Firstly, it will be necessary to calculate the reactions RA and RB.
Taking moments about B gives:
15 kN m + RA × 5 m + 10 kN × 1 m = 30 kN m
i.e. 5RA = 30 − 10 − 15 = 5
from which, RA = 5/5 = 1 kN
Resolving forces vertically gives:
RA + RB = 10 kN
i.e. 1 + RB = 10
from which, RB = 10 − 1 = 9 kN
For the span C to A, see Figure(b)
Bending moment (BM)
At x, M = 15 kN m (constant) .....(6.15)
Shearing force (SF) At x = 0, F= 0 kN ......(6.16)
For the span A to D, see Figure(c)
Bending moment (BM)
At x, M = 15 kN m + RA × (x − 2)
15 + 1(x − 2)
15 + x − 2
i.e. M = 13 + x (a straight line) ......(6.17)
At A, x = 2 m, therefore MA = 13 + 2 = 15 kN m
At D, x = 4 m, therefore MD(−) = 13 + 4
= 17 kN m
Note that MD(−) means that M is calculated to the left of D
Shearing force (SF)
At x, F = 1 kN (constant) ......(6.18)
For the span D to B, see Figure (d)
Bending moment (BM)
At x, M = 15 kN m + 1 kN m × (x − 2) − 30 kN m
= 15 + x − 2 − 30
i.e. M = x − 17 (a straight line) .......(6.19)
At D, x = 4 m, therefore MD(+) = 4 − 17 = −13 kN m
Note that MD(+) means that M is calculated just to the right of D.
At B, x = 7 m, therefore MB(−) = 7 − 17 = −10 kN m
Shearing force (SF) At x, F= −1 kN (constant) ......(6.20)
For the span B to E, see Figure (e).
In this case we will consider the right of the beam as there is only one force to the right of the section at x.
M = −10×(8 − x)= −80 + 10x (a straight line)
At x, F = 10 kN (positive as the right hand is going down, and constant) Plotting the above equations for the various spans, results in the bending moment and shearing force diagrams of Figure(f).