CHAPTER 6 SIMPLY SUPPORTED BEAMS
EXERCISE 40, Page 87
1. Determine the moment of a force of 25 N applied to a spanner at an effective length of 180 mm
from the centre of a nut.
Moment, M = force × distance
= 25 N × 0.18 m = 4.5 N m
2. A moment of 7.5 N m is required to turn a wheel. If a force of 37.5 N applied to the rim of the
wheel can just turn the wheel, calculate the effective distance from the rim to the hub of the
wheel.
Moment, M = force × distance
from which, distance from rim to hub =
moment, M 7.5 N m
= 0.2 m = 200 mm
=
force, F
37.5 N
3. Calculate the force required to produce a moment of 27 N m on a shaft, when the effective
distance from the centre of the shaft to the point of application of the force is 180 mm.
Moment, M = force × distance
from which, force =
moment, M
27 N m
= 150 N
=
dis tan ce, d 180 ×10−3 m
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EXERCISE 41, Page X89
1. Determine distance d and the force acting at the support A for the force system shown below,
when the system is in equilibrium.
Clockwise moment = anticlockwise moment
Hence,
i.e.
2.8 × d = 1 × 140
distance, d =
1×140
= 50 mm
2.8
Force at support A, R A = 1 + 2.8 = 3.8 kN
2. If the 1 kN force shown below is replaced by a force F at a distance of 250 mm to the left of R A ,
find the value of F for the system to be in equilibrium.
Clockwise moment = anticlockwise moment
Hence, if d = 50 mm from above, then
2.8 × 50 = F × 250
and
force, F =
2.8 × 50
= 0.56 kN = 560 N
250
3. Determine the values of the forces acting at A and B for the force system shown below.
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R A + R B = 20 + 30 = 50 N
At equilibrium,
(1)
Taking moments about point A gives:
clockwise moment = anticlockwise moment
20 × 20 + 30 × 50 = R B × 76
Hence,
400 + 1500 = 76 R B
i.e.
from which,
force acting at B, R B =
From equation (1),
from which,
1900
= 25 N
76
R A + 25 = 50
R A = 50 – 25 = 25 N
4. The forces acting on a beam are as shown below. Neglecting the mass of the beam, find the
value of R A and distance d when the beam is in equilibrium.
At equilibrium,
from which,
R A + 60 = 40 + 25
R A = 40 + 25 – 60 = 5 N
Taking moments about the 60 N force gives:
clockwise moment = anticlockwise moment
Hence,
25 × d + R A × 35 = 40 × (35 – 15)
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i.e.
25d + 5 × 35 = 40 × 20
i.e.
25d + 175 = 800
i.e.
from which,
25d = 800 - 175
distance, d =
800 − 175 625
=
= 25 mm
25
25
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EXERCISE 42, Page 92
1. Calculate the force R A and distance d for the beam shown below. The mass of the beam should
be neglected and equilibrium conditions assumed.
At equilibrium,
from which,
0.2 + 2.7 + 0.4 = R A + 1.3
R A = 0.2 + 2.7 + 0.4 – 1.3 = 2.0 kN
Taking moments about the 2.7 kN force gives:
clockwise moment = anticlockwise moment
Hence,
0.4 × (d + 15) + R A × 10 = 1.3 × d + 0.2 × (12 + 10)
i.e.
0.4d + 6 + 2.0 × 10 = 1.3d + 4.4
i.e.
0.4d + 6 + 20 = 1.3d + 4.4
i.e.
6 + 20 – 4.4 = 1.3d – 0.4d
and
21.6 = 0.9d
from which,
distance, d =
21.6
= 24 mm
0.9
2. For the force system shown below, find the values of F and d for the system to be in equilibrium.
At equilibrium,
1.4 + 0.7 + F = 2.3 + 0.8
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i.e.
force, F = 2.3 + 0.8 – 1.4 – 0.7 = 1.0 kN
Taking moments about the 0.7 kN force gives:
clockwise moment = anticlockwise moment
F × d + 2.3 × 12 = 0.8 × (d + 5) + 1.4 × (14 + 12)
Hence,
i.e.
