CHAPTER 6 FORCES IN STRUCTURES
EXERCISE 36, Page 78
1. Determine the internal forces in the following pin-jointed truss using a graphical method.
The spaces between the forces are shown by upper case letters in the diagram below.
In all such problems, joints with two or less unknowns should be tackled first.
The vector diagram, drawn to scale, for the whole framework is shown below.
By measurement, bc = - 2.0 kN (2-3), ca = - 3.5 kN (1-3), dc = 1.7 kN (1-2), da = 3.0 kN ( R 1 ) and
bd = 1.0 kN ( R 2 )
i.e.
R 1 = 3.0 kN, R 2 = 1.0 kN, 1-2, 1.7 kN, 1-3, - 3.5 kN, 2-3, - 2.0 kN
2. Determine the internal forces in the following pin-jointed truss using a graphical method.
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The spaces between the forces are shown by upper case letters in the diagram below.
The vector diagram, drawn to scale, for the whole framework is shown below.
By measurement, ac = + 3.0 kN (1-3), bc = - 5.2 kN (2-3), cd = - 1.5 kN (1-2), be = 6.0 kN ( H 2 ),
ed = 2.6 kN ( R 2 ) and da = - 2.6 kN ( R 1 )
i.e. R 1 = - 2.6 kN, R 2 = 2.6 kN, H 2 = 6.0 kN, 1-2, - 1.5 kN, 1-3, 3.0 kN, 2-3, - 5.2 kN
3. Determine the internal forces in the following pin-jointed truss using a graphical method.
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The spaces between the forces are shown by upper case letters in the diagram below.
The vector diagram, drawn to scale, for the whole framework is shown below.
By measurement, fa = - 7.1 kN (1-3), cf = - 1.3 kN (2-3), df = + 1.0 kN (1-2), cd = 1.0 kN ( R 2 ),
de = 5.0 kN ( R 1 ) and ea = 4.0 kN ( H1 )
i.e. R 1 = 5.0 kN, R 2 = 1.0 kN, H 1 = 4.0 kN, 1-2, 1.0 kN, 1-3, - 7.1 kN, 2-3, - 1.4 kN
4. Determine the internal forces in the following pin-jointed truss using a graphical method.
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The spaces between the forces are shown by upper case letters in the diagram below.
As there is no joint with less than 3 unknowns, taking moments about joint 2 gives:
R 1 × 12 = 2 × 9 + 4 × 6 + 6 × 3
i.e.
12 R 1 = 18 + 24 + 18 = 60
60
= 5 kN = de
12
and
R1 =
and
R 2 = 12 – 5 = 7 kN = cd
The vector diagram for the whole framework, drawn to scale, is shown below.
By measurement, bh = - 8.0 kN (4-5), hg = + 4.0 kN (4-6), ga = - 8.0 kN (3-4), df = + 8.7 kN (1-6),
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jc = - 14 kN (5-2), fg = - 2.0 kN (3-6), dj = + 12.1 kN (6-2), ef = - 10.0 kN (1-3) and
jh = - 6.0 kN (5-6)
i.e. R 1 = 5.0 kN, R 2 = 7.0 kN, 1-3, - 10.0 kN, 1-6, 8.7 kN, 3-4, - 8.0 kN, 3-6, - 2.0 kN,
4-6, 4.0 kN, 4-5, - 8.0 kN, 5-6, - 6.0 kN, 5-2, - 14.0 kN, 6-2, 12.1 kN
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EXERCISE 37, Page 84
1. Using the method of joints, determine the unknown forces for the following pin-jointed truss.
All unknown member forces are assumed to be in tension.
For joint 3 (see diagram):
Resolving vertically,
4 + F2 3 cos 60º + F13 cos 30º = 0
Resolving horizontally,
(1)
F13 sin 30º = F2 3 sin 60º
i.e.
 sin 60 
 0.866 
F13 = F2 3 
  F23 

