Bent bracket deflection by unit load method

Data

$P = 5$ kN, $a = BC = 1.0$ m, $b = AB = 1.5$ m, $EI = 8000$ kN m$^2$.

  B ------------- C
  |      a        | P = 5 kN
  |               v
  | b
  |
 ###  A (fixed)

Real bending moments

• In BC ($x$ measured from C, $0 \le x \le a$): $M = -Px$ (hogging).
• In AB ($y$ measured downward from B, $0 \le y \le b$): the load acts at a lever arm $a$, so $M = -Pa$, constant.

Vertical deflection of C

Unit vertical load at C downward: $m_{BC} = -x$, $m_{AB} = -a$.

$$\begin{aligned} \delta_v &= \frac{1}{EI}\left[\int_0^a (Px)(x)\,dx + \int_0^b (Pa)(a)\,dy\right] \\ &= \frac{P}{EI}\left[\frac{a^3}{3} + a^2 b\right] \end{aligned}$$ $$\delta_v = \frac{5}{8000}\left[0.3333 + 1.5000\right] = \frac{5\times 1.8333}{8000} = 1.146\times10^{-3}\ \text{m}$$

Horizontal deflection of C

Unit horizontal load at C acting to the right. In BC the line of action of the unit load lies along the member, so $m_{BC} = 0$. In AB, a section at depth $y$ below B is at a vertical distance $y$ from the unit load (C is level with B), so $m_{AB} = -y$. The sign is the same as that of the real moment $-Pa$, because both loads turn the column clockwise.

$$\delta_h = \frac{1}{EI}\int_0^b (Pa)(y)\,dy = \frac{Pa\,b^2}{2EI}$$

With $b^2/2 = 1.125$:

$$\delta_h = \frac{5\times1.0\times1.125}{8000} = 0.703\times10^{-3}\ \text{m}$$

The positive sign shows that C moves in the direction of the unit load (to the right). The load P, acting to the right of the column, bends the column so that its top B tilts to the right. Arm BC does not bend under the unit load, so C moves right with B.

Answer: Vertical deflection of C $= 1.15$ mm downward; horizontal deflection $= 0.70$ mm to the right.