN (downward) and $V_0 = 135$ kN (up).

Virtual systems

1. Unit horizontal load at node 3 (to the right): $H_0 = 1$ (to the left), vertical reactions zero.
2. Unit couple at node 0 (clockwise): $V_0 = 1/6$ (down), $V_3 = 1/6$ (up), $H_0 = 0$.

Bending moments ($s$ from the first node of each member; $M$, $m_1$ and $m_2$ use the same sign convention)

For the unit couple, $m_2 = 1$ in column 0-1 (the vertical reaction at 0 has no lever arm there), $m_2$ falls linearly to zero at joint 2 along the beam, and $m_2 = 0$ in column 2-3, because the only force on that side, the vertical reaction at the roller 3, acts along the column's axis.

Integrals ($\int Mm\,ds$ / stiffness, units $1/EI$)

$$\Delta_{3h} = -\frac{71\,077.5}{EI},\qquad \theta_0 = -\frac{5670}{EI}$$

The negative signs mean the displacements are opposite to the unit loads: the roller moves to the left and the hinge rotates anticlockwise.

Answer: horizontal displacement of the roller $= 71\,077.5/EI$ (kNm$^3$/EI) to the left (in the direction of the UDL); rotation of the hinge $= 5670/EI$ rad anticlockwise, with $EI$ the stiffness of column 2-3 (the other two members have $2EI$).