Shear stress distribution in an I-section

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 $\tau = VQ/(Ib)$

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

 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.