---
title: "T-beam: bending stress 15 cm above base"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:t-beam-udl-5m-bending-stress"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/t-beam-udl-5m-bending-stress
site: BCT Engineering
---

# T-beam: bending stress 15 cm above base

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2075 Asoj, 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-5m-bending-stress
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A simply supported beam of span 5m loaded with udl 4 kN/m. Determine the maximum value of bending stress 15 cm above the base of the cross section. The cross section is T-section as shown in figure. [Figure: T-section with flange 10 cm wide, web 3 cm thick, total depth 42 cm.]

## Short answer

Max sagging moment is 12.5 kNm. NA is 24.45 cm from base; at 15 cm above base (9.45 cm below NA), tension stress = My/I = 4.40 N/mm².

## Answer

**Assumption:** the figure gives only the flange width (10 cm), web thickness (3 cm) and overall depth (42 cm). The flange thickness is taken as 4 cm, flange on top, so the web is 38 cm deep.

#### Bending moment
$$
M = \frac{wL^2}{8} = \frac{4\times 5^2}{8} = 12.5\ \text{kNm (sagging, at mid-span)}
$$

#### Neutral axis (from base, cm)
$$
\bar y = \frac{3\times38\times19 + 10\times4\times40}{114+40} = \frac{2166+1600}{154} = 24.45\ \text{cm}
$$

#### Moment of inertia
$$
I = \frac{3\times38^3}{12} + 114(19-24.45)^2 + \frac{10\times4^3}{12} + 40(40-24.45)^2 = 26829.5\ \text{cm}^4 = 2.683\times10^8\ \text{mm}^4
$$

#### Stress at 15 cm above the base
Distance from the neutral axis: $y = 24.45 - 15 = 9.45$ cm $= 94.5$ mm, **below** the neutral axis, so the fibre is in tension (sagging).

$$
\sigma = \frac{My}{I} = \frac{12.5\times10^6\times 94.5}{2.683\times10^8} = 4.40\ \text{N/mm}^2
$$

**Answer: bending stress at 15 cm above the base = 4.40 N/mm² (tensile).**

## Check yourself

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

1. What is the moment for a 5 m span with 4 kN/m?
   A. 25 kNm
   B. 12.5 kNm
   C. 50 kNm
   D. 10 kNm
   Answer: B. wL^2/8 = 4 x 25/8 = 12.5 kNm. (Key point: Bending moment)

2. How is the NA position determined?
   A. At 15 cm
   B. Top of flange
   C. Mid-depth 21 cm
   D. Area-weighted centroid from base (24.45 cm)
   Answer: D. Sum(A y)/Sum(A) = 3766/154 = 24.45 cm. (Key point: Neutral axis)

3. What y is used for a point 15 cm above the base?
   A. 9.45 cm above the NA
   B. 9.45 cm below the NA
   C. 15 cm
   D. 27 cm
   Answer: B. 24.45 - 15 = 9.45 cm below NA. (Key point: Distance y)

4. Is the stress tensile or compressive and why?
   A. Compressive, it is a sagging beam
   B. Tensile, it is below the NA in a sagging beam
   C. Zero at that level
   D. Tensile, above the NA
   Answer: B. Sagging beams are in tension below the NA: 4.40 N/mm^2. (Key point: Nature of stress)

## Related in Strength of Materials

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