The maximum pull in a cable, carrying a uniformly distributed load and supported at two ends which are at the same level, is at

The natural frequency of free transverse vibrations due to uniformly distributed load acting over a simply supported shaft is (where $$\delta {\text{S}}$$ = Static deflection of simply supported shaft due to uniformly distributed load)

A

$$\frac{{0.4985}}{{\sqrt {\delta {\text{S}}} }}$$

B

$$\frac{{0.5615}}{{\sqrt {\delta {\text{S}}} }}$$

C

$$\frac{{0.571}}{{\sqrt {\delta {\text{S}}} }}$$

D

$$\frac{{0.6253}}{{\sqrt {\delta {\text{S}}} }}$$

When a uniformly distributed load, longer than the span of the girder, moves from left to right, then the maximum bending moment at mid section of span occurs when the uniformly distributed load occupies

A

Less than the left half span

B

Whole of left half span

C

More than the left half span

D

Whole span

The shear force at the ends of a simply supported beam carrying a uniformly distributed load of w per unit length is

A

Zero at its both ends

B

$${\text{w}}l$$ at one end and $$ - {\text{w}}l$$ at the other end

C

$$\frac{{{\text{w}}l}}{2}$$ at one end and $$ - \frac{{{\text{w}}l}}{2}$$ at the other end

D

$$\frac{{{\text{w}}{l^2}}}{2}$$ at one end and $$ - \frac{{{\text{w}}{l^2}}}{2}$$ at the other end

The simply supported beam 'A' of length '$$l$$' carries a central point load 'W'. Another beam 'B' is loaded with a uniformly distributed load such that the total load on the beam is 'W'. The ratio of maximum deflections between beams 'A' and 'B' is

A

$$\frac{5}{8}$$

B

$$\frac{8}{5}$$

C

$$\frac{5}{4}$$

D

$$\frac{4}{5}$$

A simply supported beam A carries a point load at its mid span. Another identical beam B carries the same load but uniformly distributed over the entire span. The ratio of the maximum deflections of the beams A and B, will be

A

$$\frac{2}{3}$$

B

$$\frac{3}{2}$$

C

$$\frac{5}{8}$$

D

$$\frac{8}{5}$$

The bending moment in the center of a simply supported beam carrying a uniformly distributed load of w per unit length is

A

Zero

B

$$\frac{{{\text{w}}{l^2}}}{2}$$

C

$$\frac{{{\text{w}}{l^2}}}{4}$$

D

$$\frac{{{\text{w}}{l^2}}}{8}$$

The shear force diagram for a simply supported beam carrying a uniformly distributed load of w per unit length, consists of

A

One right angled triangle

B

Two right angled triangles

C

One equilateral triangle

D

Two equilateral triangles

In a simply supported beam carrying a uniformly distributed load w per unit length, the point of contraflexure

A

Lies in the center of the beam

B

Lies at the ends of the beam

C

Depends upon the length of beam

D

Does not exist

The shear force in the center of a simply supported beam carrying a uniformly distributed load of ‘w’ per unit length, is

A

Zero

B

$$\frac{{{\text{w}}{l^2}}}{2}$$

C

$$\frac{{{\text{w}}{l^2}}}{4}$$

D

$$\frac{{{\text{w}}{l^2}}}{8}$$

The bending moment diagram for a simply supported beam carrying a uniformly distributed load of ‘w’ per unit length, will be

A

A horizontal line

B

A vertical line

C

An inclined line

D

A parabolic curve

The maximum pull in a cable, carrying a uniformly distributed load and supported at two ends which are at the same level, is at

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