Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 32)
32.
The principal stresses at a point are 100, 100 and-200 kgf/cm2, the octo hedral shear stress at the point is :
Discussion:
9 comments Page 1 of 1.
Umesh said:
9 years ago
Could anyone explain it?
Sss said:
9 years ago
100 (Sqrt 2).
SRIDHAR said:
9 years ago
Please describe the solution.
Shakeel said:
9 years ago
Anyone know how to solve it? Please explain it.
(2)
Jeevan said:
9 years ago
Octo hydrel shear stress= yield stress or max stress/3 x√2
To fine max stress.
Stress=( p1+p2)/2 + 1/2√{(p1-p1)^2 + 4(txy)^2}
=(100+100)/2 + 1/2 √{(100-100)^2 +4(200)^2}
= 100+200
=300.
From above formula:
= 300/3 * √2.
= 100 * √2.
So ans is option 1.
To fine max stress.
Stress=( p1+p2)/2 + 1/2√{(p1-p1)^2 + 4(txy)^2}
=(100+100)/2 + 1/2 √{(100-100)^2 +4(200)^2}
= 100+200
=300.
From above formula:
= 300/3 * √2.
= 100 * √2.
So ans is option 1.
(2)
Shashi Ranjan said:
8 years ago
@Jeevan's solution is correct (You interpreted the question well).
(1)
Anant said:
8 years ago
Octahedral shear stress = √[ (p1-p2)^2 +(p2-p3)^2 +(p3-p1)^2] /3.
(3)
Durlavi said:
5 years ago
Thanks @Anant.
(1)
Amit kushwaha said:
5 years ago
Thanks @Jeevan.
(1)
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