Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 7)
7.
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum normal stress will be
400 MPa
500 MPa
900 MPa
1400 MPa
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
58 comments Page 6 of 6.

Kaushal roy said:   8 years ago
Two different stress act on a member design should be based on max principal stress so here we find max principal stress.

Mustaf HD said:   8 years ago
Normal stress= (x+y)/2-[{(x-y)/2}cos2@]-(z*sin2θ).

Z-shear stress.
θ perpendicular means 90.
900-(-300)-0=1200.
The answer should be 1200.
(5)

Bangaram said:   7 years ago
1200+6000/2+1/2 ( (1200-600) + 4 (400) ) =1400mpa.

So it is by using formula of maximum normal stress.
(7)

Kamal Talukdar. said:   7 years ago
1200+600/2+1/2 * (1200-600)^2+4 * (400)^2.
= 1800/2+1/2 * (600)^2+4 * (400)^2,
= 900+1/2 * 360000+640000,
= 900+1/2*1000,
= 900+500.
= 1400MPa.
(27)

Yugal Chaudhari said:   6 years ago
Use @Kamal talukdar formula. It is correct.

Rashid Khan Kakar said:   4 years ago
It's subject to tow perpendicular normal stresses on oblique section.
(1)

Dadi Lavanya lahari said:   1 year ago
Thank you for explaining the answer @Sagarbankar.

Pavan dagge said:   4 weeks ago
@All.

The Thermal stress is only induced in a material when its natural expansion or contraction is restrained by external supports or boundaries.

Since the problem specifically states the steel bar is free to expand, it will undergo thermal strain (change in length) without any internal resistance. Because there is no resistance to the deformation, the internal restoring force is zero, resulting in zero stress.


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