---
title: "Cantilever with self-weight: P and moment of resistance"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:cantilever-self-weight-t-section"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cantilever-self-weight-t-section
site: BCT Engineering
---

# Cantilever with self-weight: P and moment of resistance

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2075 Chaitra, 8 marks
- Video: [Understanding Stresses in Beams](https://www.youtube.com/watch?v=f08Y39UiC-o) — The Efficient Engineer
- 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-self-weight-t-section
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A 3.0m long cantilever beam having self-weight 1.5 kN/m is subjected to a downwards point load of 'P' kN at the free end. Determine the value of 'P' and the moment of resistance of the beam. Take permissible bending stress in tension and compression as 150MPa. The cross section is shown in figure. [Figure: T-section with flange 18 cm wide and 6 cm thick on top, stem 4 cm wide and 18 cm deep.]

## Short answer

Neutral axis is 16.2 cm from base; smaller section modulus at bottom governs. Equating hogging moment (3P+6.75 kNm) to moment of resistance 78.6 kNm gives P=23.95 kN. Moment of resistance = 78.6 kNm.

## Answer

**Section:** flange 18 cm × 6 cm on top, stem 4 cm × 18 cm below; overall depth 24 cm. $\sigma_t = \sigma_c = 150\ \text{N/mm}^2$.

```text
 P
 |      w = 1.5 kN/m
 v  vvvvvvvvvvvvvvvvv
 |##################  free end
 fixed    3 m
```

#### Section properties (from base, cm)
$$
\bar y = \frac{18\times6\times21 + 4\times18\times9}{108+72} = 16.2\ \text{cm}
$$

$$
I = \frac{18\times6^3}{12} + 108(21-16.2)^2 + \frac{4\times18^3}{12} + 72(9-16.2)^2 = 8488.8\ \text{cm}^4
$$

Top fibre distance $y_t = 24 - 16.2 = 7.8$ cm; bottom $y_b = 16.2$ cm.

$$
Z_t = \frac{I}{y_t} = 1088.3\ \text{cm}^3,\qquad Z_b = \frac{I}{y_b} = 524.0\ \text{cm}^3
$$

#### Moment of resistance
A downward load causes hogging: top in tension, bottom in compression. Both have the same permissible stress, so the smaller $Z$ governs:

$$
M_r = 150\times 524.0\times10^3 = 78.6\times10^6\ \text{Nmm} = 78.6\ \text{kNm}
$$

#### Value of P
$$
M_{max} = P\times 3 + \frac{1.5\times 3^2}{2} = 3P + 6.75
$$

$$
3P + 6.75 = 78.6 \Rightarrow P = 23.95\ \text{kN}
$$

**Answer: $P = 23.95$ kN; moment of resistance $= 78.6$ kNm.**

## Check yourself

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

1. A downward load on this cantilever puts which fibre in tension?
   A. Both
   B. Bottom
   C. Neither
   D. Top
   Answer: D. Cantilever hogs, so the top is in tension and bottom in compression. (Key point: Tension side)

2. Why is the smaller Z (524 cm^3) used for the moment of resistance?
   A. It is the top fibre
   B. Equal permissible stresses, so the fibre farthest from NA governs
   C. Because tension is weaker
   D. Z is always smaller
   Answer: B. M = sigma Z, with the same sigma the smaller Z limits the moment. (Key point: Governing Z)

3. What is M_max at the fixed end?
   A. P + 4.5 kNm
   B. 3P
   C. 3P + 6.75 kNm
   D. 3P + 13.5 kNm
   Answer: C. P x 3 plus w L^2/2 = 1.5 x 9/2 = 6.75 kNm. (Key point: Max moment)

4. Solving 3P + 6.75 = 78.6 gives
   A. P = 28.6 kN
   B. P = 19.5 kN
   C. P = 23.95 kN
   D. P = 26.2 kN
   Answer: C. P = (78.6 - 6.75)/3 = 23.95 kN. (Key point: Value of P)

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