Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 1 (Q.No. 37)
37.
The tangential component of stress on an plane inclined θ° to the direction of the force, may be obtained by multiplying the normal stress by
Discussion:
13 comments Page 1 of 2.
Boyee said:
4 years ago
Normal stress = pcosθ2(a).
Cosθ2(a) *tan(a)=cos(a)sin(a) = sin(2a)/2.
So, by multiplying tan(a) to normal stess we get. P/2*sin(2a) which is tangential stress.
Cosθ2(a) *tan(a)=cos(a)sin(a) = sin(2a)/2.
So, by multiplying tan(a) to normal stess we get. P/2*sin(2a) which is tangential stress.
(5)
Vikash kirad said:
4 years ago
When tanθ multiplied by normal stress we obtain the value of tangential stress.
So, option C should be the right answer.
So, option C should be the right answer.
(3)
Rakesh Ojha said:
7 years ago
It will be D.
TS = σ/2 *sin2x = σ/2 *( 2sinxcosx).
= σ* sinxcosx.
TS = σ/2 *sin2x = σ/2 *( 2sinxcosx).
= σ* sinxcosx.
(2)
Irfan mir said:
4 years ago
It will be D in terms of the applied normal stress.
But if we need to find in terms of the normal stress produced on the same plane on which we need to find shear stress then we need to multiply by cot θ.
But if we need to find in terms of the normal stress produced on the same plane on which we need to find shear stress then we need to multiply by cot θ.
(2)
Rajni said:
8 years ago
How? I Can't get it.
Madhuri said:
7 years ago
Can Anyone give the explanation of the solution?
Dilip Yadav said:
6 years ago
The Correct answer is D.
Narayan ch.saha said:
6 years ago
The Correct answer is E. I agree with the given answer.
Mukesh said:
5 years ago
@Rakesh Ojha.
Your expression is about normal shear stress.
Your expression is about normal shear stress.
MJM GCEK said:
4 years ago
Yes, I agree with you @Rakesh Ojha.
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