---
title: "Mid-span deflection of beam, 10 kN load"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:ss-beam-deflection-mid-span-4m"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/ss-beam-deflection-mid-span-4m
site: BCT Engineering
---

# Mid-span deflection of beam, 10 kN load

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2066 Chaitra (old course), 8 marks
- Video: [(DIM) Simply supported beam with concentrated load at midspan](https://www.youtube.com/watch?v=A3KaoOJ3cRk) — Engr Pogs
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/ss-beam-deflection-mid-span-4m
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

Find deflection of the beam shown below at middle of the span. Given that: Modulus of Elasticity of the material is 20000 MPa. Section of the beam is $150\ \text{mm}\times250\ \text{mm}$. Point Load W = 10 kN. [Figure: simply supported beam with a 10 kN load W at 4 m from the left support and 3 m from the right support (span 7 m).]

## Short answer

Using the deflection formula for a simply supported beam with a point load, the mid-span (3.5 m) deflection is 17.76 mm downward.

## Answer

**Given:** simply supported beam, span $L = 7$ m = 7000 mm, load $W = 10$ kN at $a = 4$ m from the left support ($b = 3$ m from the right). $E = 20000\ \text{N/mm}^2$, section $150\times250$ mm.

```text
              W = 10 kN
                 v
 A ^-------------|-------^ B
   |<----4 m---->|<-3 m->|
   mid-span at 3.5 m
```

#### Moment of inertia
$$
I = \frac{150\times250^3}{12} = 1.953\times10^8\ \text{mm}^4
$$

$$
EI = 20000\times1.953\times10^8 = 3.906\times10^{12}\ \text{N mm}^2
$$

#### Deflection at mid-span
The mid-span point $x = 3500$ mm lies in the left segment ($x < a = 4000$ mm). For a point load, in the segment $x \le a$:

$$
y = \frac{W\,b\,x}{6EIL}\left(L^2 - b^2 - x^2\right)
$$

$$
y = \frac{10\times10^3\times3000\times3500\,(7000^2 - 3000^2 - 3500^2)}{6\times3.906\times10^{12}\times7000}
$$

$$
y = \frac{1.05\times10^{11}\times2.775\times10^7}{1.6406\times10^{17}} = 17.76\ \text{mm}
$$

**Answer: deflection at mid-span $= 17.76$ mm downward.**

## Check yourself

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

1. What is I of the 150x250 mm section?
   A. db^3/12 = 7.03x10^7 mm^4
   B. bd^2/6
   C. bd^3/12 = 1.953x10^8 mm^4
   D. bd^3/3
   Answer: C. I = 150x250^3/12 about the horizontal axis. (Key point: Moment of inertia)

2. Why is the formula for x <= a used?
   A. Because the load is at mid span
   B. Because it is nearer the right support
   C. Because x exceeds a
   D. Mid span (3.5 m) lies in the left segment, since 3.5 < 4 m
   Answer: D. The expression depends on whether the point is left or right of the load. (Key point: Which segment)

3. What formula is used for the deflection?
   A. y = W b x (L^2 - b^2 - x^2)/(6 EI L)
   B. y = W L^3/(48EI)
   C. y = W a^2 b^2/(3EIL)
   D. y = 5wL^4/(384EI)
   Answer: A. For x <= a this is the standard point-load deflection formula. (Key point: Formula)

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