Series ISSN: 2573-3168
Solving Practical Engineering Mechanics Problems
BAKHTIYAROV
Synthesis Lectures on
Mechanical Engineering
STATICS
Sayavur I. Bakhtiyarov, New Mexico Institute of Mining and Technology
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SOLVING PRACTICAL ENGINEERING MECHANICS PROBLEMS: STATICS
Engineering mechanics is one of the fundamental branches of science that is important in
the education of professional engineers of any major. Most of the basic engineering courses,
such as mechanics of materials, fluid and gas mechanics, machine design, mechatronics,
acoustics, vibrations, etc. are based on engineering mechanics courses. In order to absorb the
materials of engineering mechanics, it is not enough to consume just theoretical laws and
theorems—a student also must develop an ability to solve practical problems. Therefore, it
is necessary to solve many problems independently. This book is a part of a four-book series
designed to supplement the engineering mechanics courses. This series instructs and applies
the principles required to solve practical engineering problems in the following branches of
mechanics: statics, kinematics, dynamics, and advanced kinetics. Each book contains between
6 and 8 topics on its specific branch and each topic features 30 problems to be assigned as
homework, tests, and/or midterm/final exams with the consent of the instructor. A solution
of one similar sample problem from each topic is provided.
This first book contains seven topics of statics, the branch of mechanics concerned with
the analysis of forces acting on construction systems without an acceleration (a state of the
static equilibrium). The book targets the undergraduate students of the sophomore/junior
level majoring in science and engineering. The author welcomes all feedback/comments from
the reader. Please feel free to contact him at sayavur.bakhtiyarov@nmt.edu.
Solving Practical
Engineering
Mechanics Problems
STATICS
Sayavur I. Bakhtiyarov
Synthesis Lectures on
Mechanical Engineering
Solving Practical Engineering
Mechanics Problems: Statics
iii
Synthesis Lectures on
Mechanical Engineering
Synthesis Lectures on Mechanical Engineering series publishes 60–150 page publications pertaining to this diverse
discipline of mechanical engineering. The series presents Lectures written for an audience of researchers, industry
engineers, undergraduate and graduate students. Additional Synthesis series will be developed covering key areas
within mechanical engineering.
Solving Practical Engineering Mechanics Problems: Statics
Sayavur I. Bakhtiyarov
October 2017
Resistance Spot Welding: Fundamentals and Applications for the Automotive Industry
Menachem Kimchi and David H. Phillips
October 2017
Unmanned Aircraft Design: Review of Fundamentals
Mohammad Sadraey
September 2017
Introduction to Refrigeration and Air Conditioning Systems: Theory and Applications
Allan Kirkpatrick
September 2017
MEMS Barometers Toward Vertical Position Detection: Background Theory, System Prototyping, and Measurement Analysis
Dimosthenis E. Bolanakis
May 2017
Vehicle Suspension System Technology and Design
Avesta Goodarzi, Amir Khajepour
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Ramana M. Pidaparti
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Solving Practical Engineering Mechanics Problems: Statics
Sayavur I. Bakhtiyarov
www.morganclaypool.com
ISBN: 9781681731889 print
ISBN: 9781681731896 ebook
DOI 10.2200/S00799ED1V01Y201709MEC008
A Publication in the Morgan & Claypool Publishers series
SYNTHESIS LECTURES ON MECHANICAL ENGINEERING, #8
Series ISSN: 2573-3168 Print 2573-3176 Electronic
Solving Practical Engineering
Mechanics Problems: Statics
Sayavur I. Bakhtiyarov
New Mexico Institute of Mining and Technology
SYNTHESIS LECTURES ON MECHANICAL ENGINEERING #8
M
&C
MORGAN
& CLAYPOOL PUBLISHERS
vi
ABSTRACT
Engineering mechanics is one of the fundamental branches of science that is important in the education of professional engineers of any major. Most of the basic engineering courses, such as mechanics of materials, fluid and gas
mechanics, machine design, mechatronics, acoustics, vibrations, etc. are based on engineering mechanics courses. In
order to absorb the materials of engineering mechanics, it is not enough to consume just theoretical laws and theorems—a student also must develop an ability to solve practical problems. Therefore, it is necessary to solve many
problems independently. This book is a part of a four-book series designed to supplement the engineering mechanics
courses. This series instructs and applies the principles required to solve practical engineering problems in the following branches of mechanics: statics, kinematics, dynamics, and advanced kinetics. Each book contains between 6 and 8
topics on its specific branch and each topic features 30 problems to be assigned as homework, tests, and/or midterm/
final exams with the consent of the instructor. A solution of one similar sample problem from each topic is provided.
This first book contains seven topics of statics, the branch of mechanics concerned with the analysis of forces
acting on construction systems without an acceleration (a state of the static equilibrium). The book targets the undergraduate students of the sophomore/junior level majoring in science and engineering. The author welcomes all
feedback/comments from the reader. Please feel free to contact him at sayavur.bakhtiyarov@nmt.edu.
KEYWORDS
force, moment, torque, gravity, equilibrium, center of gravity
vii
Contents
Acknowledgments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ���������������������������� ix
1
Topic S-1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ����������������������������� 1
1.1 Determination of the Reaction Forces of Supports for Rigid Body ������������������������������������������ 1
1.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2
Topic S-2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 19
2.1 Application of the Method of Joints to Find Unknown Forces in a Plane (2D)
Truss Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.3.1 Determining Reaction Forces by Analytical Method ������������������������������������������������ 32
2.3.2 Determining the Forces in the Truss Members ���������������������������������������������������������� 33
3
Topic S-3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 37
3.1 Determination of Reaction Forces of Supports for Composite Stud ���������������������������������������� 37
3.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4
Topic S-4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 49
4.1 Determination of Reaction Forces of Supports for Composite Construction
(System of Two Bodies) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
4.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
5
Topic S-5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 67
5.1 Determination of Reaction Forces of Supports for Composite Construction
(System of Three Bodies) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
5.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
5.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
6
Topic S-6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 85
6.1 Determination of reaction forces of rods supporting rectangular plate ������������������������������������ 85
6.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
6.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
7
Topic S-7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� 105
7.1 Determination of Center of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
7.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
Author Biography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� 125
ix
Acknowledgments
The author acknowledges that this work is essentially a translation and a revision of selected problems provided
by Professor A. A. Yablonski (Collection of Problems for Course Projects in Theoretical Mechanics, 2nd ed., Vischaya
Shkola Publishers, 1972, in Russian). The author intended to introduce this unique work to western academia,
which is the product of material covered by him in many classes over a period of four decades in a number of
universities and colleges.
