Mechanical Engineering Principles

64 MECHANICAL ENGINEERING PRINCIPLES

12 mm

0.2 kN

RA

d

2.7 kN 0.4 kN

1.3 kN

10 mm 15 mm

Figure 5.16

2. For the force system shown in Figure 5.17,
   find the values ofFanddfor the system
   to be in equilibrium. [1.0 kN, 64 mm]

1.4 kN

12 mm

d

F

14 mm

5 mm

2.3 kN 0.8 kN

0.7 kN

Figure 5.17

3. For the force system shown in Figure 5.18,
   determine distanced for the forces RA
   andRBto be equal, assuming equilibrium
   conditions. [80 m]

10 N

RA RB

d 20 m20 m20 m

15 N 25 N

Figure 5.18

4. A simply supported beamABis loaded as
   shown in Figure 5.19. Determine the load
   Fin order that the reaction atAis zero.
   [36 kN]

10 kN

R 1 R 2

2 m 2 m 2 m 2 m

16 kN F

A B

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 beamPQshown
   in Figure 5.20, determine (a) the reaction
   at each support, (b) the maximum force
   which can be applied atQwithout losing
   equilibrium.

[(a)R 1 =3kN,R 2 =12 kN (b) 15.5 kN]

4 kN

R 1 R 2

1.5 m 4.0 m 1.5 m 2.0 m

6 kN 5 kN

P Q

Figure 5.20

7. A uniform beamABis 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 fromA, 7.0 m fromA and
   atB. 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 loadF which causes the beam
   to just lift off the supportB. [3.25 kN]

F 10 kN 4 kN 5 kN

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