---
title: "Maximum deflection with unsymmetrical loading"
subject: "Strength of Materials"
chapter: 8
key: "semester_4/strength_of_materials_mechanical:unsymmetrical-load-maximum-deflection-moment-area"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/unsymmetrical-load-maximum-deflection-moment-area
site: BCT Engineering
---

# Maximum deflection with unsymmetrical loading

Strength of Materials · Chapter 8: Deflection of Beams by Moment- area Method

- Asked in: year not given, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/unsymmetrical-load-maximum-deflection-moment-area
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-8

## Question

A simply supported beam AB of span 5 m carries a point load of 40 kN at C, 2 m from A. E = 200 GPa and I = 1.5 x 10^8 mm^4. Using the moment-area method find (a) the slopes at A and B, (b) the deflection under the load, (c) the position and value of the maximum deflection.

## Short answer

θ_A = 2.13×10⁻³ rad, θ_B = 1.87×10⁻³ rad. Deflection under 40 kN load = 3.20 mm. Maximum deflection = 3.29 mm at 2.35 m from A (2.65 m from B).

## Answer

$EI = 200\times10^9\times1.5\times10^{-4} = 30\times10^6\ \text{N m}^2 = 30\,000\ \text{kN m}^2$.

#### Bending moment diagram
$$
R_A = \frac{40\times3}{5} = 24\ \text{kN}, \qquad R_B = 16\ \text{kN}, \qquad M_C = 24\times2 = 48\ \text{kN m}
$$

The $M/EI$ diagram is a triangle of base 5 m and height $48/EI$, with its apex at C. Total area $= \frac12\times5\times48 = 120$ (in kN m$^2$, divided by $EI$).

```text
 A      C(2 m)            B
 /\    48
/  \_____
 triangle area = 120/EI
```

#### (a) Slopes at the supports
Deviation of B from the tangent at A (moment of the area about B). The centroid of a triangle lies at the mean position of its three corners: from A, $\bar x = (0+2+5)/3 = 2.333$ m, so its distance from B is $5 - 2.333 = 2.667$ m.

$$
t_{B/A} = 120\times2.667 = 320/EI
$$

$$
\theta_A = \frac{t_{B/A}}{L} = \frac{320}{5\,EI} = \frac{64}{EI} = \frac{64}{30000} = 2.13\times10^{-3}\ \text{rad}
$$

Similarly, for the deviation of A from the tangent at B, the centroid is $2.333$ m from A:

$$
t_{A/B} = 120\times2.333 = 280/EI \;\Rightarrow\; \theta_B = \frac{280}{5EI} = \frac{56}{EI} = 1.87\times10^{-3}\ \text{rad}
$$

#### (b) Deflection under the load C
At C, the tangent at A lies $\dfrac{2}{5}t_{B/A} = \dfrac{128}{EI}$ below the line joining the supports. The deviation of C from the tangent at A is the moment of the area from A to C about C: area $\frac12\times2\times48 = 48$, centroid $2/3$ m from C:

$$
t_{C/A} = 48\times\frac23 = \frac{32}{EI}
$$

$$
y_C = \frac{128 - 32}{EI} = \frac{96}{30000} = 3.2\times10^{-3}\ \text{m} = 3.2\ \text{mm}
$$

#### (c) Maximum deflection
The point of zero slope D lies in the longer segment CB (3 m), at a distance $x$ from B. The slope at B equals the area of the $M/EI$ diagram from B to D, where $M = R_Bx = 16x$:

$$
\theta_B = \frac12\,x\,\frac{16x}{EI} \;\Rightarrow\; 8x^2 = 56 \;\Rightarrow\; x = \sqrt7 = 2.646\ \text{m}\ (<3\text{ m, so valid})
$$

So D is 2.646 m from B, i.e. 2.354 m from A. The maximum deflection equals the deviation of B from the horizontal tangent at D:

$$
y_{max} = 8x^2\times\frac{2x}{3}\Big/EI = \frac{16x^3}{3EI} = \frac{16\times18.52}{3\times30000} = 3.29\times10^{-3}\ \text{m}
$$

**Answer: $\theta_A = 2.13\times10^{-3}$ rad, $\theta_B = 1.87\times10^{-3}$ rad; $y_C = 3.20$ mm; $y_{max} = 3.29$ mm at 2.35 m from A.**

## Check yourself

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

1. What does the M/EI diagram look like and what is its area?
   A. A rectangle with height 48/EI, area 240/EI
   B. A triangle with height 40/EI, area 100/EI
   C. A parabola with area 80/EI
   D. A triangle with base 5 m and height 48/EI, area 120/EI
   Answer: D. RA = 24 kN, MC = 48 kN m, area = ½×5×48 = 120. (Key point: M/EI diagram)

2. How is θA found?
   A. tB/A divided by L: 120×2.667/5 = 64/EI
   B. Area divided by L: 120/5 = 24/EI
   C. tB/A multiplied by L
   D. Moment of area about A divided by L
   Answer: A. tB/A = 320/EI, so θA = 320/(5EI) = 64/EI = 2.13×10⁻³ rad. (Key point: Slope at A)

3. How is yC obtained?
   A. tC/A alone, 32/EI
   B. Tangent height at C (2/5 of tB/A = 128/EI) minus tC/A = 32/EI, giving 3.2 mm
   C. tB/A alone, 320/EI
   D. The sum 128/EI + 32/EI
   Answer: B. The tangent from A is above the curve at C; deflection is the difference 96/EI. (Key point: Deflection at C)

4. Where is the point of zero slope and how is it found?
   A. At C, under the load
   B. At mid-span, 2.5 m from A
   C. In the longer segment CB, from area of M/EI between B and D equalling θB, 2.646 m from B
   D. In segment AC, at 1 m from A
   Answer: C. ½·x·16x = 56 gives x = √7 = 2.646 m, valid as less than 3 m. (Key point: Maximum deflection location)

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