1
CHAPTER
1
Topic S-1
1.1
DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR
RIGID BODY
Find reaction forces in supports of the given construction systems schematically shown in Figures 1.1‒1.30. The sizes
are in meters and the loads are shown in the table below.
Problem #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
G
10
12
8
14
16
6
10
10
12
10
4
20
25
20
5
15
kN
P
5
8
4
6
10
6
7
6
8
4
6
10
6
4
10
5
10
4
10
8
10
7
6
14
16
4
10
6
10
M
kN m
20
10
5
8
7
4
5
6
4
9
7
8
6
10
4
10
8
6
7
8
10
7
20
14
8
7
8
14
q
kN m
1
4
2
3
1
2
1
2
2
1
0.5
2
1
2
2
0.5
1
0.5
0.5
1
2
1.5
0.5
1
2.5
3
1
0
α
Degrees
30
60
60
30
45
60
45
60
30
30
45
45
30
45
60
45
45
30
45
45
30
30
30
30
45
30
45
30
15
30
2
1. PROBLEM S-1
q
α
A
B
M
G
P
2
2
Figure 1.1.
C
P
α
q
α
D
M
A
B
G
2
Figure 1.2.
1
3
1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY
P
q
α
A
B
M
G
2
2
Figure 1.3.
1
2
3
A
B
M
q
G
Figure 1.4.
α
3
4
1. PROBLEM S-1
A
4
M
C
B
p
2
α
q
D
Figure 1.5.
P
q
α
3
2
M
A
Figure 1.6.
1
B
1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY
2
B
α
P
2
q
α
M
A
Figure 1.7.
4
P
1
3
α
D
A
Figure 1.8.
M
E
F
q
G
B
α
C
5
6
1. PROBLEM S-1
1
1
2
q
B
A
M
α
α
G
P
Figure 1.9.
1
1
q
P
M
B
E
A
α
C
2
1.5
G
2.5
2α
D
Figure 1.10.
B
2α
M
q
α
A
Figure 1.11.
4
3
P
1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY
C
B
P
G
M
α
A
Figure 1.12.
P
F
D
α
q
M
α
1
C
1
1
G
Figure 1.13.
A
B
F
1
E
7
1. PROBLEM S-1
q
1
3
M
1
A
2
α
G
P
Figure 1.14.
1
2
q
G
α
4
8
M
A
Figure 1.15.
P
1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY
q
P
2
2
2
G
A
B
Figure 1.16.
4
P
1.5
2
α
B
q
G
A
Figure 1.17.
α
P
A
Figure 1.18.
α
G
M
G
α
B
α
9
10
1. PROBLEM S-1
A
1
M
q
2
2q
B
3
α
P
Figure 1.19.
2
q
M
3
A
Figure 1.20.
α
B
4
α
P
C
1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY
M
P
α
2
4
q
4
A
B
α
Figure 1.21.
α
P
M
2
A
B
α
Figure 1.22.
2
2
q
11
12
1. PROBLEM S-1
P
M
B
q
α
A
2
Figure 1.23.
3
M
B
P
2α
2
α
q
A
Figure 1.24.
C
1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY
q
P
α
B
2
2
2
M
A
Figure 1.25.
2q
P
α
3
B
2
3
A
M
2
Figure 1.26.
2
q
13
14
1. PROBLEM S-1
Figure 1.27.
A
2
M
q
B
2
P
Figure 1.28.
α
2
1.2 SAMPLE PROBLEM
2
q
M
α
2
1
2α P
B
α
A
Figure 1.29.
P
B
α
M
A
α
G
Figure 1.30.
1.2
SAMPLE PROBLEM
Define the reaction forces in support A and rod CD of the construction schematically shown in Figure 1.31.
G = 10 kN; P = 5 kN; M = 8 kN m; q = 0.5 kN/m; a = 30°. The sizes are in meters.
15
16
1. PROBLEM S-1
D
q
P
2
1
M
C
α
A
1
B
2
G
Figure 1.31.
1.3
SOLUTION
Let’s consider the equilibrium of the forces applied to the stud AB. Remove the supports: pin support at the point A,
the member CD, and the string attached to point B. The actions of the support are replaced by the appropriate reaction forces (Figure 1.32). Because the direction of the reaction force of the pin A is unknown, we need to determine
its components XA and YA. We also show the reaction SCD in member CD and the reaction S in the string, which is
by magnitude equal to P. Evenly distributed load of the intensity q is replaced by the concentrated force Q, which is
equal: Q = 2 ∙ q = 2 ∙ 0.5=1 kN and applied in the center of gravity of the distributed forces.
y
90°
YA
A
C
α
XA
1
Figure 1.32.
SCD
Q
M
B
G
2
S
1
2
x
1.3 SOLUTION
17
Three equilibrium equations can be written for the plane (2D) system of forces applied to the system:
∑ MiA = 0; ‒Q ∙ 1 ‒ G ∙ 3 + SCD ∙ 4 sin 30° ‒ M + S ∙ 6 = 0
(1.1)
∑ Yi = YA ‒ Q ‒ G + SCD cos 60° + S = 0
(1.3)
∑ Xi = 0;XA ‒ SCD cos 30° = 0
(1.2)
From Equation (1.1):
SCD =
Q ∙ 1 + G ∙ 3 + M ‒ S ∙ 6 1 ∙ 1 + 10 ∙ 3 + 8 ‒ 5 ∙ 6
=
= 4.5 kN.
4 sin 30°
4 ∙ 0.5
From Equation (1.2):
XA = SCD cos 30° = 4.5 ∙ 0.866 = 3.90 kN.
From Equation (1.3):
YA = Q + G ‒ SCD cos 60° ‒ S = 1 + 10 ‒ 4.5 ∙ 0.5 ‒ 5 = 3.75 kN.
The values of XA, YA, and SCD are positive. It means the assumed directions of these forces coincide with their true
directions.
19
CHAPTER
2
Topic S-2
2.1
APPLICATION OF THE METHOD OF JOINTS TO FIND UNKNOWN
FORCES IN A PLANE (2D) TRUSS STRUCTURE
Using a method of joints find the forces in the members 1, 2, 3, 4 and 5 of the given truss structure. The truss structures are schematically shown in Figures 2.1‒2.30. The external forces (loads) acting on the trusses are given in the
table below. The sizes are in meters.
