Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 2 (Q.No. 19)
19.
If a member carries a tensile force P on its area of cross-section A, the normal stress introduced on an inclined plane making an angle θ with its transverse plane, is
Discussion:
2 comments Page 1 of 1.
Yogesh said:
9 years ago
Pθ = (P/2) + (P/2)cos2θ,
Pθ = P/2 (1+cos2θ),
Pθ = P/2 (2cos^2(θ)),
Therefore, Pθ = P/2 cos^2(θ).
Pθ = P/2 (1+cos2θ),
Pθ = P/2 (2cos^2(θ)),
Therefore, Pθ = P/2 cos^2(θ).
(2)
Mounika reddy said:
1 decade ago
Force in the normal(perpendicular) direction = PCos0.
Area in inclined say A'.
Then A = A'Cos0, A' = A/cos0.
stress = force/area.
Pcos0/A/cos0.
pcos^20/A.
Area in inclined say A'.
Then A = A'Cos0, A' = A/cos0.
stress = force/area.
Pcos0/A/cos0.
pcos^20/A.
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