Under an axial force, the tangential component of stress on an inclined plane whose normal makes an angle θ with the direction of the force may be obtained by multiplying the normal component of stress on that plane by:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 29/01/2018 (Morning)
Detailed Explanation
Disputed key: the book's key is sin²θ, but its own working shows that sin²θ converts the applied stress P/A into the normal component, not the normal component into the tangential one. The original "A only … D only" options are shown directly as values.
Under an axial stress σ = P/A, on a plane whose normal is at θ to the load: σn = σ cos²θ and τ = σ sinθ cosθ.
\[ \frac{\tau}{\sigma_n} = \frac{\sigma\sin\theta\cos\theta}{\sigma\cos^2\theta} = \tan\theta \Rightarrow \tau = \sigma_n\tan\theta \]
(If θ is measured from the direction of the force to the plane itself, σn = σ sin²θ and the multiplier becomes cot θ.)
Why “tan θ” Is Correct
Using the same convention as the companion question (ratio of tangential to normal stress = tan θ), the tangential component equals the normal component multiplied by tan θ. We give (c) and flag the book's (d).
Why the Other Options Are Wrong
- A. sin θ: sin θ alone does not convert σn to τ.
- B. cos θ: cos θ alone does not convert σn to τ.
- D. sin² θ: sin²θ converts the axial stress P/A into σn (book's key), not σn into τ.
Key Point
τ/σn = tan θ (θ = angle of the plane's normal to the load); equal when θ = 45°.
Key Concepts for Students
- σn = σ cos²θ, τ = (σ/2) sin 2θ.
- τ = σn when θ = 45°.
- Book key (d) is flagged as disputed on this page.
Practice more: this is question 9 in Principal Stress and Principal Planes MCQs – Part 1. Try the whole set and check every answer instantly.