solutions lista 4.pdf

Prévia do material em texto

Structural and Stress
Analysis
Second Edition
by
Dr. T.H.G. Megson
Solutions Manual
S o l u t i o n s t o C h a p t e r 2 P r o b l e m s
S.2.1
(a) Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shown
in Fig. S.2.1. 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
R
C
A15 kN
10 kN
60°
u
O FIGURE S.2.1
(b) From Eq. (2.1) and Fig. S.2.1
R2 = 152 + 102 + 2 × 15 × 10 cos 60◦
which gives
R = 21.8 kN
Also, from Eq. (2.2)
tan θ = 10 sin 60
◦
15 + 10 cos 60◦
so that
θ = 23.4◦.
3
4 • Solutions Manual
S.2.2
(a) The vectors do not have to be drawn in any particular order. Fig. S.2.2 shows the
vector diagram with the vector representing the 10 kN force drawn first.
12 kN
8 kN
10 kN20 kN R
u
FIGURE S.2.2
The resultant 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 = 10 + 8 cos 60◦ − 12 cos 30◦ − 20 cos 55◦ = −7.9 kN
Then, resolving forces in the positive y direction
Fy = 8 cos 30◦ + 12 cos 60◦ − 20 cos 35◦ = −3.5 kN
The resultant R is given by
R2 = (−7.9)2 + (−3.5)2
so that
R = 8.6 kN
Also
tan θ = 3.5
7.9
which gives
θ = 23.9◦.
Solutions to Chapter 2 Problems • 5
S.2.3 Initially the forces are resolved into vertical and horizontal components as shown
in Fig. S.2.3.
FIGURE S.2.3
20.0 kN
30°
30°
45°
35.4 kN 50 kN
40 kN 80 kN
69.3 kN
35.4 kN
40 kN
34.6 kN
(�1, 1.25)
(0, 0.5)
(1.25, 0.25)
(1.0, 1.6)
O
y
Ry
Rx
x
60 kN
x
y
Then
Rx = 69.3 + 35.4 − 20.0 = 84.7 kN
Now taking moments about the x axis
Rx y¯ = 35.4 × 0.5 − 20.0 × 1.25 + 69.3 × 1.6
which gives
y¯ = 1.22 m
Also, from Fig. S.2.3
Ry = 60 + 40 + 34.6 − 35.4 = 99.2 kN
Now taking moments about the y axis
Ry x¯ = 40.0 × 1.0 + 60.0 × 1.25 − 34.6 × 1.0
so that
x¯ = 0.81 m
The resultant R is then given by
R2 = 99.22 + 84.72
from which
R = 130.4 kN
Finally
θ = tan−1 99.2
84.7
= 49.5◦.
6 • Solutions Manual
S.2.4
(a) In Fig. S.2.4(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
4 m 6 m 5 m 5 m
7 kN 8 kN
5.7 kN6.1 kN
3.5 kN 5.7 kN
A B
RBRA,V
RA,H
60° 45°
FIGURE S.2.4(a)
Taking moments about B
RA,V × 20 − 3 × 16 − 6.1 × 10 − 5.7 × 5 = 0
which gives
RA,V = 6.9 kN
Now resolving vertically
RB,V + RA,V − 3 − 6.1 − 5.7 = 0
so that
RB,V = 7.9 kN
Finally, resolving horizontally
RA,H − 3.5 − 5.7 = 0
so that
RA,H = 9.2 kN
Note that all reactions are positive in sign so that their directions are those
indicated in Fig. S.2.4(a).
(b) The loads on the cantilever beam will produce a vertical reaction and a moment
reaction at A as shown in Fig. S.2.4(b).
Resolving vertically
RA − 15 − 5 × 10 = 0
which gives
RA = 65 kN
Solutions to Chapter 2 Problems • 7
MA
RA
10 m
15 kN
5 kN/m
A B
FIGURE S.2.4(b)
Taking moments about A
MA − 15 × 10 − 5 × 10 × 5 = 0
from which
MA = 400 kN m
Again the signs of the reactions are positive so that they are in the directions
shown.
(c) In Fig. S.2.4(c) there are horizontal and vertical reactions at A and a vertical
reaction at B.
RA,H
A
B
RA,V RB
10 kN
2 m 4 m 2 m 2 m
5 m
15 kN
5 kN/m
20 kN
FIGURE S.2.4(c)
By inspection (or by resolving horizontally)
RA,H = 20 kN
Taking moments about A
RB × 8 + 20 × 5 − 5 × 2 × 9 − 15 × 6 − 10 × 2 = 0
which gives
RB = 12.5 kN
Finally, resolving vertically
RA,V + RB − 10 − 15 − 5 × 2 = 0
8 • Solutions Manual
so that
RA,V = 22.5 kN.
(d) The loading on the beam will produce vertical reactions only at the supports as
shown in Fig. S.2.4(d).
A B
RA RB
75 kN/m
8 kN/m
9 m3 m
FIGURE S.2.4(d)
Taking moments about B
RA × 12 + 75 − 8 × 12 × 6 = 0
Hence
RA = 41.8 kN
Now resolving vertically
RB + RA − 8 × 12 = 0
so that
RB = 54.2 kN.
S.2.5
(a) The loading on the truss shown in Fig. P.2.5(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 + 5 × 4 = 0
which gives
RA = 57 kN (upwards)
Now resolving vertically
RB + RA − 5 − 10 − 15 − 15 − 5 − 5 = 0
from which
RB = −2 kN (downwards).
Solutions to Chapter 2 Problems • 9
(b) The angle of the truss is tan−1(4/10)= 21.8◦. The loads on the rafters are sym-
metrically arranged and may be replaced by single loads as shown in Fig. S.2.5.
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.
FIGURE S.2.5
2000 N 8000 N
7427.9 N
1857.0 NRA,V
RA,H RB
20 m
4 m
742.7 N
2970.9 N
21.8°
21.8°
21.8°
Taking moments about B
RA,V × 20 + 742.7 × 2 − 1857.0 × 15 + 2970.9 × 2 + 7427.9 × 5 = 0
which gives
RA,V = −835.6 N (downwards).
Now resolving vertically
RB + RA,V − 1857.0 + 7427.9 = 0
from which
RB = −4735.3 N (downwards).
Finally, resolving horizontally
RA,H − 742.7 − 2970.9 = 0
so that
RA,H = 3713.6 N.
10 • Solutions Manual
S o l u t i o n s t o C h a p t e r 3 P r o b l e m s
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.
FIGURE S.3.1 (a)
A
B
C
D
E
20 m
40 m
15 m
25 m
35 m
20 m
uB
uD
uC
(b)
22.5
kN
119.6 kN
211.4 kN
314.9 kN
247.4 kN
166.4 kN
74.6 kN
Then,
θB = tan−1
(
20
35
)
= 29.7◦, θC = tan−1
(
20
25
)
= 38.7◦, θD = tan−1
(
20
15
)
= 53.1◦.
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 = 5 × 4.5 = 22.5 kN.
At a section immediately below B,
N = 22.5 + 4 × 15 cos 29.7◦ = 74.6 kN.
At a section immediately above C,
N = 74.6 + 10 × 4.5 = 119.6 kN.
At a section immediately below C,
N = 119.6 + 4 × 15 cos 38.7◦ = 166.4 kN.
At a section immediately above D,
N = 166.4 + 10 × 4.5 = 211.4 kN.
At a section immediately below D,
N = 211.4 + 4 × 15 cos 53.1◦ = 247.4 kN.
Finally, at a section immediately above E,
N = 247.4 + 15 × 4.5 = 314.9 kN.
The distribution of compressive force in the mast is shown in Fig. S.3.1(b).
Solutions to Chapter 3 Problems • 11
S.3.2 The beam support reactions have been calculated in S.2.4(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.
(a)
(b)
(c)
(d)
A
3 kN
7 kN 8 kN 5.7 kN
5.7 kN
6.1 kN
6.9 kN 7.9 kN
9.2 kN
3.5 kN
B
C D E
5.7 kN
9.2 kN 9.2 kN
�ve
A B C D E
6.9 kN
3.9 kN
2.2 kN
7.9 kN
�ve
�ve
A B C
D E
27.6 kN m
51.0 kN m
40.0 kN m
�ve
A B C D E
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 = 9.2 kN (the vertical 3 kN load has no effect on the normal force).
Then
NCD = 9.2 − 3.5 = 5.7 kN
and
NDE = 9.2 − 3.5 − 5.7 = 0
Note that NDE could have been found directly by considering forces to the right of
any section betweenD 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.
12 • Solutions Manual
At any section between A and B,
SAB = −RA,V = −6.9 kN.
At any section between B and C,
SBC = −6.9 + 3 = −3.9 kN.
At any section between C and D,
SCD = −6.9 + 3 + 6.1 = 2.2 kN.
Finally, at any section between D and E,
SDE = +RE = 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 = ME = 0.
At B,
MB = 6.9 × 4 = 27.6 kN m.
At C,
MC = 6.9 × 10 − 3 × 6 = 51.0 kN m.
At D,
MD = 6.9 × 15 − 3 × 11 − 6.1 × 5 = 40 kN m.
Alternatively, MD = 7.9× 5= 39.5 kN m; 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 horizon-
tal components of 8 and 6 kN respectively, the latter acting to the right. Resolving
horizontally, RE,H = 6 kN. Now taking moments about E
RB × 10 − 2 × 8 × 11 − 8 × 3 = 0
which gives
RB = 20 kN
Resolving vertically
RE,V + RB − 2 × 8 − 8 = 0
Solutions to Chapter 3 Problems • 13
so that
RE,V = 4 kN
A
(a)
B
5 m
x
3 m 4 m
10 kN
6 kN
2 kN/m
RB RE,V
RE,H
8 kN
3 m
C D
E
(b)
�ve
6 kN 6 kN
A B C D E
(c)
�ve
10 kN
10 kN
�ve
�ve
4 kN
4 kN 4 kN
4 kN
A B
C D
E
(d)
25 kN m
12 kN m
�ve
�ve
4 kN m
A
B C
D E
FIGURE S.3.3
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 =RE,H = 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 = +2x
14 • Solutions Manual
Therefore, when x= 0, SAB = 0 and when x= 5 m (i.e. at section just to the left of B),
SAB = 10 kN. Also, at any section between B and C a distance x from A
SBC = +2x − RB = 2x − 20
Therefore, when x= 5 m (i.e. at a section just to the right of B), SBC =−10 kN and
when x= 8 m (i.e. at a section just to the left of C), SBC =−4 kN.
Between C and D the shear force is constant and SCD = 2× 8− 20=−4 kN. Finally,
between D and E, SDE = 2× 8− 20+ 8=+4 kN (or, alternatively, SDE =+RE,V =
+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 = −2x
( x
2
)
= −x2
Then, when x= 0, MAB = 0 and when x= 5 m, MAB =−25 kN m. Note that the
distribution is parabolic and that when x= 0, (dMAB/dx)= 0.
At any section between B and C
MBC = −x2 + RB(x − 5) = −x2 + 20(x − 5)
When x= 5 m, MBC =−25 kN m and when x= 8 m, MBC =−4 kN m. 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 =RE,V × 3=+12 kN m; the
complete distribution is shown in Fig. S.3.3(d).
S.3.4 By inspection, the vertical reactions at A and B are each equal to W as shown
in Fig. S.3.4(a).
Shear force
The shear force in AB is equal to−W while that in BC=−W +W = 0. Also the shear
force in CD is equal to +W and the complete distribution is shown in Fig. S.3.4(b).
A B C D
W
W
W
W
L
4
L
2
L
4
(a)
FIGURE S.3.4(a)
Solutions to Chapter 3 Problems • 15
�ve
�ve
A B C
W
W
D
(b)
(c)
�ve
A DB C
WL
4
WL
4 FIGURES S.3.4(b) and (c)
Bending moment
From symmetry, MA =MD = 0 and MB =MC =WL/4 giving the distribution shown in
Fig. S.3.4(c).
