Draw the Shear and Moment Diagram for the beam. Indicate values at the supports and at points where change in load occurs.

 Draw the Shear and Moment Diagram for the beam. Indicate values at the supports and at points where change in load occurs.

Solution:

Let RA and RB are the vertical support reactions at A and B, and MA and MB be the bending moments at the same.

As per the Conditions of Static Equilibrium of the Structural Element:
∑Fy = 0, ∑Fx = 0, and ∑M = 0

Values at Supports:

The Shear Force at the left support A, VA = 0.5*[(-4.0)] * 1.5
or VA = -3.0 kN

The bending moment at the left support A, MA =0.5 * (-4.0) * (1.5) * ( 1.5/3)

Hence, MA = -1.5 kNm

The Shear Force at the right support B, VB = 0.5*[(4.0)] * 1.5
or VB = 3.0 kN

The bending moment at the right support B, MB = 0.5 * (-4.0) * (1.5) * (1.5/3)

Hence, MB = -1.5 kNm

Using the static equilibrium conditions,
∑Fy = 0, RA + RB = 0.5 * [4.0] * (1.5) + 4 * (2.0) + 0.5 * [4.0] * (1.5)

Hence RA + RB = 14.0 kN — eq(1)
Also, ∑M =0, hence the algebraic sum of the moments at left support is equated to zero.
Hence, RB*(5.0-1.5-1.5) = [(-1.5) – (-1.5) + (3.0)(5.0 – 1.5 – 1.5) -4.0 * (2.0) * 1.0

Hence, RB = 7.0 kN
From eq(1), RA = 7.0 kN

Shear and Moment equations:

In between A and B,
for 1.5m <= x <= 3.5m:
Shear, Vx = 10.0 – 4.0*x

Moment, Mx = VA(x-1.5) + -4.0(x -1.5)^2/2 = -2.0x2 + 10.0x – 12.0

In between B and the right end,
for 3.5m <= x <= 5.0m:
Load intensity at x, wx = 2.667x – 13.33

Shear, Vx = 1.333x*2 – 13.33x + 33.33

Moment, Mx = 0.445x3 – 6.667x2 + 33.33x – 55.55

Shear Force and Bending Moment Diagram:

distance x. Shear V Moment M
0.0. 0.0. 0.0
1.5 4.0 -1.5
3.5 3.0. -1.5
5.0 0.0 0.0

Maximum +ve Shear Force, Vmax = 4.0 kN
at x = 1.5 m from the left end.

Maximum -ve Shear Force, Vmin = -4.0 kN
at x = 3.5 m from the left end.

Maximum +ve Bending Moment, Mmax = 0.5 kN-m
at x = 2.5 m from the left end.

Maximum -ve Bending Moment, Mmin = -1.5 kN-m
at x = 3.5 m from the left end.