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.
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.
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.
Sharda said:
1 decade ago
Equivalent spring constant will be k1k2/k1+k2.
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.
Hrushikesh said:
1 decade ago
@Vishal.
Did not get you?
Did not get you?
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.
Lokesh Reddy said:
1 decade ago
k1k2/k1+k2. This is a case of series combination.
Gauri shankar said:
1 decade ago
Parallel connection: 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.
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