Question

For a symmetrical I-section beam, the shear stress is maximum at ...........

• A. top fiber
• B. bottom fiber
• C. neutral axis
• D. at the flange tips

Hint:

In all standard beams, including I-sections, shear stress is zero at extreme fibers and is maximum at the neutral axis — crucial for shear design.

Answer 1

The Correct Option is C

In a symmetrical I-section beam, the distribution of shear stress follows a parabolic pattern across the depth of the section. The shear stress is zero at both the top and bottom fibers and increases towards the center. It reaches its maximum value at the neutral axis, which is the axis passing through the centroid of the section. This is because shear stress \( \tau \) at a given depth in a beam is given by: \[ \tau = \frac{VQ}{Ib} \] Where:
- \( V \) = Shear force,
- \( Q \) = First moment of area above (or below) the point,
- \( I \) = Moment of inertia of the whole section,
- \( b \) = Width at the level where shear stress is being calculated.
At the neutral axis, the value of \( Q \) is maximum for the web, and since the web thickness is relatively small, the resulting shear stress is the highest here. \[ \text{Maximum shear stress location} = \text{Neutral Axis} \]