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 5 of 6.
L3FA said:
9 years ago
Thanks for the given explanations.
Rakesh patra said:
8 years ago
Principle stress are maximum or minimum normal stress which may be developed on a loaded body.
Ram prasad said:
8 years ago
Thanks for the given explanation.
Rishu said:
8 years ago
Thanks.
Hareshkumar said:
8 years ago
The resultant tensile stress is (1200+600)/2 =900 MPA shear stress =400 Mpa using maximum normal stress theory 1/2(900+ 900^2+4 * 400^2) = 1400Mpa.
Manikanta said:
8 years ago
Plane at which max normal stress & minimum normal stress present is called as principal plane. So, maximum principal stress is same as the maximum normal stress.
(1)
Afrooj said:
8 years ago
Thanks for all for the explanation.
Shiva said:
8 years ago
Maximum normal stress : σ 1=1200 Mpa , σ2=600 Mpa, τ= 400 Mpa.
Maximum normal stress formula= (σ1 + σ2)/2 + √(σ1 + σ2)/2 + τ 1500 Mp.
Maximum normal stress formula= (σ1 + σ2)/2 + √(σ1 + σ2)/2 + τ 1500 Mp.
(3)
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.
(36)
Anwarul haque said:
8 years ago
Fastest Method is Mohrs circle Method.
(4)
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