Problem #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
P1
P2
5
5
10
10
5
10
5
10
10
20
10
10
10
10
10
10
10
10
10
10
10
10
5
10
10
10
10
20
10
10
5
10
5
30
5
20
10
20
20
10
20
20
10
10
20
20
20
40
10
40
20
10
10
20
20
20
20
20
20
10
P3
kN
5
20
20
50
20
10
20
30
30
10
40
30
10
10
10
20
20
20
40
20
10
20
10
20
20
20
10
10
20
20
P4
P5
20
30
20
20
10
20
30
20
10
30
20
20
40
30
40
20
40
30
30
20
20
20
20
30
a
m
2
2
2
2
3
3
3
3
2
2
2
2
2
-
h
m
6
5
6
1.5
5
3.5
3.5
4
2.4
2.4
2.3
3
2.2
-
α
degrees
30
30
30
30
45
45
45
-
20
2. PROBLEM S-2
P1
P2
3a
P3
3
2
4
1
A
P4
B
5
P5
6a
Figure 2.1.
P1
3
A
4
5
P2
6a
Figure 2.2.
2
P3
1
α
P4
B
2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE
3
P2
90°
4
2
A
5
1
h
P1
B
P4
P3
6a
Figure 2.3.
P1
2a
3
2
4
B
A
1
P2
P3
5
4a
Figure 2.4.
21
22
2. PROBLEM S-2
P1
3
90°
P2
2
5
4
A
α
1
P3
B
P4
6a
Figure 2.5.
P1
5
3
2
A
P2
P3
P4
5a
Figure 2.6.
4
1
α
P5
B
2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE
45°
3
B
P2
P1
α
P3
4
2
5
1
A
3a
Figure 2.7.
P1
3
h
90°
P2
2
A
4
1
5
P3
B
4a
2a
P1
a
A
3
4
B
2
1
P3
6a
Figure 2.9.
a/2
Figure 2.8.
5
P2
23
24
2. PROBLEM S-2
B
60°
P1
3
4
α
P2
2
5
A
1
P3
a
a
P4
a
a
a
a
Figure 2.10.
B
a
a
a
P1
A
3
α
a
a
2
a
5
4
P2
1
P3
Figure 2.11.
P1
3
P2
h
90°
2
5
A
4
P3
2a
Figure 2.12.
B
1
P4
4a
2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE
P2
a
a
P3
a
a
3
5
2
α
P1
P4
4
α
4a
P5
A
B
Figure 2.13.
P3
P4
5
1
2
3
P1
4
3a
B
Figure 2.14.
A
P5
3a
P2
25
26
2. PROBLEM S-2
3/2 a
3a
P3
2h
h
P1
3
4
P2
5
2h
2
1
B
A
Figure 2.15.
P3
3a
P1
5
P2
3a
3
2
1
A
Figure 2.16.
B
4
P4
2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE
P2
P1
5
3
a
2
4
P3
A
1
h
4a
B
Figure 2.17.
3
5
h
3/2 h
P1
2
A
1
4
P2
6a
Figure 2.18.
P3
B
P4
27
28
2. PROBLEM S-2
P1
3
3/2 h
90°
4
P2
A
h
2
1
5
P3
B
P4
6a
Figure 2.19.
a/2
5
P1
a
3
2
A
4
1
P2
B
P3
6a
Figure 2.20.
P2
P1
3
a
2
A
1
P3
B
P4
6a
Figure 2.21.
4
h
5
2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE
P2
P1
45°
3
a
2
A
5
4
P3
1
B
P4
6a
Figure 2.22.
6a
3
P1
2
4
1
P4
3a
P2
P5
P3
A
B
Figure 2.23.
P2
3
5
2
A
4
1
B
6a
Figure 2.24.
P4
h
P1
P3
29
2. PROBLEM S-2
P1
P2
5
A
3
2
4
P3
B
h
30
1
6a
Figure 2.25.
5
3
2
A
h
P1
4
1
B
P3
P2
6a
Figure 2.26.
P2
P1
P3
a
3
2
A
1
5
B
5a
Figure 2.27.
4
2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE
P1
A
P2
3
B
5
4
a
2
1
P3
4a
Figure 2.28.
P2
45°
5
4
h
P1
3
2
1
A
P3
B
P4
6a
Figure 2.29.
P1
3
5
2
B
4
P2
1
P5
P4
3a
Figure 2.30.
a
A
P3
31
32
2. PROBLEM S-2
2.2
SAMPLE PROBLEM
The truss schematically is given in Figure 2.31. Define the truss forces S1, S2, S3, S4, and S5.
P1 = 58 kN; P2 = 50 kN; P3 = 85 kN.
3
a/2
4
90°
P3
a
2
5
A
1
P1
P2
6a
B
Figure 2.31.
2.3
SOLUTION
2.3.1
DETERMINING REACTION FORCES BY ANALYTICAL METHOD
Let’s consider an equilibrium of the forces applied on the truss. Remove the supports at A and B by replacing
them with appropriate reaction forces. Reaction force at point A is resolved into two components: XA and YA in the
directions of the coordinate system axes. The reaction force at the pin support B will be shown vertically up along
member BN.
The force P3 is resolved into two components P3′ and P 3″ . The magnitudes of these components:
P ′3 = P3 cos α and P 3″ = P3 sin α (Figure 2.32).
Calculate sin α and cos α:
DE
sin α = CD
=
CE
cos α = CD
=
a
2√a2 +(2a )2
a
2√a2 +(2a )2
= 0.447;
= 0.894.
Determine the magnitudes of P 3′ and P 3″:
P ′3 = 85 ∙ 0.894 = 76 kN; P 3″ = 85 ∙ 0.447 = 38 kN.
For forces applied to the 2D truss we write three equilibrium equations:
∑ MA = 0; ‒ P1 ∙ a ‒ P2 ∙ 2a ‒ P ′3 ∙ 5a + P 3″ ∙ a + RB ∙ 5a = 0;
(2.1)
2.3 SOLUTION
∑ Xi = 0; ‒ P 3″ + XA = 0;
(2.2)
∑ Yi = 0; YA ‒ P1 ‒ P2 ‒ P ′3 + RB = 0.
(2.3)
33
y
P3’
a
a/2
D
E
YA
A
XA
α
α
P3
P3”
C
x
N
P1
P2
RB
6a
B
Figure 2.32.
From Equation (2.1):
RB = P1 + P2 ∙ 2 + P ′3 ∙ 5 ‒ P 3″ = 58 + 50 ∙ 2 + 76 ∙ 5 - 38 = 100o kN.
5
5
From Equation (2.2):
XA = P 3″ = 38 kN.
From Equation (2.3):
YA = P1 + P2 + P ′3 ‒ RB = 58 + 50 + 76 ‒ 100 = 84 kN.
2.3.2
DETERMINING THE FORCES IN THE TRUSS MEMBERS
Let’s determine the forces in truss members using method of joints. In order to find the forces in the members 1, 2,
and 3 (Figure 2.33) we do a section I-I and we consider an equilibrium of the forces applied to one of the parts of
the truss (Figure 2.34).
