Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 44)
44.
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 minimum normal stress will be
Discussion:
3 comments Page 1 of 1.
Rajasekhar said:
6 years ago
@Dinesh.
You are calculating for max normal stresses min normal stress has a negative sign before root.
So, 900-500 = 400 is the correct answer.
You are calculating for max normal stresses min normal stress has a negative sign before root.
So, 900-500 = 400 is the correct answer.
(1)
DINESH said:
8 years ago
Answer 1400 MPa.
Solution:
Assume,
stress in X direction : sigma(X)= 1200 MPa,
stress in Y direction : sigma(Y)= 600 MPa,
shear stress (Ï„ XY)=400 MPa.
We know maximum normal stress formula and substitute in it.
=((1200+600)/2)+sq root of (((1200-600)/2)^2 + 400^2).
=(900+500) MPa.
=1400 MPa.
Solution:
Assume,
stress in X direction : sigma(X)= 1200 MPa,
stress in Y direction : sigma(Y)= 600 MPa,
shear stress (Ï„ XY)=400 MPa.
We know maximum normal stress formula and substitute in it.
=((1200+600)/2)+sq root of (((1200-600)/2)^2 + 400^2).
=(900+500) MPa.
=1400 MPa.
(2)
Shail said:
1 decade ago
Min normal stress:
= (1200+600)/2 - sqrt[ {(1200-600)/2}sq + {400}sq ].
= 900- sqrt[ {300}sq + {400}sq].
= 900- sqrt[90000+160000].
= 900- sqrt[250000].
= 900- 500.
= 400 MPa.
= (1200+600)/2 - sqrt[ {(1200-600)/2}sq + {400}sq ].
= 900- sqrt[ {300}sq + {400}sq].
= 900- sqrt[90000+160000].
= 900- sqrt[250000].
= 900- 500.
= 400 MPa.
(1)
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