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 1 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.
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.
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.
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.
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.
Kikani vishal said:
1 decade ago
Here springs are in parallel because ends are rigid connected. So equivalent spring constant is equal to k1+k2. That is not given in option.
Vicky said:
1 decade ago
Answer is k1+k2, as both the ends are fixed. So it is a case of spring connected in parallel. But answer is not given in option.
Prakash Joshi said:
1 decade ago
This is a case of 2 springs in series. Hence equivalent spring constant will be k1k2/(k1+K2).
C should be answer.
C should be answer.
Abin said:
1 decade ago
The equivalent spring constant should be k1k2/(k1+k2) as this is a case of springs in series.
Chandra prakash said:
9 years ago
For series = k1k2/k1+k2.
For parallel = k1 + k2.
But the case is not mentioned in question.
For parallel = k1 + k2.
But the case is not mentioned in question.
Suresh Kumar said:
9 years ago
These springs are having series connection so the answer must be k1k2/k1+k2.
(1)
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