# Determine expressions for the bending moment and shearing force distributions for the beam of Figure.

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Determine expressions for the bending moment and shearing force distributions for the beam of Figure. Hence, sketch the bending moment and shearing force diagrams.

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Firstly, it will be necessary to calculate the reactions RA and RB.

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).