S.3.5 The support reactions for the beam have been calculated in S.2.4(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.
65 kN
15 kN
�ve
A
(b)
B
(c)
400 kN m
�ve
A B
5 kN/m
15 kN
10 m
x
A B
(a)
FIGURE S.3.5
16 • Solutions Manual
Shear force
At any section a distance x, say, from B the shear force is given by
SAB = −15 − 5x
Then, when x= 0, SAB =−15 kN and when x= 10 m, SAB =−65 kN; the distribution
is linear as shown in Fig. S.3.5(b).
Bending moment
The bending moment is given by
MAB = −15x − 5x
( x
2
)
= −15x − 5x
2
2
When x= 0, MAB = 0 and when x= 10 m, MAB =−400 kN m. The distribution, shown
in Fig. S.3.5(c), is parabolic and does not have a turning value between A and B.
S.3.6 Only vertical reactions are present at the support points. Referring to
Fig. S.3.6(a) and taking moments about C
A B
RB RC
x
1 kN/m 5 kN
5 m 5 m10 m
C D
(a)
5 kN
5 kN
�ve
�ve
�ve�ve
5 kN
6.25 kN
3.75 kN
A
(b)
B C D
5.5 kN m
�ve
3.75 m
12.5 kN m
25 kN m(c)
A B C D
FIGURE S.3.6
Solutions to Chapter 3 Problems • 17
RB × 10 − 1 × 15 × 7.5 + 5 × 5 = 0
from which
RB = 8.75 kN
Now resolving vertically
RC + RB − 1 × 15 − 5 = 0
so that
RC = 11.25 kN
Shear force
The shear force at any section between A and B a distance x from A is given by
SAB = +1x
At A, where x= 0, SAB = 0 and at B where x= 5 m, SAB =+5 kN.
In BC the shear force at any section a distance x from A is given by
SBC = +1x − RB = x − 8.75 (i)
Then, when x= 5 m, SBC =−3.75 kN and when x= 15 m, SBC = 6.25 kN.
In CD the shear force is constant and given by
SCD = −5 kN (considering forces to the right of any section)
The complete distribution is shown in Fig. S.3.6(b).
Bending moment
At any section between A and B the bending moment is given by
MAB = −1x
( x
2
)
= −x
2
2
Therefore, when x= 0, MAB = 0 and when x= 5 m, MAB =−12.5 kN m. The distribu-
tion is parabolic and when x= 0, (dMAB/dx)= 0.
In BC
MBC =
(
−x2
2
)
+ RB(x − 5) = −0.5x2 + 8.75(x − 5) (ii)
When x= 5 m, MBC =−12.5 kN m and when x= 15 m, MBC =−25 kN m. The distri-
bution is parabolic and has a turning value when SBC = 0 (see Eq. (3.4)) and, from
Eq. (i), this occurs at x= 8.75 m. Alternatively, but lengthier, Eq. (ii) could be dif-
ferentiated with respect to x and the result equated to zero. When x= 8.75 m, from
Eq. (ii), MBC =−5.5 kN m.
In CD the bending moment distribution is linear, is zero at D and −25 kN m at C. The
complete distribution is shown in Fig. S.3.6(c).
18 • Solutions Manual
S.3.7 Referring to Fig. S.3.7(a) and taking moments about C
(a)
(b)
(c)
A
A
A
B
B
B
C
C
C
D
D
D
3 m 3 m
4.4 kN
2.9 m
7.4 kN
1.5 kN
16.8 kN m
0.9 kN m
5.6 kN
�ve
�ve
�ve
�ve
�ve
1.5 m
10 kN
1 kN/m
x
RA RC
FIGURE S.3.7
RA × 6 − 10 × 3 − 1 × 4.5 × 0.75 = 0
from which
RA = 5.6 kN
Resolving vertically
RC + RA − 10 − 1 × 4.5 = 0
i.e.
RC = 8.9 kN
Shear force
In AB the shear force is constant and equal to −RA =−5.6 kN.
At any section in BC a distance x from A
SBC = −RA + 10 + 1(x − 3) = 1.4 + x (i)
Therefore, when x= 3 m, SBC = 4.4 kN and when x= 6 m, SBC = 7.4 kN.
Solutions to Chapter 3 Problems • 19
In CD the shear force varies linearly from zero at D to −1× 1.5=−1.5 kN at C. The
complete distribution is shown in Fig. S.3.7(b).
Bending moment
The bending moment in AB varies linearly from zero at A to RA × 3= 5.6× 3=
16.8 kN m at B.
In BC
MBC = 5.6x − 10(x − 3) − 1(x − 3)
2
2
= −1.4x− 0.5x2 + 25.5 (ii)
so that when x= 3, MBC = 16.8 kN m and when x= 6 m, MBC =−0.9 kN m. Note that
the bending moment at C, by considering the overhang CD, should be equal to
−1× 1.52/2=−1.125 kN m; 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 = 0 is obtained by setting Eq. (ii) equal to zero and solving. This gives x= 5.9 m
so that MBC = 0 at a distance of 2.9 m from B. The complete distribution is shown in
Fig. S.3.7(c).
S.3.8 The vertical reaction at A is given by (taking moments about B)
RA × 10 − w × 10 × 5 + 10 × 2 = 0
which gives
RA = 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 = −(5w − 2) + wx (i)
From Eq. (i), SAB = 0 when x= (5w − 2)/w. This corresponds to a turning value, i.e.
a maximum value, of bending moment (see Eq. (3.4)). Therefore, for x= 10/3 m,
w= 1.2 kN/m. The maximum value of bending moment in AB is then
MAB(max) = (5 × 1.2 − 2) × 103 − 1.2 ×
(
10
3
)2
2
= 6.7 kN m.
S.3.9 Suppose that the vertical reaction at A is RA, then, taking moments about B
RAL − nw
(
L
2
)(
5L
4
)
− wL
(
L
2
)
= 0
which gives
RA = wL(5n + 4)8 (i)
20 • Solutions Manual
The shear force at any section of the beam between A and B a distance x from A is
given by
SAB = −RA + nwL2 + wx = −wL
5n + 4
8
+ nwL
2
+ wx = −nwL
8
− wL
2
+ wx (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= 2L/3. Then, from Eq. (ii)
n = 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
MAB = RAx − nw
(
L
2
)(
L
4
+ x
)
− wx
2
2
(iii)
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.3.10 Referring to Fig. S.3.10(a) the support reaction at A is given by, taking
moments about B
RA × 20 − 5 × 20 × 10 + 20 × 5 = 0
from which
RA = 45 kN
Resolving forces vertically
RB + RA − 5 × 20 − 20 = 0
which gives
RB = 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 = −RA + 5x = −45 + 5x (i)
When x= 0, SAB =−45 kN and when x= 20 m, SAB = 55 kN.
In BC, considering forces to the right of any section, SBC =−20 kN and the complete
distribution is shown in Fig. S.3.10(b).
Solutions to Chapter 3 Problems • 21
(a)
(b)
A
202.5 kN m
B C
B
�ve
�ve
�ve �ve
�ve
C
55 kN
20 kN
20 kN
RBRA
100 kN m18 m
45 kN
9 m
5 m
5 kN/m
B C
A
20 m
x
A
(c) FIGURE S.3.10
Bending moment
The bending moment in AB will be a maximum when SAB = 0, i.e. when, from Eq. (i),
x= 9 m. Also
MAB = 45x − 5x
2
2
(ii)
so that
MAB(max) = 202.5 kN m
The bending moment distribution is parabolic in AB and when x= 0, MAB = 0 and
when x= 20 m, MAB =−100 kN m.
In BC the bending moment distribution is linear and varies from zero at C to
−100 kN m at B.
Finally, from Eq. (ii), MAB changes sign, i.e. there is a point of contraflexure, at
x= 18 m (Fig. S.3.10(c)).
22 • Solutions Manual
S.3.11 The beam and its loading are shown in Fig. S.3.11.
FIGURE S.3.11
A
C
x
D
120 kN (total)
100 kN (total)
B
3.5 m2.5 m2 m
RBRA
Taking moments about B
RA × 8 − 100 × 4 − 120 × 4.75 = 0
which gives
RA = 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
SCD = −121.25 + 100 x8 +
120(x − 2)
2.5
= −217.25 + 60.5x (i)
For the bending moment to be a maximum SCD = 0 so that, from Eq. (i), x= 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
MAB(max) = 121.25 × 3.6 −
(
100
8
)
3.62
2
−
(
120
2.5
)
(3.6 − 2)2
2
= 294.1 kN m
At mid-span x= 4 m, so that
MAB = 121.25 × 4 −
(
100
8
)
42
2
−
(
120
2.5
)
(4 − 2)2
2
= 289 kN m.
S.3.12 The beam and its loading are shown in Fig. S.3.12(a)
Taking moments about B
RA × 6 − 12 × 6 × 2 × 2 = 0
from which
RA = 2 kN
Solutions to Chapter 3 Problems • 23
�ve
�ve
�ve
A
A
(a)
(b)
(c)
2 kN
B
B
4.62 kN m
3.46 m
6 m
w
4 kN
RA RB
2 kN m
A B
x
FIGURE S.3.12
Shear force
The shear force at any section a distance x from A is given by
SAB = −2 +
(
1
2
)
xw
where w= (1/3)x from similar triangles. Then
SAB = −2 + x
2
6
(i)
When x= 0, SAB =−2 kN and when x= 6 m, SAB = 4 kN. Examination of Eq. (i) shows
that SAB = 0 when x= 3.46 m and that (dSAB/dx)= 0 when x= 0. The distribution is
shown in Fig. S.3.12(b).
Bending moment
The bending moment is given by
MAB = 2x −
(
1
2
)
xw
( x
3
)
= 2x − x
3
18
(ii)
From Eq. (ii) MAB = 0 when x= 0 and x= 6 m. Also, from Eq. (i), MAB is a maximum
when x= 3.46 m. Then, from Eq. (ii)
MAB(max) = 4.62 kN m.
and the distribution is as shown in Fig. S.3.12(c).
24 • Solutions Manual
S.3.13 The arrangement is shown in Fig. S.3.13.
Self-weight w
Sling
A
B
L
x
a
FIGURE S.3.13
The same argument applies to this problem as that in Ex. 3.11 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
RA(L − a) − wL
(
L
2
− a
)
= 0
from which
RA = wL
(
L
2 − a
)
L − a
The shear force at a section a distance x from A in AB is given by
SAB = −RA + wx = −wL
(
L
2 − a
)
L − a + wx (i)
At the position of maximum bending moment in AB the shear force is zero giving
x = L
(
L
2 − a
)
L − a
Then
MAB(max) = RAL
(
L
2 − a
)
L − a −
w
2

L
(
L
2 − a
)
L − a


2
(ii)
The bending moment at B due to the cantilever overhang is −wa2/2 which is numer-
ically 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 + L2 = 0
Solutions to Chapter 3 Problems • 25
Solving gives
a = 0.29L
Alternatively, as in Ex. 3.11, 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.3.14 The truss of Fig. P.3.14 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.3.14(a).
BA C D
150 kN
560 kN m
480 kN m
140 kN
140 kN
10 kN
�ve
�ve
�ve
� ve
(a)
(b)
(c)
50 kN
60 kN
A
A
B
B
C
C
D
D
60 kN
FIGURE S.3.14
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+ 50=−10 kN and between C and D the shear force
is equal to +140 kN; the complete distribution is shown in Fig. S.3.14(b).
26 • Solutions Manual
Bending moment
The bending moment at the supports is zero. At B the bending moment is equal
to +60× 8= 480 kN m while at C the bending moment is +140× 4= 560 kN m; the
distribution is shown in Fig.S.3.14(c).
S.3.15 The support reactions have been previously calculated in S.2.5(a) and are as
shown in Fig. S.3.15(a).
