REVISION TEST 3 (Page 119)
This assignment covers the material contained in chapters 6 to 8. The marks available are
shown at the end of each question.
Problem 1. Determine the unknown internal forces in the pin-jointed truss of Figure RT3.1.
Figure RT3.1
Marks
Assuming all unknown forces are in tension and considering joint (3):
Joint 3
Resolving vertically:
upward forces = downward forces
0 = F 23 sin 30 + 2 kN
from which,
F 2 3 = 
2 kN
2 kN

= - 4.0 kN (compressive)
sin 30
0.5
3
Resolving horizontally: Forces to the left = Forces to the right
F 13 + F 23 cos 30 = 0
or
F 13 = - F 23 cos 30 = - (- 4.0)(0.866)
i.e.
F 1 3 = 3.464 kN (tension)
4
The table below gives a summary of the internal forces determined.
Member
Force (kN)
1-3
3.464
2-3
- 4.0
Total:
7
20
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Problem 2. Determine the unknown internal forces in the pin-jointed truss of Figure RT3.2.
Figure RT3.2
Marks
Assuming all unknown forces are in tension and considering joint (3):
Joint 3
Resolving vertically:
0 = 2 + F 3 4 sin 45
from which,
F 3 4 = -
i.e.
F 34 = - 2.828 kN (compressive)
Resolving horizontally:
F 13 + F 3 4 cos 45 = 0
from which,
F 13 = - F 3 4 cos 45 = - (- 2.828)(0.707)
i.e.
F 1 3 = 2.0 kN (tensile)
2
sin 45
2
2
Considering joint (4):
Joint 4
21
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Resolving vertically:
F 1 4 cos 45 + F 3 4 cos 45 = F 24 cos 30
i.e.
0.707 F 1 4 - (2.828)(0.707) = 0.866 F 24
0.707
(2.828)(0.707)
F1 4 
0.866
0.866
from which,
F2 4 
or
F 24 = 0.816 F 1 4 - 2.309
(1)
2
(2)
2
(3)
2
Resolving horizontally: F 1 4 sin 45 + F 24 sin 30 = F 3 4 sin 45
i.e.
0.707 F 1 4 + 0.5 F 24 = (- 2.828)(0.707)
from which,
F24 
or
F 24 = - 4 – 1.414 F 1 4
(2.828)(0.707) 0.707

