In the calculation of induced shear stress in helical springs, the Wahl's stress factor is used to take care of

When an open coiled helical compression spring is subjected to an axial compressive load, the maximum shear stress induced in the wire is (where D = Mean diameter of the spring coil, d = Diameter of the spring wire, K = Wahl's stress factor and W = Axial compressive load on the spring)

A

$$\frac{{{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

B

$$\frac{{2{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

C

$$\frac{{4{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

D

$$\frac{{8{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

A closely coiled helical spring is acted upon by an axial force. The maximum shear stress developed in the spring is $$\tau $$. The half of the length of the spring if cut off and the remaining spring is acted upon by the same axial force. The maximum shear stress in the spring in new condition will be

A

$$\frac{\tau }{2}$$

B

$$\tau $$

C

$$2\tau $$

D

$$4\tau $$

The design stress, which is more than the damaging stress, is the least unit stress that will render a member unfit for service before the end of its normal life. The design stress factor or factor of safety indicates the margin between design stress and the __________ stress.

A

Working

B

Damaging

C

Allowable

D

None of these

The strain energy stored in a body due to shear stress, is (where $$\tau $$ = Shear stress, C = Shear modulus and V = Volume of the body)

A

$$\frac{{\tau {\text{V}}}}{{2{\text{C}}}}$$

B

$$\frac{{2{\text{C}}}}{{\tau {\text{V}}}}$$

C

$$\frac{{{\tau ^2}{\text{V}}}}{{2{\text{C}}}}$$

D

$$\frac{{2{\text{C}}}}{{{\tau ^2}{\text{V}}}}$$

When a circular beam of diameter d is subjected to a shear force F, the maximum shear stress induced will be

A

$$\frac{{4{\text{F}}}}{{\pi {{\text{d}}^2}}}$$

B

$$\frac{{{\text{6F}}}}{{\pi {{\text{d}}^2}}}$$

C

$$\frac{{{\text{8F}}}}{{\pi {{\text{d}}^2}}}$$

D

$$\frac{{{\text{16F}}}}{{3\pi {{\text{d}}^2}}}$$

When a beam of rectangular cross-section of width b and depth d, is subjected to a shear force F, the maximum shear stress induced will be

A

$$\frac{{\text{F}}}{{{\text{bh}}}}$$

B

$$\frac{{{\text{3F}}}}{{{\text{2bh}}}}$$

C

$$\frac{{{\text{2F}}}}{{{\text{bh}}}}$$

D

$$\frac{{{\text{4F}}}}{{{\text{bh}}}}$$

If a spring is cut down into two springs, the stiffness of cut springs will be

A

Half

B

Same

C

Double

D

None of the above

The cracks in helical springs used in railway carriages usually start on the inner side of the coil because of the fact that

A

It is subjected to a higher cyclic loading than the outer side

B

It is subjected to a higher stress than the outer side

C

It is more stretched than the outer side during the manufacturing process

D

It has a lower curvature than the outer side

The stress induced in a body, when suddenly loaded, is _________ the stress induced when the same load is applied gradually.

A

Equal to

B

One-half

C

Twice

D

Four times

When a body is subjected to biaxial stress i.e. direct stresses ($${\sigma _{\text{x}}}$$) and ($${\sigma _{\text{y}}}$$) in two mutually perpendicular planes accompanied by a simple shear stress ($${\tau _{{\text{xy}}}}$$ ), then maximum shear stress is

A

$$\frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$

B

$$\frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$

C

$$\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$

D

$$\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$

In the calculation of induced shear stress in helical springs, the Wahl's stress factor is used to take care of

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