Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 3 (Q.No. 39)
39.
At a point in a strained material, if two mutually perpendicular tensile stresses of 2000 kg/cm2 and 1000 kg/cm2 are acting, then the intensity of tangential stress on a plane inclined at 15° to the axis of the minor stress will be :
Discussion:
11 comments Page 1 of 2.
Savan said:
4 years ago
σ t= (σ x - σ y)/2. X (Sin2*15).
σ t 250.
σ t 250.
Amit Singh said:
5 years ago
Shere stress = (sigma1-sigma2)/2*(sin2θ)
= (2000-1000)/2*(sin2*15)
= 250.
= (2000-1000)/2*(sin2*15)
= 250.
Subhankar said:
5 years ago
sin 30 = 1/2.
So, 2sin 30 degree = 1.
D is the correct answer.
So, 2sin 30 degree = 1.
D is the correct answer.
Suraj said:
6 years ago
Formula for finding Tangential(Shear) Stress at any plane when τ xy = 0, τ= -(6x-6y)/2*sin(2θ) = -500Sin(2*15) = 250.
Pintu said:
7 years ago
Please explain the solution properly.
Sam said:
7 years ago
1000*sin15 = 258.819.
James said:
7 years ago
@Subhajit.
Here, 2 * sin30 results in 1. The answer becomes 1000.
Here, 2 * sin30 results in 1. The answer becomes 1000.
Nanu said:
7 years ago
@ Manjunatha R.N.
On that formula, there must be: Sin(2*angle).
On that formula, there must be: Sin(2*angle).
Manjunatha R.N. said:
8 years ago
@Subhajit.
Why you have taken 30 degree?
but given is 15 degree only.
Why you have taken 30 degree?
but given is 15 degree only.
Subhajit said:
8 years ago
Ans b is correct.
sigma t =(sigma x - sigma y)/2 sin 30 degree,
=(2000-1000)/2 sin 30 degree,
=500.1/2=250 kg/cm^2.
sigma t =(sigma x - sigma y)/2 sin 30 degree,
=(2000-1000)/2 sin 30 degree,
=500.1/2=250 kg/cm^2.
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