---
title: "Shear stress distribution in an I-section"
subject: "Strength of Materials"
chapter: 5
key: "semester_4/strength_of_materials_mechanical:shear-stress-i-section-distribution"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/shear-stress-i-section-distribution
site: BCT Engineering
---

# Shear stress distribution in an I-section

Strength of Materials · Chapter 5: Transverse loading

- Asked in: year not given, 8 marks
- Video: [Shear Stresses In I Section](https://www.youtube.com/watch?v=EIIi_fWZ5R0) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/shear-stress-i-section-distribution
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-5

## Question

An I-section beam has flanges 150 mm wide and 20 mm thick, and a web 20 mm thick and 200 mm deep (overall depth 240 mm). At a certain section the shear force is 100 kN. Calculate the shear stress at the neutral axis, at the top of the web and in the flange just above the web junction, and sketch the distribution. What fraction of the shear force is carried by the web?

## Short answer

I=86.13×10⁶ mm⁴, V=100 kN. At neutral axis Q=430×10³ mm³, τ=24.96 MPa. At web top Q=330×10³ mm³, τ=19.16 MPa. In flange just above web, τ=2.55 MPa. The web carries about 92% of the shear force.

## Answer

#### Section property
Overall depth $240$ mm. Moment of inertia about the NA (flanges plus web):

$$
I = \frac{150\times240^3 - 130\times200^3}{12} = 86.13\times10^6\ \text{mm}^4
$$

(here 130 mm $= 150-20$ is the total width of the two voids on either side of the web).

#### First moments of area
- Flange area above the web junction (the junction is 100 mm from the NA): $A_f = 150\times20 = 3000$ mm$^2$, with its centroid at $110$ mm:

$$
Q_{junction} = 3000\times110 = 330\,000\ \text{mm}^3
$$

- At the NA, add the upper half of the web, $100\times20 = 2000$ mm$^2$ at $50$ mm:

$$
Q_{NA} = 330\,000 + 2000\times50 = 430\,000\ \text{mm}^3
$$

#### Shear stresses $\tau = VQ/(Ib)$
| Level | $b$ (mm) | $Q$ (mm$^3$) | $\tau$ (MPa) |
|---|---|---|---|
| Neutral axis | 20 | 430 000 | $\dfrac{100\,000\times430\,000}{86.13\times10^6\times20} = 24.96$ |
| Web, just below flange | 20 | 330 000 | $\dfrac{100\,000\times330\,000}{86.13\times10^6\times20} = 19.16$ |
| Flange, just above web | 150 | 330 000 | $\dfrac{100\,000\times330\,000}{86.13\times10^6\times150} = 2.55$ |
| Top of flange | 150 | 0 | 0 |

The stress jumps from 2.55 to 19.16 MPa at the junction because the width changes from 150 to 20 mm.

```text
 flange  |==|            0 - 2.6 MPa
         |  \
 web     |   )---> 19.2 (junction)
         |   )---> 25.0 (NA, maximum)
         |   )---> 19.2
 flange  |==|
```

#### Share of shear force
In the web the stress varies parabolically with $y$ (mm from the NA): $\tau(y) = \dfrac{V}{I\,b}\left[330\,000 + \dfrac{b\,(100^2-y^2)}{2}\right]$. Integrating $\tau\,b\,dy$ over the web (100 mm each side of the NA):

$$
V_{web} = \int \tau\,b\,dy = 92.1\ \text{kN} \approx 92\ \%\ \text{of}\ V
$$

Check: the web area is $20\times200 = 4000$ mm$^2$, so the average web shear stress is $92\,100/4000 = 23.0$ MPa, which lies between the junction value (19.16 MPa) and the NA value (24.96 MPa), as expected. The simple estimate $V/(\text{web area}) = 100\,000/4000 = 25$ MPa is close to the maximum.

**Answer: $\tau_{NA} = 24.96$ MPa (maximum), $\tau_{web\,top} = 19.16$ MPa, $\tau_{flange} = 2.55$ MPa; the web carries about 92 % of $V$.** The flanges carry the bending moment and the web carries almost all the shear.

## Check yourself

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

1. Where is the shear stress largest in the I-section?
   A. At the top of the flange
   B. At the neutral axis
   C. At the flange-web junction
   D. At the bottom fibre
   Answer: B. Q is largest at the NA (430000 mm^3) with web width 20 mm, giving 24.96 MPa. (Key point: Where maximum shear occurs)

2. Why does shear stress jump from 2.55 MPa to 19.16 MPa at the flange-web junction?
   A. Q changes suddenly
   B. The width b changes from 150 mm to 20 mm with the same Q
   C. The shear force changes
   D. I changes
   Answer: B. Q is the same (330000) just above and below, but tau = VQ/(Ib) and b falls from 150 to 20 mm. (Key point: Jump at the junction)

3. What is Q for the flange area above the junction?
   A. 150 x 20 x 110 = 330000 mm^3
   B. 150 x 20 x 100 = 300000 mm^3
   C. 150 x 20 x 120 = 360000 mm^3
   D. 20 x 100 x 50 = 100000 mm^3
   Answer: A. The flange area 3000 mm^2 has its centroid 110 mm from the NA. (Key point: Q at the junction)

4. What share of the shear force is carried by the web?
   A. About 8 %
   B. About 50 %
   C. About 92 %
   D. 100 %
   Answer: C. Integrating tau b dy over the web gives about 92 kN of the 100 kN; flanges carry bending, web carries almost all the shear. (Key point: Role of web)

## Related in Strength of Materials

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