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Solutions to Chapter 3 Problems
SOLUTIONS TO CHAPTER 3 PROBLEMS S.3.1
Fig. S.3.1(a) shows the mast with two of each set of cables; the other two cables in each set are in a plane perpendicular to the plane of the paper. A B qB 40 m
qC
74.6 kN 22.5 kN
35 m C
166.4 kN
119.6 kN 25 m
D
247.4 kN 211.4 kN
qD
15 m E 314.9 kN
20 m
20 m
(a)
(b)
FIGURE S.3.1
Then, θB ¼ tan 1
20 20 20 ¼ 29:7°, θC ¼ tan 1 ¼ 38:7°, θD ¼ tan 1 ¼ 53:1°: 35 25 15
The normal force at any section of the mast will be compressive and is the sum of the self-weight and the vertical component of the tension in the cables. Furthermore the self-weight will vary linearly with distance from the top of the mast. Therefore, at a section immediately above B, N ¼ 5 4:5 ¼ 22:5 kN: At a section immediately below B, N ¼ 22:5 þ 4 15 cos 29:7° ¼ 74:6 kN: At a section immediately above C, N ¼ 74:6 þ 10 4:5 ¼ 119:6 kN: At a section immediately below C, N ¼ 119:6 þ 4 15 cos 38:7° ¼ 166:4 kN: At a section immediately above D, N ¼ 166:4 þ 10 4:5 ¼ 211:4 kN: At a section immediately below D, N ¼ 211:4 þ 4 15 cos 53:1° ¼ 247:4 kN:
e14
Solutions to Chapter 3 Problems
e15
Finally, at a section immediately above E, N ¼ 247:4 þ 15 4:5 ¼ 314:9 kN: The distribution of compressive force in the mast is shown in Fig. S.3.1(b). S.3.2
The beam support reactions have been calculated in S.2.6(a) and are as shown in Fig. S.3.2(a); the bays of the beam have been relettered as shown in Fig. P.3.2. 3 kN
7 kN
6.1 kN
A
5.7 kN
8 kN
C
E
D
9.2 kN B (a)
3.5 kN
6.9 kN 9.2 kN
5.7 kN
7.9 kN
9.2 kN 5.7 kN ve
(b)
A
B
C
E
D
7.9 kN
A
2.2 kN C
B ve
ve D
E
D
E
3.9 kN (c)
6.9 kN A
B
C
ve 27.6 kNm 40.0 kNm (d)
51.0 kNm
FIGURE S.3.2
Normal force The normal force at any section of the beam between A and C is constant and given by NAC ¼ 9.2 kN (the vertical 3 kN load has no effect on the normal force). Then NCD ¼ 9:2 3:5 ¼ 5:7 kN
e16
Solutions Manual and NDE ¼ 9:2 3:5 5:7 ¼ 0 Note that NDE could have been found directly by considering forces to the right of any section between D and E. The complete distribution of normal force in shown in Fig. S.3.2(b).
Shear force The shear force in each bay of the beam is constant since only concentrated loads are involved. At any section between A and B, SAB ¼ RA, V ¼ 6:9 kN: At any section between B and C, SBC ¼ 6:9 þ 3 ¼ 3:9 kN: At any section between C and D, SCD ¼ 6:9 þ 3 þ 6:1 ¼ 2:2 kN: Finally, at any section between D and E, SDE ¼ þRE ¼ 7:9 kN: The complete shear force distribution is shown in Fig. S.3.2(c). Bending moment Since only concentrated loads are present it is only necessary to calculate values of bending moment at the load points. Note that MA ¼ ME ¼ 0. At B, MB ¼ 6:9 4 ¼ 27:6 kNm: At C, MC ¼ 6:9 10 3 6 ¼ 51:0 kNm: At D, MD ¼ 6:9 15 3 11 6:1 5 ¼ 40 kNm: Alternatively, MD ¼ 7.9 5 ¼ 39.5 kNm; the difference in the two values is due to rounding off errors. The complete distribution is shown in Fig. S.3.2(d). S.3.3
There will be vertical and horizontal reactions at E and a vertical reaction at B as shown in Fig. S.3.3(a). The inclined 10 kN load will have vertical and horizontal components of 8 and 6 kN respectively, the latter acting to the right. Resolving horizontally, RE,H ¼ 6 kN. Now taking moments about E
Solutions to Chapter 3 Problems 8 kN
10 kN 2 kN/m
RE,H
E A
B
x
C
RB
5m
3m
D
6 kN 4m
RE,V 3m
(a) A
B
C
D
E ve
(b)
6 kN
6 kN
4 kN
4 kN
10 kN ve C A
B
ve
D
E
ve 4 kN
4 kN 10 kN
(c)
25 kNm
ve
4 kNm D
A B
C
E ve
12 kNm
(d)
FIGURE S.3.3
RB 10 2 8 11 8 3 ¼ 0 which gives RB ¼ 20 kN Resolving vertically RE,V þ RB 2 8 8 ¼ 0 so that RE,V ¼ 4 kN
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e18
Solutions Manual
Normal force The normal force at all sections of the beam between A and D is zero since there is no horizontal reaction at B and no horizontal forces between A and D. In CD, NCD ¼ RE,H ¼ 6 kN (compressive and therefore negative); the distribution is shown in Fig. S.3.3(b). Shear force We note that over the length of the uniformly distributed load the shear force will vary linearly with distance from, say, A. Then, at any section between A and B, a distance x from A, the shear force is given by SAB ¼ þ2x Therefore, when x ¼ 0, SAB ¼ 0 and when x ¼ 5 m (i.e. at section just to the left of B), SAB ¼ 10 kN. Also, at any section between B and C a distance x from A SBC ¼ þ2x RB ¼ 2x 20 Therefore, when x ¼ 5 m (i.e. at a section just to the right of B), SBC ¼ 10 kN and when x ¼ 8 m (i.e. at a section just to the left of C), SBC ¼ 4 kN. Between C and D the shear force is constant and SCD ¼ 2 820 ¼ 4 kN. Finally, between D and E, SDE ¼ 2 820 þ 8 ¼ þ 4 kN (or, alternatively, SDE ¼ þ RE,V ¼ þ 4 kN). The complete distribution is shown in Fig. S.3.3(c). Bending moment At any section between A and B the bending moment is given by MAB ¼ 2x
x 2
¼ x 2
Then, when x ¼ 0, MAB ¼ 0 and when x ¼ 5 m, MAB ¼ 25 kNm. Note that the distribution is parabolic and that when x ¼ 0, (dMAB/dx) ¼ 0. At any section between B and C MBC ¼ x 2 þ RB ðx 5Þ ¼ x 2 þ 20ðx 5Þ When x ¼ 5 m, MBC ¼ 25 kNm and when x ¼ 8 m, MBC ¼ 4 kNm. The distribution of bending moment between B and C is parabolic but has no turning value between B and C. At D the bending moment is most simply given by MD ¼ RE,V 3 ¼ þ 12 kNm; the complete distribution is shown in Fig. S.3.3(d). S.3.4
The forces and moments acting on the beam ABC are shown in Fig. S.3.4(a). Since the beam is a cantilever it is unnecessary to calculate the support reactions if forces and moments to the right of any section of the beam are considered.
Normal force At any section between B and C the normal force is given by NBC ¼ þ10 kN ðtensionÞ
Solutions to Chapter 3 Problems 1 5 kN
1 5 kN
7 kNm
A
B
10 kN
18 kNm
x 2.5 m
C
10 kN
18 kNm 1.3 m
(a) 20 kN
20 kN 10 kN +ve
Normal force (b)
56.6 kN 39.1 kN 24.1 kN +ve
15 kN Shear force (c)
181.0 kNm
61.4 kNm 43.4 kNm –ve
18.0 kNm Bending moment (d)
FIGURE S.3.4
At any section between A and B the normal force is given by NAB ¼ þ10 þ 10 ¼ þ20 kN ðtensionÞ The normal force diagram is therefore as shown in Fig. S.3.4(b).
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e20
Solutions Manual Shear force At any section between A and B the shear force is given by SAB ¼ þ15 þ 15 þ 7ð3:8 xÞ which gives SAB ¼ 56:6 7x When x ¼ 0, SAB ¼ þ 56.6 kN and when x ¼ 2.5 m, SAB ¼ þ 39.1 kN. At any section between B and C the shear force is given by SBC ¼ þ15 þ 7ð3:8 xÞ so that SBC ¼ 41:6 7x When x ¼ 2.5 m, SBC ¼ þ 24.1 kN and when x ¼ 3.8 m, SBC ¼ þ 15 kN. The shear force diagram is therefore as shown in Fig. S.3.4(c). Bending moment The bending moment at any section between B and C is given by 7 MBC ¼ 15ð3:8 xÞ ð3:8 xÞ2 18 2 which gives MBC ¼ 125:5 þ 41:6x 3:5x 2 When x ¼ 3.8 m, MBC ¼ 18.0 kNm and when x ¼ 2.5 m, MBC ¼ 43.4 kNm and the distribution is parabolic. At any section between A and B the bending moment is given by 7 MAB ¼ 15ð3:8 xÞ 15ð2:5 xÞ ð3:8 xÞ2 18 18 2 which gives MAB ¼ 181:0 þ 56:6x 3:5x 2 When x ¼ 2.5 m, MAB ¼ 61.4 kNm and when x ¼ 0, MAB ¼ 181.0 kNm and the distribution is parabolic. The complete bending moment distribution is then as shown in Fig. S.3.4(d).
S.3.5
Again, as in P.3.4, it is unnecessary to calculate the support reactions if forces and moments to the right of any section are considered.
Normal force There are no horizontal forces applied to the beam so that the normal force in both AB and CD is zero. The normal force in BC is due to the component, resolved parallel to BC, of the force at C produced by the distributed load on CD, that is
Solutions to Chapter 3 Problems
e21
NBC ¼ þ4 1 cos 60° ¼ þ2 kN ðtensionÞ The normal force diagram is therefore as shown in Fig. S.3.5(a).
