A close coil helical spring when subjected to a moment M having its axis along the axis of the helix
A. It is subjected to pure bending
B. Its mean diameter will decrease
C. Its number of coils will increase
D. All the above
Answer: Option A
Solution (By Examveda Team)
A close coil helical spring is designed primarily to resist axial loads, but when a moment is applied along the axis of the helix, it behaves differently.In this case, the axial moment causes each coil to bend rather than twist or compress.
This results in pure bending of the spring wire, without torsion or axial compression/stretching.
Option B: Its mean diameter will decrease – Not true in this scenario; there is no significant radial compression due to moment.
Option C: Its number of coils will increase – This is incorrect; the number of coils is fixed and does not change with loading.
Therefore, only Option A is correct, as the spring is subjected to pure bending under the applied moment.
This Question Belongs to Civil Engineering >> Theory Of Structures
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Comments (2)
A. $$\frac{2}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{5}{8}$$
D. $$\frac{8}{5}$$
Principal planes are subjected to
A. Normal stresses only
B. Tangential stresses only
C. Normal stresses as well as tangential stresses
D. None of these
A. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$
B. $$\frac{{\text{I}}}{{\text{M}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$
C. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{F}}}{{\text{Y}}}$$
D. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{Y}}}{{\text{F}}}$$
A. $$\frac{{\text{M}}}{{\text{T}}}$$
B. $$\frac{{\text{T}}}{{\text{M}}}$$
C. $$\frac{{2{\text{M}}}}{{\text{T}}}$$
D. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
Correct option. B
Its answer should be all of the above options as number of coils will increase along with reduced mean dia