F14
0.5
0.5
Equating equations (1) and (2) gives:
0.816 F 1 4 - 2.309 = - 4 – 1.414 F 1 4
i.e.
(0.816 + 1.414)F 1 4 = - 4 + 2.309
from which,
and
2.230 F 1 4 = - 1.691
F 1 4 = 
1.691
= - 0.758 kN (compressive)
2.230
Substituting equation (3) into equation (1) gives:
F 24 = 0.816(- 0.758) - 2.309
i.e.
F 24 = - 2.928 kN (compressive)
2
The table below provides a summary of the internal forces determined.
Member
Force (kN)
1-3
2.0
1-4
- 0.758
2-4
- 2.928
3-4
- 2.828
Total:
12
22
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Problem 3. A beam of length 3 m is simply supported at its ends. A clockwise couple of 4 kN m is
placed at a distance of 1 m from the left hand support. (a) Determine the end reactions.
(b) If the beam now carries an additional downward load of 12 kN at a distance of 1 m from the
right hand support, sketch the bending moment and shearing force diagrams.
Marks
(a)
(R 1  3) + 4 = 0
Taking moments about B:
from which,
R1 = -
Resolving forces vertically:
hence,
4
= - 1.333 kN (acting downwards)
3
1
R1 + R 2 = 0
R 2 = - R 1 = +1.333 kN
1
(b)
i.e.
Taking moments about B:
(R 1  3) + 4 = (12  1)
from which,
3 R 1 = 12 – 4 = 8
8
= 2.667 kN
3
i.e.
R1 =
Resolving forces vertically:
R 1 + R 2 = 12
hence,
R 2 = 12 - R 1 = 12 – 2.667
R 2 = 9.333 kN
2
2
23
© John Bird & Carl Ross Published by Taylor and Francis
The bending moment and shearing force diagrams are shown below.
3
Total:
9
Problem 4. A beam of length 4 m is simply supported at its right extremity and at 1 m from the left
extremity. If the beam is loaded with a downward load of 2 kN at its left extremity and with another
downward load of 10 kN at a distance of 1 m from its right extremity, sketch its bending moment
and shearing force diagrams.
Marks
Taking moments about B:
from which,
(R 1  3) = (2  4) + (10  1) = 18
R 1 = 6 kN
Resolving forces vertically:
1
R 1 + R 2 = 2 + 10
from which,
R 2 = 12 - R 1 = 12 – 6
i.e.
R 2 = 6 kN
1
24
© John Bird & Carl Ross Published by Taylor and Francis
The shearing force and bending moment diagrams can now be obtained and are shown
below.
5
Total:
7
Problem 5. (a) Find the second moment of area and radius of gyration about the axis XX for the
beam section shown in Figure RT3.3. (b) Determine the position of the centroid of the section.
(c) Calculate the second moment of area and radius of gyration about an axis through the centroid
parallel to axis XX.
Figure RT3.3
25
© John Bird & Carl Ross Published by Taylor and Francis
Marks
The I-section is divided into three rectangles, A, B and C and their centroids denoted
by C A , C B and C C respectively, as shown below.
(a) For rectangle A:
The second moment of area about C A (an axis through C A parallel to XX)
=
Using the parallel axis theorem:
bd 3
(6.0)(1.0)3
=
= 0.5 cm 4
12
12
1
I XX = 0.5 + AH 2
where A = (6.0)(1.0) = 6.0 cm2 and H = 10.5 cm
I XX = 0.5 + 6.0(10.5) 2 = 662 cm 4
Hence,
2
For rectangle B:
The second moment of area about C B (an axis through C B parallel to XX)
=
Using the parallel axis theorem:
For rectangle C:
bd 3
(2.0)(8.0)3
=
= 85.33 cm 4
12
12
I XX = 85.33 + (8.0)(2.0)(6.0)
I XX
2
= 661.33 cm 4
bd 3
(10.0)(2.0)3
=
=
= 26.67 cm 4
3
3
1
2
1
Total second moment of area for the I-section about axis XX,
I XX = 662 + 661.33 + 26.67 = 1350 cm4
Total area of I-section = (6.0)(1.0) + (2.0)(8.0) + (10.0)(2.0) = 42 cm2
Radius of gyration, k XX =
I XX
=
area
 1350 

 = 5.67 cm
 42 
2
2
2
26
© John Bird & Carl Ross Published by Taylor and Francis
(b) The centroid of the I-section will lie on the axis of symmetry, shown as SS in the
above diagram. Using a tabular approach:
Part Area (a cm 2 )
Distance of centroid
Moment about XX
from XX (i.e. y cm)
(i.e. ay cm 2 )
A
6
10.5
63
B
16
6.0
96
C
20
1.0
20
a
 ay = 179
= A = 42
Ay =
 ay , from which, y =
 ay
A
=
179
= 4.26 cm
42
Thus the centroid is positioned on the axis of symmetry 4.26 cm from axis XX.
(c) From the parallel axis theorem:
7
I XX = I CC + AH 2
1350 = I CC + (42)(4.26) 2
i.e.
= I CC + 762.2
from which, second moment of area about axis CC, I CC = 1350 – 762.2
= 587.8 cm 4
Radius of gyration, k CC =
3
ICC
587.8
= 3.74 cm

area
42
2
Total:
25
TOTAL MARKS FOR REVISION TEST 3: 60
27
© John Bird & Carl Ross Published by Taylor and Francis