FIGURE S.3.15
10 kN
30 kN
5 kN
A
A
A
B
B
B
C
C
C
D
D
D
E
E
E
F
F
F
G
G
G
H
H
H
15 kN
15 kN
5 kN
5 kN
15 kN 5 kN 5 kN
2 kN57 kN
7 kN
12 kN
(a)
(b)
(c)
�ve
�ve
�ve
100 kNm
40 kNm
20 kNm
10 kNm
76 kNm
Shear force
In AB the shear force is equal to +5 kN.
In BC the shear force is equal to +5 + 10 = +15 kN.
Solutions to Chapter 3 Problems • 27
In CD the shear force is equal to +5 + 10 + 15 = +30 kN.
In DE the shear force is equal to +5 + 10 + 15 + 15 − 57 = −12 kN.
In EF the shear force is equal to+5+10+15+15−57+5 = −7 kN or, more simply,
considering forces to the right of any section, −5 − 2 = −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.3.15(b).
Bending moment
Since only concentrated loads are involved it is only necessary to calculate values of
bending moment at the load points. Then
MA = MG = MH = 0
MB = −5 × 2 = −10 kN m
MC = −5 × 4 − 10 × 2 = −40 kN m
MD = −5 × 6 − 10 × 4 − 15 × 2 = −100 kN m
ME = −5 × 12 − 2 × 8 = −76 kN m
MF = −5 × 4 = −20 kN m
The complete distribution is shown in Fig. S.3.15(c).
S.3.16 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.3.16.
Consider CB
The shear force in CB is constant and equal to +3 kN (if viewed in the direction BA).
The bending moment is zero at C and varies linearly to −3× 2=−6 kN m 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=−9 kN m at A. The torque is constant and equal to
6 kN m.
28 • Solutions Manual
FIGURE S.3.16
A
3 kN
C 2 m
B
3 m
3 kN
6 kN m Equivalent
loading on BA
S.3.17 From Fig. P.3.17 the torque in CB is constant and equal to −300 N m. In BA
the torque is also constant and equal to −300− 100=−400 N m.
S.3.18 Referring to Fig. S.3.18(a)
(b) A B C
1500 Nm
1000 Nm
�ve
(a)
A
B C
500 Nm
x
1 Nm/mm
FIGURE S.3.18
the torque in CB at any section a distance x from C is given by
TCB = 1x (T is in N m when x is in mm)
Then, when x= 0, TCB = 0 and when x= 1 m, TCB = 1000 N m.
In BA the torque is constant and equal to 1× 1000+ 500= 1500 N m.
The complete distribution is shown in Fig. S.3.18(b).
Solutions to Chapter 4 Problems • 29
S.3.19 From symmetry the support reactions are equal and are each 2400 N m. Then,
referring to Fig. S.3.19(a)
FIGURE S.3.19
400 N m
400 N m
400 N m
400 N m
2 N m/mm
0.5 m 0.5 m1.0 m
A
A
(b)
(a)
x
B
B
C
C
D
D
2400 N m
1400 N m
1400 N m
�ve
�ve
�ve �ve
1000 N m
1000 N m
2400 N m
the torque at any section a distance x from D is given by
TDC = 400 + 2x (T is in N m when x is in mm)
Therefore, when x= 0, TDC = 400 N m and when x= 0.5 m, TDC = 1400 N m.
In CB the torque is given by
TCB = 400 + 2x − 2400 = 2x − 2000
and when x= 0.5 m, TCB =−1000 N m. Note that TCB = 0 when x= 1.0 m.
The remaining distribution follows from antisymmetry and is shown in Fig. S.3.19(b).
The maximum value of torque is 1400 N m and occurs at C and B.
S o l u t i o n s t o C h a p t e r 4 P r o b l e m s
S.4.1 Initially the support reactions are calculated. Only vertical reactions are present
so that, referring to Fig. S.4.1(a), and taking moments about E
RB × 18 − 30 × 24 + 60 × 6 = 0
i.e.
RB = 20 kN (upwards)
30 • Solutions Manual
Now resolving vertically,
RE + RB − 30 − 60 = 0
which gives
RE = 70 kN (upwards)
A
30 kN 60 kNRB RE
B
G H J K
C
5 � 6 m
8 m
u
D E F
FIGURE S.4.1(a)
All members are assumed to be in tension as shown in Fig. S.4.1(a) and for simplicity
the forces in the members are designated by their joint letters. Also the diagonals of
the truss are each 10 m long so that cos θ = 0.6 and sin θ = 0.8.
By inspection
HC = 0, BG = −20 kN and EK = −70 kN
Joint A:
Resolving vertically,
AG sin θ − 30 = 0
i.e.
AG = +37.5 kN
Resolving horizontally,
AB + AG cos θ = 0
i.e.
AB = −37.5 × 0.6 = −22.5 kN
Joint B:
Resolving horizontally,
BC − BA = 0
Solutions to Chapter 4 Problems • 31
i.e.
BC = BA = −22.5 kN
Joint G:
Resolving vertically,
GC sin θ + GB + GA sin θ = 0
i.e.
GC × 0.8 − 20 + 37.5 × 0.8 = 0
GC = −12.5 kN
Resolving horizontally,
GH + GC cos θ − GA cos θ = 0
i.e.
GH − 12.5 × 0.6 − 37.5 × 0.6 = 0
GH = 30 kN
Joint H:
By inspection (or resolving horizontally)
HJ = HG = 30 kN
Joint C:
Resolving vertically,
CJ sin θ + CG sin θ = 0
i.e.
CJ × 0.8 − 12.5 × 0.8 = 0
CJ = 12.5 kN
Joint J:
Resolving vertically,
JD + JC sin θ = 0
i.e.
JD + 12.5 × 0.8 = 0
JD = −10 kN
32 • Solutions Manual
Resolving horizontally,
JK − JH − JC cos θ = 0
i.e.
JK − 30 − 12.5 × 0.6 = 0
JK = 37.5 kN
Joint D:
Resolving vertically,
DK sin θ + DJ = 0
i.e.
DK × 0.8 − 10 = 0
DK = 12.5 kN
Resolving horizontally,
DE + DK cos θ − DC = 0
i.e.
DE + 12.5 × 0.6 + 37.5 = 0
DE = −45 kN
Joint F:
Resolving vertically,
FK sin θ − 60 = 0
i.e.
FK × 0.8 − 60 = 0
FK = 75 kN
Resolving horizontally,
FE + FK cos θ = 0
i.e.
FE + 75 × 0.6 = 0
FE = −45 kN
Solutions to Chapter 4 Problems • 33
To check the forces in the members JK and DE take a section cutting JK, KD and DE
as shown in Fig. S.4.1(b).
FIGURE S.4.1(b)
and (c)
KKJ
ED E F DE D
DJ
KJ K
F
KD
60 kN 60 kN
RE � 70 kN RE � 70 kN
E
(b) (c)
Taking moments about K,
ED × 8 − 60 × 6 = 0
ED = 45 kN
Taking moments about D,
KJ × 8 − 60 × 12 + 70 × 6 = 0
KJ = 37.5 kN
To check the force in the member JD take a section cutting JK, JD and CD as shown
in Fig. S.4.1(c).
Resolving vertically,
DJ + 70 − 60 = 0
DJ = −10 kN
S.4.2 Referring to Fig. S.4.2 the reactions at A and B are each 15 kN from symmetry.
C
15 kN
10 kN
10 kN
10 kN
15 kN
4 mE
4 �
 2 m
G
J
H
A P F
60�u
B
30�
FIGURE S.4.2
34 • Solutions Manual
Also sin θ = 4/8= 0.5 so that θ = 30◦, the remaining angles follow. By inspection (or
by resolving vertically) BJ=−15 kN. Further, by inspection, FB= 0. All members are
assumed to be in tension.
Joint A:
Resolving vertically,
AC sin θ + 15 = 0
i.e.
AC × 0.5 + 15 = 0
AC = −30 kN
Resolving horizontally,
AP + AC cos θ = 0
i.e.
AP − 30 × 0.866 = 0
AP = 26.0 kN
Joint C:
Resolving parallel to CP,
CP + 10 cos 30◦ = 0
i.e.
CP = −8.7 kN
Resolving parallel to CE,
CE − CA − 10 sin 30◦ = 0
i.e.
CE + 30 − 10 × 0.5 = 0
CE = −25 kN
Joint P:
Resolving vertically,
PE cos 30◦ + PC cos 30◦ = 0
Solutions to Chapter 4 Problems • 35
i.e.
PE = −PC = 8.7 kN
Resolving horizontally,
PF + PE cos 60◦ − PC cos 60◦ − PA = 0
i.e.
PF + 8.7 × 0.5 + 8.7 × 0.5 − 26.0 = 0
PF = 17.3 kN
Joint F:
Resolving vertically,
FE cos 30◦ + FH cos 30◦ = 0
i.e.
FE = −FH
Resolving horizontally,
FH cos 60◦ − FE cos 60◦ − FP = 0 (FB = 0)
FH × 0.5 + FH × 0.5 − 17.3 = 0
FH = −FE = 17.3 kN
Joint G:
Resolving parallel to GH,
GH + 10 cos 30◦ = 0
i.e.
GH = −8.7 kN
Resolving parallel to GE,
GJ − GE − 10 cos 60◦ = 0
i.e.
GJ − GE − 10 × 0.5 = 0 (i)
Joint H:
Resolving perpendicularly to HJ,
HE cos 30◦ + HG cos 30◦ = 0
36 • Solutions Manual
i.e.
HE = −HG = 8.7 kN
Resolving parallel to HJ,
HJ − HF+ HG cos 60◦ − HE cos 60◦ = 0
i.e.
HJ − 17.3 − 8.7 × 0.5 − 8.7 × 0.5 = 0
HJ = 26.0 kN
Joint J:
Resolving vertically,
JG cos 60◦ + JH cos 30◦ + JB = 0
i.e.
JG × 0.5 + 26.0 × 0.866 − 15 = 0
JG = −15.0 kN
Substituting for JG in Eq. (i) gives
GE = −20.0 kN
S.4.3 From Fig. P.4.3, by inspection, there will be a horizontal reaction of 4 kN acting
to the right at A. Taking moments about B,
RA,V × 12 − 40 × 10 − 40 × 8 = 0
i.e.
RA,V = 60 kN
Resolving vertically,
RB + 60 − 40 − 40 = 0
i.e.
RB = 20 kN
First take a section cutting the members EG, EH and FH as shown in Fig. S.4.3(a).
Note that tan θ = 1.5/2 so that θ = 36.9◦
Solutions to Chapter 4 Problems • 37
FIGURE S.4.3
EG
EH
FH H 2 m
1.5 m
20 kN
G B
u
4 kN 4 kN
40 kN
60 kN
A C EC
EF
FH
F
(a) (b)
Resolving vertically,
EH sin 36.9◦ + 20 = 0
i.e.
EH = −33.3 kN
Taking moments about H,
EG × 1.5 + 20 × 6 = 0
i.e.
EG = −80 kN
Resolving horizontally,
FH + EH cos 36.9◦ + EG = 0
i.e.
FH − 33.3 cos 36.9◦ − 80 = 0
FH = 106.6 kN
Now take a section cutting EC, EF and FH as shown in Fig. S.4.3(b)
Resolving vertically,
EF − 40 + 60 = 0
EF = −20 kN
S.4.4 There will be horizontal and vertical reactions at A and a vertical reaction at B.
Then, resolving horizontally,
RA,H − 2 × 6 cos 30◦ = 0
38 • Solutions Manual
i.e.
RA,H = 10.4 kN
Taking moments about B,
RA,V × 12 − 5 × 36 × 6 − 6 × 3 = 0
i.e.
RA,V = 91.5 kN
From symmetry at K, the forces in the members KE and KG are equal. Then, assum-
ing they are tensile and resolving vertically,
2 × KE cos 30◦ + 36 = 0
i.e.
KE = −20.8 kN (=KG)
Knowing KE (and KG) we can now take a section cutting through KG, EG, EF and
DF as shown in Fig. S.4.4.
