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
-100 MPa
250 MPa
300 MPa
400 MPa
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Rashid Khan Kakar said:   3 years ago
150+1/2* √(300^2+4*200^2) = 400.

AMANDEEP Singh mattu said:   3 years ago
Formula = 300 /2 +1/2 under root (300)^2+4(200)^2 = 500.
300/2 = 150, 1/2 = 0.5,
500*.5 = 250 + 150 = 400 max normal stress.

Neto said:   3 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.

Ramana said:   4 years ago
Relation is shear stress = (1/2) * normal stress.

Sahil said:   4 years ago
By solving the above formula 300/2 +underroot 300 ka square bla bla we got 650, not 400.

Basha said:   5 years ago
Please explain the answer.

Numankakar said:   5 years ago
Please explain the formula.

Shubham jagtap said:   5 years ago
Maximum normal stress = 300/2+ √ (300/2)^2+200^2.

Kanji dodiya said:   6 years ago
Max normal stress.
= 300/2 + 1/2 √[(300^2 +4*200^2)],
= 150+250,
= 400.

SATYABRAT PATRA said:   7 years ago
Max normal stress.

&sigme;1 =300/2+ √[(300/2)^2+200^2].
= 400 MPa.


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