 sin 30 
 0.5 
i.e.
F13 = 1.732 F2 3
(2)
Substituting equation (2) into equation (1) gives:
4 + F2 3 (0.5) + 1.732 F2 3 (0.866) = 0
F2 3 (0.5 + 1.5) = - 4
or
F23 = - 2.0 kN (compression)
and
From equation (2),
(3)
F13 = 1.732(- 2.0)
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F1 3 = - 3.46 kN (compression)
i.e.
(4)
For joint 1 (see diagram):
Resolving horizontally,
F12  F13 cos 60
i.e.
F12  ( 3.46)(0.5)  1.73kN
Resolving vertically,
R1  F13 sin 60  0
R1  F13 (0.866)
or
and from equation (4),
R1  (3.46)(0.866)
i.e.
R1  3.0kN
Resolving vertically,
R1  R 2  4
Hence, from equation (6),
R 2  4  3  1.0kN
(5)
(6)
Summarising, R 1 = 3.0 kN, R 2 = 1.0 kN, 1-2, 1.73 kN, 1-3, - 3.46 kN, 2-3, - 2.0 kN
2. Using the method of joints, determine the unknown forces for the following pin-jointed truss.
For joint 3 (see diagram):
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F13 cos30  F23 cos 60  0
Resolving vertically,
i.e.
(0.866)F13  (0.5)F23  0
Resolving horizontally,
F13 sin 30  F23 sin 60  6
i.e.
0.5F13  0.866F23  6
F13  1.732F23  12
i.e.
(1)
(2)
Substituting equation (2) into equation (1) gives:
(0.866)(1.732F23  12)  (0.5)F23  0
i.e.
1.5F23  10.392  0.5F23  0
and
2F23  10.392
10.392
= - 5.2 kN
2
i.e.
F23 = 
From equation (2),
F13  1.732(5.2)  12
i.e.
F1 3 = 3.0 kN
(3)
(4)
For joint 1 (see diagram):
Resolving vertically,
F13 sin 60  R1  0
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or
R1  (0.866)F13
From equation (4),
R1  (0.866)(3.0)
R1 = - 2.6 kN
i.e.
Resolving vertically (overall),
R1  R 2  0
R 2  R1  2.6 kN
i.e.
Resolving horizontally gives:
F13 cos 60  F12  0
or
F12  0.5F13
From equation (4),
F12  0.5(3.0)  1.5kN
Resolving horizontally (overall),
Summarising,
(5)
(6)
H1  6kN
R 1 = -2.61 kN, R 2 = 2.61 kN, H 2 = 6.0 kN, 1-2, - 1.5 kN, 1-3, 3.0 kN,
2-3, - 5.2 kN
3. Using the method of joints, determine the unknown forces for the following pin-jointed truss.
For joint 3 (see diagram):
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Resolving horizontally,
F13 sin 45  4  F23 sin 45
(0.707)F13  4  (0.707)F23
i.e.
4
0.707
and
F13  F23 
or
F13  F23 - 5.657
Resolving vetically,
6  F13 cos 45  F23 cos 45  0
i.e.
6
 F13  F23  0
cos 45
and
8.485  F13  F23  0
F13  F23  8.485
or
(1)
(2)
Substituting equation (1) into equations (2) gives:
F23  5.657  F23  8.485
2F23  2.828
i.e.
F23  1.414kN
and
(3)
Substituting equation (3) into equations (1) gives:
F1 3 -1.414 - 5.657 = - 7.07 kN
For joint 2 (see diagram):
Resolving vertically,
R 2  F23 sin 45  0
or
R 2  (0.707)F23
(4)
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From equation (3),
R 2  (0.707)(1.414)  1.0kN
Resolving horizontally,
F23 cos 45  F12  0
i.e.
F12  (0.707)F23
From equation (3),
F12  (0.707)( 1.414) =1.0 kN
Resolving vertically (overall),
R1  R 2 = 6 or R1  6  R 2  5.0kN
Resolving horizontally (overall),
H1  4kN
Summarising, R 1 = 5.0 kN, R 2 = 1.0 kN, H 1 = 4.0 kN 1-2, 1.0 kN, 1-3, - 7.07 kN,
2-3, - 1.41 kN
4. Using the method of joints, determine the unknown forces for the following pin-jointed truss.
As there are no joints which have two or less unknowns, it will be necessary to calculate R 1 and
R2 .
Taking moments about joint 2 gives: R 1 × 12 = 2 × 9 + 4 × 6 + 6 × 3
= 18 + 24 + 18 = 60
R1 
i.e.
60
= 5 kN
12
Resolving vertically (overall),
R 1 + R 2 = 2 + 4 + 6 = 12
Hence,
R 2 = - R 1 + 12 = 12 – 5
i.e.
R 2 = 7 kN
(1)
(2)
For joint 1 (as there are only 2 unknowns here) - see diagram:
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Resolving vertically,
R1  F13 sin 30  0
F13  
i.e.
From equation (1),
Resolving horizontally,
F13  
R1
0.5
R1
5