It is advisable to consider an equilibrium of that part of the truss where less computations are needed. Also,
it is better if the selected each equilibrium equation contains only one unknown force. It will allow for the determination of each force independently from forces acting on other members. We will assume that all members are in
extension. Then a negative sign in answer will indicate that the member is in compression.
34
2. PROBLEM S-2
y
4
II
I
3
III
P3’
P3
2
YA
5
P3”
1
XA
x
N
P1
Figure 2.33.
P2
II
S3
I
3/2 a
B
P3’
F
P3”
G
S1
III
P3
β
S2
H
RB
I
2a
I
RB
B
Figure 2.34.
To define the force S1 we will write a moment equation about the point where the lines of actions of the forces
S2 and S3 are crossing:
∑ MiF = 0; ‒S1 ∙ 1.5a ‒ P ′3 ∙ 2a ‒ P 3″ ∙ 0.5a + RB ∙ 2a.
From here:
S1 =
RB ∙ 2 ‒ P ′3 ∙ 2 ‒ P 3″ ∙ 0.5 100 ∙ 2 ‒ 76 ∙ 2 ‒ 38 ∙ 0.5
=
= 19.3 kN.
1.5
1.5
To find S2 we will project the forces on axes Ay:
∑ Yi = 0; ‒S2 cos β ‒ P ′3 + RB = 0,
FG =
where cos β = FH
1.5a
√a2 + (1.5a)2
= 0.832. From here:
2.3 SOLUTION
RB ‒ P ′3
100 ‒ 76
=
=28.8 kN.
cos β
0.832
S2=
To define the force S2 we will write a moment equation about the point where the lines of actions of the forces
S1 and S3 are crossing:
∑ MiH = 0; S3 ∙ 1.5a ‒ P ′3 ∙ 3a + P 3″ ∙ a + RB ∙ 3a = 0.
From here:
S3=
P ′3 ∙ 3a + P 3″ ‒ RB ∙ 3a 76 ∙ 3 ‒ 38 ‒ 100 ∙ 3
=
= -73.3 kN.
1.5
1.5
In order to find the forces in the member 4 we do a section II-II and we consider an equilibrium of the forces
applied the left side of the truss (Figure 2.35).
II
L
K
YA
S6
α
V
XA
P1
Figure 2.35.
S4
S7
II
An equilibrium equation:
∑ MiV = 0; ‒YA ∙ a ‒ S4 ∙ VK = 0,
where VK = VL ∙ cos α = a · 0.894.
From here:
S4 =
YA ∙ a
84
=‒
= ‒94 kN.
0.894a
0.894
In order to find the force S5 we do a section III-III and we consider an equilibrium of the forces applied the
right side of the truss (Figure 2.36).
35
36
2. PROBLEM S-2
III
P3’
S8
α
P3
P3”
S5
M
S9
III
2a
Figure 2.36.
Equilibrium equation:
∑ MiM = 0; P ′3 ∙ 2a + S5 ∙ 2a + P 3″ ∙ 2a = 0.
From here:
S5 = ‒
P ′3 ∙ 2 + P 3″
76 ∙ 2 + 38
=‒
= ‒ 95 kN.
2
2
37
CHAPTER
3
Topic S-3
3.1
DETERMINATION OF REACTION FORCES OF SUPPORTS FOR
COMPOSITE STUD
Determine the reaction forces and the forces in the interim pins of the composite stud. The studs and acting forces
are schematically shown in Figures 3.1‒3.30. All sizes are in meters. The external forces (loads) acting on the studs
are given in the following table.
Problem #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
P1
6
7
12
8
9
10
4
5
10
8
10
7
8
10
14
11
8
10
9
12
6
12
8
9
9
6
10
7
5
18
P2
kN
10
4
18
10
15
7
6
4
7
14
5
12
12
7
9
15
12
15
13
6
10
20
14
15
15
12
16
10
13
7
P3
5
9
14
13
6
5
8
12
16
14
-12
M1
25
20
36
30
32
18
20
16
24
26
12
15
24
30
26
40
20
35
25
32
30
30
25
28
24
26
32
26
30
16
kN m
M2
25
28
20
18
28
24
35
-
q
kN/m
0.8
1
1.4
1.8
0.9
1.5
1.2
1
1.5
1.6
1.2
1
1.4
0.8
1.5
2
1
1.2
1.5
1.4
1.3
2
-
38
3. PROBLEM S-3
P1
M1
A
B
60°
C
q
P2
2.0
2.0
2.0
2.0
2.0
F
E
D
2.0
4.0
Figure 3.1.
q
P1
A
P2
B
2.0
M1
E
60°
2.0
D
C
2.0
F
2.0
2.0
45°
2.5
3.5
Figure 3.2.
q
P1
A
B
45°
2.0
2.0
D
C
3.0
P2
M1
E
2.0
2.5
F
60°
2.5
2.0
Figure 3.3.
P1
P2
60°
A
45°
B
2.0
Figure 3.4.
2.0
P3
M1
1.5
D
E
M2
C
2.5
2.0
1.5
2.5
2.0
F
3.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE STUD 39
q
P1
A
E
B
1.5
2.0
M1
F 60°
D
C
60°
P2
3.5
2.0
2.5
3.0
1.5
Figure 3.5.
q
P1
A
P2
M1
60°
B
2.0
2.0
E
D
C
2.0
2.0
3.5
F
2.0
30°
2.5
Figure 3.6.
P1
A
P2
M1
45°
M2
C
B
2.0
2.0
P3
E
F
30°
D
1.5
1.5
2.5
2.5
2.0
2.0
Figure 3.7.
P1
A
60°
q
P2
M1
D
C
F
E
B
1.5
Figure 3.8.
2.0
2.5
2.0
2.5
2.0
30°
3.5
40
3. PROBLEM S-3
P1
q
P2
60°
A
P3
M1
B
D
C
2.0
2.0
2.0
2.0
2.0
E
45°
1.5
2.0
F
2.5
Figure 3.9.
P1
M1
A
q
P2
F
30°
C
B
45°
2.5
2.5
D
1.5
2.0
E
2.5
2.0
3.0
Figure 3.10.
P1
A
P2
M1
60°
C
B
2.5
3.5
1.5
P3
M2
E
D
3.0
1.5
30°
2.0
2.0
Figure 3.11.
P1
A
P2
q
D 60°
45°
B
2.0
Figure 3.12.