10.4 kN
36 kN 36 kN
36 kN
91.5 kN
A
D DF
EF
K
EC KGEG
FIGURE S.4.4
Resolving vertically,
EF cos 30◦ + KG cos 30◦ + 3 × 36 − 91.5 = 0
i.e.
EF × 0.866 − 20.8 × 0.866 + 108 − 91.5 = 0
EF = 1.7 kN
Taking moments about E,
DF × 1.5 tan 60◦ − KG × 1.5 tan 60◦ − 36 × 1.5+ 36 × 3 − 91.5 × 4.5
+ 10.4 × 1.5 tan 60◦ = 0
Solutions to Chapter 4 Problems • 39
i.e.
DF × 2.6 + 20.8 × 2.6 − 54 + 108 − 411.8 + 10.4 × 2.6 = 0
DF = 106.4 kN
Resolving horizontally,
EG + EF cos 60◦ + KG cos 60◦ + DF + 10.4 = 0
i.e.
EG + 1.7 × 0.5 − 20.8 × 0.5 + 106.4 + 10.4 = 0
EG = −107.3 kN
S.4.5 The equation of the parabola which passes through the upper chord points
is y= kx2. When x= 9 m, y= 7 m so that 7= k × 81 which gives k= 0.086. At A,
when x= 3 m, y= 0.086×9= 0.77 m so that BA= 7−0.77= 6.23 m and when x= 6 m,
y= 0.086 × 36= 3.1 m and CD= 7 − 3.1= 3.9 m.
From symmetry, the vertical reactions at the supports are both equal to 3.5 kN. The
truss is now cut through AD, BD and BC as shown in Fig. S.4.5.
u1 D
DA
DB
CB
C
3.9 m
u2
2 kN 3.5 kN FIGURE S.4.5
From Fig. S.4.5, tan θ1 = (6.23− 3.9)/3= 0.78 so that θ1 = 37.8◦. Also tan θ2 = 3.9/3 =
1.3, i.e. θ2 = 52.4◦. Taking moments about D,
CB × 3.9 − 3.5 × 3 = 0
i.e.
CB = 2.7 kN
40 • Solutions Manual
Resolving vertically,
DA sin θ1 − DB sin θ2 − 2 + 3.5 = 0
i.e.
0.61 DA − 0.79 DB + 1.5 = 0
DA − 1.3 DB + 2.46 = 0 (i)
Now resolving horizontally,
DA cos θ1 + DB cos θ2 + CB = 0
i.e.
0.79 DA + 0.61 DB + 2.7 = 0
or
DA + 0.77 DB + 3.42 = 0 (ii)
Eq. (i) − Eq. (ii)
−2.07 DB − 0.96 = 0
DB = −0.5 kN
From Eq. (i) or Eq. (ii)
DA = −3.1 kN
S.4.6 In Fig. S.4.6 the horizontal through A meets the diagonal DE at I, the length
AI= 3 m.
B (3, 1)
T
J
D (1, 1)
y
H (3, 0) G (2, 0)
7.5 kN 5 kN
F (1, 0) x
C (2, 1)
I 1 m
E (0, 0)
45�
A
(3.5, 0.5)
0.5 m 1 m 1 m 1 m
FIGURE S.4.6
Solutions to Chapter 4 Problems • 41
Taking moments about A
T × JA − 7.5 × 1.5 − 5 × 3.5 = 0
i.e.
T × 3 sin 45◦ − 11.25 − 17.5 = 0
T = 13.6 kN
The coordinates of each joint, referred to the xy axes shown in Fig. S.4.6, are now
calculated and inserted in Fig. S.4.6.
Joint E:
x; tEF(xF − xE) + tED(xD − xE) = 0
tEF(1 − 0) + tED(1 − 0) = 0
i.e.
tEF + tED = 0 (i)
y; tEF(yF − yE) + tED(yD − yE) − 5 = 0
tEF(0 − 0) + tED(1 − 0) − 5 = 0
i.e.
tED = 5
Substituting in Eq. (i),
tEF = −5
Joint F:
x; tDE(xE − xD) + tDC(xC − xD) + tDF(xF − xD) + T cos 45◦ = 0
tDE(0 − 1) + tDC(2 − 1) + tDF(1 − 1) + 13.6 cos 45◦ = 0
i.e.
tDC − tDE + 9.6 = 0
Substituting for tDE from the above,
tDC = −4.6
y; tDE(yE − yD) + tDC(yC − yD) + tDF(yF − yD) + T sin 45◦ = 0
tDE(0 − 1) + tDC(1 − 1) + tDF(0 − 1) + 13.6 sin 45◦ = 0
42 • Solutions Manual
i.e.
−tDE − tDF + 9.6 = 0
Substituting for tDE from the above
tDF = 4.6
Now, instead of writing down the equations in terms of x and y initially, we shall insert
the coordinates directly.
Joint F:
x; tFE(0 − 1) + tFD(1 − 1) + tFC(2 − 1) + tFG(2 − 1) = 0
−tFE + tFC + tFG = 0 (ii)
y; tFE(0 − 0) + tFD(1 − 0) + tFC(1 − 0) + tFG(0 − 0) = 0
tFD + tFC = 0
tFC = −tFD = −4.6
Then, from Eq. (ii),
tFG = −0.4
Joint C:
x; tCB(3 − 2) + tCF(1 − 2) + tCD(1 − 2) + tCG(2 − 2) = 0
tCB − tCF − tCD = 0
Substituting the values of tCF and tCD
tCB = −9.2
y; tCB(1 − 1) + tCF(0 − 1) + tCD(1 − 1) + tCG(0 − 1) = 0
−tCF − tCG = 0
tCG = 4.6
Joint G:
x; tGB(3 − 2) + tGH(3 − 2) + tGF(1 − 2) + tGC(2 − 2) = 0
tGB + tGH − tGF = 0 (iii)
Solutions to Chapter 4 Problems • 43
y; tGB(1 − 0) + tGH(0 − 0) + tGF(0 − 0) + tGC(1 − 0) − 7.5 = 0
tGB + tGC − 7.5 = 0
Then
tGB = 2.9
Substituting in Eq. (iii) for tGB and tGF gives tGH = −3.3.
Joint B:
x; tBG(2 − 3) + tBA(3.5 − 3) + tBC(2 − 3) + tBH(3 − 3) = 0
−tBG + 0.5 tBA − tBC = 0 (iv)
Substituting in Eq. (iv) for tBG and tBC gives tBA = −12.6
y; tBG(0 − 1) + tBA(0.5 − 1) + tBC(1 − 1) + tBH(0 − 1) = 0
−tBG − 0.5 tBA − tBH = 0 (v)
Substituting in Eq. (v) for tBG and tBA gives tBH = 3.4
Joint H:
x; tHA(3.5 − 3) + tHB(3 − 3) + tHG(2 − 3) = 0
0.5 tHA − tHG = 0
tHA = −6.6
The tension coefficients are now multiplied by the length of each member to obtain
the force in each member:
Member Length (m) Tension coefficient Force (kN)
ED 1.414 5 7.1
EF 1 −5 −5
DC 1 −4.6 −4.6
DF 1 4.6 4.6
FG 1 −0.4 −0.4
FC 1.414 −4.6 −6.5
CB 1 −9.2 −9.2
CG 1 4.6 4.6
GB 1.414 2.9 4.1
GH 1 −3.3 −3.3
BA 0.707 −12.6 −8.9
BH 1 3.4 3.4
HA 0.707 −6.6 −4.7
44 • Solutions Manual
S.4.7 Truss of P.4.1: The support reactions have been calculated in P.4.1 and are as
shown in Fig. S.4.7(a); the spaces between the members and forces are now numbered.
2
7 8
965
1 4 3
70 kN20 kN30 kN 60 kN
10
11 12
A
G H J K
FB C D E
(a)
1
4
2
5
6 9 11
107, 8
(b) 123
FIGURES S.4.7(a)
and (b)
Starting at joint A and moving in a clockwise sense round joint A the vector 12 is
drawn vertically downwards and is equal to 30 kN to a suitable scale. The vector 25 is
drawn parallel to AG and intersects the vector 51, which is horizontal, at 5. The vector
25, equal to 37.5 kN, acts away from A so that AG is in tension. The vector 51, equal
to 22.5 kN, acts towards A so that AB is in compression.
Now moving in a clockwise sense round joint B the vector 41 is drawn vertically
upwards to represent 20 kN. The vector 15 has already been drawn and the vector
56 is vertical so that the point 6 is located by the intersection of this vector with the
horizontal vector 64. Then, BG=−20 kN (vector 56) and BC=−22.5 kN (vector 64).
At joint H the vector 72 is horizontal as is the vector 28. Further, the vector 87 is
vertical but since 72 and 28 are both horizontal the vector 87 must be zero in length
and the points 7 and 8 therefore coincide. From the force polygon HG= 30.0 kN (72)
and HJ= 30.0 kN (28), HC= 0.
Solutions to Chapter 4 Problems • 45
We now move to joint C where the point 9 is located by drawingthe vectors 89, parallel
to CJ, and 94 horizontally through 4. Then, CJ= 12.5 kN (89), CG=−12.5 kN (67)
and CD=−37.5 (94).
The point 10 is found by drawing the vectors 2 10 (horizontal) and 9 10 (vertical).
JK= 37.5 kN (2 10) and JD=−10.0 kN (9 10).
Point 11 is found by drawing the vectors 11 4 horizontally and 10 11 parallel to DK.
Then, DK= 12.5 kN (10 11) and DE=−45 kN (11 4).
The point 3 is located by drawing the vector 23 vertically downwards and equal to
60 kN. The point 12 is positioned at the intersection of the vertical through 11 and
the horizontal through 3. Then, KE=−70 kN (11 12), EF=−45 kN (12 3) and KF=
75 kN (12 2).
Truss of P.4.2: The reactions have been calculated in S.4.2 and are shown in
Fig. S.4.7(c); The spaces between the loads and members are now numbered.
A B
C
E
G
H
FP
J
15 kN 15 kN
10 kN
10 kN
10 kN
(c) 1
2
3
4 5
6
7
8
9
11
12
10
2
3
1, 12
4
6
7
10
11
9 5
8
(d) FIGURES S.4.7(c) and (d)
Starting at joint A the vector 12 is drawn vertically upwards to represent the 15 kN
reaction (Fig. S.4.7(d)). The point 6 is located at the intersection of 16 (horizontal)
and 26 (parallel to AC). Then, AC=−30 kN (26) and AP= 26 kN (61).
Moving to joint C the vector 23 is drawn to represent the 10 kN load and the point 7 is
found by drawing 37 (parallel to AC) and 76 (parallel to CP). CP=−8.7 kN (76) and
CE=−25 kN (37).
46 • Solutions Manual
At P, 78 is drawn parallel to PE and 81 drawn parallel to PF, hence the point 8. Then,
PE= 8.7 kN (78) and PF= 17.3 kN (81).
We now cannot move to joints E, F, G or H since there are more than two unknowns
at each of these joints. Therefore moving to joint B, the vector 51 is drawn vertically
upwards to represent the 15 kN reaction. Now 1 12 is horizontal and 12 5 is vertical so
that the points 1 and 12 must coincide; Then, FB= 0 (1 12) and BJ=−15 kN (12 5).
The point 11 is found by drawing 11 5 parallel to GJ and 12 11 parallel to JH. This gives
JH= 26 kN (12 11) and GJ=−15 kN (11 5). Considering joint G the point 4 is located
by drawing the vector 45 vertically downwards to represent the 10 kN load. Then, point
10 is the intersection of 10 4 (parallel to EG) and 11 10 (parallel to HG). This gives
EG=−20 kN (10 4) and GH=−8.7 kN (11 10). Finally point 9 is the intersection of
12 9 (parallel to HF) and 89 (parallel to EF) or the intersection of either of these
two with 10 9 (parallel to EH). Therefore, EH= 8.7 kN (9 10), EF=−17.3 kN (89),
HF= 17.3 kN (12 9).
