In this tutorial we will:
Consider this example of a simply supported beam with three point loads.
Reactions R1 and R2 are calculated as follows (upward forces +ve}.
Σ vertical forces = 0
Gives R1 + R2 - 2 - 8 - 6 = 0 Thus R1 + R2 = 16 kN --------- (i)
Σ of moments about left hand end = 0
Clockwise moments are +ve
Gives (2 x 2) + (8 x 4) + (6 x 6) - (R2 x 8) = 0
Thus (8 x R2) = 4 +32 +36 = 72 Gives R2 = 9 kN
Substitute R2 in (i) Gives R1 = 7 kN
We will now calculate shearing forces V and bending moments M along the beam using free body diagrams from datum x = 0 .
Firstly note that M = 0 at x = 0 and at x = 8 since points of simple support cannot sustain bending moments.
Now consider free body diagrams for sections of the beam between each point load, say at x=1, x=3. x=5, x=7 and take Σ vertical forces = 0 and Σ moments = 0 for each section.
Firstly consider the section from x = 0 to x = 1. Assume that we do not know the signs of V and M.
Σ vertical forces = 0
±V + 7 = 0 Thus V (appears to be) = -7 kN
However, the sign convention for shearing forces on an element of the beam is:
We must reverse the sign of V so in this case V is +ve thus V = + 7 kN
Σ moments = 0 (at position x = 1)
R1 x 1) ± M = 0
Gives ( 7 x 1) ± M = 0 Thus M (appears to be) = -7 kNm
However, the sign convention for +ve and -ve bending moments on an element of the beam is:
We must reverse the sign of M So in this case M is +ve thus M = +7 kNm
The free body diagram is as follows showing +ve bending and +ve shear.
Note that the conflict regarding sign conventions can be avoided if the direction of the x axis is reversed and the datum point for taking sections is from the right hand end of the beam. On balance I prefer using the conventional direction for the x axis.
Sections from x = 0 to x = 3, x = 5 and x = 7 follow below.
At x = 3 V = 5 kN M = 19 kNm
At x = 5 V = -3 kN M = 21 kNm (note V changes sign from + to - )
At x = 7 V = -9 kN M = 9kNm
Note that V is constant between the point loads, i.e.
for
0 < x < 2 V = 7 kN
for
2 < x < 4 V = 5 kN
for
4 < x < 6 V = -3 kN
for
6 < x < 8 V = -9 kN
At positions of point loads and reaction forces V is indeterminate, i.e. there is a step change.
We plot values for V on a shearing force diagram
Before we construct the equivalent bending moment diagram consider the changes in M and values of V for sections along the beam.
By calculation:
at x = 2 M = 14 kNm
at x = 4 M = 24 kNm
at x = 6 M = 18 kNm
Section of
beam ( Δx = 1 m) |
ΔM (kNm) | ΔM/Δx | V (kN) |
x = 0 to 1 | +7 | +7 | +7 |
x = 1 to 2 | +7 | +7 | +7 |
x = 2 to 3 | +5 | +5 | +5 |
x = 3 to 4 | +5 | +5 | +5 |
x = 4 to 5 | -3 | -3 | -3 |
x = 5 to 6 | -3 | -3 | -3 |
x = 6 to 7 | -9 | -9 | -9 |
x = 7 to 8 | -9 | -9 | -9 |
We can now construct the bending moment diagram for the beam in this example noting that the slope of M is constant between the point loads and equals V (as shown in the above table).
I welcome feedback at:
alistair@alistairstutorials.co.uk