Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 38)
38.
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 shear stress will be
Discussion:
9 comments Page 1 of 1.
Rashid Khan Kakar said:
4 years ago
400 is the right answer.
(2)
Karan said:
6 years ago
Max.shear stress = & radic(a-b/2)^2+c^2
Where,a=tensile stress along X-direction
b=tensile stress along Y-direction
c=shear stress.
Given,a=300 mpa,
b=0 mpa.
c=200 mpa.
The answer is 250 mpa.
Where,a=tensile stress along X-direction
b=tensile stress along Y-direction
c=shear stress.
Given,a=300 mpa,
b=0 mpa.
c=200 mpa.
The answer is 250 mpa.
(1)
Debattam Dey said:
2 years ago
The general Formula of principal stress is (P1)=Px+Py/2+ √(Px-Py/2)²+q² .
Maximum shear stress i.e. Pt(max) which will be √ (Px-Py/2)²+q².
Here given, Py=0, Px=300MPa and q=200MPa. [Px=Major principal stress, Py=Minor principal stress and q=Shear stress].
so, √(Px/2)²+q²,
= √(300/2)²+200²,
= 250MPa (Ans).
Maximum shear stress i.e. Pt(max) which will be √ (Px-Py/2)²+q².
Here given, Py=0, Px=300MPa and q=200MPa. [Px=Major principal stress, Py=Minor principal stress and q=Shear stress].
so, √(Px/2)²+q²,
= √(300/2)²+200²,
= 250MPa (Ans).
(1)
Anand said:
1 decade ago
Can anyone explain please?
Rajat said:
1 decade ago
Max. shear stress = 0.5*square root [{(300-0)/2}square + 200sqr].
Pratik said:
9 years ago
Max shear stress = (tensile stress ^2 - shear stress ^2)/2.
Thulasya naik said:
9 years ago
Max.shear stress = 0.5 * square root[300^2 + 4 * 200^2].
= 250 MPA.
= 250 MPA.
Rahil said:
9 years ago
Please, explain properly.
BIPLAB said:
9 years ago
Square root [{(300 - 0)/2}square + 200sqr] = square root[150^2 + 200^2] = 250.
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