2 kN A
B
(a) 14 kN
2 kN
+ve
Normal force C
D
14 kN +ve
A
3.46 kN B
Shear force 4 kN
+ve
+ve (b)
C
D
20 kNm
–ve
A
6 kNm 6 kNm
Bending moment
–ve B 2 kNm 2 kNm –ve C D (c)
FIGURE S.3.5
Shear force Since there is a uniformly distributed load on CD the shear force distribution will be linear and vary from zero at D to þ 4 1 ¼ þ 4 kN at C. In BC the shear force is equal to the component, perpendicular to BC, of the load at C from CD, that is SBC ¼ þ4 1 sin 60° ¼ þ3:46 kN In AB the shear force at any section is given by SAB ¼ þ4 1 þ 10 ¼ þ14 kN The distribution of shear force is then as shown in Fig. S.3.5(b).
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Solutions Manual Bending moment The bending moment at any section a distance x, say, from D is given by MCD ¼ 4
x2 2
When x ¼ 0, MCD ¼ 0 and when x ¼ 1 m, MCD ¼ 2 kNm and the distribution is parabolic. In BC and AB there are no applied distributed loads so that the bending moment distributions will be linear. Then MC ¼ 2 kNm ðalready calculatedÞ MB ¼ 4 1 1:5 ¼ 6 kNm MA ¼ 4 1 2:5 10 1 ¼ 20 kNm The complete distribution is then as shown in Fig. S.3.5(c). S.3.6
As in the previous two problems it is unnecessary to calculate the support reactions. D A
B –ve
3.54kN
C
_ve
5kN
5kN
3.54kN
Normal force
(a) 5.54kN 3.54kN 3.54kN
D Shear force
+ve A
B
C
(b) 8.08 kNm
3.44 kNm D
A
B
C
(c)
FIGURE S.3.6
Normal force At any section between D and C the normal force is given by NDC ¼ 5 kN
Solutions to Chapter 3 Problems
e23
At any section between A and C the normal force is given by NAC ¼ 5 cos 45° ¼ 3:54 kN The normal force distribution is therefore as shown in Fig. S.3.6(a). Shear force By inspection SDC ¼ 0. At any section between B and C the shear force is given by SBC ¼ þ5 sin 45° ¼ þ3:54 kN At any section between A and B, a distance x, say, from B, the shear force is given by SAB ¼ þ5 sin 45° þ 2x ¼ þ3:54 þ 2x When x ¼ 0, SAB ¼ þ 3.54 kN and when x ¼ 1 m, SAB ¼ þ 5.54 kN. The complete distribution is then as shown in Fig. S.3.6(b). Bending moment The bending moment at any section between D and C is zero. In BC the bending moment distribution is linear and varies from zero at C to 5 sin 45° 1 ¼ 3:54 kNm at B. At any section in AB a distance x, say, from B the bending moment is given by MAB ¼ 5 sin 45°ð1 þ xÞ 2
x2 2
that is MAB ¼ 3:54 3:54x x 2 When x ¼ 0, MAB ¼ 3.54 kNm and when x ¼ 1 m, MAB ¼ 8.08 kNm and the distribution is parabolic. The complete distribution is then as shown in Fig. S.3.6(c). S.3.7
By inspection, the vertical reactions at A and B are each equal to W as shown in Fig. S.3.7(a). W A
W B
C
D W
W L 4 (a)
FIGURE S.3.7(a)
L 2
L 4
e24
Solutions Manual Shear force The shear force in AB is equal to W while that in BC ¼ W þ W ¼ 0. Also the shear force in CD is equal to þ W and the complete distribution is shown in Fig. S.3.7(b). W ve A
B
C
B
C
D
ve (b) W A
D
ve
(c)
WL 4
WL 4
FIGURE S.3.7(b) AND (c)
Bending moment From symmetry, MA ¼ MD ¼ 0 and MB ¼ MC ¼ WL/4 giving the distribution shown in Fig. S.3.7(c). S.3.8
The support reactions for the beam have been calculated in S.2.10(b). However, in this case, if forces and moments to the right of any section are considered, the calculation of the support reactions is unnecessary.
Shear force At any section a distance x, say, from B the shear force is given by SAB ¼ 15 5x Then, when x ¼ 0, SAB ¼ 15 kN and when x ¼ 10 m, SAB ¼ 65 kN; the distribution is linear as shown in Fig. S.3.8(b). Bending moment The bending moment is given by MAB ¼ 15x 5x
x 2
¼ 15x
5x 2 2
Solutions to Chapter 3 Problems
e25
15 kN 5 kN/m
A
B 10 m
(a)
x
A
B 15 kN ve
(b)
65 kN 400 kN m
ve
(c)
B
A
FIGURE S.3.8
When x ¼ 0, MAB ¼ 0 and when x ¼ 10 m, MAB ¼ 400 kNm. The distribution, shown in Fig. S.3.8(c), is parabolic and does not have a turning value between A and B. S.3.9
Only vertical reactions are present at the support points. Referring to Fig. S.3.9(a) and taking moments about C RB 10 1 15 7:5 þ 5 5 ¼ 0 from which RB ¼ 8:75 kN Now resolving vertically RC þ RB 1 15 5 ¼ 0 so that RC ¼ 11:25 kN
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Solutions Manual 5 kN
1 kN/m
A
B
x
C
D RC
RB 5m
10 m
5m
(a) 6.25 kN 5 kN ve A
ve
B
ve
3.75 kN
(b) A
D
C
ve
5 kN
B
5 kN
C
D
5.5 kNm ve 12.5 kNm 3.75 m (c)
25 kNm
FIGURE S.3.9
Shear force The shear force at any section between A and B a distance x from A is given by SAB ¼ þ1x At A, where x ¼ 0, SAB ¼ 0 and at B where x ¼ 5 m, SAB ¼ þ 5 kN. In BC the shear force at any section a distance x from A is given by SBC ¼ þ1x RB ¼ x 8:75 Then, when x ¼ 5 m, SBC ¼ 3.75 kN and when x ¼ 15 m, SBC ¼ 6.25 kN. In CD the shear force is constant and given by SCD ¼ 5 kN ðconsidering forces to the right of any sectionÞ The complete distribution is shown in Fig. S.3.9(b).
(i)
Solutions to Chapter 3 Problems
e27
Bending moment At any section between A and B the bending moment is given by MAB ¼ 1x
x 2
¼
x2 2
Therefore, when x ¼ 0, MAB ¼ 0 and when x ¼ 5 m, MAB ¼ 12.5 kNm. The distribution is parabolic and when x ¼ 0, (dMAB/dx) ¼ 0. In BC
x 2 MBC ¼ þ RB ðx 5Þ ¼ 0:5x 2 þ 8:75ðx 5Þ 2
(ii)
When x ¼ 5 m, MBC ¼ 12.5 kNm and when x ¼ 15 m, MBC ¼ 25 kNm. The distribution is parabolic and has a turning value when SBC ¼ 0 (see Eq. (3.4)) and, from Eq. (i), this occurs at x ¼ 8.75 m. Alternatively, but lengthier, Eq. (ii) could be differentiated with respect to x and the result equated to zero. When x ¼ 8.75 m, from Eq. (ii), MBC ¼ 5.5 kNm. In CD the bending moment distribution is linear, is zero at D and 25 kNm at C. The complete distribution is shown in Fig. S.3.9(c). S.3.10 Referring to Fig. S.3.10(a) and taking moments about C RA 6 10 3 1 4:5 0:75 ¼ 0 from which RA ¼ 5:6 kN Resolving vertically RC þ RA 10 1 4:5 ¼ 0 i.e. RC ¼ 8:9 kN Shear force In AB the shear force is constant and equal to RA 5 5.6 kN. At any section in BC a distance x from A SBC ¼ RA þ 10 þ 1ðx 3Þ ¼ 1:4 þ x
(i)
Therefore, when x ¼ 3 m, SBC 5 4.4 kN and when x 5 6 m, SBC 5 7.4 kN. In CD the shear force varies linearly from zero at D to 1 1.5 ¼ 1.5 kN at C. The complete distribution is shown in Fig. S.3.10(b). Bending moment The bending moment in AB varies linearly from zero at A to RA 3 ¼ 5.6 3 ¼ 16.8 kNm at B.
e28
Solutions Manual 10 kN 1 kN/m A C
B RA
x
D RC
3m
1.5 m
3m
(a) 7.4 kN 4.4 kN ve
D
C 1.5 kN
ve (b)
ve
B
A
5.6 kN A
0.9 kNm
B 2.9 m
C
D ve
ve
(c)
16.8 kNm
FIGURE S.3.10
In BC MBC ¼ 5:6x 10ðx 3Þ
1ðx 3Þ2 ¼ 1:4x 0:5x 2 þ 25:5 2
(ii)
so that when x ¼ 3, MBC ¼ 16.8 kNm and when x ¼ 6 m, MBC ¼ 0.9 kNm. Note that the bending moment at C, by considering the overhang CD, should be equal to 1 1.52/2 ¼ 1.125 kNm; the discrepancy is due to rounding off errors. We see from Eq. (i) that there is no turning value of bending moment in BC and that the slope of the bending moment diagram at D is zero. Also, the value of x at which MBC ¼ 0 is obtained by setting Eq. (ii) equal to zero and solving. This gives x ¼ 5.9 m so that MBC ¼ 0 at a distance of 2.9 m from B. The complete distribution is shown in Fig. S.3.10(c).
