---
title: "Cantilever: slope and deflection at free end"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:cantilever-slope-deflection-50kn-40kn"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cantilever-slope-deflection-50kn-40kn
site: BCT Engineering
---

# Cantilever: slope and deflection at free end

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2073 Shrawan, 8 marks
- Video: [Slope and Deflection of Cantilever Beam | Point Load at Free End | Strength of Materials](https://www.youtube.com/watch?v=qUBLJHWvXuQ) — EASY SCIENCE PREP
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/cantilever-slope-deflection-50kn-40kn
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

Determine the slope and deflection at the free end of the cantilever beam shown in figure below. [Figure: cantilever fixed at A; 50 kN at B (2 m from A); 40 kN at the free end C (3 m from B); $E = 200\ \text{kN/mm}^2$; cross-section rectangular 5 cm wide $\times$ 10 cm deep.]

## Short answer

Using superposition of standard cantilever cases, EI=833.3 kNm². Slope at free end = 0.72 rad. Deflection at free end = 2520 mm downward.

## Answer

**Given:** $AB = 2$ m with 50 kN at B; $BC = 3$ m with 40 kN at free end C; total length 5 m. $E = 200\ \text{kN/mm}^2 = 2\times10^5\ \text{N/mm}^2$.

$$
I = \frac{50\times100^3}{12} = 4.167\times10^6\ \text{mm}^4,\qquad EI = 8.333\times10^{11}\ \text{N mm}^2 = 833.3\ \text{kNm}^2
$$

```text
 fixed       50kN        40kN
 |A----2m----B-----3m-----C
```

Use superposition with standard cantilever results (point load $P$ at distance $a$ from the fixed end: $\theta = Pa^2/2EI$, $y = Pa^3/3EI$ at the load point, and the portion beyond the load stays straight).

#### Slope at C
$$
\theta_C = \frac{40\times10^3\times(5000)^2}{2EI} + \frac{50\times10^3\times(2000)^2}{2EI} = 0.600 + 0.120 = 0.72\ \text{rad}
$$

#### Deflection at C
- Due to 40 kN at C: $\dfrac{40\times10^3\times5000^3}{3EI} = 2000$ mm
- Due to 50 kN at B: $\dfrac{50\times10^3\times2000^3}{3EI} = 160$ mm at B, plus slope $0.12$ rad $\times$ 3000 mm $= 360$ mm further to C

$$
y_C = 2000 + 160 + 360 = 2520\ \text{mm}
$$

**Answer: slope at free end = 0.72 rad; deflection = 2520 mm (2.52 m) downward.** These values are far beyond the small-deflection limit because the section (50 × 100 mm) is very flexible for these loads; the formulas still give the theoretical result.

## Check yourself

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

1. Which method is used to combine the two loads?
   A. Superposition of standard cantilever results
   B. Moment-area only for UDL
   C. Castigliano with unit load
   D. Averaging
   Answer: A. Each load is treated separately and effects added. (Key point: Method)

2. Slope at the load point of a cantilever for load P at a from fixed end:
   A. Pa^3/(3EI)
   B. Pa^2/(3EI)
   C. Pa/EI
   D. Pa^2/(2EI)
   Answer: D. theta = Pa^2/2EI. (Key point: Slope formula)

3. Why add 0.12 rad x 3000 mm to the deflection from the 50 kN load?
   A. Because of shear
   B. The portion BC remains straight with slope at B, adding slope x length
   C. Because BC bends
   D. Because of self weight
   Answer: B. Beyond the load the cantilever is unloaded and straight, so extra deflection = theta_B x 3000 mm = 360 mm. (Key point: Beyond the 50 kN load)

4. What is the slope at C?
   A. 0.60 rad
   B. 0.12 rad
   C. 0.72 rad
   D. 0.52 rad
   Answer: C. 0.600 rad from 40 kN plus 0.120 rad from 50 kN. (Key point: Slope result)

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## Related in other subjects

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