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
Discussion:
57 comments Page 6 of 6.
Kaushal roy said:
7 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:
7 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.
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.
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.
= 1800/2+1/2 * (600)^2+4 * (400)^2,
= 900+1/2 * 360000+640000,
= 900+1/2*1000,
= 900+500.
= 1400MPa.
(25)
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:
12 months ago
Thank you for explaining the answer @Sagarbankar.
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