Solutions to Chapter 3 Problems
e29
S.3.11 The vertical reaction at A is given by (taking moments about B) RA 10 w 10 5 þ 10 2 ¼ 0 which gives RA ¼ 5w 2 The position of the maximum sagging bending moment in AB is most easily found by determining the shear force distribution in AB. Then, at any section a distance x from A SAB ¼ ð5w 2Þ þ wx
(i)
From Eq. (i), SAB ¼ 0 when x ¼ (5w2)/w. This corresponds to a turning value, i.e. a maximum value, of bending moment (see Eq. (3.4)). Therefore, for x ¼ 10/3 m, w ¼ 1.2 kN/m. The maximum value of bending moment in AB is then 2 10 10 3 ¼ 6:7 kNm: MAB ðmax Þ ¼ ð5 1:2 2Þ 1:2 2 3 S.3.12 Suppose that the vertical reaction at A is RA, then, taking moments about B L 5L L wL ¼0 RA L ¼ nw 2 4 2 which gives RA ¼
wLð5n þ 4Þ 8
(i)
The shear force at any section of the beam between A and B a distance x from A is given by L 5n þ 4 L L L SAB ¼ RA þ nw þ wx ¼ wL þ nw þ wx ¼ nw w þ wx 2 8 2 8 2
(ii)
The shear force is zero at a position of maximum bending moment and in this case occurs at L/3 from the right hand support, i.e. when x ¼ 2L/3. Then, from Eq. (ii) n¼
4 3
A point of contraflexure occurs at a section where the bending moment changes sign, i.e. where the bending moment is zero. At a section of the beam a distance x from A the bending moment is given by L L wx 2 þx (iii) MAB ¼ RA x nw 2 4 2 Substituting in Eq. (iii) for RA from Eq. (i) and the calculated value of n and equating to zero gives a quadratic equation in x whose factors are L/3 and L. Clearly the required value is L/3. S.3.13 Referring to Fig. S.3.13(a) the support reaction at A is given by, taking moments about B RA 20 5 20 10 þ 20 5 ¼ 0
e30
Solutions Manual 20 kN
5 kN/m
A
B
RA
C
RB
x 20 m
5m
(a) 55 kN
ve A
C
B ve
ve
20 kN (b) 45 kN 18 m
100 kNm
9m ve B
A
C
ve
(c)
202.5 kNm
FIGURE S.3.13
from which RA ¼ 45 kN Resolving forces vertically RB þ RA 5 20 20 ¼ 0 which gives RB ¼ 75 kN Shear force The shear force at any section of the beam between A and B a distance x from A is given by SAB ¼ RA þ 5x ¼ 45 þ 5x
(i)
Solutions to Chapter 3 Problems
e31
When x ¼ 0, SAB ¼ 45 kN and when x ¼ 20 m, SAB ¼ 55 kN. In BC, considering forces to the right of any section, SBC ¼ 20 kN and the complete distribution is shown in Fig. S.3.13(b). Bending moment The bending moment in AB will be a maximum when SAB ¼ 0, i.e. when, from Eq. (i), x ¼ 9 m. Also MAB ¼ 45x
5x 2 2
(ii)
so that MAB ðmax Þ ¼ 202:5 kNm The bending moment distribution is parabolic in AB and when x ¼ 0, MAB ¼ 0 and when x ¼ 20 m, MAB ¼ 100 kNm. In BC the bending moment distribution is linear and varies from zero at C to 100 kNm at B. Finally, from Eq. (ii), MAB changes sign, i.e. there is a point of contraflexure, at x ¼ 18 m (Fig. S.3.13(c)). 120 kN (total) 100 kN (total) B
A C RA
D
x 2m
RB 2.5 m
3.5 m
FIGURE S.3.14
S.3.14 The beam and its loading are shown in Fig. S.3.14. Taking moments about B RA 8 100 4 120 4:75 ¼ 0 which gives RA ¼ 121:25 kN Suppose that the maximum bending moment occurs in the bay CD. The shear force in CD, at any section a distance x from A is given by x 120ðx 2Þ ¼ 217:25 þ 60:5x SCD ¼ 121:25 þ 100 þ 8 2:5
(i)
e32
Solutions Manual For the bending moment to be a maximum SCD ¼ 0 so that, from Eq. (i), x ¼ 3.6 m. This value of x is within CD so that the assumption that the bending moment is a maximum in CD is correct. Then 100 3:62 120 ð3:6 2Þ2 ¼ 294:1 kNm MAB ðmax Þ ¼ 121:25 3:6 2 2 8 2:5 At mid-span x ¼ 4 m, so that MAB ¼ 121:25 4
100 42 120 ð4 2Þ2 ¼ 289 kNm: 2 2 8 2:5
S.3.15 The beam and its loading are shown in Fig. S.3.15(a) Taking moments about B 1 RA 6 6 2 2 ¼ 0 2 from which RA ¼ 2 kN
2 kN/m w A
B x
RA
RB 6m
(a) 4 kN 3.46 m ve A B
ve (b)
2 kN B
A ve
(c)
FIGURE S.3.15
4.62 kNm
Solutions to Chapter 3 Problems
e33
Shear force The shear force at any section a distance x from A is given by 1 xw SAB ¼ 2 þ 2 where w 5 (l/3)x from similar triangles. Then SAB ¼ 2 þ
x2 6
(i)
When x 5 0, SAB ¼ 2 kN and when x ¼ 6 m, SAB ¼ 4 kN. Examination of Eq. (i) shows that SAB ¼ 0 when x 5 3.46 m and that (dSAB/dx) 5 0 when x 5 0. The distribution is shown in Fig. S.3.15(b). Bending moment The bending moment is given by 1 x x3 MAB ¼ 2x xw ¼ 2x 18 2 3
(ii)
From Eq. (ii) MAB 5 0 when x ¼ 0 and x ¼ 6 m. Also, from Eq. (i), MAB is a maximum when x ¼ 3.46 m. Then, from Eq. (ii) MAB ðmax Þ ¼ 4:62 kNm: and the distribution is as shown in Fig. S.3.15(c). S.3.16 The arrangement is shown in Fig. S.3.16. Self-weight w Sling A
x
B a L
FIGURE S.3.16
The same argument applies to this problem as that in Ex. 3.14 in that the optimum position for the sling is such that the maximum sagging bending moment in AB will be numerically equal to the hogging bending moment at B. Then, taking moments about B L RA ðL aÞ wL a ¼ 0 2
e34
Solutions Manual from which RA ¼ wL
L a 2
La
The shear force at a section a distance x from A in AB is given by L a SAB ¼ RA þ wx ¼ wL 2 þ wx La
(i)
At the position of maximum bending moment in AB the shear force is zero giving L a x¼L 2 La Then
2 L 32 w L a 5 4 2 MAB ðmax Þ ¼ RA L La La 2 L a 2
(ii)
The bending moment at B due to the cantilever overhang is wa2/2 which is numerically equal to the maximum value of MAB. Therefore substituting for RA in Eq. (ii) and adding the two values of bending moment (their sum is zero) we obtain 2a2 4aL þ L2 ¼ 0 Solving gives a ¼ 0:29L Alternatively, as in Ex. 3.14, the relationship of Eq. (3.7) could have been used. Then, the area of the shear force diagram between A and the point of maximum sagging bending moment is equal to minus half the area of the shear force diagram between this point and B. S.3.17 The truss of Fig. P.3.17 is treated in exactly the same way as though it were a simply supported beam. The support reactions are calculated by taking moments and resolving forces and are as shown in Fig. S.3.17(a). Shear force Between A and B the shear force is constant and equal to 60 kN. Between B and C the shear force is equal to 60 þ 50 ¼ 10 kN and between C and D the shear force is equal to þ 140 kN; the complete distribution is shown in Fig. S.3.17(b). Bending moment The bending moment at the supports is zero. At B the bending moment is equal to þ 60 8 ¼ 480 kNm while at C the bending moment is þ 140 4 ¼ 560 kNm; the distribution is shown in Fig. S.3.17(c).
Solutions to Chapter 3 Problems
(a)
A
B
C
60 kN
50 kN
150 kN
e35
D
140 kN 140 kN ve
A
B
C
D
ve ve (b)
10 kN
60 kN A
B
C
D
ve
480 kNm (c)
560 kNm
FIGURE S.3.17
S.3.18 The support reactions have been previously calculated in S.2.11(a) and are as shown in Fig. S.3.18(a). Shear force In AB the shear force is equal to þ 5 kN. In BC the shear force is equal to þ 5 þ 10 ¼ þ 15 kN. In CD the shear force is equal to þ 5 þ 10 þ 15 ¼ þ 30 kN. In DE the shear force is equal to þ 5 þ 10 þ 15 þ 1557 ¼ 12 kN. In EF the shear force is equal to þ 5 þ 10 þ 15 þ 1557 þ 5 ¼ 7 kN or, more simply, considering forces to the right of any section, 52 ¼ 7 kN. In FG the shear force is equal to 5 kN while in GH the shear force is zero. The complete distribution is shown in Fig. S.3.18(b). Bending moment Since only concentrated loads are involved it is only necessary to calculate values of bending moment at the load points. Then
e36
Solutions Manual 5 kN
10 kN 15 kN 15 kN 5 kN
A
B
C
D
(a)
5 kN
E
F
57 kN
G
H
2 kN
30 kN
15 kN ve
5 kN
E B
A
C
F 7 kN
(b)
G
H
ve
D
5 kN
12 kN
100 kNm 76 kNm
40 kNm
ve 20 kNm
10 kNm (c)
A
B
C
D
E
F
FIGURE S.3.18
MA ¼ MG ¼ MH ¼ 0 MB ¼ 5 2 ¼ 10 kNm MC ¼ 5 4 10 2 ¼ 40 kNm MD ¼ 5 6 10 4 15 2 ¼ 100 kNm ME ¼ 5 12 2 8 ¼ 76 kNm MF ¼ 5 4 ¼ 20 kNm The complete distribution is shown in Fig. S.3.18(c).