 10kN
0.5
0.5
(3)
F16  F13 cos 30 = 0
i.e.
F16  (0.866)F13
(4)
From equation (4),
F16  (0.866)(10)  8.66kN
(5)
For joint 3 (see diagram):
Resolving horizontally,
F13 cos30  F36 cos30  F34 cos30
i.e.
F36  F13  F34
(6)
From equation (3),
F36  10  F34
(7)
Resolving vertically,
F13 sin 30  F36 sin 30  2  F34 sin 30
or
F13  F36 
2
 F3 4
0.5
(8)
Substituting equation (7) into equation (8) gives:
F13  10  F34  4  F34
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or
F13  10  4  2F34
From equation (3),
10  10  4  2F34
i.e.
2F34  16
from which,
F34  8kN
From equations (3) and (6),
F36  10  8  10  8  2kN
(9)
(10)
For joint 4 (see diagram):
Resolving horizontally,
F34  F45
(11)
From equation (10),
F45  8kN
(12)
Resolving vertically,
4  F46  F34 cos 60  F45 cos 60  0
(13)
From equations (10) and (12), 4  F46  8cos 60  8cos 60  0
i.e.
4  F46  4  4  0
from which,
F46  4  4  4  4kN
(14)
For joint 5 (see diagram):
Resolving horizontally,
F45 cos30  F56 cos30  F52 cos30
i.e.
F56  F52  F45
(15)
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From equation (12),
F56  F52  8
(16)
Resolving vertically,
F45 sin 30  6  F56 sin 30  F52 sin 30
(17)
From equations (12) and (16),
8(0.5)  6  (F52  8)(0.5)  F52 (0.5)
– 4 = 6 + 4 + F5 2
i.e.
i.e.
F52  4  6  4  14kN
(18)
From equation (16),
F56  14  8  6kN
(19)
For joint 2 (see diagram):
Resolving horizontally,
F26  F52 cos30
i.e.
F26  F52 cos30  (14)(0.866)
i.e.
F26  12.1kN
Summarising, R 1 = 5.0 kN, R 2 = 7.0 kN, 1-3, - 10.0 kN, 1-6, 8.7 kN, 3-4, - 8.0 kN,
3-6, - 2.0 kN, 4-6, 4.0 kN, 4-5, - 8.0 kN, 5-6, - 6 kN, 5-2, - 14 kN, 2-6, 12.1 kN
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EXERCISE 38, Page 85
1. Determine the internal member forces of the following truss, by the method of sections.
To calculate the reactions:
Let l = horizontal distance between joint (1) and joint (2), and let h = the perpendicular distance of
joint (3) from the base.
l = a + b where a =
h
h
= 0.577 h and b =
= 1.732 h
tan 60
tan 30
Taking moments about joint (2) gives:
R1  l  4  b  4 1.732h
i.e.
R1 (a  b)  6.928h
i.e.
R1 (0.577h  1.732h)  6.928h
from which,
R1 =
6.928h
6.928h 6.928


