Mechanical Engineering - Machine Design - Discussion
Discussion Forum : Machine Design - Section 2 (Q.No. 41)
41.
A shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will be
Discussion:
7 comments Page 1 of 1.
Herat Sinroja said:
5 years ago
(T)max= √((80-0)/2)^2 + (30)^2),
= 50 N/mm^2.
(T) = (σ)y/2 = 280/2 = 140 N/mm^2,
FOS = 140/50 = 2.8.
= 50 N/mm^2.
(T) = (σ)y/2 = 280/2 = 140 N/mm^2,
FOS = 140/50 = 2.8.
(4)
Lakshman said:
6 years ago
Thank you all.
(1)
Manoj said:
1 decade ago
How did you get it?
Saurav said:
1 decade ago
Allowable shear stress = 6(sigma)yt/2 = 140.
So,
140/f.s. = sqrt((6x/2)^2+tau^2).
Solving, f.s. = 2.8.
So,
140/f.s. = sqrt((6x/2)^2+tau^2).
Solving, f.s. = 2.8.
Shahid said:
10 years ago
I m getting 4.66 By using this formula.
Tau (Max) = (Sigma) yt/(2*F.S).
Tau (Max) = (Sigma) yt/(2*F.S).
Vishwas Patel said:
10 years ago
Max principle stress = (sigma)/2 + sqrt[(sigma/2)^2 + (tau)^2].
= 40*10^6 + 50*10^6.
= 90*10^6.
Min. principle stress = (sigma)/2 - sqrt[(sigma/2)^2 + (tau)^2].
= 40*10^6 - 50*10^6.
= -10*10^6.
Now according to max shear stress theory:
(Max stress - Min stress)/2 = (Strength at yield point)/2* F.S.
Hence F.S. = 2.8.
= 40*10^6 + 50*10^6.
= 90*10^6.
Min. principle stress = (sigma)/2 - sqrt[(sigma/2)^2 + (tau)^2].
= 40*10^6 - 50*10^6.
= -10*10^6.
Now according to max shear stress theory:
(Max stress - Min stress)/2 = (Strength at yield point)/2* F.S.
Hence F.S. = 2.8.
Balakrishnan said:
9 years ago
It is very useful for my placement perspective process.
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