G
H
Solutions to Chapter 3 Problems 1kN B
D
4kN 2kN 1m
RA,H
A
C RA,V
E RE,V
5kN 4 x Im
(a) 5kN 1kN
1kN
5kN
+ve Shear force
–ve 4kN
4kN
(b) 2kNm –ve
1kNm
Bending moment
+ve 3kNm 5kNm
(c)
FIGURE S.3.19
S.3.19 Referring to Fig. S.3.19(a) and resolving horizontally RA,H ¼ 3 kN Taking moments about E RA,V 4 þ 1 1 þ 2 1 5 3 4 1 ¼ 0 which gives RA, V ¼ 4 kN Resolving forces vertically RE,V þ 4 5 4 ¼ 0 so that RE,V ¼ 5 kN Shear force At any section between A and C the shear force is given by SAC ¼ 4 kN At any section between C and D the shear force is given by SCD ¼ 4 þ 5 ¼ þ1 kN
e37
e38
Solutions Manual At any section between D and E the shear force is given by SDE ¼ 4 þ 5 þ 4 ¼ þ5 kN The complete distribution is then as shown in Fig. S.3.19(b). Bending moment The horizontal reaction at A together with the horizontally applied loads may be regarded as producing a couple equal to 1 1 ¼ 1 kNm at A acting in a clockwise sense together with a couple equal to 2 1 ¼ 2 kNm at E also acting in a clockwise sense. Since no distributed loads are applied the bending moment distribution will be linear between A and C, C and D and D and E. Then MA MC MD ME
¼ þ1 kNm ¼ þ1 þ 4 1 ¼ þ5 kNm ¼ þ1 þ 4 3 5 2 ¼ þ3 kNm ¼ þ1 þ 4 4 5 3 ¼ 2 kNm
Note that ME is equal to the magnitude of the couple at E. The complete distribution is then as shown in Fig. S.3.19(c). S.3.20 Referring to Fig. S.3.20(a), the weight per davit is 7.5 kN vertically downwards at C. D
5 1.
m
B
C
7.5 kN 2m
A (a)
FIGURE S.3.20(a)
Now, referring to Fig. S.3.20(b), the shear force SCB at a section where the radius is inclined at an angle θ to the horizontal is given by SCB þ 7:5 sin θ ¼ 0 MCB SCB θ 7.5 kN (b)
FIGURE S.3.20(b)
Solutions to Chapter 3 Problems
e39
so that SCB ¼ 7:5 sin θ: When θ ¼ 0, SCB ¼ 0, when θ ¼ 90°, SCB ¼ 7.5 kN and when θ ¼ 180°, SCB ¼ 0. Also, in BA, SBA ¼ 0. The bending moment at any section between C and B is given by MCB þ 7:5ð1:5 1:5 cos θÞ ¼ 0 so that MCB ¼ 11:25ð cos θ 1Þ When θ ¼ 0, MCB ¼ 0, when θ ¼ 90°, MCB ¼ 11.25 kNm and when θ ¼ 180°, MCB ¼ 22.5 kNm. In BA, the bending moment is constant and equal to 22.5 kNm. S.3.21 With the head of the engine at mid-span, the load on a truss is as shown in Fig. S.3.21(a). 100 kN/m 25 kN/m A
D B 6m
x 4m
C RD 2m
(a)
FIGURE S.3.21(a)
Taking moments about A RD 12 100 4 8 25 2 11 ¼ 0, which gives RD ¼ 312:5 kN Then, at any section between D and C, a distance x from D SDC ¼ 312:5 25x When x ¼ 0, SD ¼ 312.5 kN, and when x ¼ 2 m, SC ¼ 262.5 kN. At any section between C and B, a distance x from D SCB ¼ 312:5 25 2 100ðx 2Þ i.e. SCB ¼ 462:5 100x When x ¼ 2 m, SC ¼ 262.5 kN and when x ¼ 6 m, SB ¼ 137.5 kN. Note that SCB ¼ 0 when x ¼ 4.625 m. In BA, the shear force is constant and equal to 137.5 kN and the complete
e40
Solutions Manual shear force distribution is as shown in Fig. S.3.21(b). The bending moment at any section between D and C is given by MDC ¼ RD x
25 2 x ¼ 312:5x 12:5x 2 2
Then, when x ¼ 0, MD ¼ 0 and when x ¼ 2 m, MC ¼ 575 kNm. Note that there is no turning value for MDC between D and C. At any section between C and B, the bending moment is given by MCB ¼ 312:5x 25 2ðx 1Þ
100 ðx 2Þ2 2
i.e. MCB ¼ 462:5x 50x 2 150 When x ¼ 2 m, MC ¼ 575 kNm (as before) and when x ¼ 6 m, MB ¼ 825 kNm. Further, dMCB ¼ 462:5 100x ¼ 0 for a turning value dx i.e. when x ¼ 4.625 m. Substituting this value of x gives MCBð max Þ ¼ 919:5 kNm: Between B and A, the bending moment distribution is linear and MA ¼ 0. The complete distribution of bending moment is as shown in Fig. S.3.21(c). 262.5 kN
312.5 kN Shear force
+ A
B
D C
– 137.5 kN
137.5 kN
4.625 m
(b)
FIGURE S.3.21(b)
A
B
C
D
575 kNm 825 kNm (c)
FIGURE S.3.21(c)
919.5 kNm
Bending moment
Solutions to Chapter 3 Problems
e41
S.3.22 In this problem it is unnecessary to calculate the support reactions. Also, the portion CB of the cantilever is subjected to shear and bending while the portion BA is subjected to shear, bending and torsion as shown in Fig. S.3.22.
A 3m
3 kN
3 kN
B 6 kNm
C
Equivalent loading on BA
2m
FIGURE S.3.22
Consider CB The shear force in CB is constant and equal to þ 3 kN (if viewed in the direction BA). The bending moment is zero at C and varies linearly to 3 2 ¼ 6 kNm at B. The torque in CB is everywhere zero. Consider AB The shear force in AB is constant and equal to 3 kN. The bending moment varies linearly from zero at B to 3 3 ¼ 9 kNm at A. The torque is constant and equal to 6 kNm. S.3.23 From Fig. P.3.23 the torque in CB is constant and equal to 300 Nm. In BA the torque is also constant and equal to 300100 ¼ 400 Nm. S.3.24 Referring to Fig. S.3.24(a) the torque in CB at any section a distance x from C is given by TCB ¼ 1x ðT is in Nm when x is in mmÞ Then, when x ¼ 0, TCB ¼ 0 and when x ¼ 1 m, TCB ¼ 1000 Nm. In BA the torque is constant and equal to 1 1000 þ 500 ¼ 1500 Nm. The complete distribution is shown in Fig. S.3.24(b).
e42
Solutions Manual B
1 Nm/mm
A
C
500 Nm x
(a) 1500 Nm 1000 Nm ve
(b)
A
B
C
FIGURE S.3.24
S.3.25 From symmetry the support reactions are equal and are each 2400 Nm. Then, referring to Fig. S.3.25(a) 2 Nm/mm
B 2400 Nm
A 400 Nm
C 2400 Nm x 1.0 m
0.5 m
D 400 Nm
0.5 m
(a) 1400 Nm 1000 Nm
B
A
ve
D
ve
ve
400 Nm
ve
C
400 Nm 1000 Nm (b)
1400 Nm
FIGURE S.3.25
the torque at any section a distance x from D is given by TDC ¼ 400 þ 2x ðT is in Nm when x is in mmÞ Therefore, when x ¼ 0, TDC ¼ 400 Nm and when x ¼ 0.5 m, TDC ¼ 1400 Nm. In CB the torque is given by TCB ¼ 400 þ 2x 2400 ¼ 2x 2000
Solutions to Chapter 3 Problems
e43
and when x ¼ 0.5 m, TCB ¼ 1000 Nm. Note that TCB ¼ 0 when x ¼ 1.0 m. The remaining distribution follows from antisymmetry and is shown in Fig. S.3.25(b). The maximum value of torque is 1400 Nm and occurs at C and B. S.3.26 The weight of the concrete slab is 22 3 0.15 1.5 ¼ 14.85 kN producing a uniformly distributed load on each joist of 14.85/(2 1.5) ¼ 4.95 kN/m. The weight of each joist is 17 9.81/1000 ¼ 0.17 kN/m. Then the total uniformly distributed load per joist ¼ 4.95 þ 0.17 ¼ 5.12 kN/m. The applied torques at C and E on the beam are therefore given by TC,E ¼ 5:12 1:52 þ 0:2 1:5 ¼ 6:1 kNm and are shown in Fig. S.3.26(a). 6.1 kNm B E 6.1 kNm
6.1 kNm
C
A 6.1 kNm (a)
FIGURE S.3.26(a)
From symmetry, the reaction torques at A and B are each 6.1 kNm so that the distribution of torque along the beam is as shown in Fig. S.3.26(b).
6.1 kNm + E A
B
C –
(b)
6.1 kNm
FIGURE S.3.26(b)
S.3.27 With a single load W positioned at B as shown in Fig. S.3.27(a), the reactions at A and D are 3W/4 and W/4, respectively. The shear force diagram for this case is then as shown in Fig. S.3.27(b). With a single load W positioned at C, the shear force diagram is as shown in Fig. S.3.27(c).
e44
Solutions Manual Superimposing Figs S.3.27(b) and (c) gives the complete shear force diagram for the beam shown in Fig. S.3.27(d). The bending moment distributions corresponding to these two separate cases are shown in Figs S.3.27(e) and (f), respectively. Superimposing the two diagrams gives the complete bending moment distribution for the beam shown in Fig. S.3.27(g).
