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Solutions to Chapter 2 Problems
Solutions Manual
SOLUTIONS TO CHAPTER 2 PROBLEMS S.2.1
By a. b. c. d.
inspection: Translation parallel to BA. Translation parallel to BD, clockwise rotation. No translation, possible clockwise rotation. The force F at C may be resolved into two components as shown in Fig. S.2.1(a). The force system is then equivalent to a force parallel to AD of 2F Fcos 45° ¼ 1.293F and a force of 0.707F parallel to AB both acting at the centre of the block together with an anticlockwise torque as shown in Fig. S.2.1(b). The resultant of the two forces then acts at an angle α to the direction of AD given by
tan α ¼
0:707F ¼ 0:547 1:293F
which gives α ¼ 28:7° 2F A
B
A
B
0.707F
F 45º D (a)
45º C Fcos 45º
Fsin 45º
α D (b)
C 1.293F
FIGURE S.2.1
e1
e2
Solutions Manual
S.2.2
a. Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shown in Fig. S.2.2. Parallel vectors AC and BC are then drawn to intersect at C. The resultant is the vector OC which is 21.8 kN at an angle of 23.4° to the 15 kN force.
B
C
10 kN R
60°
q 15 kN
O
A
FIGURE S.2.2
b. From Eq. (2.1) and Fig. S.2.2 R 2 ¼ 152 þ 102 þ 2 15 10 cos 60° which gives R ¼ 21:8 kN Also, from Eq. (2.2) tan θ ¼
10 sin 60° 15 þ 10 cos 60°
so that θ ¼ 23:4°: S.2.3
a. The vectors do not have to be drawn in any particular order. Fig. S.2.3 shows the vector diagram with the vector representing the 10 kN force drawn first. The resultants R is then equal to 8.6 kN and makes an angle of 23.9° to the negative direction of the 10 kN force. b. Resolving forces in the positive x direction Fx ¼ 10 þ 8 cos 60° 12 cos 30° 20 cos 55° ¼ 7:9 kN Then, resolving forces in the positive y direction Fy ¼ 8 cos 30° þ 12 cos 60° 20 cos 35° ¼ 3:5 kN The resultant R is given by R 2 ¼ ð7:9Þ2 þ ð3:5Þ2
Solutions to Chapter 2 Problems
e3
12 kN
8 kN
20 kN
R
q
10 kN
FIGURE S.2.3
so that R ¼ 8:6 kN Also tan θ ¼
3:5 7:9
which gives θ ¼ 23:9°: S.2.4
Since the joints at A, B, and C are hinged and the load is applied at joint B, the forces in AB and BC will be purely axial. Assume that the force in AB is tension and in BC is compression (if a wrong assumption is made the answer will be negative). 1.5 m
A
FAB
B
10 kN 3m
FCB α
C
FIGURE S.2.4(a)
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Solutions Manual Referring to Fig. S.2.4(a), the hinge at B is in equilibrium under the action of the horizontal force FAB, the inclined force FCB, and the 10 kN load. Also α ¼ tan 1
1:5 ¼ 26:6 3
Resolving forces vertically at B FCB cos 26:6° 10 ¼ 0, which gives FCB ¼ 11:26 kN ðcompressionÞ Now resolving forces horizontally at B FAB FCB sin 26:6° ¼ 0 from which FAB ¼ 5:0 kN ðtensionÞ Alternatively, FAB can be found by resolving forces at B in a direction perpendicular to CB, i.e. FAB cos 26:6° 10 sin 26:6° ¼ 0, which again gives FAB ¼ 5:0 kN ðtensionÞ Since the hinge at B is in equilibrium under the action of the three forces acting at B, it may be represented by a triangle of forces as shown in Fig. S.2.4(b), which is constructed as d
26.6°
11.2 kN 10 kN
f
e 5.0 kN
FIGURE S.2.4(b)
Solutions to Chapter 2 Problems
e5
follows. Draw to a suitable scale the vector df vertically downwards to represent the 10 kN load. Then draw fe and ed parallel to AB and CB, respectively. The point of intersection e completes the triangle of forces. The vector fe represents the force FAB in magnitude and direction, while the vector ed represents the force FCB in magnitude and direction. Note that the direction of the forces as depicted by the arrows in Fig. S.2.4(b) is their action on the hinge at B. S.2.5
Label the spaces between the forces as shown in Fig. S.2.5(a). Note that the reaction at the support B has been previously calculated. Draw cd to represent the 5 kN load downwards and to the right parallel to the load. Then draw de to the same scale to represent the 10 kN load upwards and to the right. Now draw ef, fg, and gh to represent the 0.7 kN reaction at B, the downward 2 kN load, and the 3 kN load, respectively. The force diagram is completed by the vector hc, which represents the reaction at A in magnitude and direction as shown in Fig. S.2.5(b).
