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:
58 comments Page 1 of 6.
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.
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.
Dadi Lavanya lahari said:
1 year ago
Thank you for explaining the answer @Sagarbankar.
Rashid Khan Kakar said:
4 years ago
It's subject to tow perpendicular normal stresses on oblique section.
(1)
Yugal Chaudhari said:
6 years ago
Use @Kamal talukdar formula. It is correct.
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.
(27)
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)
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.
Z-shear stress.
θ perpendicular means 90.
900-(-300)-0=1200.
The answer should be 1200.
(5)
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.
Anwarul haque said:
8 years ago
Fastest Method is Mohrs circle Method.
(4)
Sagar bankar said:
8 years ago
Maximum normal stress : σ 1=1200 Mpa , σ2=600 Mpa, τ= 400 Mpa.
Maximum normal stress =(σ 1+σ2/2) +(1/2) √[(σ 1-σ2)^2 +4(T)^2].
=(1200+600/2) +(1/2) √ [(1200-600)^2 +4(400)^2],
=(1800/2)+(1000/2),
=1400 MPa.
Maximum normal stress =(σ 1+σ2/2) +(1/2) √[(σ 1-σ2)^2 +4(T)^2].
=(1200+600/2) +(1/2) √ [(1200-600)^2 +4(400)^2],
=(1800/2)+(1000/2),
=1400 MPa.
(37)
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