W A RA
C
B L/4
L/2
L/4
D RD (a)
W/4 (b)
+ − 3W 4
+
3W (c) 4
−
W 4
W +
(d)
− W +
(e)
3WL/16
(f) +
3WL/16 (g)
+ WL/4
FIGURE S.3.27(a)–(g)
WL 4
Fig. S.3.1(a) shows the mast with two of each set of cables; the other two cables in each set are in a plane perpendicular to the plane of the paper. A B qB 40 m
qC
74.6 kN 22.5 kN
35 m C
166.4 kN
119.6 kN 25 m
D
247.4 kN 211.4 kN
qD
15 m E 314.9 kN
20 m
20 m
(a)
(b)
FIGURE S.3.1
Then, θB ¼ tan 1
20 20 20 ¼ 29:7°, θC ¼ tan 1 ¼ 38:7°, θD ¼ tan 1 ¼ 53:1°: 35 25 15
The normal force at any section of the mast will be compressive and is the sum of the self-weight and the vertical component of the tension in the cables. Furthermore the self-weight will vary linearly with distance from the top of the mast. Therefore, at a section immediately above B, N ¼ 5 4:5 ¼ 22:5 kN: At a section immediately below B, N ¼ 22:5 þ 4 15 cos 29:7° ¼ 74:6 kN: At a section immediately above C, N ¼ 74:6 þ 10 4:5 ¼ 119:6 kN: At a section immediately below C, N ¼ 119:6 þ 4 15 cos 38:7° ¼ 166:4 kN: At a section immediately above D, N ¼ 166:4 þ 10 4:5 ¼ 211:4 kN: At a section immediately below D, N ¼ 211:4 þ 4 15 cos 53:1° ¼ 247:4 kN:
e14
Solutions to Chapter 3 Problems
e15
Finally, at a section immediately above E, N ¼ 247:4 þ 15 4:5 ¼ 314:9 kN: The distribution of compressive force in the mast is shown in Fig. S.3.1(b). S.3.2
The beam support reactions have been calculated in S.2.6(a) and are as shown in Fig. S.3.2(a); the bays of the beam have been relettered as shown in Fig. P.3.2. 3 kN
7 kN
6.1 kN
A
5.7 kN
8 kN
C
E
D
9.2 kN B (a)
3.5 kN
6.9 kN 9.2 kN
5.7 kN
7.9 kN
9.2 kN 5.7 kN ve
(b)
A
B
C
E
D
7.9 kN
A
2.2 kN C
B ve
ve D
E
D
E
3.9 kN (c)
6.9 kN A
B
C
ve 27.6 kNm 40.0 kNm (d)
51.0 kNm
FIGURE S.3.2
Normal force The normal force at any section of the beam between A and C is constant and given by NAC ¼ 9.2 kN (the vertical 3 kN load has no effect on the normal force). Then NCD ¼ 9:2 3:5 ¼ 5:7 kN
e16
Solutions Manual and NDE ¼ 9:2 3:5 5:7 ¼ 0 Note that NDE could have been found directly by considering forces to the right of any section between D and E. The complete distribution of normal force in shown in Fig. S.3.2(b).
Shear force The shear force in each bay of the beam is constant since only concentrated loads are involved. At any section between A and B, SAB ¼ RA, V ¼ 6:9 kN: At any section between B and C, SBC ¼ 6:9 þ 3 ¼ 3:9 kN: At any section between C and D, SCD ¼ 6:9 þ 3 þ 6:1 ¼ 2:2 kN: Finally, at any section between D and E, SDE ¼ þRE ¼ 7:9 kN: The complete shear force distribution is shown in Fig. S.3.2(c). Bending moment Since only concentrated loads are present it is only necessary to calculate values of bending moment at the load points. Note that MA ¼ ME ¼ 0. At B, MB ¼ 6:9 4 ¼ 27:6 kNm: At C, MC ¼ 6:9 10 3 6 ¼ 51:0 kNm: At D, MD ¼ 6:9 15 3 11 6:1 5 ¼ 40 kNm: Alternatively, MD ¼ 7.9 5 ¼ 39.5 kNm; the difference in the two values is due to rounding off errors. The complete distribution is shown in Fig. S.3.2(d). S.3.3
There will be vertical and horizontal reactions at E and a vertical reaction at B as shown in Fig. S.3.3(a). The inclined 10 kN load will have vertical and horizontal components of 8 and 6 kN respectively, the latter acting to the right. Resolving horizontally, RE,H ¼ 6 kN. Now taking moments about E
Solutions to Chapter 3 Problems 8 kN
10 kN 2 kN/m
RE,H
E A
B
x
C
RB
5m
3m
D
6 kN 4m
RE,V 3m
(a) A
B
C
D
E ve
(b)
6 kN
6 kN
4 kN
4 kN
10 kN ve C A
B
ve
D
E
ve 4 kN
4 kN 10 kN
(c)
25 kNm
ve
4 kNm D
A B
C
E ve
12 kNm
(d)
FIGURE S.3.3
RB 10 2 8 11 8 3 ¼ 0 which gives RB ¼ 20 kN Resolving vertically RE,V þ RB 2 8 8 ¼ 0 so that RE,V ¼ 4 kN
e17
e18
Solutions Manual
Normal force The normal force at all sections of the beam between A and D is zero since there is no horizontal reaction at B and no horizontal forces between A and D. In CD, NCD ¼ RE,H ¼ 6 kN (compressive and therefore negative); the distribution is shown in Fig. S.3.3(b). Shear force We note that over the length of the uniformly distributed load the shear force will vary linearly with distance from, say, A. Then, at any section between A and B, a distance x from A, the shear force is given by SAB ¼ þ2x Therefore, when x ¼ 0, SAB ¼ 0 and when x ¼ 5 m (i.e. at section just to the left of B), SAB ¼ 10 kN. Also, at any section between B and C a distance x from A SBC ¼ þ2x RB ¼ 2x 20 Therefore, when x ¼ 5 m (i.e. at a section just to the right of B), SBC ¼ 10 kN and when x ¼ 8 m (i.e. at a section just to the left of C), SBC ¼ 4 kN. Between C and D the shear force is constant and SCD ¼ 2 820 ¼ 4 kN. Finally, between D and E, SDE ¼ 2 820 þ 8 ¼ þ 4 kN (or, alternatively, SDE ¼ þ RE,V ¼ þ 4 kN). The complete distribution is shown in Fig. S.3.3(c). Bending moment At any section between A and B the bending moment is given by MAB ¼ 2x
x 2
¼ x 2
Then, when x ¼ 0, MAB ¼ 0 and when x ¼ 5 m, MAB ¼ 25 kNm. Note that the distribution is parabolic and that when x ¼ 0, (dMAB/dx) ¼ 0. At any section between B and C MBC ¼ x 2 þ RB ðx 5Þ ¼ x 2 þ 20ðx 5Þ When x ¼ 5 m, MBC ¼ 25 kNm and when x ¼ 8 m, MBC ¼ 4 kNm. The distribution of bending moment between B and C is parabolic but has no turning value between B and C. At D the bending moment is most simply given by MD ¼ RE,V 3 ¼ þ 12 kNm; the complete distribution is shown in Fig. S.3.3(d). S.3.4
The forces and moments acting on the beam ABC are shown in Fig. S.3.4(a). Since the beam is a cantilever it is unnecessary to calculate the support reactions if forces and moments to the right of any section of the beam are considered.
Normal force At any section between B and C the normal force is given by NBC ¼ þ10 kN ðtensionÞ
Solutions to Chapter 3 Problems 1 5 kN
1 5 kN
7 kNm
A
B
10 kN
18 kNm
x 2.5 m
C
10 kN
18 kNm 1.3 m
(a) 20 kN
20 kN 10 kN +ve
Normal force (b)
56.6 kN 39.1 kN 24.1 kN +ve
15 kN Shear force (c)
181.0 kNm
61.4 kNm 43.4 kNm –ve
18.0 kNm Bending moment (d)
FIGURE S.3.4
At any section between A and B the normal force is given by NAB ¼ þ10 þ 10 ¼ þ20 kN ðtensionÞ The normal force diagram is therefore as shown in Fig. S.3.4(b).
e19
e20
Solutions Manual Shear force At any section between A and B the shear force is given by SAB ¼ þ15 þ 15 þ 7ð3:8 xÞ which gives SAB ¼ 56:6 7x When x ¼ 0, SAB ¼ þ 56.6 kN and when x ¼ 2.5 m, SAB ¼ þ 39.1 kN. At any section between B and C the shear force is given by SBC ¼ þ15 þ 7ð3:8 xÞ so that SBC ¼ 41:6 7x When x ¼ 2.5 m, SBC ¼ þ 24.1 kN and when x ¼ 3.8 m, SBC ¼ þ 15 kN. The shear force diagram is therefore as shown in Fig. S.3.4(c). Bending moment The bending moment at any section between B and C is given by 7 MBC ¼ 15ð3:8 xÞ ð3:8 xÞ2 18 2 which gives MBC ¼ 125:5 þ 41:6x 3:5x 2 When x ¼ 3.8 m, MBC ¼ 18.0 kNm and when x ¼ 2.5 m, MBC ¼ 43.4 kNm and the distribution is parabolic. At any section between A and B the bending moment is given by 7 MAB ¼ 15ð3:8 xÞ 15ð2:5 xÞ ð3:8 xÞ2 18 18 2 which gives MAB ¼ 181:0 þ 56:6x 3:5x 2 When x ¼ 2.5 m, MAB ¼ 61.4 kNm and when x ¼ 0, MAB ¼ 181.0 kNm and the distribution is parabolic. The complete bending moment distribution is then as shown in Fig. S.3.4(d).
S.3.5
Again, as in P.3.4, it is unnecessary to calculate the support reactions if forces and moments to the right of any section are considered.
Normal force There are no horizontal forces applied to the beam so that the normal force in both AB and CD is zero. The normal force in BC is due to the component, resolved parallel to BC, of the force at C produced by the distributed load on CD, that is
Solutions to Chapter 3 Problems
e21
NBC ¼ þ4 1 cos 60° ¼ þ2 kN ðtensionÞ The normal force diagram is therefore as shown in Fig. S.3.5(a).