D 5 kN
10 kN
C
E
G
H RA
3 kN
F RB = 0.7 kN
2 kN
(a)
FIGURE S.2.5(a)
e f
c
g
RA = 9.6 kN
h 10.8° (b)
d
FIGURE S.2.5(b)
S.2.6
Referring to Fig. S.2.6 the resultant, R, of the force system is given by R ¼ 5 þ 4 8 þ 10 ¼ 11 kN
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Solutions Manual
a y
5 kN R
4 kN
10 kN
30°
(0,1.0)
8 kN x 1.0
m
0.5
0.7 m
5m
FIGURE S.2.6(a)
Then Rx ¼ 11 cos 60° ¼ 5:5 kN, Ry ¼ 11 sin 60° ¼ 9:5 kN: Now taking moments about the line of action of the 5 kN force Ra ¼ 4 1 8 6:5 þ 10 2 25 ¼ 11a from which a ¼ 1:32 m: 1.3
2m
y 5 kN
R
a (0,1.0) 60°
FIGURE S.2.6(b)
b
c
f
e d
x
Solutions to Chapter 2 Problems
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In Fig. S.2.6(b) the resultant, R, of the force system crosses the x axis at the point e. Then fe ¼ fd ed. Also fd ¼ bc ¼ 1.32 cos 30° ¼ 1.14 m and ed ¼ cd tan 30° ¼ bf tan 30° ¼ (af ab) tan 30° ¼ (1 1.32 sin 30°) tan 30° ¼ 0.2 m. Therefore, fe ¼ 1.14 0.2 ¼ 0.94 m. S.2.7
Referring to Fig. S.2.7 the resultant vertical force FV is given by FV ¼ 10 þ 8 þ 3 ¼ 21 kN Taking moments about the centre of the cylinder FV x ¼ 10 1:25 þ 8 0:75 3 1:0 so that 21x ¼ 15:5 – x
10 kN
8 kN
3 kN
FV 0.5 m
0.75 m
1.0 m
FIGURE S.2.7
which gives x ¼ 0:738 m The force system is then equivalent to a vertically downward force of 21 kN acting through the centre of the cylinder together with a torque T ¼ 210.738¼15.5 kNm (anticlockwise). Alternatively, and more directly T ¼ 10 1:25 þ 8 0:75 3 1:0 ¼ 15:5 kNm ðanticlockwiseÞ The bending moment at the built in end is given by M ¼ 21 2:5 ¼ 52:5 kNm S.2.8
Referring to Fig. S.2.7, the total vertical downward load on the end of the cantilever is given by FV ¼ 10 þ 8 þ 3 ¼ 21 kN
e8
Solutions Manual The line of action has already been calculated in S.2.7 and is a distance of 0.738 m from the centre of the cylinder cross section. Therefore, the loading system may be replaced by an anticlockwise torque given by T ¼ 21 0:738 ¼ 15:5 kNm together with a vertically downward load of 21 kN acting through the cylinder cross section as shown in Fig. S.2.8.
