Question

A cantilever beam 2.5m long carries a uniformly distributed load 50 kn per meter over length of 1.5m from the free end. The value of maximum shear force and bending moment for the beam will be respectively;

Answer 1

diagram.

It may also be observed that a constant shear force produces a uniform change in the bending moment, resulting in straight line in the moment diagram. If no shear force exists along a certain portion of a beam, then it indicates that there is no change in moment takes place. It may also further observe that dm/dx= F therefore, from the fundamental theorem of calculus the maximum or minimum moment occurs where the shear is zero. In order to check the validity of the bending moment diagram, the terminal conditions for the moment must be satisfied. If the end is free or pinned, the computed sum must be equal to zero. If the end is built in, the moment computed by the summation must be equal to the one calculated initially for the reaction. These conditions must always be satisfied.

Answer 2

Answer:

Maximum shear force = 75 KN.

Maximum bending moment = 103.75 KN-m

Explanation:

The given data :-

Type of beam - cantilever beam.

Uniformly distributed load of 50 KN/m.

Length of beam = 2.5 m.

Given that load is distributed to 1.5 m from the free end.

Let the reaction of beam at fixed end $R_{a}$

Net vertical load of beam = 1.5 * 50 = 75 KN ( downward )

Sum of vertical force = 0

$R_{a}$  =75 KN ( upward)

The maximum shear force = 75 KN between fixed end to a distance of 1 m from fixed end.

The moment reaction at fixed end $M_{a}$  = 135 KN-m ( anticlockwise )

Let a cross-section x-x from the fixed end at a distance of x.

Moment equation is,

$M_{x-x} = - 135 +(75 * x) - ( 50 *(x-1)*\frac{x-1}{2} )$

For maximum bending moment

$\frac{dM_{x-x} }{dx} = 0$

we get x = 0.5

putting the value of x in moment equation

M = -135 +37.5 - 6.25 = 103.75 KN-m ( anticlockwise )