2.0
M1
C
4.0
4.0
2.0
2.0
E
3.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE STUD 41
P1
M1
45°
A
E 60°
C
B
2.0
1.5
P3
q
P2
D
2.5
1.5
4.0
2.5
2.0
Figure 3.13.
P1
A
P2
30°
q
M1
E
B
2.0
2.5
D
C
2.5
2.0
30°
5.0
2.0
Figure 3.14.
P1
M1
A
P2
60°
D
B
3.5
M2
P3
45°
C
2.0
2.0
E
2.5
2.0
1.5
2.5
Figure 3.15.
P1
P2
q
C
A
B
45°
3.0
Figure 3.16.
M1
2.0
60°
D
3.5
3.5
E
2.0
2.0
42
3. PROBLEM S-3
P1
P2
M1
A
B
M2
45°
C
2.0
3.5
2.5
D
2.0
E
2.5
3.5
Figure 3.17.
P1
P2
M1
30°
C
60°
B
A
2.5
P3
2.0
D
1.5
3.5
E
2.5
2.0
2.0
Figure 3.18.
P2
M1 P1
A
q
B
60°
C
30°
3.0
2.0
2.0
2.5
D
E
5.0
1.5
Figure 3.19.
P2
P1
45° A
M2
q
B
D
C
2.0
Figure 3.20.
2.0
3.0
2.0
4.0
E
3.0
3.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE STUD 43
P1
q
M1
60° A
P2
E
B
D
C
2.0
4.0
2.0
F
3.0
30°
1.5
2.0
1.5
Figure 3.21.
P1
A
45°
M1
q
B
P2
M2
D
F
E
C
2.0
2.5
2.5
2.0
3.5
2.0
1.5
Figure 3.22.
P1
q
M1
A
60° F
B
30°
2.0
P2
3.0
D
C
2.5
2.5
E
3.0
1.5
1.5
Figure 3.23.
q
P1
C
A
30°
D
Figure 3.24.
F
E
B
4.0
P2
M1
2.0
3.0
1.5
2.0
1.5
45°
2.0
44
1. PROBLEM S-1
P1
P2
60°
A
C
45°
D
B
1.5
P3
M2
M1
F
E
2.5
1.5
2.0
2.0
2.5
1.5
2.5
Figure 3.25.
q
P1
P3
P2
A
M1
C
45°
60°
D
E
B
2.0
2.0
2.5
F
1.5
2.5
2.0
1.5
2.0
Figure 3.26.
q
M1
A
E
C
30°
F
D
B
4.0
P2
P1
2.0
2.5
2.0
1.5
2.0
2.0
Figure 3.27.
P2
P1
q
A
D
B
2.0
Figure 3.28.
1.5
E
C
60°
2.0
M1
P3
3.5
2.0
3.0
2.0
3.3 SOLUTION
q
C
A
60°
E
D
B
5.0
P2
P1
M1
2.5
2.5
45
45°
2.0
2.0
2.0
Figure 3.29.
P2
P1
45°
M1
B
P1
60°
E
C
30°
2.5
1.5
3.0
D
2.0
2.0
3.5
Figure 3.30.
3.2
SAMPLE PROBLEM
A composite stud schematically shown in Figure 3.31. Determine the reaction forces and the forces in the interim
pins if P1 = 12 kN; P2 = 20 kN; M = 50 kNm; and q = 2 kN/m.
P1
A
B
M
60°
C
4.0
4.0
4.0
P2
q
E
D
4.0
2.0
2.0
Figure 3.31.
3.3
SOLUTION
A composite stud consists of a system of the simple studs connected via pins. Therefore, we will consider the system
of the forces in equilibrium applied to each simple stud. The forces in the pins connecting these studs will be taking
into the account.
First, we will consider the stud DE (Figure 3.32), as a number of unknown forces (RE, XD, and YD) applied
to this stud is equal to the number of equilibrium equations:
∑MiD = 0; RE ∙ 4 ‒ P2 ∙ 2 cos 30° ‒ Q1 ∙ 1 = 0,
where Q1 = q ∙ 2 = 2 ∙ 2 = 4 kN.
(3.1)
46
1. PROBLEM S-1
∑Xi = 0; XD ‒ P2 cos 60° = 0,
(3.2)
∑Yi = 0; YD ‒ Q1 ‒ P2 cos 30° + RE = 0.
From Equation (3.1):
RE =
(3.3)
P2 ∙ 2 cos 30° + Q1 ∙ 1 20 ∙ 2 ∙ √ 2 + 4 ∙ 1
=
= 9.66 kN.
4
4
3
From Equation (3.2):
XD = P2 cos 60° = 20 ∙ 0.5 = 10 kN.
From Equation (3.3):
YD = Q1 + P2 cos 30° ‒ RE = 4 + 20 ∙ √32 ‒ 9.66 = 11.6 kN.
P2
YD
60°
XD
D
Q1
1.0 1.0
RE
E
2.0
Figure 3.32.
For the stud BD (Figure 3.33):
∑ MiB = 0; ‒M + RC ∙ 4 ‒ Q2 ∙ 6 ‒ YD′ ∙ 8 = 0,
(3.4)
∑ Xi = 0; XB ‒ XD′ = 0;
(3.5)
where Q2 = q ∙ 4 = 2 ∙ 4 = 8 kN.
∑ Yi = 0; YB + RC ‒ Q2 - YD′ = 0.
From Equation (3.4):
RC =
M + Q2 ∙ 6 + YD′ ∙ 8 50 + 8 ∙ 6 + 11.6 ∙ 8
=
= 47.7 kN.
4
4
From Equation (3.5):
XB = XD′ = 10 kN.
From Equation (3.6):
YB = RC + Q2 + YD′ = 47.7 + 8 + 11.6 = 28.1 kN.
(3.6)
47
YB
M
XB
B
RC
X D’
C
4.0
Q2
D
2.0
2.0
YD’
Figure 3.33.
For the stud AB (Figure 3.34):
∑ MiA = 0; MA ‒ P1 ∙ 4 ‒ YB′ ∙ 4 = 0;
(3.7)
∑ Yi = 0; YA ‒ P1 ‒ YB′ = 0.
(3.9)
∑ Xi = 0; XA ‒ XB′ = 0;
YA
A
MA
(3.8)
P1
XA
B
4.0
4.0
XB’
YB’
Figure 3.34.
From Equation (3.7):
MA = P1 ∙ 4 + YB′ ∙ 8 = 12 · 4 ‒ 28.1 · 8 = -177 kNm.
From Equation (3.8):
XA = XB′ = 10 kN.
From Equation (3.9):
YA = P1 + YB′ = 12 ‒ 28.1 = -16.1 kN.
