---
title: "Bent bracket deflection by unit load method"
subject: "Mechanics of Solids"
chapter: 5
key: "semester_5/mechanics_of_solids:bracket-deflection-unit-load"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/bracket-deflection-unit-load
site: BCT Engineering
---

# Bent bracket deflection by unit load method

Mechanics of Solids · Chapter 5: Deflections and Slope of Statically Determinate and Indeterminate Structures

- Asked in: year not given, 8 marks
- Video: [Strain Energy - Problem No 10 (Deflection under a point load in a Frame)](https://www.youtube.com/watch?v=YjJUPYwtdX8) — Stan Academy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/bracket-deflection-unit-load
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-5

## Question

A bracket ABC is fixed at A. Member AB is vertical, 1.5 m long, and member BC is horizontal, 1.0 m long, running from the top of AB to the right [C is the free end]. A vertical load of 5 kN acts downward at C. Both members have the same $EI = 8000$ kN m$^2$. Using the unit load method (bending energy only) find the vertical and horizontal deflection of C.

## Short answer

Apply unit loads at C in the vertical and horizontal directions. Integrate real bending moments times virtual moments over members AB and BC. Vertical deflection = 1.15 mm downward; horizontal deflection = 0.70 mm to the right.

## Answer

#### Data

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

```text
  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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the real bending moment in vertical member AB?
   A. Zero
   B. -P x, varying
   C. -P a, constant
   D. -P b, constant
   Answer: C. The 5 kN load acts at lever arm a = BC = 1.0 m from AB, so M = -Pa is constant along AB. (Key point: Real bending moments)

2. Which formula gives the vertical deflection of C?
   A. P a^3/(3EI)
   B. P b^3/(3EI)
   C. P a b^2/(2EI)
   D. (P/EI)(a^3/3 + a^2 b)
   Answer: D. Both members contribute: BC gives Pa^3/3EI and AB gives Pa^2 b/EI, total 1.146 mm. (Key point: Vertical deflection formula)

3. Why does a horizontal unit load at C give m = 0 in BC?
   A. Its line of action passes along BC, so it causes no bending there
   B. BC is rigid
   C. P is vertical
   D. BC carries no real load
   Answer: A. The unit load acts along the axis of BC, so only AB bends (m = -y). (Key point: Horizontal unit load)

4. What is the horizontal deflection of C and its direction?
   A. 1.15 mm to the left
   B. 0.70 mm to the right
   C. 0.70 mm to the left
   D. 1.15 mm to the right
   Answer: B. delta_h = P a b^2/(2EI) = 5 x 1.125/8000 = 0.703 mm, positive, so in the direction of the unit load (right). (Key point: Horizontal deflection value)

## Related in Mechanics of Solids

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- [Castigliano's theorem and unit load method](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Acastigliano-unit-load-theorems) — Mechanics of Solids, chapter 5
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