Frame rotation and vertical deflection at free end

Given data

Frame fixed at the base A. Left column: 4 m high ($2EI$), with 50 kN horizontal (to the right) 1 m above the base. Beam: 6 m ($1.5EI$) with UDL 20 kN/m (down). Right column: $EI$, hanging 2 m from the beam end, with 30 kN horizontal (to the left) at its free end. Required: rotation and vertical deflection at the free end D.

        20 kN/m (1.5EI, 6 m)
   B ==================== C
   |                      |  EI, 2 m
   |                      D <-- 30 kN
   |  2EI, 3 m
   |
   |--> 50 kN (1 m above A)
   |  1 m
   A (fixed)

Virtual systems

• Rotation: unit couple (clockwise) at D.
• Vertical deflection: unit load (downward) at D.

The frame is a cantilever, so $M$ and $m$ are found from the free end D without reactions. $s$ is measured from D along DC, from C along CB, and from B down BA. All moments below are the clockwise moments, about the section, of the forces between D and the section; $M$ and $m$ use the same convention.

Bending moments (kNm)

Every load (the 30 kN to the left at D, the UDL and the 50 kN to the right on the column) gives a clockwise moment, so $M$ has the same sign in every member.

Integrals $\int Mm\,ds$ / stiffness

Both totals are positive, so the free end moves in the sense of the virtual loads:

$$\theta_D = \frac{1512.5}{EI}\ \text{(clockwise)},\qquad \Delta_{Dv} = \frac{7275}{EI}\ \text{(downward)}$$

Answer: rotation at the free end $= 1512.5/EI$ rad (clockwise); vertical deflection $= 7275/EI$ m (downward), with $EI$ in kNm$^2$.