S.4.8 The x, y and z coordinates of each joint are, O(0, 0, 0), A(−3.5, −4, 9),
B(6.5, −4, 9) and C(1, 8, 9). Then,
x; tOA(−3.5 − 0) + tOB(6.5 − 0) + tOC(1 − 0) + 5 = 0
i.e.
tOA − 1.86 tOB − 0.29 tOC − 1.43 = 0 (i)
y; tOA(−4 − 0) + tOB(−4 − 0) + tOC(8 − 0) + 40 = 0
i.e.
tOA + tOB − 2 tOC − 10 = 0 (ii)
z; tOA(9 − 0) + tOB(9 − 0) + tOC(9 − 0) = 0
i.e.
tOA + tOB + tOC = 0 (iii)
Eq. (ii) − Eq. (iii)
−3 tOC − 10 = 0
i.e.
tOC = −3.33
Eq. (i) − Eq. (ii)
−2.86 tOB + 1.71 tOC + 8.57 = 0
Solutions to Chapter 4 Problems • 47
Substituting for tOC gives
tOB = 1.01
From Eq. (iii)
tOA = 2.32
The length of member OA is given by
LOA =
√
(xA − xO)2 + (yA − yO)2 + (zA − zO)2
=
√
(−3.5 − 0)2 + (−4 − 0)2 + (9 − 0)2 = 10.45 m
Similarly,
LOB = 11.8 m and LOC = 12.08 m
Then
OA = 2.32 × 10.45 = 24.2 kN, OB = 1.01 × 11.8 = 11.9 kN,
OC = −3.33 × 12.08 = −40.2 kN.
S.4.9 Take the origin of axes at A and assume the axes system shown in Fig. S.4.9.
A
x
z y
25 kN
25 kN
D
B
C
45�
45�
FIGURE S.4.9
The coordinates of each joint are as follows: A(0, 0, 0), B(−2.5, 0, 4), C(0, 3, 4) and
D(2.5, 0, 4). Then,
x; tAB(−2.5 − 0) + tAC(0 − 0) + tAD(2.5 − 0) + 25 cos 45◦ − 25 cos 45◦ = 0
i.e.
tAB − tAD = 0 (i)
48 • Solutions Manual
y; tAB(0 − 0) + tAC(3 − 0) + tAD(0 − 0) + 2 × 25 cos 45◦ = 0
i.e.
tAC = −11.79 (ii)
z; tAB(4 − 0) + tAC(4 − 0) + tAD(4 − 0) + 25 = 0
i.e.
tAB + tAC + tAD + 6.25 = 0 (iii)
Substituting in Eq. (iii) from Eqs (i) and (ii)
tAB = 2.77 = tAD
Now
LAB = LAD =
√
42 + 2.52 = 4.72 m and LAC =
√
42 + 32 = 5.0 m.
Then,
AB = 2.77 × 4.72 = 13.1 kN = AD, AC = −11.79 × 5 = −59.0 kN.
S.4.10 Referring to Fig. P.4.10 choose an origin of axes at E with the x axis parallel to
EF, the y axis parallel to EB and the z axis vertical. The coordinates of the joints are
then E(0, 0, 0), B(0, 4, 0), C(0, 4, 3), F(3, 0, 0), A(3, 4, 0) and D(3, 4, 3).
Joint E:
x; tEB(0 − 0) + tEC(0 − 0) + tEF(3 − 0) + 3 = 0
i.e.
tEF = −1 (i)
y; tEB(4 − 0) + tEC(4 − 0) + tEF(0 − 0) = 0
i.e.
tEB + tEC = 0 (ii)
z; tEB(0 − 0) + tEC(3 − 0) + tEF(0 − 0) + 9 = 0
i.e.
tEC = −3
Therefore, from Eq. (ii),
tEB = 3
Solutions to Chapter 4 Problems • 49
Joint F:
x; tFE(0 − 3) + tFB(0 − 3) + tFA(3 − 3) + tFD(3 − 3) = 0
i.e.
tFE + tFB = 0 (iii)
Therefore, from Eq. (i),
tFB = 1
y; tFE(0 − 0) + tFB(4 − 0) + tFA(4 − 0) + tFD(4 − 0) = 0
i.e.
tFA + tFD + 1 = 0 (iv)
z; tFE(0 − 0) + tFB(0 − 0) + tFA(0 − 0) + tFD(3 − 0) + 6 = 0
i.e.
tFD = −2
Then, from Eq. (iv)
tFA = 1
From Fig. P.4.10, LEB =LFA = 4 m, LEF = 3 m and LEC =LFB =LFD = 5 m. Then
EF = −1 × 3 = −3 kN, EC = −3 × 5 = −15 kN, EB = 3 × 4 = 12 kN,
FB = 1 × 5 = 5 kN, FA = 1 × 4 = 4 kN, FD = −2 × 5 = −10 kN.
S.4.11 With the origin of axes at joint D as shown in Figs S.4.7(a) and (b) the coordi-
nates of the joints are D(0, 0, 0), E(4, 0, 0), A(−2, 4, −4), B(6, 4, −4), C(2, −4, −4).
Further, the analysis can only begin at joint D since there are more than three unknowns
at every other joint.
FIGURE S.4.11
4 m
A, B C
D, E
z z
y 80 kN
4 m
4 m
2 m
A
40 kN D Ex
C B
2 m 2 m 2 m
(a) (b)
Joint D:
x; tDE(4 − 0) + tDA(−2 − 0) + tDC(2 − 0) − 40 = 0
50 • Solutions Manual
i.e.
2 tDE − tDA + tDC − 20 = 0 (i)
y; tDE(0 − 0) + tDA(4 − 0) + tDC(−4 − 0) = 0
i.e.
tDA − tDC = 0 (ii)
z; tDE(0 − 0) + tDA(−4 − 0) + tDC(−4 − 0) = 0
i.e.
tDA + tDC = 0 (iii)
Eqs (ii) and (iii) can only be satisfied if tDA = tDC = 0. Then, from Eq. (i) tDE = 10.
Joint E:
x; tED(0 − 4) + tEA(−2 − 4) + tEB(6 − 4) + tEC(2 − 4) = 0
i.e.
−40 − 6 tEA + 2 tEB − 2 tEC = 0 (iv)
y; tED(0 − 0) + tEA(4 − 0) + tEB(4 − 0) + tEC(−4 − 0) − 80 = 0
i.e.
4 tEA + 4 tEB − 4 tEC − 80 = 0 (v)
z; tED(0 − 0) + tEA(−4 − 0) + tEB(−4 − 0) + tEC(−4 − 0) = 0
i.e.
−4 tEA − 4 tEB − 4 tEC = 0 (vi)
Adding Eqs (v) and (vi) gives
−8 tEC − 80 = 0
i.e.
tEC = −10
Also, Eq. (v) − Eq. (vi) gives
tEA + tEB − 10 = 0 (vii)
Substituting in Eq. (iv) for tEC gives
−20 − 6 tEA + 2 tEB = 0 (viii)
Solutions to Chapter 5 Problems • 51
2×Eq. (vii) − Eq. (viii) gives
tEA = 0
Then, from Eq. (vii) tEB = 10
From Fig. S.4.11, LED = 4 m, LEB =LDA =LEC =LDC =
√
22 + 42 + 42 = 6 m. Then,
AD= 0, DC= 0, DE= 10× 4= 40 kN, AE= 0, CE= − 10× 6= − 60 kN, BE= 10×
6= 60 kN.
S o l u t i o n s t o C h a p t e r 5 P r o b l e m s
S.5.1 Referring to Fig. S.5.1 where H is the tension in the portion CD of the cable and
taking moments about A
RA,H
RA,V
A
B
H
C
a
b
5 kN
5 kN
1.5 m
FIGURE S.5.1
H × 1.5 − 5 × 5 − 5 × 2.5 = 0
i.e.
H = 25 kN = TCD
Resolving horizontally
RA,H − H = 0
i.e.
RA,H = 25 kN
From symmetry,
RA,V = 10 kN
The tension in AB is equal to the resultant of RA,V and RA,H,
i.e.
TAB =
√
102 + 252 = 26.9 kN
52 • Solutions Manual
From Fig. S.5.1,
tan α = 10
25
= 0.4
Then
α = 21.8◦
Also the vertical distance of B below A is 2.5 tanα= 1.0 m. Then
tan β = 0.5
2.5
= 0.2
so that
β = 11.3◦
Now resolving horizontally at C
TCB cos 11.3◦ = TCD = 25 kN
i.e.
TCB = 25.5 kN
The tensions in the remaining parts of the cable follow from symmetry.
S.5.2 From Fig. S.5.2
C
B
D
0.7 m
0.5 m
A1 kN
2 kN
RA,V
RA,H
RD,V
RD,H
a
g
FIGURE S.5.2
resolving vertically
RA,V + RD,V = 1 + 2 = 3 kN (i)
Taking moments about C
RD,H × 0.5 − RD,V × 2 = 0
i.e.
RD,H − 4 RD,V = 0 (ii)
Taking moments about A
RD,H × 0.7 + RD,V × 6 − 1× 2 − 2× 4 = 0
Solutions to Chapter 5 Problems • 53
i.e.
RD,H + 8.57 RD,V − 14.29 = 0 (iii)
Eq. (ii) − Eq. (iii)
−12.57 RD,V + 14.29 = 0
i.e.
RD,V = 1.14 kN
From Eq. (ii)
RD,H = 4.56 kN = RA,H
From Eq. (i)
RA,V = 1.86 kN
Now
tan α = RA,V
RA,H
= 1.86
4.56
= 0.41
i.e.
α = 22.19◦
Therefore the sag of B relative to A is = 2 tan α= 0.81 m.
Further
γ = tan−1 (1.2 − 0.81)
2
= 11.03◦
At A the resultant of RA,H and RA,V is equal to the tension TAB,
i.e.
TAB =
√
4.562 + 1.862 = 4.9 kN
Also
TDC =
√
4.562 + 1.142 = 4.7 kN
At B
TBC cos 11.03◦ − TBA cos 22.19◦ = 0
i.e.
TBC = 4.6 kN.
S.5.3 Suppose the horizontal and vertical components of reaction at A and E are
RA,H, RA,V and RE,H, RE,V, respectively. Then, taking moments about B,
RA,V × 4 − RA,H × 2.6 = 0
54 • Solutions Manual
i.e.
1.54 RA,V − RA,H = 0 (i)
Now taking moments about E
RA,V × 18 − RA,H × 0.5 − 3× 14 − 5× 9 − 4× 4 = 0
i.e.
36 RA,V − RA,H − 206 = 0 (ii)
Eq. (i) − Eq. (ii)
−34.46 RA,V + 206 = 0
i.e.
RA,V = 5.98 kN
From Eq. (i)
RA,H = 9.21 kN
Then
TAB =
√
5.982 + 9.212 = 10.98 kN
Suppose α is the angle AB makes with the horizontal, then
α = tan−1 2.6
4
= 33.02◦
Also let β be the angle BC makes with the horizontal. Then, resolving horizontally
at B
TBC cosβ = TAB cosα = 10.98 cos 33.02◦ = 9.21 kN
Resolving vertically at B
TBC sin β + 3 = TAB sin α = 10.98 sin 33.02◦ = 5.98 kN
i.e.
TBC sin β = 2.98
Then
tan β = 2.98
9.21
= 0.324
i.e.
β = 17.93◦
and
TBC = 9.68 kN
Solutions to Chapter 5 Problems • 55
The sag of C below A is equal to 2.6+ 5 tan β = 4.22 m.
Let γ be the angle CD makes with the horizontal. Then, resolving horizontally at C,
TCD cos γ = TCB cosβ = 9.68 cos 17.93◦ = 9.21 kN
Resolving vertically
TCD sin γ = 5 − TCB sin β = 2.02
Then
γ = tan−1 2.02
9.21
= 12.37◦
and
TCD = 9.21cos γ = 9.431 kN
The sag of D below A is equal to 4.22− 5 tan γ = 3.13 m.
Resolving horizontally
RE,H = RA,H = 9.21 kN
and resolving vertically
RE,V = 12 − RA,V = 6.02 kN
Then
TED =
√
6.022 + 9.212 = 11.0 kN.