2 kN A
B
(a) 14 kN
2 kN
+ve
Normal force C
D
14 kN +ve
A
3.46 kN B
Shear force 4 kN
+ve
+ve (b)
C
D
20 kNm
–ve
A
6 kNm 6 kNm
Bending moment
–ve B 2 kNm 2 kNm –ve C D (c)
FIGURE S.3.5
Shear force Since there is a uniformly distributed load on CD the shear force distribution will be linear and vary from zero at D to þ 4 1 ¼ þ 4 kN at C. In BC the shear force is equal to the component, perpendicular to BC, of the load at C from CD, that is SBC ¼ þ4 1 sin 60° ¼ þ3:46 kN In AB the shear force at any section is given by SAB ¼ þ4 1 þ 10 ¼ þ14 kN The distribution of shear force is then as shown in Fig. S.3.5(b).
e22
Solutions Manual Bending moment The bending moment at any section a distance x, say, from D is given by MCD ¼ 4
x2 2
When x ¼ 0, MCD ¼ 0 and when x ¼ 1 m, MCD ¼ 2 kNm and the distribution is parabolic. In BC and AB there are no applied distributed loads so that the bending moment distributions will be linear. Then MC ¼ 2 kNm ðalready calculatedÞ MB ¼ 4 1 1:5 ¼ 6 kNm MA ¼ 4 1 2:5 10 1 ¼ 20 kNm The complete distribution is then as shown in Fig. S.3.5(c). S.3.6
As in the previous two problems it is unnecessary to calculate the support reactions. D A
B –ve
3.54kN
C
_ve
5kN
5kN
3.54kN
Normal force
(a) 5.54kN 3.54kN 3.54kN
D Shear force
+ve A
B
C
(b) 8.08 kNm
3.44 kNm D
A
B
C
(c)
FIGURE S.3.6
Normal force At any section between D and C the normal force is given by NDC ¼ 5 kN
Solutions to Chapter 3 Problems
e23
At any section between A and C the normal force is given by NAC ¼ 5 cos 45° ¼ 3:54 kN The normal force distribution is therefore as shown in Fig. S.3.6(a). Shear force By inspection SDC ¼ 0. At any section between B and C the shear force is given by SBC ¼ þ5 sin 45° ¼ þ3:54 kN At any section between A and B, a distance x, say, from B, the shear force is given by SAB ¼ þ5 sin 45° þ 2x ¼ þ3:54 þ 2x When x ¼ 0, SAB ¼ þ 3.54 kN and when x ¼ 1 m, SAB ¼ þ 5.54 kN. The complete distribution is then as shown in Fig. S.3.6(b). Bending moment The bending moment at any section between D and C is zero. In BC the bending moment distribution is linear and varies from zero at C to 5 sin 45° 1 ¼ 3:54 kNm at B. At any section in AB a distance x, say, from B the bending moment is given by MAB ¼ 5 sin 45°ð1 þ xÞ 2
x2 2
that is MAB ¼ 3:54 3:54x x 2 When x ¼ 0, MAB ¼ 3.54 kNm and when x ¼ 1 m, MAB ¼ 8.08 kNm and the distribution is parabolic. The complete distribution is then as shown in Fig. S.3.6(c). S.3.7
By inspection, the vertical reactions at A and B are each equal to W as shown in Fig. S.3.7(a). W A
W B
C
D W
W L 4 (a)
FIGURE S.3.7(a)
L 2
L 4
e24
Solutions Manual Shear force The shear force in AB is equal to W while that in BC ¼ W þ W ¼ 0. Also the shear force in CD is equal to þ W and the complete distribution is shown in Fig. S.3.7(b). W ve A
B
C
B
C
D
ve (b) W A
D
ve
(c)
WL 4
WL 4
FIGURE S.3.7(b) AND (c)
Bending moment From symmetry, MA ¼ MD ¼ 0 and MB ¼ MC ¼ WL/4 giving the distribution shown in Fig. S.3.7(c). S.3.8
The support reactions for the beam have been calculated in S.2.10(b). However, in this case, if forces and moments to the right of any section are considered, the calculation of the support reactions is unnecessary.
Shear force At any section a distance x, say, from B the shear force is given by SAB ¼ 15 5x Then, when x ¼ 0, SAB ¼ 15 kN and when x ¼ 10 m, SAB ¼ 65 kN; the distribution is linear as shown in Fig. S.3.8(b). Bending moment The bending moment is given by MAB ¼ 15x 5x
x 2
¼ 15x
5x 2 2
Solutions to Chapter 3 Problems
e25
15 kN 5 kN/m
A
B 10 m
(a)
x
A
B 15 kN ve
(b)
65 kN 400 kN m
ve
(c)
B
A
FIGURE S.3.8
When x ¼ 0, MAB ¼ 0 and when x ¼ 10 m, MAB ¼ 400 kNm. The distribution, shown in Fig. S.3.8(c), is parabolic and does not have a turning value between A and B. S.3.9
Only vertical reactions are present at the support points. Referring to Fig. S.3.9(a) and taking moments about C RB 10 1 15 7:5 þ 5 5 ¼ 0 from which RB ¼ 8:75 kN Now resolving vertically RC þ RB 1 15 5 ¼ 0 so that RC ¼ 11:25 kN
e26
Solutions Manual 5 kN
1 kN/m
A
B
x
C
D RC
RB 5m
10 m
5m
(a) 6.25 kN 5 kN ve A
ve
B
ve
3.75 kN
(b) A
D
C
ve
5 kN
B
5 kN
C
D
5.5 kNm ve 12.5 kNm 3.75 m (c)
25 kNm
FIGURE S.3.9
Shear force The shear force at any section between A and B a distance x from A is given by SAB ¼ þ1x At A, where x ¼ 0, SAB ¼ 0 and at B where x ¼ 5 m, SAB ¼ þ 5 kN. In BC the shear force at any section a distance x from A is given by SBC ¼ þ1x RB ¼ x 8:75 Then, when x ¼ 5 m, SBC ¼ 3.75 kN and when x ¼ 15 m, SBC ¼ 6.25 kN. In CD the shear force is constant and given by SCD ¼ 5 kN ðconsidering forces to the right of any sectionÞ The complete distribution is shown in Fig. S.3.9(b).
(i)
Solutions to Chapter 3 Problems
e27
Bending moment At any section between A and B the bending moment is given by MAB ¼ 1x
x 2
¼
x2 2
Therefore, when x ¼ 0, MAB ¼ 0 and when x ¼ 5 m, MAB ¼ 12.5 kNm. The distribution is parabolic and when x ¼ 0, (dMAB/dx) ¼ 0. In BC
x 2 MBC ¼ þ RB ðx 5Þ ¼ 0:5x 2 þ 8:75ðx 5Þ 2
(ii)
When x ¼ 5 m, MBC ¼ 12.5 kNm and when x ¼ 15 m, MBC ¼ 25 kNm. The distribution is parabolic and has a turning value when SBC ¼ 0 (see Eq. (3.4)) and, from Eq. (i), this occurs at x ¼ 8.75 m. Alternatively, but lengthier, Eq. (ii) could be differentiated with respect to x and the result equated to zero. When x ¼ 8.75 m, from Eq. (ii), MBC ¼ 5.5 kNm. In CD the bending moment distribution is linear, is zero at D and 25 kNm at C. The complete distribution is shown in Fig. S.3.9(c). S.3.10 Referring to Fig. S.3.10(a) and taking moments about C RA 6 10 3 1 4:5 0:75 ¼ 0 from which RA ¼ 5:6 kN Resolving vertically RC þ RA 10 1 4:5 ¼ 0 i.e. RC ¼ 8:9 kN Shear force In AB the shear force is constant and equal to RA 5 5.6 kN. At any section in BC a distance x from A SBC ¼ RA þ 10 þ 1ðx 3Þ ¼ 1:4 þ x
(i)
Therefore, when x ¼ 3 m, SBC 5 4.4 kN and when x 5 6 m, SBC 5 7.4 kN. In CD the shear force varies linearly from zero at D to 1 1.5 ¼ 1.5 kN at C. The complete distribution is shown in Fig. S.3.10(b). Bending moment The bending moment in AB varies linearly from zero at A to RA 3 ¼ 5.6 3 ¼ 16.8 kNm at B.
e28
Solutions Manual 10 kN 1 kN/m A C
B RA
x
D RC
3m
1.5 m
3m
(a) 7.4 kN 4.4 kN ve
D
C 1.5 kN
ve (b)
ve
B
A
5.6 kN A
0.9 kNm
B 2.9 m
C
D ve
ve
(c)
16.8 kNm
FIGURE S.3.10
In BC MBC ¼ 5:6x 10ðx 3Þ
1ðx 3Þ2 ¼ 1:4x 0:5x 2 þ 25:5 2
(ii)
so that when x ¼ 3, MBC ¼ 16.8 kNm and when x ¼ 6 m, MBC ¼ 0.9 kNm. Note that the bending moment at C, by considering the overhang CD, should be equal to 1 1.52/2 ¼ 1.125 kNm; the discrepancy is due to rounding off errors. We see from Eq. (i) that there is no turning value of bending moment in BC and that the slope of the bending moment diagram at D is zero. Also, the value of x at which MBC ¼ 0 is obtained by setting Eq. (ii) equal to zero and solving. This gives x ¼ 5.9 m so that MBC ¼ 0 at a distance of 2.9 m from B. The complete distribution is shown in Fig. S.3.10(c).
