Frame vertical deflection at D

Given data and assumptions

Frame: A(0,0) at the base of the left column, B(0,6), C(6,6), and D at the foot of the right column, hanging from C (free end at D). Column height 6 m, beam BC = 6 m with UDL 4 kN/m (down), 15 kN horizontal at B (taken to the right). Section 25 cm $\times$ 45 cm, $E = 150$ kN/mm$^2 = 1.5\times10^8$ kN/m$^2$. Since a vertical deflection is asked for at D, D is taken as a free end; the frame is then stable only if A is fixed, so it is treated as a cantilever frame from A (a hinge at A with D free would be a mechanism).

$$I = \frac{0.25\times0.45^3}{12} = 1.898\times10^{-3}\ \text{m}^4,\qquad EI = 284\,766\ \text{kNm}^2$$

Virtual load

Unit vertical load (down) at D. $s$ is measured from D up the right column, from C along CB, and from B down the left column ($EI$ constant).

Details: in CB, $M = 4s^2/2 = 2s^2$ and $m = s$, so $\int_0^6 2s^3\,ds = 648$. In BA, $M = 4\times6\times3 + 15s = 72 + 15s$ and $m = 6$, so $\int_0^6 6(72+15s)\,ds = 6(432 + 270) = 4212$. (The UDL and the 15 kN load both bend the left column in the same sense as the unit load.)

$$\Delta_{Dv} = \frac{648 + 4212}{EI} = \frac{4860}{EI} = \frac{4860}{284\,766} = 17.07\ \text{mm}$$

Answer: vertical deflection at D $= 4860/EI = 17.07$ mm downward.