point loads
1. Calculate the force RA and distance d for the beam shown in Figure 5.16. The mass of the beam should be neglected and equilibrium conditions assumed.
12 mm 0.2 kN RA d 2.7 kN 0.4 kN 1.3 kN 15 mm 10 mm Figure 5.16
2. For the force system shown in Figure 5.17, find the values ofF andd for the system to be in equilibrium. [1.0 kN, 64 mm] 1.4 kN 12 mm d F 14 mm 5 mm 0.8 kN 2.3 kN 0.7 kN Figure 5.17
3. For the force system shown in Figure 5.18, determine distance d for the forces RA andRB to be equal, assuming equilibrium
conditions. [80 m] 10 N RA RB 20 m 20 m d 20 m 15 N 25 N Figure 5.18
4. A simply supported beamABis loaded as shown in Figure 5.19. Determine the load F in order that the reaction atAis zero.
[36 kN] 10 kN R1 R2 2 m 2 m 2 m 2 m 16 kN F B A Figure 5.19
5. A uniform wooden beam, 4.8 m long, is supported at its left-hand end and also at
3.2 m from the left-hand end. The mass of the beam is equivalent to 200 N acting vertically downwards at its centre. Deter- mine the reactions at the supports.
[50 N, 150 N] 6. For the simply supported beamP Qshown in Figure 5.20, determine (a) the reaction at each support, (b) the maximum force which can be applied atQwithout losing equilibrium. [(a)R1=3 kN, R2 =12 kN (b) 15.5 kN] 4 kN R1 R2 4.0 m 1.5 m 1.5 m 2.0 m 6 kN 5 kN Q P Figure 5.20
7. A uniform beamAB is 12 m long and is supported at distances of 2.0 m and 9.0 m fromA. Loads of 60 kN, 104 kN, 50 kN and 40 kN act vertically downwards at A, 5.0 m from A, 7.0 m from A and at B. Neglecting the mass of the beam, determine the reactions at the supports.
[133.7 kN, 120.3 kN] 8. A uniform girder carrying point loads is shown in Figure 5.21. Determine the value of load F which causes the beam to just lift off the supportB. [3.25 kN]
4 kN 5 kN 10 kN F 2 m 4 m 3 m 2 m 4 m RA RB Figure 5.21
5.4
Simply supported beams with
couples
The procedure adopted here is a simple extension to Section 5.3, but it must be remembered that the units
of a couple are in: N m, N mm, kN m, etc, unlike that of a force. The method of calculating reactions on beams due to couples will now be explained with the aid of worked problems.
Problem 11. Determine the end reactions for the simply supported beam of
Figure 5.22, which is subjected to an anti-clockwise couple of 5 N m applied at mid-span. 5 kN m 1.5 m 1.5 m RA RB B C A Figure 5.22
Taking moments aboutB:
Now the reaction RA exerts a clockwise moment about B given by: RA × 3 m. Additionally, the couple of 5 kN m is anti-clockwise and its moment is 5 kN m regardless of where it is placed.
Clockwise moments aboutB
= moments aboutanti-clockwiseB
i.e. RA×3 m=5 kN m (5.1)
from which, RA= 5
3 kN
or RA =1.667 kN (5.2) Resolving forces vertically gives:
Upward forces=downward forces
i.e. RA+RB =0 (5.3)
It should be noted that in equation (5.3) the 5 kN m couple does not appear; this is because it is a couple and not a force. From equations (5.2) and (5.3),
RB = −RA=−1.667 kN
i.e. RB acts in the opposite direction to RA, so thatRB andRA also form a couple that resists the 5 kN m couple.
Problem 12. Determine the end reactions for the simply supported beam of
Figure 5.23, which is subjected to an
anti-clockwise couple of 5 kN m at the pointC 5 kN m 2 m 1 m RA A B C RB Figure 5.23
Taking moments aboutB gives:
RA×3 m=5 kN m (5.4)
from which, RA= 5
3 kN
or RA =1.667 kN
Resolving forces vertically gives:
i.e. RA+RB =0
from which, RB = −RA=−1.667 kN It should be noted that the answers for the reactions are the same for Problems 11 and 12, thereby proving by induction that the position of a couple on a beam, simply supported at its ends, does not affect the values of the reactions.
Problem 13. Determine the reactions for the simply supported beam of Figure 5.24.
10 kN m 8 kN m 6 kN m 1 m 1 m 1 m 1 m A B C D E RA RB Figure 5.24
Taking moments aboutB gives:
RA×4 m+8 kN m=10 kN m+6 kN m
i.e. 4RA=10+6−8=8
from which, RA =
8
Resolving forces vertically gives: RA+RB =0
from which, RB = −RA=−2 kN
Problem 14. Determine the reactions for the simply supported beam of Figure 5.25.
1 m 1 m 1 m 1 m 10 kN m 8 kN m 6 kN RA RB D A C E B Figure 5.25
Taking moments aboutB gives:
RA ×4 m+8 kN m+6 kN× 1 m = 10 kN m i.e. 4RA=10−8−6= −4 from which, RA = − 4 4 = −1 kN (acting downwards) Resolving forces vertically gives:
RA+RB +6=0
from which, RB = −RA−6= −(−1)−6
i.e. RB =1−6=−5 kN
(acting downwards)
Now try the following exercise
Exercise 28 Further problems on simply