Calculations on moment
QUESTION 1
A beam was simply supported at both ends and carries loads that’s uniformly distributed
together with concentrated loads, as shown in figure 6.0 below
2KN/M
6KN
4m
2KN
2m
2m
RA
RB
Fig 6.0
Find the reactions at the two ends.
Solution
Making reference to fig 6.0 above all the loads on the beam are acting downward.
The uniform distributed loading is 2k/m.
Area of u.d.l = 2 x 4
Area of u.d.l = 8KN
The load would act half way from total length of the u.d.l i.e. ½ x 4
Concentrated loads are 6KN and 2KN.
8KN
2m
6KN
2m
2KN
2m
RA
Figure 6.1 free body diagram of figure 6.0
Taking moment about RA.
Clockwise moment = Anticlockwise moment
8 x 2+ 6 x 4+2 x 6 = RB x 8
16 +24 +12 = 8 RB
RB `= 52 / 8
RB = 6.5KN
To find reaction at RA
Since downward force=upward force
RA + RB = 8 + 6 +2
RA+ RB = 16KN
2m
RB
RB = 5.2KN
RA = 16 - 5.2
RA = 9.5KN
Question 2
A beam is supported at point A on fixed support and at B with on a roller support, its
subjected to loading as shown in figure 6.1, find the reaction at the supports.
10KN
2M
20KN
10KN/M
2M
3.0M
2M
Fig 6.2
Loading system
Area of u.d.l = 10x3
Area of u.d.l = 30KN
10KN
20KN
30KN
RH
RA
RVA
2M
2M
2M
RB
1.5M
1.5M
Fig 6.3 a free body diagram of fig 6.2
Taking Moment about RvA
Anticlockwise moment =clockwise moment
RB x 6 = 10 x 2+20 x 4 +7.5 x 30
6 RB = 20 +80 + 225
RB = 325 / 6
RB = 54.17KN
RH is horizontal reaction it is zero because no other force acting in horizontal direction.
RH = 0
Taking moment about RB
Anticlockwise moment = Clockwise moment
10 x 4+20 x 2 = RvA x 6 + 30 x1.5
40+40 = RvA x 6 + 45
80 - 45 = 6RVA
35 = 6RvA
35/6 = RvA
5.83KN = RvA
RvA = 5.83KN
To check for the balance of reaction at the supports
Downward forces = upward forces
10KN + 20KN + 30KN = RvA + RB
60KN = 5.83 + 54.17
60KN = 60KN
i.e. the beam is in equilibrium