S.5.4 Consider half the cable as shown in Fig. S.5.4, it is immaterial which half.
40 m
D
B
H
C
10 kN/m
RB,H
RB,V
FIGURE S.5.4
Taking moments about B
H × D − 10 × 40
2
2
= 0
i.e.
H = 8000
D
56 • Solutions Manual
Resolving horizontally
RB,H − H = 0
i.e.
RB,H = 8000D
Resolving vertically
RB,V = 10 × 40 = 4000 kN
Now
Tmax =
√
R2B,V + R2B,H =
√
4002 +
(
8000
D
)2
But
Tmax = 1000 kN =
√
4002 +
(
8000
D
)2
which gives
D = 8.73 m.
S.5.5 Consider half the cable shown in Fig. S.5.5; the cable is symmetrical about the
mid-span point.
H C
D
B
d
10 kN
RB,H
RB,V
17 m 34 m
36 N/m
3 m
FIGURE S.5.5
Resolving vertically
RB,V = 0.036× 51 + 10 = 11.84 kN
Taking moments about C
RB,H × 3 − 11.84× 51 + 10× 17 + 0.036× 51
2
2
= 0
which gives
RB,H = 129.0 kN
Then
Tmax =
√
11.842 + 129.02 = 129.5 kN
Solutions to Chapter 5 Problems • 57
Now taking moments about D
11.84 × 34 − 129.0 d − 0.036 × 34
2
2
= 0
i.e.
d = 2.96 m.
S.5.6 The cable is symmetrical about the mid-span point. Then, referring to Fig. S.5.6
and taking moments about B
H × 8 − 24 × 40
2
2
= 0
24 kN/m
8 m
B
H
40 m
RB,V
RB,H
Tmax
a
C
FIGURE S.5.6
i.e.
H = 2400 kN = RB,H
Resolving vertically
RB,V = 24 × 40 = 960 kN
The angle the suspension cable makes with the horizontal at the top of each tower is
given by
tan α = 960
2400
= 0.4
i.e.
α = 21.8◦
Since the cable passes over frictionless pulleys and there is no bending moment at the
base of a tower the anchor cable must be inclined at the same angle to the horizontal
as the suspension cable, i.e. 21.8◦.
Also
Tmax =
√
24002 + 9602 = 2584.9 kN
58 • Solutions Manual
The vertical force on a tower is caused by the tension in the suspension cable and the
tension in the anchor cable, i.e.
Vertical force = 2 × 2584.9 sin 21.8◦ = 1919.9 kN.
S.5.7 The suspension cable is shown in Fig. S.5.7. Since the towers are of differ-
ent height the lowest point in the cable will not be at mid-span. The position of the
lowest point may be found by either using Eq. (5.17) directly or by working from first
principles. We shall adopt the latter approach.
A
C
B
25 kN/m
120 m
2.5 m
7.5 m
H
L1 L2
FIGURE S.5.7
Taking moments about B for CB
H × 10 = 25L
2
2
2
i.e.
H = 5L
2
2
4
(i)
Now taking moments about A for CA
H × 7.5 = 25L
2
1
2
i.e.
H = 5L
2
1
3
(ii)
Equating Eqs (i) and (ii) gives
L1 = 0.866L2
Then, since L1 +L2 = 120 m,
L2 = 64.31 m
From Eq. (i)
H = 5169.7 kN
Solutions to Chapter 5 Problems • 59
H is equal to the horizontal component of the maximum tension in the suspension
cable which must occur at B since the span CB is greater than the span CA. The
vertical component of the maximum tension is equal to 25× 64.31= 1607.8 kN. Then
Tmax =
√
5169.72 + 1607.82 = 5413.9 kN
We must now investigate the tension in the anchor cable since this will be different
to the tension in the suspension cable. There is no resultant horizontal force on the
top of a tower since the saddles are on rollers. Therefore
TAC cos 55◦ = H = 5169.7 kN
i.e.
TAC = 9013.1 kN
The maximum tension in the cable is therefore 9013.1 kN and the maximum stress is
σmax = 9013.1 × 10
3(
πD2
4
) = 240
Then
D = 218.7 mm
The vertical load on the tallest tower= 1607.8+ 9013.1 sin 55◦ = 8990.9 kN.
S.5.8 The central half of the cable has its cross-sectional area reduced to
0.8× 0.08= 0.064 m2. Considering one half of the cable as shown in Fig. S.5.8 the
critical points are C where the tension is a maximum and B where the cross-sectional
area is reduced.
A
B
C
40 m
Tmax
TB
100 m100 m
H
w kN/m
a
FIGURE S.5.8
Consider the portion AC and suppose that the load intensity is w kN/m. Taking
moments about C
H × 40 − w × 200
2
2
= 0
i.e.
H = 500 w
60 • Solutions Manual
H is equal to the horizontal component of the maximum tension in the cable at C. The
vertical component is equal to 200w. Therefore
Tmax =
√
(200w)2 + (500w)2 = 538.5w
The maximum allowable stress in the cable is 500 N/mm2. Therefore
538.5w = 500 × 103 × 0.08
which gives
w = 74.3 kN/m
The horizontal component of the tension in the cable at B is equal to H , the vertical
component is equal to 100w. Then
TB =
√
(100w)2 + (500w)2 = 509.9w
Therefore
509.9w = 500× 103 × 0.064
from which
w = 62.8 kN/m
The maximum allowable value of w is therefore 62.8 kN/m.
At the top of the towers the inclination of the cable to the horizontal is given by
α = tan−1 200w
500w
= 21.8◦.
S.5.9 The suspension bridge is supported by two cables; the load/cable is therefore
12.5 kN/m.
(a) Consider the right hand half of the cable shown in Fig. S.5.9.
12.5 kN/m
25 m
B
H
125 m
RB,V
RB,H
C
FIGURE S.5.9
Solutions to Chapter 5 Problems • 61
Taking moments about B
H × 25 − 12.5 × 125
2
2
= 0
i.e.
H = 3906.25 kN
Resolving horizontally
RB,H = H = 3906.25 kN
Resolving vertically
RB,V = 12.5 × 125 = 1562.5 kN
Then
Tmax =
√
(3906.25)2 + (1562.5)2 = 4207.16 kN
Therefore the required area of cross-section of each cable is
A = 4207.16 × 10
3
800
= 5259.0 mm2.
(b) (i) The load in the anchor cable is equal to 4207.16 kN and the overturningforce is
given by
Overturning force = RB,H − 4207.16 cos 45◦ = 3906.25 − 2974.9 = 931.3 kN.
(ii) The horizontal components of the maximum tension in the cable and the tension
in the anchor cable are equal and opposite since the cable passes over a saddle
resting on rollers, i.e.
TAC cos 45◦ = 3906.25
so that
TAC = 5524.3 kN
There is zero overturning force on the tower.
S.5.10 As in P.5.7 we need, initially, to determine the position of the lowest point of
the cable.
For CB, taking moments about B
H × 16 = 5L
2
2
2
i.e.
H = 5L
2
2
32
(i)
Similarly, for CA and taking moments about A
H = 5L
2
1
24
(ii)
62 • Solutions Manual
A
C
B
5 kN/m
80 m
12 m
16 m
H
L1 L2
RB,V
RB,H
FIGURE S.5.10
Equating Eqs (i) and (ii) gives
L1 = 0.866L2
Therefore, since L1 +L2 = 80,
L2 = 42.87 m
Then, from Eq. (i)
H = 287.16 kN = RB,H
Resolving vertically
RB,V = 5 × 42.87 = 214.35 kN
The maximum tension in the cable is given by
Tmax =
√
(287.16)2 + (214.35)2 = 358.3 kN
The horizontal component of the tension in the anchor cable is equal to the horizontal
component of the maximum tension in the suspension cable since the cable passes
over a saddle on rollers. Therefore
TAC cos 45◦ = 287.16
i.e.
TAC = 406.1 kN
The vertical thrust on the tallest tower is equal to the sum of the vertical components
of the maximum tension in the suspension cable and the tension in the anchor cable,
i.e.
Vertical thrust= 214.35+ 287.16= 501.5 kN.
Solutions to Chapter 6 Problems • 63
S o l u t i o n s t o C h a p t e r 6 P r o b l e m s
S.6.1 Referring to Fig. S.6.1 and taking moments about B
y
10 m
x
B
C
A
RA,H
RA,V RB,V
RB,H
20 kN/m
FIGURE S.6.1
RA,V × 20− 20× 10× 15= 0
i.e.
RA,V = 150 kN
Now taking moments about C
RA,H × 10−RA,V × 10+ 20× 10
2
2
= 0
i.e.
RA,H = 50 kN
With the origin of axes at the centre of the arch the equation of the arch is
x2 + y2 = 102
Therefore, when x=−5 m, y=√102 − 52 = 8.66 m.
The bending moment at this point is given by
M = RA,V × 5 − RA,H × 8.66 − 20 × 5
2
2
= 150 × 5 − 50 × 8.66 − 250 = 67 kN m.
S.6.2 For the origin of axes shown in Fig. S.6.2 the equation of the arch is
x2 + y2 = 122
Then, when y = 4 m, x = 11.31 m and when y = 6 m, x = 10.39 m.
64 • Solutions Manual
3 m 2 m
20 kN
D
A
RA,V
RB,V
RB,H
RA,H
8 m
6 m
12
m
y
x
C
f B
FIGURE S.6.2
Taking moments about B
RA,V × 21.7 − RA,H × 2 − 20× 8.39 = 0
i.e.
10.85RA,V − RA,H − 83.9 = 0 (i)
Now taking moments about C
RA,V × 11.31 − RA,H × 8 = 0
i.e.
1.41RA,V − RA,H = 0 (ii)
Eq. (i)−Eq. (ii)
9.44RA,V − 83.9 = 0
i.e.
RA,V = 8.89 kN
From Eq. (ii)
RA,H = 12.53 kN
The radius at D makes an angle with the vertical given by
φ = sin−1 3
12
= 14.48◦
Then
NF = 12.53 cosφ + 8.89 sinφ = 14.35 kN (compression)
and
SF = 8.89 cosφ − 12.53 sinφ = 5.48 kN
The vertical distance of D above A = (√122 − 32) − 4 = 7.62 m. Then the bending
moment at D is given by
BM = 8.89 × 8.31 − 12.53 × 7.62 = −21.6 kN m (hogging).
Solutions to Chapter 6 Problems • 65
S.6.3 Take the origin of axes at the crown C of the arch as shown in Fig. S.6.3; the
equation of the arch is then
y = k x2
When x = −7 m, y = 3 m so that k = 3/72 = 0.0612.
A
10 m 7 m
y
x
D
RB,V
RB,H
RA,V
RA,H
3 m
4 m
40 kN/m
B
C
FIGURE S.6.3
Therefore, when x = 10 m, y = 6.12 m (at A). Also, when x = −3 m, y = 0.55 m (D).
Taking moments about C for AC
RA,V × 10 − RA,H × 6.12 = 0
i.e.
RA,V − 0.612RA,H = 0 (i)
Now taking moments about B
RA,V × 17 − RA,H × 3.12 − 40 × 7
2
2
= 0
i.e.
RA,V − 0.184RA,H − 57.65 = 0 (ii)
Then, Eq. (i) − Eq. (ii) gives −0.428RA,H + 57.65 = 0
i.e.
RA,H = 134.7 kN
From Eq. (i)
RA,V = 82.4 kN
66 • Solutions Manual
The bending moment at D is then given by
MD = 82.4 × 13 − 134.7 × 5.57 − 40 × 3
2
2
= 140.9 kN m (sagging).
S.6.4 For the parabolic part of the arch take the origin of axes at C as shown in Fig.
S.6.4; the equation of this part of the arch is then y = kx2. When x = 9 m, y = 5 m so
that k = 0.0617. Then at D where x = 5 m, y = 1.54 m and the vertical height of D
above A is 5 − 1.54 = 3.46 m.