Torque = 15.5 kNm 21 kN
FIGURE S.2.8
S.2.9
Initially the forces are resolved into vertical and horizontal components as shown in Fig. S.2.9. Then Rx ¼ 69:3 þ 35:4 20:0 ¼ 84:7 kN Now taking moments about the x axis Rx y ¼ 35:4 0:5 20:0 1:25 þ 69:3 1:6 which gives y ¼ 1:22 m Also, from Fig. S.2.9 Ry ¼ 60 þ 40 þ 34:6 35:4 ¼ 99:2 kN Now taking moments about the y axis Ry x ¼ 40:0 1:0 þ 60:0 1:25 34:6 1:0 so that x ¼ 0:81 m
Solutions to Chapter 2 Problems y
(1.0, 1.6) x
69.3 kN 30° 80 kN 40 kN
( 1, 1.25)
20.0 kN
40 kN
Rx
50 kN
35.4 kN
30°
e9
34.6 kN
y
45° (0, 0.5)
35.4 kN
Ry
(1.25, 0.25) x
O 60 kN
FIGURE S.2.9
The resultant R is then given by R 2 ¼ 99:22 þ 84:72 from which R ¼ 130:4 kN Finally θ ¼ tan 1 S.2.10
99:2 ¼ 49:5°: 84:7
a. In Fig. S.2.10(a) the inclined loads have been resolved into vertical and horizontal components. The vertical loads will generate vertical reactions at the supports A and B while the horizontal components of the loads will produce a horizontal reaction at A only since B is a roller support. 3 kN
7 kN
8 kN 6.1 kN
60°
A RA,H
3.5 kN
5.7 kN 45°
B
5.7 kN
RA,V
RB,V 4m
6m
5m
FIGURE S.2.10(a)
Taking moments about B RA,V 20 3 16 6:1 10 5:7 5 ¼ 0
5m
e10
Solutions Manual which gives RA,V ¼ 6:9 kN Now resolving vertically RB,V þ RA,V 3 6:1 5:7 ¼ 0 so that RB,V ¼ 7:9 kN Finally, resolving horizontally RA,H 3:5 5:7 ¼ 0 so that RA,H ¼ 9:2 kN Note that all reactions are positive in sign so that their directions are those indicated in Fig. S.2.10(a). b. The loads on the cantilever beam will produce a vertical reaction and a moment reaction at A as shown in Fig. S.2.10(b). 5 kNm MA A
B 15 kN
RA 10 m
FIGURE S.2.10(b)
Resolving vertically RA 15 5 10 ¼ 0 which gives RA ¼ 65 kN Taking moments about A MA 15 10 5 10 5 ¼ 0 from which MA ¼ 400 kNm Again the signs of the reactions are positive so that they are in the directions shown. c. In Fig. S.2.10(c) there are horizontal and vertical reactions at A and a vertical reaction at B. By inspection (or by resolving horizontally) RA,H ¼ 20 kN
Solutions to Chapter 2 Problems
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20 kN
5m 10 kN
15 kN 5 kN/m
A RA,H
B RB
RA,V 2m
4m
2m
2m
FIGURE S.2.10(c)
Taking moments about A RB 8 þ 20 5 5 2 9 15 6 10 2 ¼ 0 which gives RB ¼ 12:5 kN Finally, resolving vertically RA,V þ RB 10 15 5 2 ¼ 0 so that RA,V ¼ 22:5 kN: d. The loading on the beam will produce vertical reactions only at the supports as shown in Fig. S.2.10(d). 75 kN/m 8 kN/m A
B RB
RA 3m
9m
FIGURE S.2.10(d)
Taking moments about B RA 12 þ 75 8 12 6 ¼ 0 Hence RA ¼ 41:8 kN
e12
Solutions Manual Now resolving vertically RB þ RA 8 12 ¼ 0 so that RB ¼ 54:2 kN:
S.2.11
