Mechanical Engineering - Machine Design - Discussion

Discussion Forum : Machine Design - Section 4 (Q.No. 6)
6.
A circular shaft can transmit a torque of 5 kN-m. If the torque is reduced to 4 kN-m, then the maximum value of bending moment that can be applied to the shaft is
1 kN-m
2 kN-m
3 kN-m
4 kN-m
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

Manish said:   5 years ago
@All.

Acc. to question there should be some relation between bending moment and torque, we can't equate it to equivalent torque, as it states that it can transfer torque of 5kN-m and if this torque is reduced, then the value of bending moment is what, this means there should lie a value of bending moment when the torque is 5kNm and when we reduce the torque to 4kNm what will be the value of bending moment. so it can't be equated to equivalent torque equation, in another way if we equate the max. bending equals to the max. shear, then we can get a linear relation between bending moment and torque. as T=2M, so this can be used in this question, so the answer would have been 2kNm.

At T=5,M=2.5
at T=4, M=2
(1)

Gaurav said:   1 decade ago
How 3 kN-m?

Engineer Abdul baqi said:   1 decade ago
In this equation if you people said me the using formula only send me using formula of this equation.

Zuhaib said:   1 decade ago
Send me formula dear.

Getaneh said:   10 years ago
What is the formula?

Sayad nasim said:   10 years ago
Hello dear one, formula is:

T(equivalent) = sqrt root (T^2+M^2).

=> 5^2 = sqrt root (4^2+M^2).

=> M = 3.

Sayad nasim said:   10 years ago
T (equivalent) = sqrt root (M^2+T^2).

Tulu said:   9 years ago
We know that

T = √ T^2 + √ M^2.

From this formula, we found that M = 3Knm.

Shankar B C said:   8 years ago
Nice explanations. Thanks.

Jayram chauhan said:   8 years ago
Thanks for the explanation.


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