Related Questions

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 $$\frac{{0.6253}}{{\sqrt {\delta {\text{S}}} }}$$
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 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 $$\frac{4}{5}$$
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
A simply supported beam which carries a uniformly distributed load has two equal overhangs. To have maximum B.M. produced in the beam least possible, the ratio of the length of the overhang to the total length of the beam, is
A 0.207
B 0.307
C 0.407
D 0.508
The bending moment of a cantilever beam of length '$$l$$' and carrying a uniformly distributed load of 'w' per unit length is __________ at the fixed end.
A $$\frac{{{\text{w}}l}}{4}$$
B $$\frac{{{\text{w}}l}}{2}$$
C $${\text{w}}l$$
D $$\frac{{{\text{w}}{l^2}}}{2}$$
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
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 $$\frac{{{\text{w}}{l^2}}}{8}$$
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

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