---
title: "Simply supported T-beam: max bending stresses"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:t-beam-6m-100kn-10kn"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/t-beam-6m-100kn-10kn
site: BCT Engineering
---

# Simply supported T-beam: max bending stresses

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2081 Baisakh, 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-6m-100kn-10kn
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A simply supported T-beam of 6 m span is subjected to the load as shown in figure. Calculate maximum bending stresses in tension and compression. [Figure: simply supported beam AB of span 6 m with a 100 kN point load at 3 m from A and a 10 kN load at 3 m from B (i.e. at the middle region as drawn, 3 m + 3 m); cross-section T-shape with flange 200 mm wide and 50 mm thick, web 50 mm thick, total depth 200 mm below the flange as marked.]

## Short answer

Maximum compressive bending stress at the top fibre: 127.2 N/mm². Maximum tensile bending stress at the bottom fibre: 236.1 N/mm².

## Answer

**Reading of the figure:** simple span $L = 6$ m. The 100 kN and 10 kN loads both act at 3 m from the supports, i.e. at mid-span, so together they form a 110 kN central load. The T-section has a flange 200 mm $\times$ 50 mm and a web 50 mm thick extending 200 mm below the flange (total depth 250 mm). The flange is on top.

#### Maximum bending moment
$$
M_{max} = \frac{WL}{4} = \frac{110\times6}{4} = 165.0\ \text{kN·m} = 1.6500\times10^{8}\ \text{N·mm}
$$

#### Section properties (distances from the bottom of the web)
$$
A = 200(50) + 50(200) = 20000\ \text{mm}^2
$$

$$
\bar y = \frac{(200\times50)(225) + (50\times200)(100)}{20\,000} = 162.50\ \text{mm} \quad(\text{from the bottom}), \qquad H - \bar y = 87.50\ \text{mm}\ (\text{from the top})
$$

$$
I = \left[\frac{200(50)^3}{12} + 10\,000(225-162.50)^2\right] + \left[\frac{50(200)^3}{12} + 10\,000(162.50-100)^2\right] = 1.1354\times10^{8}\ \text{mm}^4
$$

#### Bending stresses ($\sigma = My/I$)
$$
\sigma_{compression}\ (\text{top}) = \frac{1.6500\times10^{8}\times87.50}{1.1354\times10^{8}} = 127.16\ \text{N/mm}^2
$$

$$
\sigma_{tension}\ (\text{bottom}) = \frac{1.6500\times10^{8}\times162.50}{1.1354\times10^{8}} = 236.15\ \text{N/mm}^2
$$

**Answer:** maximum compressive stress $= 127.2$ N/mm² (top of the flange); maximum tensile stress $= 236.1$ N/mm² (bottom of the web).

## Check yourself

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

1. Since both loads act at mid-span, what maximum moment results?
   A. WL/4 with W = 110 kN, giving 165 kN·m
   B. WL/4 with W = 100 kN
   C. WL/8 with W = 110 kN
   D. WL/2 with W = 110 kN
   Answer: A. A central 110 kN load gives 110 x 6/4 = 165 kN·m. (Key point: Combining loads)

2. Where is the neutral axis of the T-section measured?
   A. At the top of the flange
   B. At 162.5 mm from the bottom of the web
   C. At 87.5 mm from the bottom
   D. At mid-depth
   Answer: B. Taking area moments about the bottom gives 162.5 mm. (Key point: Centroid)

3. For a sagging beam with the flange on top, which fibre is in tension?
   A. Top of the flange
   B. Bottom of the web
   C. Neither
   D. Both
   Answer: B. Sagging puts the bottom in tension, top in compression. (Key point: Which fibre is in tension)

4. Which pair is correct?
   A. Compression 127.2, tension 236.1 N/mm^2
   B. Both 181.7 N/mm^2
   C. Compression 87.5, tension 162.5 N/mm^2
   D. Compression 236.1, tension 127.2 N/mm^2
   Answer: A. sigma = My/I with y = 87.5 mm (top) and 162.5 mm (bottom). (Key point: Stress values)

## Related in Strength of Materials

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