= 3 kN
0.577h  1.732h 2.309h 2.309
Resolving vertically gives:
Hence, since R1  3kN
(1)
R1  R 2 = 4
R 2 = 1 kN
(2)
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Resolving vertically gives:
R1  F13 sin 60  0
i.e.
F13  
R1
sin 60
From equation (1),
F1 3 
3
= - 3.46 kN
0.866
Resolving horizontally,
F13 cos 60  F12  0
i.e.
F12  F13 cos 60
From equation (3),
F1 2 = - 3.46 × 0.5 = 1.73 kN
(3)
Consider section (2) (as shown):
Resolving horizontally gives: F12  F23 cos30  0
i.e.
Summarising,
F23 = 
F1 2
1.73

= - 2.0 kN
cos 30
0.866
(4)
R 1 = 3.0 kN, R 2 = 1.0 kN, 1-2, 1.73 kN, 1-3, - 3.46 kN, 2-3, - 2.0 kN
2. Determine the internal member forces of the following truss, by the method of sections.
Let l = horizontal distance between joint (1) and joint (2), and let h = the perpendicular distance of
joint (3) from the base, and l = a + b as in previous Problem.
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l = a + b where a =
h
h
= 0.577 h and b =
= 1.732 h
tan 60
tan 30
Taking moments about joint (1) gives:
R1  l  6  h  0
i.e.
R1 = 
6h
6h
6h
= - 2.6 kN


l
a b
 0.577 1.732  h
Resolving vertically gives:
(1)
R1  R 2 = 0
Hence, since R1  2.6 kN
R 2 = 2.6 kN
(2)
Consider section (1), as shown:
Resolving vertically gives:
F13 sin 60  R1  0
i.e.
F13  
R1
sin 60
From equation (1),
F1 3 
2.6
= 3.0 kN
0.866
Resolving horizontally,
F13 cos 60  F12  0
i.e.
F12  F13 cos 60
From equation (3),
F1 2 = - 3.0 × 0.5 = - 1.5 kN
(3)
(4)
Consider section (2) (as shown):
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Resolving horizontally gives:
By inspection, H 2 = 6 kN
and
6 + F23 cos30  F12  0
and from equation (4),
F23 =
Summarising,
 F1 2  6   1.5  6

= - 5.2 kN
cos 30
0.866
R 1 = - 2.61 kN, R 2 = 2.61 kN, H 2 = 6 kN, 1-2, - 1.5 kN, 1-3, 3.0 kN,
2-3, - 5.2 kN
3. Determine the internal member forces of the following truss, by the method of sections.
Let l = horizontal distance between joint (1) and joint (2).
Taking moments about joint (1) gives:
l
l
R1  l  H1  0  6   4 
2
2
i.e.
R1 = 3 + 2 = 5.0 kN
(1)
Hence,
R 2  6.0  R1  6.0  5.0  1kN
(2)
By inspection,
H1  4kN
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Consider section (1), as shown:
Resolving vertically gives:
F13 sin 45  R1  0
i.e.
F13  
R1
sin 45
From equation (1),
F1 3 
5.0
= - 7.07 kN
0.707
Resolving horizontally,
H1  F12  F13 cos 45  0
i.e.
F12  H1  0.707F13
From equation (3),
F1 2 = 4  0.707  7.07  = 1.0 kN
(3)
(4)
Consider section (2) (as shown):
Resolving horizontally gives:
F23 cos 45  F12  0
and from equation (4),
Summarising,
F23 =
 F1 2
1.0