Solutions to Chapter 3 Problems
e29
S.3.11 The vertical reaction at A is given by (taking moments about B) RA 10 w 10 5 þ 10 2 ¼ 0 which gives RA ¼ 5w 2 The position of the maximum sagging bending moment in AB is most easily found by determining the shear force distribution in AB. Then, at any section a distance x from A SAB ¼ ð5w 2Þ þ wx
(i)
From Eq. (i), SAB ¼ 0 when x ¼ (5w2)/w. This corresponds to a turning value, i.e. a maximum value, of bending moment (see Eq. (3.4)). Therefore, for x ¼ 10/3 m, w ¼ 1.2 kN/m. The maximum value of bending moment in AB is then 2 10 10 3 ¼ 6:7 kNm: MAB ðmax Þ ¼ ð5 1:2 2Þ 1:2 2 3 S.3.12 Suppose that the vertical reaction at A is RA, then, taking moments about B L 5L L wL ¼0 RA L ¼ nw 2 4 2 which gives RA ¼
wLð5n þ 4Þ 8
(i)
The shear force at any section of the beam between A and B a distance x from A is given by L 5n þ 4 L L L SAB ¼ RA þ nw þ wx ¼ wL þ nw þ wx ¼ nw w þ wx 2 8 2 8 2
(ii)
The shear force is zero at a position of maximum bending moment and in this case occurs at L/3 from the right hand support, i.e. when x ¼ 2L/3. Then, from Eq. (ii) n¼
4 3
A point of contraflexure occurs at a section where the bending moment changes sign, i.e. where the bending moment is zero. At a section of the beam a distance x from A the bending moment is given by L L wx 2 þx (iii) MAB ¼ RA x nw 2 4 2 Substituting in Eq. (iii) for RA from Eq. (i) and the calculated value of n and equating to zero gives a quadratic equation in x whose factors are L/3 and L. Clearly the required value is L/3. S.3.13 Referring to Fig. S.3.13(a) the support reaction at A is given by, taking moments about B RA 20 5 20 10 þ 20 5 ¼ 0
e30
Solutions Manual 20 kN
5 kN/m
A
B
RA
C
RB
x 20 m
5m
(a) 55 kN
ve A
C
B ve
ve
20 kN (b) 45 kN 18 m
100 kNm
9m ve B
A
C
ve
(c)
202.5 kNm
FIGURE S.3.13
from which RA ¼ 45 kN Resolving forces vertically RB þ RA 5 20 20 ¼ 0 which gives RB ¼ 75 kN Shear force The shear force at any section of the beam between A and B a distance x from A is given by SAB ¼ RA þ 5x ¼ 45 þ 5x
(i)
Solutions to Chapter 3 Problems
e31
When x ¼ 0, SAB ¼ 45 kN and when x ¼ 20 m, SAB ¼ 55 kN. In BC, considering forces to the right of any section, SBC ¼ 20 kN and the complete distribution is shown in Fig. S.3.13(b). Bending moment The bending moment in AB will be a maximum when SAB ¼ 0, i.e. when, from Eq. (i), x ¼ 9 m. Also MAB ¼ 45x
5x 2 2
(ii)
so that MAB ðmax Þ ¼ 202:5 kNm The bending moment distribution is parabolic in AB and when x ¼ 0, MAB ¼ 0 and when x ¼ 20 m, MAB ¼ 100 kNm. In BC the bending moment distribution is linear and varies from zero at C to 100 kNm at B. Finally, from Eq. (ii), MAB changes sign, i.e. there is a point of contraflexure, at x ¼ 18 m (Fig. S.3.13(c)). 120 kN (total) 100 kN (total) B
A C RA
D
x 2m
RB 2.5 m
3.5 m
FIGURE S.3.14
S.3.14 The beam and its loading are shown in Fig. S.3.14. Taking moments about B RA 8 100 4 120 4:75 ¼ 0 which gives RA ¼ 121:25 kN Suppose that the maximum bending moment occurs in the bay CD. The shear force in CD, at any section a distance x from A is given by x 120ðx 2Þ ¼ 217:25 þ 60:5x SCD ¼ 121:25 þ 100 þ 8 2:5
(i)
e32
Solutions Manual For the bending moment to be a maximum SCD ¼ 0 so that, from Eq. (i), x ¼ 3.6 m. This value of x is within CD so that the assumption that the bending moment is a maximum in CD is correct. Then 100 3:62 120 ð3:6 2Þ2 ¼ 294:1 kNm MAB ðmax Þ ¼ 121:25 3:6 2 2 8 2:5 At mid-span x ¼ 4 m, so that MAB ¼ 121:25 4
100 42 120 ð4 2Þ2 ¼ 289 kNm: 2 2 8 2:5
S.3.15 The beam and its loading are shown in Fig. S.3.15(a) Taking moments about B 1 RA 6 6 2 2 ¼ 0 2 from which RA ¼ 2 kN
2 kN/m w A
B x
RA
RB 6m
(a) 4 kN 3.46 m ve A B
ve (b)
2 kN B
A ve
(c)
FIGURE S.3.15
4.62 kNm
Solutions to Chapter 3 Problems
e33
Shear force The shear force at any section a distance x from A is given by 1 xw SAB ¼ 2 þ 2 where w 5 (l/3)x from similar triangles. Then SAB ¼ 2 þ
x2 6
(i)
When x 5 0, SAB ¼ 2 kN and when x ¼ 6 m, SAB ¼ 4 kN. Examination of Eq. (i) shows that SAB ¼ 0 when x 5 3.46 m and that (dSAB/dx) 5 0 when x 5 0. The distribution is shown in Fig. S.3.15(b). Bending moment The bending moment is given by 1 x x3 MAB ¼ 2x xw ¼ 2x 18 2 3
(ii)
From Eq. (ii) MAB 5 0 when x ¼ 0 and x ¼ 6 m. Also, from Eq. (i), MAB is a maximum when x ¼ 3.46 m. Then, from Eq. (ii) MAB ðmax Þ ¼ 4:62 kNm: and the distribution is as shown in Fig. S.3.15(c). S.3.16 The arrangement is shown in Fig. S.3.16. Self-weight w Sling A
x
B a L
FIGURE S.3.16
The same argument applies to this problem as that in Ex. 3.14 in that the optimum position for the sling is such that the maximum sagging bending moment in AB will be numerically equal to the hogging bending moment at B. Then, taking moments about B L RA ðL aÞ wL a ¼ 0 2
e34
Solutions Manual from which RA ¼ wL
L a 2
La
The shear force at a section a distance x from A in AB is given by L a SAB ¼ RA þ wx ¼ wL 2 þ wx La
(i)
At the position of maximum bending moment in AB the shear force is zero giving L a x¼L 2 La Then
2 L 32 w L a 5 4 2 MAB ðmax Þ ¼ RA L La La 2 L a 2
(ii)
The bending moment at B due to the cantilever overhang is wa2/2 which is numerically equal to the maximum value of MAB. Therefore substituting for RA in Eq. (ii) and adding the two values of bending moment (their sum is zero) we obtain 2a2 4aL þ L2 ¼ 0 Solving gives a ¼ 0:29L Alternatively, as in Ex. 3.14, the relationship of Eq. (3.7) could have been used. Then, the area of the shear force diagram between A and the point of maximum sagging bending moment is equal to minus half the area of the shear force diagram between this point and B. S.3.17 The truss of Fig. P.3.17 is treated in exactly the same way as though it were a simply supported beam. The support reactions are calculated by taking moments and resolving forces and are as shown in Fig. S.3.17(a). Shear force Between A and B the shear force is constant and equal to 60 kN. Between B and C the shear force is equal to 60 þ 50 ¼ 10 kN and between C and D the shear force is equal to þ 140 kN; the complete distribution is shown in Fig. S.3.17(b). Bending moment The bending moment at the supports is zero. At B the bending moment is equal to þ 60 8 ¼ 480 kNm while at C the bending moment is þ 140 4 ¼ 560 kNm; the distribution is shown in Fig. S.3.17(c).
Solutions to Chapter 3 Problems
(a)
A
B
C
60 kN
50 kN
150 kN
e35
D
140 kN 140 kN ve
A
B
C
D
ve ve (b)
10 kN
60 kN A
B
C
D
ve
480 kNm (c)
560 kNm
FIGURE S.3.17
S.3.18 The support reactions have been previously calculated in S.2.11(a) and are as shown in Fig. S.3.18(a). Shear force In AB the shear force is equal to þ 5 kN. In BC the shear force is equal to þ 5 þ 10 ¼ þ 15 kN. In CD the shear force is equal to þ 5 þ 10 þ 15 ¼ þ 30 kN. In DE the shear force is equal to þ 5 þ 10 þ 15 þ 1557 ¼ 12 kN. In EF the shear force is equal to þ 5 þ 10 þ 15 þ 1557 þ 5 ¼ 7 kN or, more simply, considering forces to the right of any section, 52 ¼ 7 kN. In FG the shear force is equal to 5 kN while in GH the shear force is zero. The complete distribution is shown in Fig. S.3.18(b). Bending moment Since only concentrated loads are involved it is only necessary to calculate values of bending moment at the load points. Then
e36
Solutions Manual 5 kN
10 kN 15 kN 15 kN 5 kN
A
B
C
D
(a)
5 kN
E
F
57 kN
G
H
2 kN
30 kN
15 kN ve
5 kN
E B
A
C
F 7 kN
(b)
G
H
ve
D
5 kN
12 kN
100 kNm 76 kNm
40 kNm
ve 20 kNm
10 kNm (c)
A
B
C
D
E
F
FIGURE S.3.18
MA ¼ MG ¼ MH ¼ 0 MB ¼ 5 2 ¼ 10 kNm MC ¼ 5 4 10 2 ¼ 40 kNm MD ¼ 5 6 10 4 15 2 ¼ 100 kNm ME ¼ 5 12 2 8 ¼ 76 kNm MF ¼ 5 4 ¼ 20 kNm The complete distribution is shown in Fig. S.3.18(c).
