The shear stress at a distance y above the neutral axis (NA) of a prismatic beam due to a shear force V is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2012
Detailed Explanation
Consider a slice of beam dx long and cut it horizontally at level y. The difference in bending force on the part above the cut, between the two faces of the slice, must be balanced by horizontal shear along the cut.
\[ \tau\,b\,dx = \frac{dM}{I}\,A\bar y \Rightarrow \tau = \frac{dM}{dx}\cdot\frac{A\bar y}{I\,b} = \frac{V A\bar y}{I\,b} \]
Why “VAȳ/(Ib)” Is Correct
The formula is τ = VAȳ/(Ib), where Aȳ is the first moment of the area beyond the level about the neutral axis, I is for the whole section, and b is the width at the level considered.
Why the Other Options Are Wrong
- A. VAȳb/I: b belongs in the denominator — a wider level spreads the shear and lowers τ.
- B. Vb/(AȳI): This is upside down; units do not give a stress.
- D. VI/(bAȳ): Also inverted.
Key Point
τ = VAȳ/(Ib) (Jourawski / Zhuravskii formula).
Key Concepts for Students
- Aȳ is called the first moment of area Q.
- τ is inversely proportional to the width b at that level.
- Shear flow q = τb = VAȳ/I (used for bolts/welds in built-up beams).
Practice more: this is question 45 in Bending and Shear Stress MCQs – Part 3. Try the whole set and check every answer instantly.