a. The loading on the truss shown in Fig. P.2.11(a) produces only vertical reactions at the support points A and B; suppose these reactions are RA and RB respectively and that they act vertically upwards. Then, taking moments about B RA 10 5 16 10 14 15 12 15 10 5 8 þ 5 4 ¼ 0 which gives RA ¼ 57 kN ðupwardsÞ Now resolving vertically RB þ RA 5 10 15 15 5 5 ¼ 0 from which RB ¼ 2 kN ðdownwardsÞ: b. The angle of the truss is tan1(4/10) ¼ 21.8°. The loads on the rafters are symmetrically arranged and may be replaced by single loads as shown in Fig. S.2.11. These, in turn, may be resolved into horizontal and vertical components and will produce vertical reactions at A and B and a horizontal reaction at A. Taking moments about B RA,V 20 þ 742:7 2 1857:0 15 þ 2970:9 2 þ 7427:9 5 ¼ 0 which gives RA,V ¼ 835:6 N ðdownwardsÞ: Now resolving vertically RB þ RA,V 1857:0 þ 7427:9 ¼ 0 7427.9 N 2000 N
8000 N 21.8°
21.8°
RA,H RA,V
21.8°
FIGURE S.2.11
4m
2970.9 N
742.7 N
RB
1857.0 N 20 m
Solutions to Chapter 2 Problems
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from which RB ¼ 4735:3 N ðdownwardsÞ: Finally, resolving horizontally RA,H 742:7 2970:9 ¼ 0 so that RA,H ¼ 3713:6 N: S.2.12 Referring to Fig. P.2.12 the weight of the vertical part of the column is equal to 5 500 ¼ 2500 N and the weight of the semi-circular part of the column is equal to π 0.6 500 ¼ 942.5 N. Therefore, the total vertical load on the base of the column is 2500 þ 942.5 ¼ 3442.5 N. The bending moment in the yz plane is given by Mx ¼ 942:5 0:6 ¼ 527:8 Nm and the bending moment in the xz plane is given by My ¼ 600 2:5 þ 250 5:6 ¼ 2900 Nm Therefore, the resultant bending moment is given by pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M ¼ 29002 þ 527:82 ¼ 2947:6 Nm acting in a vertical plane at an angle to the yz plane given by tan1 (2900/527.8) ¼ 79.7°. Finally, the torque on the base of the column is 250 0.6 ¼ 150 Nm.
SOLUTIONS TO CHAPTER 2 PROBLEMS S.2.1
By a. b. c. d.
inspection: Translation parallel to BA. Translation parallel to BD, clockwise rotation. No translation, possible clockwise rotation. The force F at C may be resolved into two components as shown in Fig. S.2.1(a). The force system is then equivalent to a force parallel to AD of 2F Fcos 45° ¼ 1.293F and a force of 0.707F parallel to AB both acting at the centre of the block together with an anticlockwise torque as shown in Fig. S.2.1(b). The resultant of the two forces then acts at an angle α to the direction of AD given by
tan α ¼
0:707F ¼ 0:547 1:293F
which gives α ¼ 28:7° 2F A
B
A
B
0.707F
F 45º D (a)
45º C Fcos 45º
Fsin 45º
α D (b)
C 1.293F
FIGURE S.2.1
e1
e2
Solutions Manual
S.2.2
a. Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shown in Fig. S.2.2. Parallel vectors AC and BC are then drawn to intersect at C. The resultant is the vector OC which is 21.8 kN at an angle of 23.4° to the 15 kN force.
