---
title: "Horizontal shear stress variation, SF 50 kN"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:horizontal-shear-stress-variation-t-section"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/horizontal-shear-stress-variation-t-section
site: BCT Engineering
---

# Horizontal shear stress variation, SF 50 kN

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2067 Asar (old course), 10 marks
- Video: [Problem on maximum shear stress for T section beam/ Strength of Materials](https://www.youtube.com/watch?v=3pqyKqGxUIU) — Prof.Dr.Pravin Patil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/horizontal-shear-stress-variation-t-section
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

Determine the variation of horizontal shear stress for the following section where shear force is 50 KN. [Figure: T-section in cm: flange 5 cm wide and 2 cm thick (bottom of the figure), stem 1 cm wide and 5 cm high, overall as marked: 5 cm height, 1 cm, 2 cm, 5 cm.]

## Short answer

Maximum horizontal shear stress is 107.0 N/mm² at the neutral axis. In the stem it varies parabolically, dropping abruptly from 106.9 to 21.4 N/mm² at the flange junction due to width change.

## Answer

**Section (in cm):** an inverted T. Flange 5 cm wide × 2 cm thick at the bottom; stem 1 cm wide × 5 cm high above it; overall height 7 cm. $F = 50$ kN.

```text
     +-+   <- top of stem
     | |
     | |  5
     | |
 +---+-+---+
 |         |  2
 +---------+
     5
```

#### Neutral axis (from base)
$$
\bar y = \frac{5\times2\times1 + 1\times5\times4.5}{10+5} = \frac{32.5}{15} = 2.167\ \text{cm}
$$

#### Moment of inertia
$$
I = \frac{5\times2^3}{12} + 10(1-2.167)^2 + \frac{1\times5^3}{12} + 5(4.5-2.167)^2 = 54.58\ \text{cm}^4
$$

#### Shear stress $\tau = \dfrac{F\,A\bar y}{I\,b}$
The top of the stem is 4.833 cm above the NA. The junction of stem and flange (2 cm from the base) lies 0.167 cm below the NA, since the NA is at 2.167 cm.

| Level | $A\bar y$ (cm³) | $b$ (cm) | $\tau$ (N/mm²) |
|---|---|---|---|
| Top of stem | 0 | 1 | 0 |
| Neutral axis (2.167 cm) | $1\times4.833\times4.833/2 = 11.68$ | 1 | 107.0 |
| Just above the flange (web) | $1\times5\times2.333 = 11.67$ | 1 | 106.9 |
| Just below the junction (flange) | 11.67 | 5 | 21.4 |
| Bottom of flange | 0 | 5 | 0 |

Example: $\tau_{NA} = \dfrac{50\times10^3\times11.68\times10^3}{54.58\times10^4\times10} = 107.0\ \text{N/mm}^2$.

#### Variation
- In the stem the stress varies **parabolically** from 0 at the top to the maximum at the NA, then stays almost constant down to the junction.
- At the junction it drops suddenly from 106.9 to 21.4 N/mm² because the width changes from 1 cm to 5 cm.
- In the flange it falls parabolically to zero at the bottom.

**Answer: $\tau_{max} = 107.0\ \text{N/mm}^2$ at the neutral axis; average stress on the stem $= 100\ \text{N/mm}^2$, so $\tau_{max}/\tau_{avg} \approx 1.07$.**

## Check yourself

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

1. Which formula gives shear stress at a level?
   A. tau = F A ybar/(I b)
   B. tau = M y/I
   C. tau = F/A everywhere
   D. tau = F b/(I A ybar)
   Answer: A. Horizontal shear stress uses the first moment A ybar of the area above the level, I and the width b there. (Key point: Formula)

2. Where is the maximum shear stress and what is its value?
   A. At the top of the stem, 107.0 N/mm^2
   B. At the neutral axis, 107.0 N/mm^2
   C. At the bottom of the flange, 21.4 N/mm^2
   D. At the junction in the flange, 107 N/mm^2
   Answer: B. A ybar is greatest at the NA and the width there is 1 cm, giving 107.0 N/mm^2. (Key point: Maximum stress location)

3. Why does stress drop suddenly from 106.9 to 21.4 N/mm^2 at the junction?
   A. The shear force changes there
   B. The material changes
   C. The width changes from 1 cm to 5 cm while A ybar is nearly the same
   D. The moment of inertia changes
   Answer: C. With A ybar nearly equal, tau is inversely proportional to b, so a 5 times wider flange gives about 1/5 the stress. (Key point: Jump at junction)

4. How does stress vary in the stem from the free top?
   A. Linearly from maximum at top to zero
   B. Constant throughout
   C. Zero at the NA and maximum at top
   D. Parabolically from zero at the top to the maximum near the NA
   Answer: D. A ybar grows quadratically with depth in a rectangular strip, giving a parabola. (Key point: Shape of variation)

5. What is tau_max/tau_avg for the stem?
   A. About 1.07
   B. 1.5
   C. 2.0
   D. 0.5
   Answer: A. tau_avg = 50 kN/(1x5 cm^2) = 100 N/mm^2, so the ratio is 107/100. (Key point: Average stress)

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