---
title: "T-beam: maximum tensile and compressive stress"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:t-beam-udl-40kn-30cm-tension"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/t-beam-udl-40kn-30cm-tension
site: BCT Engineering
---

# T-beam: maximum tensile and compressive stress

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2068 Chaitra, 8 marks
- Video: [Bending stresses in beams: 14: Numerical Problem (unsymmetrical T/Tee section)](https://www.youtube.com/watch?v=4USC2tRoHBk) — Mech PM
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/t-beam-udl-40kn-30cm-tension
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A horizontal beam 4m long simply supported at ends carries a uniformly distributed load of 30KN/m over the whole span along with a concentrated load of 40KN at its mid span. The beam is of T-section with web 30cm$\times$3cm and flange 18cm $\times$4cm making overall depth of 34cm. Find the maximum tensile and compressive stresses if the flange is at the top and horizontal.

## Short answer

Maximum tensile stress is 122.5 N/mm² at the bottom of the web; maximum compressive stress is 62.2 N/mm² at the top of the flange.

## Answer

**Section:** flange 18 cm × 4 cm on top, web 3 cm × 30 cm; overall depth 34 cm.

#### Maximum bending moment (mid-span, sagging)
$$
M = \frac{wL^2}{8} + \frac{PL}{4} = \frac{30\times4^2}{8} + \frac{40\times4}{4} = 60 + 40 = 100\ \text{kNm}
$$

#### Neutral axis (from base, cm)
$$
\bar y = \frac{3\times30\times15 + 18\times4\times32}{90+72} = \frac{1350+2304}{162} = 22.56\ \text{cm}
$$

Top fibre distance $y_t = 34 - 22.56 = 11.44$ cm; bottom $y_b = 22.56$ cm.

#### Moment of inertia
$$
I = \frac{3\times30^3}{12} + 90(15-22.56)^2 + \frac{18\times4^3}{12} + 72(32-22.56)^2 = 18406\ \text{cm}^4 = 1.8406\times10^8\ \text{mm}^4
$$

#### Stresses
With the flange at the top and a sagging beam, top = compression, bottom = tension.

$$
\sigma_c = \frac{100\times10^6\times114.4}{1.8406\times10^8} = 62.2\ \text{N/mm}^2
$$

$$
\sigma_t = \frac{100\times10^6\times225.6}{1.8406\times10^8} = 122.5\ \text{N/mm}^2
$$

**Answer: maximum tensile stress = 122.5 N/mm² (bottom of web); maximum compressive stress = 62.2 N/mm² (top of flange).**

## Check yourself

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

1. What is the maximum bending moment for UDL 30 kN/m over 4 m plus 40 kN at mid span?
   A. 60 kNm at mid span
   B. 100 kNm at mid span
   C. 40 kNm at mid span
   D. 160 kNm at mid span
   Answer: B. M = wL^2/8 + PL/4 = 60 + 40 = 100 kNm. (Key point: Maximum bending moment)

2. How is the neutral axis position found?
   A. At the mid-depth, 17 cm from the base
   B. At the flange-web junction
   C. Take moments of areas about the base and divide by total area, giving 22.56 cm from the base
   D. Using the web area only
   Answer: C. The T-section is unsymmetrical, so the centroid is y = sum(A y)/sum(A) = 22.56 cm. (Key point: Neutral axis)

3. With the flange on top and a sagging beam, which fibre carries the larger stress and of what kind?
   A. Top of flange, compression, 122.5 N/mm^2
   B. Bottom of web, compression, 62.2 N/mm^2
   C. Top of flange, tension, 62.2 N/mm^2
   D. Bottom of web, tension, 122.5 N/mm^2
   Answer: D. Bottom is farther (225.6 mm) from the NA and is in tension under sagging, so it has the larger stress. (Key point: Tension and compression side)

4. Which stress formula gives the compressive stress at the top?
   A. sigma = M y_t / I with y_t = 114.4 mm, giving 62.2 N/mm^2
   B. sigma = M y_b / I with y_b = 225.6 mm
   C. sigma = M / (A y)
   D. sigma = M I / y_t
   Answer: A. Bending stress is My/I with y the distance from the NA to the top fibre, 11.44 cm. (Key point: Stress formula)

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