To verify the obtained results we need to make sure that any equation of equilibrium written for the forces
applied to the whole construction is valid. For example:
∑Yi = 0; YA ‒ P1 + RC ‒ Q2 ‒ Q1 ‒ P2 cos 30° + RE = 0.
49
CHAPTER
4
Topic S-4
4.1
DETERMINATION OF REACTION FORCES OF SUPPORTS FOR
COMPOSITE CONSTRUCTION (SYSTEM OF TWO BODIES)
Determine the reaction forces and the forces in the interim pin of the composite construction. The constructions
and the acting forces are schematically shown in Figures 4.1‒4.30. All sizes are in meters. The external forces (loads)
acting on the constructions are given in the table below.
Problem #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
P1
6
5
8
10
12
14
16
12
14
8
15
15
7
5
6
8
9
7
6
7
8
5
14
10
11
15
11
12
10
9
kN
P2
8
10
12
8
6
10
8
6
10
11
15
16
18
16
17
6
10
13
10
15
14
12
9
10
M
kNm
25
26
33
25
27
18
20
28
26
29
28
15
30
24
31
26
27
35
32
30
34
36
28
33
18
36
30
35
29
q
kN/m
0.8
1.1
1.3
1
0.9
1.4
1
1.4
0.9
1
1.5
1.1
0.9
1.5
0.8
1.1
0.8
1.1
0.8
1.2
1.5
1.2
1.3
1
1.4
1.5
1.1
1.3
1.5
50
4. PROBLEM S-4
90°
P1
C
3.5
q
2
M
A
B
2
4
Figure 4.1.
1.5
C
P1
P2
M
B
A
2
Figure 4.2.
2
4
1.5
90°
4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)
P1
M
C
1
30°
P2
2
3.0
q
A
B
5
Figure 4.3.
M
3
q
P1
B
A
1
Figure 4.4.
C
1
4
q
M
P1
B
30°
C
4
4
A
Figure 4.5.
D
4
4
51
52
4. PROBLEM S-4
q
P2
P1
C
45°
B
45°
2
1
2
2
2
A
Figure 4.6.
1
P1
3
C
M
q
2
D
P2
60°
A
3
B
3
Figure 4.7.
3
3
q
1
C
30°
M
2
P1
A
B
3
Figure 4.8.
P2
4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)
q
P1
45°
C
2.5
M
A
B
3
2
2
Figure 4.9.
1
M
C
q
P1
3
90°
A
B
2
Figure 4.10.
1.5
1.5
53
54
4. PROBLEM S-4
1
2
1
q
P1
D
30°
2
C
1
2
45°
P2
M
B
A
Figure 4.11.
P1
60°
2
1
2
M
1
C
P2
2
4
q
B
Figure 4.12.
A
3
D
4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)
q
P2
B
60°
C
M
4
P1
90°
A
D
2
2
2
3
2
Figure 4.13.
C
M
5.0
B
q
A
Figure 4.14.
4
90°
55
56
4. PROBLEM S-4
2
C
2
q
M
B
60°
A
30°
3
P1
45°
P2
3
3
Figure 4.15.
A
P2
P1
90°
2
q
60°
D
C
1
Figure 4.16.
4
M
B
1
3
4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)
3
P2
3
P1
45°
2
C
A
M
2
q
B
Figure 4.17.
2
2
M
30°
C
2
2
B
q
2
P1
A
60°
P2
Figure 4.18.
57
58
4. PROBLEM S-4
A
P2
q
2
30°
2
2
P1
M
C
60°
2
3
B
Figure 4.19.
D
P1
2
2
C
B
2
P2
A
60°
4
Figure 4.20.
q
3
M
4
4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)
D
P1
2
C
2
3
M
q
2
P2
B
A
60°
4
4
Figure 4.21.
3
M
P1
P2
q
C
45°
A
D
1
2
B
1
30°
1
Figure 4.22.
P1
45°
A
M
B
1
Figure 4.23.
q
P2
2
1
D
C
1.5
2
1
59
60
4. PROBLEM S-4
D
P1
4
q
P2
M
60°
C
4
B
2
5
4
2.5
2.5
A
Figure 4.24.
q
2
B
P2
3
P1
1
M
90°
A
D
45°
1
C
1.5
1.5
Figure 4.25.
P1
30°
F
2
Figure 4.26.
M
A
4
P2
C
B
2
D
60°
2
2
q
30°
4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)
D
4
q
B
F
M
A
C
60°
P1
2
4
2
Figure 4.27.
C
M
A
45°
3
2
P1
q
B
D
60°
2
3
P2
Figure 4.28.
q
P1
A
B
4
Figure 4.29.
30°
M
C
2
2
P2
45°
D
2
2
P2
61
62
4. PROBLEM S-4
2
P2
4
P1
A
60°
30°
q
C
M
B
2
2
Figure 4.30.
4.2
SAMPLE PROBLEM
A composite structure is schematically shown in Figure 4.31. Determine the reaction forces and the forces in the
interim pin if P1 = 10 kN; P2 = 12 kN; M = 25 kNm; q = 2 kN/m; and α = 60°.
P1
α
C
M
P2
4
q
A
3
B
3
2
2
Figure 4.31.
4.3
SOLUTION
First, we will consider the system of the forces in equilibrium applied to the entire construction (Figure 4.32). To
simplify a calculation of the moments of the force P1 we will resolve it into two components P1′ and P1″ as:
P 1′ = P1 cos α = 10 ∙ 0.5 = 5 kN,
4.3 SOLUTION
63
P 1″ = P1 sin α = 10 ∙ 0.866 = 8.66 kN.
The equilibrium equations will be written as:
∑ MiA = 0; P1′ ∙ 4 + P1″ ∙ 3 ‒ Q ∙ 2 ‒ M ‒ P2 ∙ 5 + YB ∙ 7 = 0,
(4.1)
∑ Yi = 0; ‒P1″ + YA ‒ P2 + YB = 0,
(4.2)
∑ Xi = 0; XA + XB ‒ P 1′ + Q = 0.
(4.3)
where = q ∙ 4 = 2 ∙ 4 = 8 kN.
From Equation (4.1):
YB =
‒P 1′ ∙ 4 ‒ P 1″ ∙ 3 + Q ∙ 2 + M + P2 ∙ 5 ‒5 ∙ 4 ‒ 8.66 ∙ 3 + 8 ∙ 2 + 25 + 12 ∙ 5
=
= 7.86 kN.
7
7
From Equation (4.2):
YA = P 1″ + P2 ‒ YB = 8.66 + 12 ‒ 7.86 = 12.8 kN.