G
H
Solutions to Chapter 3 Problems 1kN B
D
4kN 2kN 1m
RA,H
A
C RA,V
E RE,V
5kN 4 x Im
(a) 5kN 1kN
1kN
5kN
+ve Shear force
–ve 4kN
4kN
(b) 2kNm –ve
1kNm
Bending moment
+ve 3kNm 5kNm
(c)
FIGURE S.3.19
S.3.19 Referring to Fig. S.3.19(a) and resolving horizontally RA,H ¼ 3 kN Taking moments about E RA,V 4 þ 1 1 þ 2 1 5 3 4 1 ¼ 0 which gives RA, V ¼ 4 kN Resolving forces vertically RE,V þ 4 5 4 ¼ 0 so that RE,V ¼ 5 kN Shear force At any section between A and C the shear force is given by SAC ¼ 4 kN At any section between C and D the shear force is given by SCD ¼ 4 þ 5 ¼ þ1 kN
e37
e38
Solutions Manual At any section between D and E the shear force is given by SDE ¼ 4 þ 5 þ 4 ¼ þ5 kN The complete distribution is then as shown in Fig. S.3.19(b). Bending moment The horizontal reaction at A together with the horizontally applied loads may be regarded as producing a couple equal to 1 1 ¼ 1 kNm at A acting in a clockwise sense together with a couple equal to 2 1 ¼ 2 kNm at E also acting in a clockwise sense. Since no distributed loads are applied the bending moment distribution will be linear between A and C, C and D and D and E. Then MA MC MD ME
¼ þ1 kNm ¼ þ1 þ 4 1 ¼ þ5 kNm ¼ þ1 þ 4 3 5 2 ¼ þ3 kNm ¼ þ1 þ 4 4 5 3 ¼ 2 kNm
Note that ME is equal to the magnitude of the couple at E. The complete distribution is then as shown in Fig. S.3.19(c). S.3.20 Referring to Fig. S.3.20(a), the weight per davit is 7.5 kN vertically downwards at C. D
5 1.
m
B
C
7.5 kN 2m
A (a)
FIGURE S.3.20(a)
Now, referring to Fig. S.3.20(b), the shear force SCB at a section where the radius is inclined at an angle θ to the horizontal is given by SCB þ 7:5 sin θ ¼ 0 MCB SCB θ 7.5 kN (b)
FIGURE S.3.20(b)
Solutions to Chapter 3 Problems
e39
so that SCB ¼ 7:5 sin θ: When θ ¼ 0, SCB ¼ 0, when θ ¼ 90°, SCB ¼ 7.5 kN and when θ ¼ 180°, SCB ¼ 0. Also, in BA, SBA ¼ 0. The bending moment at any section between C and B is given by MCB þ 7:5ð1:5 1:5 cos θÞ ¼ 0 so that MCB ¼ 11:25ð cos θ 1Þ When θ ¼ 0, MCB ¼ 0, when θ ¼ 90°, MCB ¼ 11.25 kNm and when θ ¼ 180°, MCB ¼ 22.5 kNm. In BA, the bending moment is constant and equal to 22.5 kNm. S.3.21 With the head of the engine at mid-span, the load on a truss is as shown in Fig. S.3.21(a). 100 kN/m 25 kN/m A
D B 6m
x 4m
C RD 2m
(a)
FIGURE S.3.21(a)
Taking moments about A RD 12 100 4 8 25 2 11 ¼ 0, which gives RD ¼ 312:5 kN Then, at any section between D and C, a distance x from D SDC ¼ 312:5 25x When x ¼ 0, SD ¼ 312.5 kN, and when x ¼ 2 m, SC ¼ 262.5 kN. At any section between C and B, a distance x from D SCB ¼ 312:5 25 2 100ðx 2Þ i.e. SCB ¼ 462:5 100x When x ¼ 2 m, SC ¼ 262.5 kN and when x ¼ 6 m, SB ¼ 137.5 kN. Note that SCB ¼ 0 when x ¼ 4.625 m. In BA, the shear force is constant and equal to 137.5 kN and the complete
e40
Solutions Manual shear force distribution is as shown in Fig. S.3.21(b). The bending moment at any section between D and C is given by MDC ¼ RD x
25 2 x ¼ 312:5x 12:5x 2 2
Then, when x ¼ 0, MD ¼ 0 and when x ¼ 2 m, MC ¼ 575 kNm. Note that there is no turning value for MDC between D and C. At any section between C and B, the bending moment is given by MCB ¼ 312:5x 25 2ðx 1Þ
100 ðx 2Þ2 2
i.e. MCB ¼ 462:5x 50x 2 150 When x ¼ 2 m, MC ¼ 575 kNm (as before) and when x ¼ 6 m, MB ¼ 825 kNm. Further, dMCB ¼ 462:5 100x ¼ 0 for a turning value dx i.e. when x ¼ 4.625 m. Substituting this value of x gives MCBð max Þ ¼ 919:5 kNm: Between B and A, the bending moment distribution is linear and MA ¼ 0. The complete distribution of bending moment is as shown in Fig. S.3.21(c). 262.5 kN
312.5 kN Shear force
+ A
B
D C
– 137.5 kN
137.5 kN
4.625 m
(b)
FIGURE S.3.21(b)
A
B
C
D
575 kNm 825 kNm (c)
FIGURE S.3.21(c)
919.5 kNm
Bending moment
Solutions to Chapter 3 Problems
e41
S.3.22 In this problem it is unnecessary to calculate the support reactions. Also, the portion CB of the cantilever is subjected to shear and bending while the portion BA is subjected to shear, bending and torsion as shown in Fig. S.3.22.
A 3m
3 kN
3 kN
B 6 kNm
C
Equivalent loading on BA
2m
FIGURE S.3.22
Consider CB The shear force in CB is constant and equal to þ 3 kN (if viewed in the direction BA). The bending moment is zero at C and varies linearly to 3 2 ¼ 6 kNm at B. The torque in CB is everywhere zero. Consider AB The shear force in AB is constant and equal to 3 kN. The bending moment varies linearly from zero at B to 3 3 ¼ 9 kNm at A. The torque is constant and equal to 6 kNm. S.3.23 From Fig. P.3.23 the torque in CB is constant and equal to 300 Nm. In BA the torque is also constant and equal to 300100 ¼ 400 Nm. S.3.24 Referring to Fig. S.3.24(a) the torque in CB at any section a distance x from C is given by TCB ¼ 1x ðT is in Nm when x is in mmÞ Then, when x ¼ 0, TCB ¼ 0 and when x ¼ 1 m, TCB ¼ 1000 Nm. In BA the torque is constant and equal to 1 1000 þ 500 ¼ 1500 Nm. The complete distribution is shown in Fig. S.3.24(b).
e42
Solutions Manual B
1 Nm/mm
A
C
500 Nm x
(a) 1500 Nm 1000 Nm ve
(b)
A
B
C
FIGURE S.3.24
S.3.25 From symmetry the support reactions are equal and are each 2400 Nm. Then, referring to Fig. S.3.25(a) 2 Nm/mm
B 2400 Nm
A 400 Nm
C 2400 Nm x 1.0 m
0.5 m
D 400 Nm
0.5 m
(a) 1400 Nm 1000 Nm
B
A
ve
D
ve
ve
400 Nm
ve
C
400 Nm 1000 Nm (b)
1400 Nm
FIGURE S.3.25
the torque at any section a distance x from D is given by TDC ¼ 400 þ 2x ðT is in Nm when x is in mmÞ Therefore, when x ¼ 0, TDC ¼ 400 Nm and when x ¼ 0.5 m, TDC ¼ 1400 Nm. In CB the torque is given by TCB ¼ 400 þ 2x 2400 ¼ 2x 2000
Solutions to Chapter 3 Problems
e43
and when x ¼ 0.5 m, TCB ¼ 1000 Nm. Note that TCB ¼ 0 when x ¼ 1.0 m. The remaining distribution follows from antisymmetry and is shown in Fig. S.3.25(b). The maximum value of torque is 1400 Nm and occurs at C and B. S.3.26 The weight of the concrete slab is 22 3 0.15 1.5 ¼ 14.85 kN producing a uniformly distributed load on each joist of 14.85/(2 1.5) ¼ 4.95 kN/m. The weight of each joist is 17 9.81/1000 ¼ 0.17 kN/m. Then the total uniformly distributed load per joist ¼ 4.95 þ 0.17 ¼ 5.12 kN/m. The applied torques at C and E on the beam are therefore given by TC,E ¼ 5:12 1:52 þ 0:2 1:5 ¼ 6:1 kNm and are shown in Fig. S.3.26(a). 6.1 kNm B E 6.1 kNm
6.1 kNm
C
A 6.1 kNm (a)
FIGURE S.3.26(a)
From symmetry, the reaction torques at A and B are each 6.1 kNm so that the distribution of torque along the beam is as shown in Fig. S.3.26(b).
6.1 kNm + E A
B
C –
(b)
6.1 kNm
FIGURE S.3.26(b)
S.3.27 With a single load W positioned at B as shown in Fig. S.3.27(a), the reactions at A and D are 3W/4 and W/4, respectively. The shear force diagram for this case is then as shown in Fig. S.3.27(b). With a single load W positioned at C, the shear force diagram is as shown in Fig. S.3.27(c).
e44
Solutions Manual Superimposing Figs S.3.27(b) and (c) gives the complete shear force diagram for the beam shown in Fig. S.3.27(d). The bending moment distributions corresponding to these two separate cases are shown in Figs S.3.27(e) and (f), respectively. Superimposing the two diagrams gives the complete bending moment distribution for the beam shown in Fig. S.3.27(g).
W A RA
C
B L/4
L/2
L/4
D RD (a)
W/4 (b)
+ − 3W 4
+
3W (c) 4
−
W 4
W +
(d)
− W +
(e)
3WL/16
(f) +
3WL/16 (g)
+ WL/4
FIGURE S.3.27(a)–(g)
WL 4