= - 1.41 kN
cos 45 0.707
R 1 = 5.0 kN, R 2 = 1.0 kN, H 1 = 4.0 kN, 1-2, 1.0 kN, 1-3, - 7.07 kN,
2-3, - 1.41 kN
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4. Determine the internal member forces of the following truss, by the method of sections.
Firstly, R1 and R 2 have to be determined.
Taking moments about joint (2) gives:
R1 12  2  9  4  6  6  3  60 kN
from which,
R1 
60
= 5.0 kN
12
Resolving vertically gives:
i.e.
(1)
R1  R 2  2  4  6
R 2  12  R1  7.0kN
(2)
Consider section (1), as shown:
h
= tan 30º from which, h = a tan 30º = 0.577 a
a
(3)
Taking moments about joint (3) gives:
F16  h  R1  a
From equations (3) and (4),
F16 
(4)
R1  a
R1
5.0


0.577a 0.577 0.577
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F16  8.7kN
i.e.
(5)
Resolving horizontally gives:
F16  (F34  F36 ) cos30  0
F3 4  
Hence,
From equations (5) and (6), F3 4  
F16
 F36
cos 30
8.7
 F36
0.866
F34  10.0  F36
i.e.
(6)
(7)
Resolving vertically gives:
R1  F34 sin 30  2  F36 sin 30
(8)
and from equation (1),
5.0  0.5F34  2  0.5F36
(9)
and from equation (7),
5.0  0.5  10  F36   2  0.5F36
i.e.
5.0  5.0  0.5F36  2  0.5F36
from which,
5.0  5.0  2  F36
i.e.
F36  2.0kN
From equation (7),
F34  10.0  2.0  8.0kN
(10)
(11)
Consider section (2), as shown:
Resolving vertically,
F13 sin 30  R1  0
from which,
F13  
R1
5.0

 10.0kN
sin 30
0.5
(12)
(13)
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Consider section (3), as shown:
Taking moments about joint (5) gives:
R 2  a  F26  h
F26 =
i.e.
7.0  a 7.0  a
7.0


= 12.1 kN
h
0.577a 0.577
Resolving vertically gives:
(14)
R 2  F45 sin 30  6  F56 sin 30
7.0  0.5F45  6  0.5F56
i.e.
67
 F56
0.5
i.e.
F45 
or
F45  2  F56
Resolving horizontally gives:
F26  F56 cos30  F45 cos30  0
From equations (14) and (15),
12.1  0.866F56  0.866F45  0
F45  
i.e.
12.1
 F56
0.866
(15)
(16)
(17)
Equating equations (15) and (17) gives:
2  F56 = 14.0  F56
i.e.
2F56 = - 14.0 + 2
i.e.
F56 = - 6.0 kN
From equation (15),
F45  2  6.0  8.0kN
(18)
(19)
Consider section (4), as shown:
110
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Resolving horizontally gives:
F26  F25 cos30  0
i.e.
F25  
F26
12.1

 14.0kN
cos 30
0.866
(20)
Consider section (5), as shown:
Resolving vertically gives:
4  F46  F34 cos 60  F45cos 60  0
From equations (11) and (19),
4  F46  8.0cos 60  8.0cos 60  0
i.e.
F46  8.0cos 60  8.0cos 60  4
i.e.
F46  4.0kN
Summarising, R 1 = 5.0 kN, R 2 = 7.0 kN,
3-6, - 2.0 kN,
(21)
1-3, - 10.0 kN, 1-6, 8.7 kN, 3-4, - 8.0 kN,
4-6, 4.0 kN, 4-5, - 8.0 kN,
5-6, - 6.0 kN,
5-2, - 14.0 kN,
6-2, 12.1 kN
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© John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 39, Page 85
Answers found from within the text of the chapter, pages 61 to 69.
EXERCISE 40, Page 85
1. (b) 2. (a) 3. (c) 4. (c) 5. (b) 6. (a)
112
© John Bird & Carl Ross Published by Taylor and Francis