Equation (4.3) contains two unknowns. It is impossible to define those unknowns. We can only determine
the relationship between them.
y
P1”
P1
α
C
P’
2
M
P2
YA
A
3
2
Q
YB
XA
3
2
2
B
XB
x
Figure 4.32.
Let’s consider an equilibrium of the forces applied to the right side of the construction (Figure 4.33):
∑ MiC = 0; ‒M ‒ P2 ∙ 2 + XB ∙ 4 + YB ∙ 4 = 0,
(4.4)
64
4. PROBLEM S-4
∑ Xi = 0; XB + XC = 0,
(4.5)
∑Yi = 0; YC ‒ P2 + YB = 0.
(4.6)
From Equation (4.4):
XB =
M + P2 ∙ 2 ‒ YB ∙ 4 25 + 12 ∙ 2 ‒ 7.86 ∙ 4
=
= 4.39 kN.
4
4
From Equation (4.5):
XC = XB = ‒4.39 kN.
From Equation (4.6):
YC = P2 ‒ YB = 12 ‒ 7.86 = 4.14 kN.
From Equation (4.3):
XA = ‒XB + P 1′ ‒ Q = ‒4.39 + 5 ‒ 8 = ‒7.39 kN.
YC
XC
C
M
4
P2
YB
XB
2
2
B
x
Figure 4.33.
To verify the obtained results we need to make sure that any equation of equilibrium written for the forces
applied to the entire construction is valid (Figure 4.31). For example:
∑ MiB = 0; P 1′ ∙ 4 + P 1″ ∙ 10 ‒ Q ∙ 2 ‒ YA ∙ 7 ‒ M + P2 ∙ 2 = 5 ∙ 4 + 8.66 ∙ 10 ‒ 8 ∙ 2 ‒ 12.8 ∙ 7 ‒ 25 + 12 ∙ 2 =
130.6 ‒ 130.6 = 0.
The results of simulations are shown in the following table.
4.3 SOLUTION
XA
-7.39
YA
12.8
Forcers, kN
XB
YB
4.39
7.86
XC
-4.39
YC
4.14
65
67
CHAPTER
5
Topic S-5
5.1
DETERMINATION OF REACTION FORCES OF SUPPORTS FOR
COMPOSITE CONSTRUCTION (SYSTEM OF THREE BODIES)
Determine the reaction forces and the forces in the interim pin of the composite construction. The constructions
and the acting forces are schematically shown in Figures 5.1‒5.30. All sizes are in meters. The external forces (loads)
acting on the constructions are given in the following table.
Problem #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
P1
6
11
9
10
8
10
16
13
11
12
8
12
15
10
12
13
7
9
12
11
8
14
13
10
15
12
9
6
8
15
kN
P2
M1
8
12
14
15
17
14
16
8
7
9
14
-
25
34
20
30
22
28
36
25
29
34
28
36
30
35
32
26
23
29
33
38
36
28
32
35
40
37
29
25
30
34
kNm
M2
30
34
37
21
34
-
q
kN/m
0.8
1
1.1
1.4
1
1.2
0.9
1.3
1.5
1
1.4
1.2
1.5
1.1
0.8
1
1.2
68
5. PROBLEM S-5
B
90°
P1
C
3.5
M1
2
q
A
4
D
2
Figure 5.1.
1.5
C
P1
P2
M1
A
Figure 5.2.
2
1.5
90°
B
D
2
4
5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)
P1
M1
B
1.5
1
1
C
30°
P2
2
3.0
q
A
D
2.5
Figure 5.3.
P2
1
M
A
B
2
90°
2
1.5
D
P1
E
60°
C
2
2
3
Figure 5.4.
P1
1
B
60°
D
3
1.5
3.0
3.0
q
M1
A
Figure 5.5.
1.5
P2
C
E
69
70
5. PROBLEM S-5
2
1
45°
2
M1
E
D
2
1.5
P1
C
B
1.5
60°
P2
2
A
Figure 5.6.
P1
M1
C
D
2
3.5
B
2
2
45°
1
E
1
M2
A
Figure 5.7.
P2
1
1
2
E
3
Figure 5.8.
M1
C
M2
2.5
A
B
45°
2
2
F
3
D
5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)
2
2
2
B
M2
1.5
M1
E
D
1
P1
2
C
A
Figure 5.9.
P1
q
M1
1.5
1.5
3.0
2.5
B
A
Figure 5.10.
C
60°
D
E
1
1.5
71
72
5. PROBLEM S-5
P1
60°
C
1
1
1
2
D
M1
2
3.0
q
A
B
E
1.5
1.5
Figure 5.11.
P1
P2
60°
A
B
1
E
1
2
1
3.0
1
C
Figure 5.12.
F
M1
D
1
1
5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)
B
M1
1.5
P1
2
1.5
2
M2
1.5
D
2.5
4.5
C
E
A
Figure 5.13.
P2
60°
1.5
C
1.5
D
A
B
2
Figure 5.14.
M1
2
1.5
P1
E
3
F
2
73
74
5. PROBLEM S-5
q
B
E
45°
P1
3.5
2
90°
P1
M1
C
A
1
1
D
3.5
2.5
F
2.5
2.5
Figure 5.15.
C
2
M1
B
4
3.0
q
A
Figure 5.16.
1.5
P1
D
5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)
M1
A
B
D
P1
3
2
3.0
M2
C
2
Figure 5.17.
M1
C
1
B
P1
q
4
60°
5
A
D
Figure 5.18.
P2
M1
B
1
Figure 5.19.
C
60°
1
2
90°
P1
A
E
D
1.5
1.5
1.5
1.5
75
76
5. PROBLEM S-5
3
M1
B
P1
3
2
45°
D
q
2
C
A
E
Figure 5.20.
A
E
C
B
2
3
2
D
2
Figure 5.21.
2
q
M1
60°
P1
5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)
2
C
60°
P2
D
P1
45°
2
1.5
A
M1
3
3
B
2
Figure 5.22.
C
90°
B
45°
3.0
q
A
Figure 5.23.
3
D
1
45°
q
3
P1
2
M1
E
77
78
5. PROBLEM S-5
A
q
P1
2
M1
2
B 1
E
1
2
1
45°
2
2
D
C
Figure 5.24.
2.5
D
3
1
M1
2
P1
q
A
2
2
45°
B
Figure 5.25.
E
C
30°
5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)
q
E
1
D
C
1.5
2
2
3.0
1
P1
1.5
A
M1
B
Figure 5.26.
A
2
q
M1
D
3
E
45°
2
3
1
B
2
C
Figure 5.27.
P1
79
80
5. PROBLEM S-5
E
2
45°
2
P1
1
A
2
3
M1
C
q
2
B
D
Figure 5.28.
