Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 37)
37.
A composite shaft consisting of two stepped portions having spring constants k1 and k2 is held between two rigid supports at the ends. Its equivalent spring constant is
k1+k2
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
29 comments Page 2 of 3.

Koribilli hemanth said:   9 years ago
The given answer is wrong.

The right answer is K1 + K2, as we know that any shaft whether stepped or normal if it fixed rigidly between the two supports, it acts like a parallel shaft. Because if apply torque anywhere on the shaft which is fixed between two rigid supports the angle of deformation of both the shafts is same.

Karthik said:   9 years ago
Which is the correct answer? Please tell me.

Swapnil Shinde said:   9 years ago
As the both ends are fixed (then torque could be applied at the junction of the compound shaft ) hence it forms the parallel combination, hence answer will be K1 + K2.

And if one end is free (then torque will be applied at the free end) hence it forms the series combination , and then the answer will be C.

Hope it will clear the doubts.

Dev verma said:   9 years ago
True Answer is option C.

V.raju said:   9 years ago
When springs in series the spring constant is k1k2 / k1 + k2.

Bindusara said:   1 decade ago
Since the compound shaft is fixed rigidly at both ends the case is similar to that of springs connected in parallel so the equivalent spring constant is K1 + K2.

Yella said:   10 years ago
It becomes parallel case so equivalent stiffness is k1 + k2.

Subbu said:   10 years ago
How it is k1k2? The units are not even matching.

Baburao said:   10 years ago
Please rectify the wrong answers. Provide answer with description.

Vitthal said:   10 years ago
What is mean by rigidly connected?


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