Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 5 (Q.No. 5)
5.
A shaft of diameter D is subjected to a twisting moment (T) and a bending moment (M). If the maximum bending stress is equal to maximum shear stress developed, then M is equal to
Discussion:
11 comments Page 1 of 2.
Hamza said:
4 years ago
@Srilakshmi.
How MZp=TZs?
How MZp=TZs?
Humayoun said:
4 years ago
For maximum bending and shear stress M=T/√2.
For simple bending and shear stress M=T/2.
For simple bending and shear stress M=T/2.
Hardik said:
6 years ago
@Imran right.
Simran said:
7 years ago
I agree with you @Imran.
SRILAKSHMI said:
8 years ago
Bending stress= shear stress.
MY/I=TR/J
Now cross multiply
MYJ=TRI
Now MZp=TZs
M(π*D^3/16)=T(π*D^3/32)
M=T(π*D^3/32)/(π*D^3/16)
M=T/2.
MY/I=TR/J
Now cross multiply
MYJ=TRI
Now MZp=TZs
M(π*D^3/16)=T(π*D^3/32)
M=T(π*D^3/32)/(π*D^3/16)
M=T/2.
(2)
Shiva said:
8 years ago
@Sandeep.
in last step J=2I;
M=(I/2I)*T;;
M=T/2.
in last step J=2I;
M=(I/2I)*T;;
M=T/2.
Siva Kumar G said:
9 years ago
@Sandeep.
Calculate well from the last second step. You will get T/2.
Calculate well from the last second step. You will get T/2.
Sandeep said:
9 years ago
I=((φ)*(D^4))/64 and,
J=((φ)*(D^4))/32, for a circle.
NOW,
Bending moment = Shear stress.
or, (M/I) * y = (T/J) * r.
or, M = (I/J) * T * (r/y) [since, r=Ymax]
or, M = (I/J) * T.
so, M = 2 T.
J=((φ)*(D^4))/32, for a circle.
NOW,
Bending moment = Shear stress.
or, (M/I) * y = (T/J) * r.
or, M = (I/J) * T * (r/y) [since, r=Ymax]
or, M = (I/J) * T.
so, M = 2 T.
Imran said:
9 years ago
@Gaurav, Thanks for your answer.
But I think that answer is for simple bending and shear stress not for Max bending and shear stresses.
But I think that answer is for simple bending and shear stress not for Max bending and shear stresses.
Boss said:
9 years ago
The answer is 2T.
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