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.
Harish said:
1 decade ago
Can you solve this sum again clearly?
Mona said:
1 decade ago
I need the solution urgently need. Can you please solve again?
Shahrukh ansari said:
1 decade ago
Max. normal stress is equal to max. principal stress.
Nabin said:
1 decade ago
LET
T.S1=1200 and T. S2=600 and S.S=400 normal stress = T.S1+T.S2-S.S.
Then 1200+600-400 = 1400. So maximum normal stress = 1400 Mpa.
T.S1=1200 and T. S2=600 and S.S=400 normal stress = T.S1+T.S2-S.S.
Then 1200+600-400 = 1400. So maximum normal stress = 1400 Mpa.
Pradeep yadav said:
10 years ago
@Suresh you have any formula for normal stress in this manner.
Mithlesh said:
10 years ago
Its due to normal stress formula.
Satyam jha said:
10 years ago
Principal stress are maximum and minimum normal stress which may be developed on a loaded bodyNormal Stress = (Shear Stress*sin2theta)+((Fx+Fy)/2)+(((Fx-Fy)/2)cos2theata).
In this Case theta = 90 because they are at right angle.
Other all values are given on calculation we get answer A)400.
Reference: Strength of materials by R.K.Rajput.
Chapter 2, Page NO.95 Eq.2.8..
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)].
In this Case theta = 90 because they are at right angle.
Other all values are given on calculation we get answer A)400.
Reference: Strength of materials by R.K.Rajput.
Chapter 2, Page NO.95 Eq.2.8..
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)].
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