1.0 × d + 27.6 = 0.8d + 4 + 36.4
i.e.
d – 0.8d = 4 + 36.4 – 27.6
and
0.2d = 12.8
from which,
distance, d =
12.8
= 64 mm
0.2
3. For the force system shown below, determine distance d for the forces R A and R B to be equal,
assuming equilibrium conditions.
R A + R B = 10 + 15 + 25 = 50 N
For equilibrium,
Hence, if R A = R B
then R A = R B =
50
= 25 N
2
Taking moments about the R A gives:
clockwise moment = anticlockwise moment
Hence,
15 × 20 + 25 × (20 + 20 + 20) = R B × (20 + 20) + 10 × d
i.e.
i.e.
and
from which,
300 + 1500 = 25 × 40 + 10d
1800 = 1000 + 10d
10d = 1800 – 1000 = 800
distance, d =
800
= 80 m
10
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4. A simply supported beam AB is loaded as shown below. Determine the load F in order that the
reaction at A is zero.
If R1 = 0, then taking moments about R 2 gives:
clockwise moment = anticlockwise moment
F × 2 = 16 ×2 + 10 × (2 + 2)
i.e.
i.e.
2F = 32 + 40 = 72
from which,
load, F =
72
= 36 kN
2
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. Determine the reactions at the supports.
The beam is shown above.
Taking moments about the left-hand support gives:
clockwise moment = anticlockwise moment
i.e.
200 × 2.4 = R B × 3.2
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RB =
from which,
For equilibrium,
200 × 2.4
= 150 N
3.2
R A + R B = 200
R A = 200 - R B
Hence,
= 200 – 150 = 50 N
6. For the simply supported beam PQ shown below, determine (a) the reaction at each support,
(b) the maximum force which can be applied at Q without losing equilibrium.
(a) Taking moments about the left-hand support gives:
clockwise moment = anticlockwise moment
i.e. 4 × 1.5 + 6 × (1.5 + 4.0) + 5 × (1.5 + 4.0 + 1.5 + 2.0) = R 2 × (1.5 + 4.0 + 1.5)
i.e.
6 + 33 + 45 = 7 R 2
R2 =
from which,
6 + 33 + 45 84
=
= 12 kN
7
7
R1 + R 2 = 4 + 6 + 5 = 15 kN
For equilibrium,
R1 + 12 = 15
Hence,
R 1 = 15 – 12 = 3 kN
from which,
(b) Let the force at Q be R Q
Taking moments about R 2 gives:
clockwise moment = anticlockwise moment
i.e.
R Q × 2.0 = 6 × 1.5 + 4 × 5.5
i.e.
2 R Q = 9 + 22 = 31
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RQ =
from which,
31
= 15.5 kN
2
7. A uniform beam AB is 12.0 m long and is supported at distances of 2.0 m and 9.0 m from A.
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.
The beam is shown above.
R1 + R 2 = 60 + 104 + 50 + 40 = 254 kN
For equilibrium,
(1)
Taking moments about A gives: clockwise moments = anticlockwise moments
5.0 × 104 + 7.0 × 50 + 12.0 × 40 = 2.0 × R1 + 9.0 × R 2
i.e.
i.e.
520 + 350 + 480 = 2 R1 + 9 R 2
i.e.
1350 = 2 R1 + 9 R 2
(2)
and from above
254 = R1 + R 2
(1)
Equation (1) × 2 gives:
508 = 2R1 + 2R 2
(3)
Equation (2) – equation (3) gives:
842 = 7R 2
842
= 120.3 kN
7
from which,
R2 =
From equation (1),
R 1 = 254 - R 2 = 254 – 120.3 = 133.7 kN
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8. A uniform girder carrying point loads is shown below. Determine the value of load F which
causes the beam to just lift off the support B.