B
C
10 kN R
60°
q 15 kN
O
A
FIGURE S.2.2
b. From Eq. (2.1) and Fig. S.2.2 R 2 ¼ 152 þ 102 þ 2 15 10 cos 60° which gives R ¼ 21:8 kN Also, from Eq. (2.2) tan θ ¼
10 sin 60° 15 þ 10 cos 60°
so that θ ¼ 23:4°: S.2.3
a. The vectors do not have to be drawn in any particular order. Fig. S.2.3 shows the vector diagram with the vector representing the 10 kN force drawn first. The resultants R is then equal to 8.6 kN and makes an angle of 23.9° to the negative direction of the 10 kN force. b. Resolving forces in the positive x direction Fx ¼ 10 þ 8 cos 60° 12 cos 30° 20 cos 55° ¼ 7:9 kN Then, resolving forces in the positive y direction Fy ¼ 8 cos 30° þ 12 cos 60° 20 cos 35° ¼ 3:5 kN The resultant R is given by R 2 ¼ ð7:9Þ2 þ ð3:5Þ2
Solutions to Chapter 2 Problems
e3
12 kN
8 kN
20 kN
R
q
10 kN
FIGURE S.2.3
so that R ¼ 8:6 kN Also tan θ ¼
3:5 7:9
which gives θ ¼ 23:9°: S.2.4
Since the joints at A, B, and C are hinged and the load is applied at joint B, the forces in AB and BC will be purely axial. Assume that the force in AB is tension and in BC is compression (if a wrong assumption is made the answer will be negative). 1.5 m
A
FAB
B
10 kN 3m
FCB α
C
FIGURE S.2.4(a)
e4
Solutions Manual Referring to Fig. S.2.4(a), the hinge at B is in equilibrium under the action of the horizontal force FAB, the inclined force FCB, and the 10 kN load. Also α ¼ tan 1
1:5 ¼ 26:6 3
Resolving forces vertically at B FCB cos 26:6° 10 ¼ 0, which gives FCB ¼ 11:26 kN ðcompressionÞ Now resolving forces horizontally at B FAB FCB sin 26:6° ¼ 0 from which FAB ¼ 5:0 kN ðtensionÞ Alternatively, FAB can be found by resolving forces at B in a direction perpendicular to CB, i.e. FAB cos 26:6° 10 sin 26:6° ¼ 0, which again gives FAB ¼ 5:0 kN ðtensionÞ Since the hinge at B is in equilibrium under the action of the three forces acting at B, it may be represented by a triangle of forces as shown in Fig. S.2.4(b), which is constructed as d
26.6°
11.2 kN 10 kN
f
e 5.0 kN
FIGURE S.2.4(b)
Solutions to Chapter 2 Problems
e5
follows. Draw to a suitable scale the vector df vertically downwards to represent the 10 kN load. Then draw fe and ed parallel to AB and CB, respectively. The point of intersection e completes the triangle of forces. The vector fe represents the force FAB in magnitude and direction, while the vector ed represents the force FCB in magnitude and direction. Note that the direction of the forces as depicted by the arrows in Fig. S.2.4(b) is their action on the hinge at B. S.2.5
Label the spaces between the forces as shown in Fig. S.2.5(a). Note that the reaction at the support B has been previously calculated. Draw cd to represent the 5 kN load downwards and to the right parallel to the load. Then draw de to the same scale to represent the 10 kN load upwards and to the right. Now draw ef, fg, and gh to represent the 0.7 kN reaction at B, the downward 2 kN load, and the 3 kN load, respectively. The force diagram is completed by the vector hc, which represents the reaction at A in magnitude and direction as shown in Fig. S.2.5(b).
D 5 kN
10 kN
C
E
G
H RA
3 kN
F RB = 0.7 kN
2 kN
(a)
FIGURE S.2.5(a)
e f
c
g
RA = 9.6 kN
h 10.8° (b)
d
FIGURE S.2.5(b)
S.2.6