C
y
B
RB,H
RB,V
RA,H
RA,V
A
D
x
4 m
9 m 3 m
5 m
18 kN/m30 kN/m
a
FIGURE S.6.4
Taking moments about C
RA,V × 9 − RA,H × 5 − 30 × 9
2
2
= 0
i.e.
1.8RA,V − RA,H − 243 = 0 (i)
Now taking moments about B
RA,V × 12 − 30 × 9 × 7.5 − 18 × 3
2
2
= 0
i.e.
RA,V = 175.5 kN
Substituting in Eq. (i)
RA,H = 72.9 kN
Solutions to Chapter 6 Problems • 67
The slope of the arch at D is dy/dx = 0.1234x. Therefore when x = 5 m, the slope =
0.617 = tanα, i.e. α = 31.67◦. Then, at D
NF = 175.5 sinα + 72.9 cosα − 30 × 4 sinα = 91.2 kN (compression)
SF = 175.5 cosα − 72.9 sinα − 30 × 4 cosα = 9.0 kN
BM = 175.5 × 4 − 72.9 × 3.46 − 30 × 4
2
2
= 209.8 kN m.
S.6.5 Suppose that the vertical and horizontal support reactions at B are RB,V and
RB,H, respectively. Then, taking moments about A
RB,V × 12 − 10 × 7.5 − 10 × 9 − 10 × 10.5 = 0
i.e.
RB,V = 22.5 kN
Now taking moments about C
RB,H × 3 − 22.5 × 6 + 10 × 1.5 + 10 × 3 + 10 × 4.5 = 0
i.e.
RB,H = 15 kN
Normal force:
In BD
NF = 22.5 cos 45◦ + 15 cos 45◦ = 26.5 kN (compression)
In DE
NF = 26.5 − 10 cos 45◦ = 19.4 kN (compression)
In EFC
NF = 15 kN (compression)
Shear force:
In BD
SF = 22.5 sin 45◦ − 15 sin 45◦ = 5.3 kN
In DE
SF = 5.3 − 10 sin 45◦ = −1.77 kN
In EF
SF = 22.5 − 10 − 10 = 2.5 kN
68 • Solutions Manual
In FC
SF = 2.5 − 10 = −7.5 kN
Bending moment:
At B
BM = 0
At D
BM = 22.5 × 1.5 − 15 × 1.5 = 11.25 kN m (sagging)
At E
BM = 22.5 × 3 − 15 × 3 − 10 × 1.5 = 7.5 kN m (sagging)
At F
BM = 22.5 × 4.5 − 15 × 3 − 10 × 3 − 10 × 1.5 = 11.25 kN m (sagging)
At C
BM = 0
All distributions are linear.
S.6.6 Referring to Fig. S.6.6 the calculated dimensions are as shown. Taking moments
about D
RA,H
RD,V
RD,H
RA,V
A
B
C
D
5 kN
15 kN
10 kN
0.75 m
0.75 m
1.3 m
1.3 m
1.5 m
60°
30°
FIGURE S.6.6
RA,V × 2.05 − RA,H × 3.55 + 5 × 0.75 + 15 × 1.5 = 0
Solutions to Chapter 7 Problems • 69
i.e.
RA,V − 1.73RA,H + 12.8 = 0 (i)
Taking moments about B
RA,V × 1.3 − RA,H × 0.75 = 0
i.e.
RA,V − 0.58RA,H = 0 (ii)
Then, Eq. (i) − Eq. (ii) gives
RA,H = 11.13 kN
Then, from Eq. (ii)
RA,V = 6.46 kN
Now resolving horizontally
RD,H = 11.13 − 15 = −3.87 kN (acting to the right)
and resolving vertically
RD,V = 6.46 + 5 + 10 = 21.46 kN
The bending moment at C is given by
MC = 3.87 × 1.5 = 5.81 kN m
Also the bending moment at D is zero. The bending moment varies linearly in DC
and is drawn on the left hand side of the member.
S o l u t i o n s t o C h a p t e r 7 P r o b l e m s
S.7.1 In this problem there are two limiting criteria, one of stress and one of change in
length; we shall consider the maximum allowable stress criterion first. From Eq. (7.1)
σ(max) = 150 = 5000 × 10
3[(
π
4
)
(3002 − d2)] (i)
where d is the internal diameter of the column. Solving Eq. (i) gives
d = 218.1 mm
70 • Solutions Manual
The shortening of the column must not exceed 2 mm. Therefore the strain in the
column, from Eq. (7.4), is limited to 2/(3× 103)= 0.00067. Then, from Eqs (7.1)
and (7.7)
0.00067 = 5000× 10
3[
200 000× (π4 ) (3002 − d2)] (ii)
Solving Eq. (ii) gives
d = 205.6 mm
The maximum allowable internal diameter of the column is therefore 205.6 mm.
S.7.2 The strain in the girder is given by Eq. (7.4), i.e.
ε = L − L0
L0
= αT = 0.00005 × 30 = 0.0015
Therefore, from Eq. (7.7), the stress is given by
σ = 0.0015 × 180 000 = 270 N/mm2.S.7.3 From Eq. (7.1)
σ = 150 = 10 000 × 10
3(
πD2
4
)
from which the required diameter of the column is D= 291.3 mm.
The shortening, δ, of the column is then, from Eqs (7.7) and (7.4), given by
δ = 150
200 000
× 3 × 103 = 2.25 mm
From Eq. (7.12) the lateral strain is 0.3× (150/200 000)= 0.000225. Then the increase
in diameter is 0.000225× 291.3= 0.066 mm.
S.7.4 The temperature rise required to produce an extension of 1.5 mm is given by
T = L − L0
αL0
= 1.5
0.000012 × 2 × 103 = 62.5
◦
The effective strain in the member when it cools is αT = 0.000012× 62.5= 0.00075
(or 1.5/2× 103). The corresponding stress is then, from Eq. (7.7)
σ = 200 000 × 0.00075 = 150 N/mm2
Suppose that the section is of side a and b and suppose that the longitudinal strain
is ε. The lateral strain is then, from Section 7.8, νε= 0.3ε. The sides of the section
Solutions to Chapter 7 Problems • 71
are therefore reduced in length by 0.3εa and 0.3εb. The percentage change in cross-
sectional area is therefore given by
%change = [ab − (a − 0.3εa)(b − 0.3εb)] × 100
ab
This reduces to
%change = 2 × 0.3εab × 100
ab
= 2 × 0.3ε × 100
since ε2 is negligibly small. Then
%change = 2 × 0.3 × 0.00075 × 100 = 0.045%.
S.7.5 The required area of cross section is, from Eq. (7.1), given by
A = 100 × 10
3
155
= 645.2 mm2
Reference to steel tables shows that two equal angles, 50× 50× 5 mm, each having an
18 mm diameter bolt hole, have sufficient area of cross section.
S.7.6 The weight of the cable is= (π × 7.52/4)× 25× 103 × 7850× 9.81/109 = 85.05 N
and the total load on the cable is therefore= 85.05+ 5× 103 = 5085.05 N.
The tensile stress in the cable at its point of support is then, from Eq. (7.1)
σ = 5085.05(
π × 7.524
) = 115.1 N/mm2
Consider the cable shown in Fig. S.7.6. The weight of a length h is equal to ρAh where
ρ is the density of the cable and A its area of cross section. The stress in the cable at
the section at the top of the length h is then ρh from Eq. (7.1).
h
dh L
FIGURE S.7.6
The extension of the elemental length δh due to the self-weight of the cable is then,
from Eqs (7.4) and (7.7), given by
ext =
(
ρh
E
)
δh
72 • Solutions Manual
The extension of the complete cable due to self-weight is therefore
ext =
∫ L
0
(
ρh
E
)
dh = ρL
2
2E
= 7850 × 9.81(25 × 10
3)2
2 × 200 000 × 109 = 0.12 mm
The extension due to the load is, from Eq. (7.28), given by
ext (load) = 5 × 10
3 × 25 × 103(
π × 7.524
)
× 200 000
= 14.15 mm
The total extension is then 14.15+ 0.12= 14.27 mm.
S.7.7 The concentrated loads applied to the chimney have previously been calculated
in S.3.1 and are 36.0 kN at a height of 15 m, 46.8 kN at a height of 25 m and 52.1 kN at
a height of 35 m. The self-weight of the chimney is
SW = 40 × 0.15 × 2500 × 9.81 × 10−3 = 147.15 kN
The total force on the chimney base is then 147.15+ 36.0+ 46.8+ 52.1= 282.05 kN
The maximum stress is then, from Eq. (7.1)
σ = 282.05 × 10
3
0.15 × 106 = 1.9 N/mm
2
From S.7.6 the shortening due to the self-weight of the chimney is given by
shortening (SW) = 2500 × 9.81(40 × 10
3)2
2 × 20 000 × 109 = 0.98 mm
From Eq. (7.28) the shortening due to the loads is
shortening (loads) = (52.1 × 35 + 46.8 × 25 + 36 × 15) × 10
6
0.15 × 106 × 20 000 = 1.18 mm
total shortening = 0.98 + 1.18 = 2.16 mm.
S.7.8 At any section a distance x from the top of the column the cross-sectional area
is given by
A = a
[
b2 + x(b1 − b2)h
]
Then, from Eq. (7.28), the shortening of an element δx is
δ� = Pδ x
AE
Solutions to Chapter 7 Problems • 73
The total shortening of the column is then
� =
∫ h
0
Pdx
AE
Substituting for A
� =
(
P
aE
)∫ h
0
dx[
b2 + x(b1−b2)h
]
i.e.
� =
(
P
aE
)[
h
(b1 − b2)
]
loge
[
b2 + x(b1 − b2)h
]h
0
from which
� =
[
Ph
aE(b1 − b2)
]
loge
(
b1
b2
)
.
S.7.9 Assume that all the members are in tension. Then, using the method of joints
(Section 4.6):
Joint C
Resolving vertically
CB cos 30◦ + 20 = 0
i.e.
CB = −23.1 kN
Resolving horizontally
CD + CB cos 60◦ = 0
i.e.
CD = 11.6 kN
Joint D
Resolving perpendicularly to DB
DA cos 30◦ − DC cos 30◦ = 0
i.e.
DA = DC = 11.6 kN
Resolving parallel to DB
DB + DC cos 60◦ + DA cos 60◦ = 0
i.e.
DB = −11.6 kN
74 • Solutions Manual
Joint B
Resolving horizontally
BA + BD cos 60◦ − BC cos 60◦ = 0
i.e.
BA = −5.1 kN
From Eqs (7.27) and (7.29)
20 × 10
3�
2
=
(
3 × 103
2 × 205 000
)[
(23.12 + 11.62 + 5.12)
200
+ (2 × 11.6
2)
100
]
× 106
from which
� = 4.5 mm
Note that the negative signs indicating compression disappear in the above equation.
S.7.10 Assuming all members are in tension and, using the method of joints (note that
the truss member forces, except CB and DA, may be obtained by inspection)
CB = −141.4 kN, CD = 100 kN, BA = −100 kN, BD = 100 kN,
DA = −141.4 kN , DE = 200 kN
Then, from Eqs (7.27) and (7.29)
100 × 10
3�
2
= (141.42 × 2√2 + 1002 × 2 + 1002 × 2 + 1002 × 2 + 141.42 × 2√2
+ 2002 × 2) × 10
9
2 × 1200 × 205 000
i.e.
� = 10.3 mm.
S.7.11 Suppose that the loads in the bars are P1, P2 and P3. For vertical equilibrium
of the system
P1 + P2 + P3 = P (i)
Taking moments about, say, bar 1
P2a + P32a − P3a2 = 0
i.e.