At equilibrium, R A + R B = F + 10 + 4 + 5
When the beam is just lifting off of the support B, then R B = 0, hence R A = (F + 19)kN.
Taking moments about the left-hand end:
clockwise moments = anticlockwise moments
i.e.
(10 × 2) + (4 × 6) + (5 × 9) = (R A × 4) + (R B × 11)
i.e.
20 + 24 + 45 = (F + 19) × 4 + (0)
i.e.
89
= (F + 19)
4
from which,
F = 22.25 – 19 = 3.25 kN
i.e. the value of force F which causes the beam to just lift off the support B is 3.25 kN
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EXERCISE 43, Page 94
1. Determine the reactions acting on the simply supported beam shown below.
Taking moments about B:
clockwise moments = anticlockwise moments
RA × 5 + 5 = 0
i.e.
Hence,
5 RA = - 5
and
R A = - 1 kN
Resolving vertically:
upward forces = downward forces
i.e.
RA+ RB= 0
R B = - R A = - (- 1) = 1 kN
i.e.
2. Determine the reactions acting on the simply supported beam shown below.
Taking moments about B:
clockwise moments = anticlockwise moments
i.e.
Hence,
RA × 5 + 5 = 0
5 RA = - 5
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and
R A = - 1 kN
Resolving vertically:
upward forces = downward forces
i.e.
RA+ RB= 0
i.e.
R B = - R A = - (- 1) = 1 kN
3. Determine the reactions acting on the simply supported beam shown below.
Resolving vertically:
upward forces = downward forces
i.e.
RA+ RB= 0
i.e.
RB= - RA
Taking moments about B:
clockwise moments = anticlockwise moments
i.e.
Hence,
R A × 8 + 10 = 6 + 12
8 R A = 6 + 12 – 10 = 8
and
R A = 1 kN
Also,
R B = - R A = - 1 kN
4. Determine the reactions acting on the simply supported beam shown below.
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Resolving vertically:
upward forces = downward forces
i.e.
RA+ RB= 0
i.e.
RA= - RB
Taking moments about R A :
clockwise moments = anticlockwise moments
10 = R B × 5 + 10
i.e.
Hence,
5 RB = 0
and
RB = 0
Also,
RA= - RB = 0
5. Determine the reactions acting on the simply supported beam shown below.
Taking moments about R B :
clockwise moments = anticlockwise moments
i.e.
Hence,
10 + R A × 3 = 10
3 RA = 0
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and
RA = 0
Resolving vertically:
upward forces = downward forces
i.e.
RA+ RB= 0
i.e.
RB = - RA = 0
6. Determine the reactions acting on the simply supported beam shown below.
Resolving vertically:
upward forces = downward forces
RA+ RB= 8
i.e.
i.e.
RB= 8 - RA
Taking moments about R B :
clockwise moments = anticlockwise moments
i.e.
Hence,
R A × 6 = 8 × 4 + 10
6 R A = 32 + 10 = 42
and
R A = 7 kN
Also,
R B = 8 - R A = 8 – 7 = 1 kN
7. Determine the reactions acting on the simply supported beam shown below.
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Resolving vertically:
upward forces = downward forces
i.e.
RA+ RB= 0
i.e.
RB= - RA
Taking moments about R B :
clockwise moments = anticlockwise moments
R A × 6 + 12 = 10
i.e.
Hence,
6 R A = 10 - 12 = - 2
1
kN = - 333.3 N
3
and
RA = −
Also,
R B = - R A = 333.3 N
EXERCISE 44, Page 95
Answers found from within the text of the chapter, pages 86 to 94.
EXERCISE 45, Page 95
1. (a) 2. (c) 3. (a) 4. (d) 5. (a) 6. (d) 7. (c) 8. (a) 9. (d) 10. (c) 11. (c)
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