Referring to Fig. S.2.6 the resultant, R, of the force system is given by R ¼ 5 þ 4 8 þ 10 ¼ 11 kN
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Solutions Manual
a y
5 kN R
4 kN
10 kN
30°
(0,1.0)
8 kN x 1.0
m
0.5
0.7 m
5m
FIGURE S.2.6(a)
Then Rx ¼ 11 cos 60° ¼ 5:5 kN, Ry ¼ 11 sin 60° ¼ 9:5 kN: Now taking moments about the line of action of the 5 kN force Ra ¼ 4 1 8 6:5 þ 10 2 25 ¼ 11a from which a ¼ 1:32 m: 1.3
2m
y 5 kN
R
a (0,1.0) 60°
FIGURE S.2.6(b)
b
c
f
e d
x
Solutions to Chapter 2 Problems
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In Fig. S.2.6(b) the resultant, R, of the force system crosses the x axis at the point e. Then fe ¼ fd ed. Also fd ¼ bc ¼ 1.32 cos 30° ¼ 1.14 m and ed ¼ cd tan 30° ¼ bf tan 30° ¼ (af ab) tan 30° ¼ (1 1.32 sin 30°) tan 30° ¼ 0.2 m. Therefore, fe ¼ 1.14 0.2 ¼ 0.94 m. S.2.7
Referring to Fig. S.2.7 the resultant vertical force FV is given by FV ¼ 10 þ 8 þ 3 ¼ 21 kN Taking moments about the centre of the cylinder FV x ¼ 10 1:25 þ 8 0:75 3 1:0 so that 21x ¼ 15:5 – x
10 kN
8 kN
3 kN
FV 0.5 m
0.75 m
1.0 m
FIGURE S.2.7
which gives x ¼ 0:738 m The force system is then equivalent to a vertically downward force of 21 kN acting through the centre of the cylinder together with a torque T ¼ 210.738¼15.5 kNm (anticlockwise). Alternatively, and more directly T ¼ 10 1:25 þ 8 0:75 3 1:0 ¼ 15:5 kNm ðanticlockwiseÞ The bending moment at the built in end is given by M ¼ 21 2:5 ¼ 52:5 kNm S.2.8
Referring to Fig. S.2.7, the total vertical downward load on the end of the cantilever is given by FV ¼ 10 þ 8 þ 3 ¼ 21 kN
e8
Solutions Manual The line of action has already been calculated in S.2.7 and is a distance of 0.738 m from the centre of the cylinder cross section. Therefore, the loading system may be replaced by an anticlockwise torque given by T ¼ 21 0:738 ¼ 15:5 kNm together with a vertically downward load of 21 kN acting through the cylinder cross section as shown in Fig. S.2.8.
Torque = 15.5 kNm 21 kN
FIGURE S.2.8
S.2.9
Initially the forces are resolved into vertical and horizontal components as shown in Fig. S.2.9. Then Rx ¼ 69:3 þ 35:4 20:0 ¼ 84:7 kN Now taking moments about the x axis Rx y ¼ 35:4 0:5 20:0 1:25 þ 69:3 1:6 which gives y ¼ 1:22 m Also, from Fig. S.2.9 Ry ¼ 60 þ 40 þ 34:6 35:4 ¼ 99:2 kN Now taking moments about the y axis Ry x ¼ 40:0 1:0 þ 60:0 1:25 34:6 1:0 so that x ¼ 0:81 m
Solutions to Chapter 2 Problems y
(1.0, 1.6) x
69.3 kN 30° 80 kN 40 kN
( 1, 1.25)
20.0 kN
40 kN
Rx
50 kN
35.4 kN
30°
e9
34.6 kN
y
45° (0, 0.5)
35.4 kN
Ry
(1.25, 0.25) x
O 60 kN
FIGURE S.2.9
The resultant R is then given by R 2 ¼ 99:22 þ 84:72 from which R ¼ 130:4 kN Finally θ ¼ tan 1 S.2.10
99:2 ¼ 49:5°: 84:7
a. In Fig. S.2.10(a) the inclined loads have been resolved into vertical and horizontal components. The vertical loads will generate vertical reactions at the supports A and B while the horizontal components of the loads will produce a horizontal reaction at A only since B is a roller support. 3 kN
7 kN
8 kN 6.1 kN
60°
A RA,H
3.5 kN
5.7 kN 45°
B
5.7 kN
RA,V
RB,V 4m
6m
5m
FIGURE S.2.10(a)
Taking moments about B RA,V 20 3 16 6:1 10 5:7 5 ¼ 0
5m
e10
