Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 8 (Q.No. 31)
31.
A body is subjected to a direct tensile stress of 300 MPa in one plane accompanied by a simple shear stress of 200 MPa. The maximum normal stress will be
Discussion:
12 comments Page 1 of 2.
Sunil said:
9 years ago
I think A is the right answer.
Neeraj said:
9 years ago
D is right answer 300/2+ √ [300^2 + 200^2] = 400.
SATYABRAT PATRA said:
8 years ago
Max normal stress.
&sigme;1 =300/2+ √[(300/2)^2+200^2].
= 400 MPa.
&sigme;1 =300/2+ √[(300/2)^2+200^2].
= 400 MPa.
Kanji dodiya said:
8 years ago
Max normal stress.
= 300/2 + 1/2 √[(300^2 +4*200^2)],
= 150+250,
= 400.
= 300/2 + 1/2 √[(300^2 +4*200^2)],
= 150+250,
= 400.
Shubham jagtap said:
7 years ago
Maximum normal stress = 300/2+ √ (300/2)^2+200^2.
Numankakar said:
6 years ago
Please explain the formula.
Basha said:
6 years ago
Please explain the answer.
Sahil said:
6 years ago
By solving the above formula 300/2 +underroot 300 ka square bla bla we got 650, not 400.
Ramana said:
6 years ago
Relation is shear stress = (1/2) * normal stress.
Neto said:
4 years ago
= 300/2 + 1/2 * √300*2 + 4*(200)*2
= 150 + 1/2 *√300*2 + 4*(40000)
= 150 + 1/2*√300*2+ 160000
= 150 + 1/2 *√300*2 + 400*2
= 150 + (700/2)
= 150 + 350
= 500.
So, the answer is 500 and here none of the option is correct.
= 150 + 1/2 *√300*2 + 4*(40000)
= 150 + 1/2*√300*2+ 160000
= 150 + 1/2 *√300*2 + 400*2
= 150 + (700/2)
= 150 + 350
= 500.
So, the answer is 500 and here none of the option is correct.
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