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
Discussion:
29 comments Page 2 of 3.
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.
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.
Suresh Kumar said:
9 years ago
These springs are having series connection so the answer must be k1k2/k1+k2.
(1)
V.raju said:
9 years ago
When springs in series the spring constant is k1k2 / k1 + k2.
Bhurmal bamne said:
10 years ago
They are parallel connections.
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?
K s pal said:
1 decade ago
Actually both springs have same magnitude of either compression or elongation [due to fixed] hence connected in parallel. So k1+k2 will be equivalent.
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