Solutions Manual which gives RA,V ¼ 6:9 kN Now resolving vertically RB,V þ RA,V 3 6:1 5:7 ¼ 0 so that RB,V ¼ 7:9 kN Finally, resolving horizontally RA,H 3:5 5:7 ¼ 0 so that RA,H ¼ 9:2 kN Note that all reactions are positive in sign so that their directions are those indicated in Fig. S.2.10(a). b. The loads on the cantilever beam will produce a vertical reaction and a moment reaction at A as shown in Fig. S.2.10(b). 5 kNm MA A
B 15 kN
RA 10 m
FIGURE S.2.10(b)
Resolving vertically RA 15 5 10 ¼ 0 which gives RA ¼ 65 kN Taking moments about A MA 15 10 5 10 5 ¼ 0 from which MA ¼ 400 kNm Again the signs of the reactions are positive so that they are in the directions shown. c. In Fig. S.2.10(c) there are horizontal and vertical reactions at A and a vertical reaction at B. By inspection (or by resolving horizontally) RA,H ¼ 20 kN
Solutions to Chapter 2 Problems
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20 kN
5m 10 kN
15 kN 5 kN/m
A RA,H
B RB
RA,V 2m
4m
2m
2m
FIGURE S.2.10(c)
Taking moments about A RB 8 þ 20 5 5 2 9 15 6 10 2 ¼ 0 which gives RB ¼ 12:5 kN Finally, resolving vertically RA,V þ RB 10 15 5 2 ¼ 0 so that RA,V ¼ 22:5 kN: d. The loading on the beam will produce vertical reactions only at the supports as shown in Fig. S.2.10(d). 75 kN/m 8 kN/m A
B RB
RA 3m
9m
FIGURE S.2.10(d)
Taking moments about B RA 12 þ 75 8 12 6 ¼ 0 Hence RA ¼ 41:8 kN
e12
Solutions Manual Now resolving vertically RB þ RA 8 12 ¼ 0 so that RB ¼ 54:2 kN:
S.2.11
a. The loading on the truss shown in Fig. P.2.11(a) produces only vertical reactions at the support points A and B; suppose these reactions are RA and RB respectively and that they act vertically upwards. Then, taking moments about B RA 10 5 16 10 14 15 12 15 10 5 8 þ 5 4 ¼ 0 which gives RA ¼ 57 kN ðupwardsÞ Now resolving vertically RB þ RA 5 10 15 15 5 5 ¼ 0 from which RB ¼ 2 kN ðdownwardsÞ: b. The angle of the truss is tan1(4/10) ¼ 21.8°. The loads on the rafters are symmetrically arranged and may be replaced by single loads as shown in Fig. S.2.11. These, in turn, may be resolved into horizontal and vertical components and will produce vertical reactions at A and B and a horizontal reaction at A. Taking moments about B RA,V 20 þ 742:7 2 1857:0 15 þ 2970:9 2 þ 7427:9 5 ¼ 0 which gives RA,V ¼ 835:6 N ðdownwardsÞ: Now resolving vertically RB þ RA,V 1857:0 þ 7427:9 ¼ 0 7427.9 N 2000 N
8000 N 21.8°
21.8°
RA,H RA,V
21.8°
FIGURE S.2.11
4m
2970.9 N
742.7 N
RB
1857.0 N 20 m
Solutions to Chapter 2 Problems
e13
from which RB ¼ 4735:3 N ðdownwardsÞ: Finally, resolving horizontally RA,H 742:7 2970:9 ¼ 0 so that RA,H ¼ 3713:6 N: S.2.12 Referring to Fig. P.2.12 the weight of the vertical part of the column is equal to 5 500 ¼ 2500 N and the weight of the semi-circular part of the column is equal to π 0.6 500 ¼ 942.5 N. Therefore, the total vertical load on the base of the column is 2500 þ 942.5 ¼ 3442.5 N. The bending moment in the yz plane is given by Mx ¼ 942:5 0:6 ¼ 527:8 Nm and the bending moment in the xz plane is given by My ¼ 600 2:5 þ 250 5:6 ¼ 2900 Nm Therefore, the resultant bending moment is given by pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M ¼ 29002 þ 527:82 ¼ 2947:6 Nm acting in a vertical plane at an angle to the yz plane given by tan1 (2900/527.8) ¼ 79.7°. Finally, the torque on the base of the column is 250 0.6 ¼ 150 Nm.