Beams - bending moments and shearing forces

cantilever with uniform distributed load

In this tutorial we will:

investigate shearing forces and bending moments for a cantilever beam with uniformly distributed load w
establish general relationships between V, M and w

Consider this example of a cantilever beam 8 m long with uniformly distributed load w kN/m

A uniformly distributed load is equivalent to a point load at the load centre. In this case the load centre is the centroid of a rectangle in the x-y plane.

The equivalent point load is (8 m) x (1 kN/m) = 8 kN

Reaction force R1 and reaction moment M_re are calculated by Σ forces = 0 and Σ moments = 0.

Σ forces = 0 (upward forces +ve)

R1 - 8 = 0 Thus R1 = 8 kN

Σ moments = 0 (take moments about the fixed support at the left hand end - clockwise moments +ve)

(8 x 4) ± M_re = 0 Thus M = -32 kNm (anti-clockwise moment)

We will now calculate shearing forces V and bending moments M along the beam using free body diagrams from datum x = 0.

Firstly consider points x = 0 and x = 8

at x = 0 there is a step change from V = 0 to V = ±8 kN. The mode of shear at the fixed support is +ve thus V = 8 kN. The absolute value of the bending moment M at the fixed support is 32 kNm (equal to reaction moment M_re ). The mode of bending is negative thus M = -32 kNm.

at x = 8 (the right hand end of the beam) shear force and bending moment cannot be sustained thus V = 0 and M = 0

Now consider, say, free body diagrams for sections of the beam from x = 0 to x = 2, x = 4, and x = 6 and take Σ forces = 0 and Σ moments = 0 at these points.

Assume that we do not know the signs of V and M

Σ forces = 0

8 -2 + V = 0 thus V (appears to) = -6 kNm but change sign according to convention.

Thus V = 6 kN

Σ moments = 0

(8 x 2) - (2 x 1) -32 + M = 0 thus M (appears to) = +18 kNm but change sign according to convention.

Thus M = -18 kNm

The free body diagram is as follows showing negative bending and positive shear

Sections at x = 4 and x = 6 follow below.

V = 4 kN M = -8 kNm

V = 2 kN M = -2 kNm

The above values for V and M plus calculated values for x = 1 x = 3 x = 5 x = 7 are tabulated below.

x (m)	V (kN)	M (kNm)
0	8	-32
1	7	-24.5
2	6	-18
3	5	-12.5
4	4	-8
5	3	-4.5
6	2	-2
7	1	-0.5
8	0	0

Before constructing the shearing force diagram consider changes (ΔV) in V along the beam recalling that the distributed load is w kN/m.

Section of beam ( Δx = 1 m)	ΔV (kN)
x = 0 to 1	-1
x = 1 to 2	-1
x = 2 to 3	-1
x = 3 to 4	-1
x = 4 to 5	-1
x = 5 to 6	-1
x = 6 to 7	-1
x = 7 to 8	-1

Below is the shearing force diagram with slope V(x) = -1 = --w

shearing force diagram

Before constructing the bending moment diagram consider changes ( ΔM) in M along the beam

Section of beam ( Δx = 1 m)	ΔM
x = 0 to 1	7.5
x = 1 to 2	6.5
x = 2 to 3	5.5
x = 3 to 4	4.5
x = 4 to 5	3.5
x = 5 to 6	2.5
x = 6 to 7	1.5
x = 7 to 8	0.5

Note that the change in ΔM between each interval Δx is -1

Using the calculated values of M at specific points along the beam we can construct an approximate bending moment diagram as follows.

The diagram is approximate because we have linked the calculated values of M at specific points with straight lines of slope ΔM/Δx whereas M is really a quadratic function of x. We will now find this function using the general relationship .d²M/dx² = -w.

We use the general model below for a beam length L with uniformly distributed load w.

From previous work above we know that the load centre of the distributed load occurs at x = L/2.

Thus the reaction moment (at point x = 0) M_re = -wL²/2

The corresponding bending moment M at x = 0 is -wL²/2 (sign convention for negative bending)

At M = L we know that M = 0 and dm/dx = 0

Now use the general relation d²M/dx² = -w and then integrate twice to find M as a function of x. These are general integrals (no defined limits) so we must include constants of integration.