P2 + 2P3 = 3P2 (ii)
Solutions to Chapter 7 Problems • 75
From Eq. (7.28) the extensions of the bars are P1L/AE, P2L/AE and P3L/AE where A
is the cross-sectional area of each bar and E is Young’s modulus. The displaced shape
of the system is shown in Fig. S.7.11.
a
P1 P2
P
P3
a
2
a
2
FIGURE S.7.11
Then, since the term L/AE is the same for each bar, the extension of each bar is directly
proportional to the load in the bar. The geometry of Fig. S.7.11 gives
P2 − P1
P3 − P1 =
1
2
(iii)
Rearranging Eq. (iii)
2P2 − P1 − P3 = 0 (iv)
Adding Eqs (i) and (iv)
P2 = P3
Substituting for P2 in Eq. (ii) gives
P3 = 7P12
Finally, from Eq. (i)
P1 = P12 .
S.7.12 Eqs (7.38) may be used directly to determine the stresses in the steel bar and
alloy cylinder. The areas of cross section are
Ast = π × 20
2
4
= 314.16 mm2, Aall = π(25
2 − 202)
4
= 176.71 mm2
Then
σst = 50 × 10
3 × 200 000
(314.16 × 200 000 + 176.71 × 70 000)
i.e.
σsteel = 132.9 N/mm2
76 • Solutions Manual
Similarly
σall = 50 × 10
3 × 70 000
(314.16 × 200 000 + 176.71 × 70 000)
i.e.
σalloy = 46.5 N/mm2
From Eq. (7.37) the shortening of the column is
δ = 50 × 10
3 × 200
(314.16 × 200 000 + 176.71 × 70 000)
i.e.
δ = 0.13 mm
From Eq. (7.30) the strain energy stored in the column is
U =
200
[(
132.92×314.16
200 000
)
+
(
46.52×176.71
70 000
)]
2
i.e.
U = 3320.3 Nmm = 3.3 Nm.
S.7.13 This problem is very similar to P.7.12 and so the same equations may be used
directly to obtain a solution. From Eqs (7.38)
σtim = 1000 × 10
3 × 15 000
[(100 × 200)15 000 + (2 × 200 × 10)200 000]
i.e.
σtimber = 13.6 N/mm2
The allowable stress in the timber is 55/3= 18.3 N/mm2 so that the timber size is
satisfactory. Also
σst = 1000 × 10
3 × 200 000
[(100 × 200)15 000 + (2 × 200 × 10)200 000]
i.e.
σsteel = 181.8 N/mm2
The allowable stress in the steel is 380/2= 190 N/mm2 so that the size of the steel plates
is satisfactory.
Solutions to Chapter 7 Problems • 77
S.7.14 If the steel portion of the bar were disconnected from the aluminium part the
separate parts would take up the positions shown in Fig. S.7.14 where
δS = αSTL1, δA = αATL2 (i)
L1
dS
d
dA
AlSteel
L2
FIGURE S.7.14
Suppose that the connected parts of the bar take up the position shown so that the
junction of the two parts has suffered a displacement δ. Then
extension of steel= δS − δ
extension of aluminium = δA + δ
These extensions produce tensile stresses in the steel and aluminium which must be
equal since their areas of cross section are the same. Therefore
σ = ES(δS − δ)
L1
= EA(δA + δ)
L2
(ii)
Rearranging Eq. (ii) gives
δ =
(
ESδS
L1
)
−
(
EAδA
L2
)
(
ES
L1
)
+
(
EA
L2
) (iii)
Now substituting for δ in the first of Eqs (ii) (or the second) and also for δS and δA
from Eq. (i) and rearranging gives
σ = T(αSL1 + αAL2)(
L1
ES
+ L2EA
) .
S.7.15 The cross-sectional areas of the steel tube and copper bar are, respectively,
Ast = π(36
2 − 302)
4
= 311.02 mm2, Ac = π × 25
2
4
= 490.9 mm2
Then, from Eqs (7.49)
σst = 80(0.000006 − 0.00001) × 490.9 × 100 000 × 200 000(490.9 × 100 000 + 311.02 × 200 000)
78 • Solutions Manual
i.e.
σsteel = −28.3 N/mm2
The stress in this case is negative so that, unlike the steel reinforcement in the concrete
column of Fig. 7.28, the steel is in tension. This is logical since the coefficient of linear
expansion of copper is greater than that of steel. Again, from Eqs (7.49)
σc = 80(0.000006 − 0.00001) × 311.02 × 100 000 × 200 000(490.9 × 100 000 + 311.02 × 200 000)
i.e.
σcopper = −17.9 N/mm2
The concrete in the column of Fig. 7.28 is in tension; here the negative answer indicates
that the copper tube is in compression; again an expected result.
S.7.16 The first part of this question is identical in form to P.7.13. Therefore, we can
substitute areas of cross section, etc. directly into Eqs (7.38).
Ast = π × 75
2
4
= 4417.9 mm2, Aal = π(100
2 − 752)
4
= 3436.1 mm2
From Eqs (7.38)
σst = 10
6 × 200 000
(4417.9 × 200 000 + 3436.1 × 80 000)
i.e.
σsteel = 172.6 N/mm2 (compression)
σal = 10
6 × 80 000
(4417.9 × 200 000 + 3436.1 × 80 000)
i.e.
σaluminium = 69.1 N/mm2 (compression)
Due to the decrease in temperature in which no change in length is allowed the strain
in the steel is αstT and the strain in the aluminium is αalT . Therefore, due to the
decrease in temperature
σst = Est αstT = 200 000 × 0.000012 × 150 = 360.0 N/mm2 (tension)
σal = Eal αalT = 80 000 × 0.000005 × 150 = 60.0 N/mm2 (tension)
The final stresses in the steel and aluminium are then
σsteel (total) = 360.0 − 172.6 = 187.4 N/mm2 (tension)
σaluminium (total) = 60.0 − 69.1 = −9.1 N/mm2 (compression).
Solutions to Chapter 8 Problems • 79
S.7.17 The cross-sectional areas of the bolt and sleeve are
AB = π × 15
2
4
= 176.7 mm2, AS = π(30
2 − 202)
4
= 392.7 mm2
From equilibrium the compressive load in the sleeve must be equal to the tensile load
in the bolt, therefore the stresses in the bolt and sleeve due to the tightening of the
nut are
σB = 10 × 10
3
176.7
= 56.6 N/mm2 (tension)
σS = 10 × 10
3
392.7
= 25.5 N/mm2 (compression)
When the external tensile load is applied the tensile stress in the bolt will be increased
while the compressive stress in the sleeve will be reduced. Equations (7.38) apply
in which Young’s modulus is the same for the bolt and sleeve so that, due to the
5 kN load
σB = 5 × 10
3
(176.7 + 392.7) = 8.8 N/mm
2 (tension) = σS
The final stresses are then
σB = 56.6 + 8.8 = 65.4 N/mm2 (tension)
σS = 25.5 − 8.8 = 16.7 N/mm2 (compression).
S.7.18 The pressure of the water in the pipe is given by
p = 120 × 1000 × 9.81 × 10−3 = 1177.2 kN/m2
Therefore, from Eq. (7.63), the minimum wall thickness is given by
tmin = 1177.2 × 10
3 × 1 × 103
2 × 20 × 106 = 29.4 mm.
S.7.19 From Eq. (7.68) the required shell thickness is given by
0.8 treq = 0.75 × 3 × 10
3
4 × 80
i.e.
treq = 8.8 mm.
80 • Solutions Manual
S o l u t i o n s t o C h a p t e r 8 P r o b l e m s
S.8.2 From Eq. (7.8) Young’s modulus E is equal to the slope of the stress–strain
curve. Then, since stress= load/area and strain= extension/original length.
E= slope of the load-extension curve multiplied by (original length/area of cross
section).
From the results given the slope of the load-extension curve 	 402.6 kN/mm. Then
E	 402.6 × 10
3 × 250(
π×252
4
) 	 205 000 N/mm2
From Eq. (11.4) the modulus of rigidity is given by
G = TL
θJ
Therefore the slope of the torque-angle of twist (in radians) graph multiplied by (L/J)
is equal to G. From the results given the slope of the torque-angle of twist graph is
	12.38 kNm/rad. Therefore
G	 12.38 × 10
6 × 250(
π × 254
32
) 	 80 700 N/mm2
Having obtained E and G the value of Poisson’s ratio may be found from Eq. (7.21), i.e.
ν =
(
E
2G
)
− 1	 0.27
Finally, the bulk modulus K may be found using either of Eqs (7.22) or (7.23). From
Eq. (7.22)
K 	 E
3(1 − 2ν) 	 148 500 N/mm
2.
S.8.3 Suppose that the actual area of cross section of the material is A and that the
original area of cross section is Ao. Then, since the volume of the material does not
change during plastic deformation
AL=AoLo
where L and Lo are the actual and original lengths of the material respectively. The
strain in the material is given by
ε = L − Lo
Lo
= Ao
A
− 1 (i)
from the above. Suppose that the material is subjected to an applied load P. The
actual stress is then given by σ =P/A while the nominal stress is given by σnom =P/Ao.
Therefore, substituting in Eq. (i) for A/Ao
ε = σ
σnom
− 1
Solutions to Chapter 9 Problems • 81
Then
σnom(1 + ε) = σ = Cεn
or
σnom = Cε
n
1 + ε (ii)
Differentiating Eq. (ii) with respect to ε and equating to zero gives
dσnom
dε
= nC(1 + ε)ε
n−1 − Cε n
(1 + ε)2 = 0
i.e.
n(1 + ε)ε n−1 − εn = 0
Rearranging gives
ε = n
(1 − n) .
S.8.4 Substituting in Eq. (8.1) from Table P.8.4
104
5 × 104 +
105
106
+ 10
6
24 × 107 +
107
12 × 107 = 0.39< 1
Therefore fatigue failure is not probable.
S o l u t i o n s t o C h a p t e r 9 P r o b l e m s
S.9.1 From symmetry the support reactions are equal and are each 5w kN. At each
support the bending moment is −w× 22/2=−2w kN m. At mid-span the bending
moment is 5w× 3−w× 52/2= 2.5w kN m which is therefore the maximum. Using the
method of Section 9.6
Iz = 200 × 340
3
12
− 185 × 300
3
12
= 2.39× 108 mm4.
Then, substituting in Eq. (9.9)
150 = 2.5w × 10
6 × 170
2.39 × 108
from which
w = 84.3 kN/m.
S.9.2 The maximum bending moment occurs at the built in end of the cantilever and is
equal to 13 300× 2.5= 33 250 Nm. The second moment of area of the beam section is
82 • Solutions Manual
230× 3003/12= 5.175× 108 mm4. The maximum direct stress due to bending is then,
from Eq. (9.9), given by
σmax = 33 250× 10
3 × 150
5.175 × 108 = 9.6 N/mm
2.
S.9.3 The arrangement of the floor and joists is shown in Fig. S.9.3.
d m
d m
4 m
FIGURE S.9.3
If the joists are spaced a distance d m apart then each joist carries a uniformly dis-
tributed load of intensity 16d kN/m. Since each joist is a simply supported beam
the maximum bending moment is, from Ex. 3.7, equal to 16d × 42/8 = 32d kN m.
The second moment of area of each joist is, from Section 9.6, 110× 3003/12 =
2.475 × 108 mm4. Then, from Eq. (9.9)
7 = 32d × 10
6 × 150
2.475 × 108
from which
d= 0.36 m.
S.9.4 The mast is shown in Fig. S.9.4.
At any section a distance h m from the top of the mast the diameter is given by
d = 100 +
(
h
15
)
× 150
i.e.
d = 100 + 10h mm
The maximum direct stress due to bending at this section is, from Eq. (9.9)
σmax =
Ph× 103 ×
(
d
2
)
(
πd4
64
) = Ph × 103 × 32
πd3
(i)
Solutions to Chapter 9 Problems • 83
100 mm
h
15 m
250 mm FIGURE S.9.4
This maximum direct stress will be greatest when dσmax/dh= 0. Therefore, substituting
for d in Eq. (i) and differentiating
0 =
(
32 × 103P
π
)
(100 + 10h)3 − 3h(100 + 10h)2 × 10
(100 + 10h)6
i.e.
100 + 10h = 30h
which gives
h = 5 m
i.e. the