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.
Nepolepn pradhan said:
1 decade ago
Let,tensile stress on one plane(t.s1) = 1200mpa.
Tensile stress on another plane(t.s2) = 600mpa.
Shear stress (s.s) = 400mpa.
Now using major principal stress formula = [(t.s1+t.s2)/2]+sqrt[{(t.s1-t.s2)*(t.s1-t.s2)}/2 +(s.s*s.s)].
Tensile stress on another plane(t.s2) = 600mpa.
Shear stress (s.s) = 400mpa.
Now using major principal stress formula = [(t.s1+t.s2)/2]+sqrt[{(t.s1-t.s2)*(t.s1-t.s2)}/2 +(s.s*s.s)].
Kaushl said:
1 decade ago
Normal stress & principal stress are not same, then how we can use the formula of max principal stress for finding normal stress.
Amol Kamale said:
1 decade ago
Yes, this formula applicable for finding Max. Normal stress in all types of questions.
Rohit said:
1 decade ago
@Amol is this formula applicable for finding Max. Normal stress in all types of questions?
Amol Kamale said:
1 decade ago
Assume stress in X direction(X) = 1200.
Stress in Y direction(Y) = 600.
Shear stress(S) = 400.
Max. Normal stress=(X+Y)/2+Sqrt(((X-Y)/2)square+S square).
=(1200+600)/2 + sqrt(300 square + 400 square).
=1400 mpa.
Stress in Y direction(Y) = 600.
Shear stress(S) = 400.
Max. Normal stress=(X+Y)/2+Sqrt(((X-Y)/2)square+S square).
=(1200+600)/2 + sqrt(300 square + 400 square).
=1400 mpa.
ABHI BANSAL said:
1 decade ago
By using mohr's circle we can easily find out the maximum principle stress.
Umesh said:
1 decade ago
Use max. Principal stress formula.
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