P1
B
1.5
A
M1
1
60°
D
1.5
2
P2
1
C
3
Figure 5.29.
E
1
5.3 SOLUTION
81
P1
A
2
60°
B
1
2
60°
1
C
2
E
2
D
M1
q
Figure 5.30.
5.2
SAMPLE PROBLEM
A composite structure is schematically shown in Figure 5.31. Determine the reaction forces and the forces in the
interim pins if P1 = 10 kN; P2 = 15 kN; M = 40 kNm; and q = 1.6 kN/m.
2
B
P1
C
2
M
4
A
60°
2
q
P2
2
2
D
2
Figure 5.31.
5.3
SOLUTION
First, we will consider the system of the forces in equilibrium applied to the stud BC (Figure 5.32). The equations
of the equilibrium are:
∑ MiB = 0; ‒P2 ∙ 2 sin 60° + YC ∙ 4 = 0,
(5.1)
82
5. PROBLEM S-5
∑ MiC = 0; ‒YB ∙ 4 + P2 ∙ 2 sin 60° = 0,
(5.2)
∑ Xi = 0; XB ‒ P2 ∙ cos 60° ‒ XC = 0.
(5.3)
From Equation (5.1):
YC =
P2 ∙ 2 sin 60° 15 ∙ 2 ∙ √32
= 6.5 kN.
=
4
4
From Equation (5.2):
YB =
P2 ∙ 2 sin 60°
= YC = 6.5 kN.
4
P2
XB
B
60°
2
C
XC
2
YB
YC
Figure 5.32.
The components XB and XC can be determined from the equations of equilibrium of the forces applied to the
element CD. Let’s consider the equilibrium of the forces applied to the element CD (Figure 5.33):
∑ Yi = 0; YD ‒ YC′ = 0,
(5.4)
∑ Xi = 0; XC′ + XD = 0.
(5.6)
∑ MiC = 0; M + XD ∙ 4 + YD ∙ 2 = 0,
(5.5)
From Equation (5.4):
YD = YC′ = 6.5 kN.
From Equation (5.5):
XD=
M + YD ∙ 2 40 + 6.5 ∙ 2
=
= ‒13.3 kN.
4
4
From Equation (5.6):
XC′ = ‒XD = 13.3 kN.
From Equation (5.3):
XB = P2 ∙ cos 60° + XC = 15 ∙ 0.5 + 13.3 = 20.8 kN.
.
5.3 SOLUTION
83
2
XC’
C
YC ’
4
M
YD
XD
D
Figure 5.33.
Finally, we will consider an equilibrium of the forces applied to the element AB (Figure 5.34):
∑ Xi = 0; XA + Q ‒ XB′ = 0,
(5.7)
∑ Yi = 0; YA ‒ P1 ‒ Y B′ = 0,
(5.8)
where Q = q ∙ 2 = 1.6 ∙ 2 = 3.2 kN.
∑ MiB = 0; MA + XA ∙ 2 ‒ YA ∙ 4 + P1 ∙ 2 + Q ∙ 1 = 0.
From Equation (5.7):
XA = Q + XB′ = ‒3.2 + 20.8 = 17.6 kN.
From Equation (5.8):
YA = P1 + YB′ = 10 + 6.5 = 16.5 kN.
From Equation (5.9):
MA = ‒XA ∙ 2 + YA ∙ 4 ‒ P1 ∙ 2 ‒ Q ∙ 1 = ‒17.6 ∙ 2 + 16.5 ∙ 4 ‒ 10 ∙ 2 ‒ 3.2 ∙ 1 = 7.6 kNm.
(5.9)
5. PROBLEM S-5
YB’
4
XB’
2
B
YA
XA
A
Q
1
P1
1
84
MA
Figure 5.34.
To verify the obtained results we need to make sure that any equation of equilibrium written for the forces
applied to the entire construction is valid (Figure 5.31). For example:
∑ Xi = 0; XA + Q ‒ P2 ∙ cos 60° + XD = 0.
17.6 + 3.2 ‒ 15 ∙ 0.5 ‒ 13.3 = 20.8 ‒ 20.8 = 0.
∑ Yi = 0; YA ‒ P1 ‒ P2 ∙ cos 30° + YD =0.
16.5 ‒ 10 ‒ 15
√3
2
+ 6.5 = 23 ‒ 23 = 0.
The results of simulations are shown in the table below.
Moment, kNm
MA
7.6
XA
17.6
YA
16.5
XB
20.8
Forcers, kN
YB
XC
6.5
13.3
YC
6.5
XD
-13.3
YD
6.5
85
CHAPTER
6
Topic S-6
6.1
DETERMINATION OF REACTION FORCES OF RODS SUPPORTING
RECTANGULAR PLATE
Determine the reaction forces in rods supporting a thin horizontal rectangular plate of weight G under action
of force P applied along the side AB. The constructions and the acting forces are schematically shown in Figures
6.1‒6.30. All sizes (in meters) and the external forces (loads) acting on the plate are given in the table below.
Problem #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
G
20
28
18
28
32
22
20
30
36
20
8
33
24
33
35
14
20
33
20
33
28
16
20
28
30
20
33
30
30
20
kN
P
a
b
25
30
20
30
35
25
25
35
40
25
10
35
25
35
40
15
25
35
25
35
30
20
25
30
35
25
35
25
35
25
8
5.5
4
7
8
9
4
5.5
6
8.5
3
6
9.5
6
7
4
8
6
4
6
7
5.5
4
7
6
8
6
6
5.5
4
2.5
5
4.5
4
4
2.5
5
5.5
6
2.5
2.5
5.5
2.5
5.5
5
3.5
2.5
5.5
5
5.5
4
3
5
4
5
2.5
5.5
5
5
5
m
c
d
3.5
3.5
3.5
4
4
4.5
3
4
4
3.5
1.5
3
3.5
4
4
3
3.5
4
4
4
4
3
3
4
4
3.5
3
3
4
4
1
1
1.5
2
1.5
1
1
1.5
1
1.5
-
86
6. PROBLEM S-6
d
6
B
b
c
3
P
A
5
4
1
2
Figure 6.1.
P
b
6
5
B
2
1
Figure 6.2.
c
A
3
4
6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE
b
B
P
A
c
5
6
4
2
3
1
d
Figure 6.3.
6
c
P
A
B
4
5
1
b
Figure 6.4.
3
2
87
88
6. PROBLEM S-6
b
P
B
5
6
3
1
c
A
4
2
Figure 6.5.
P
B
A
6
5
c
2
1
3
b
Figure 6.6.
4
6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE
6
d
B
5
4
P
3
c
A
1
2
b
Figure 6.7.
P
A
c
B
4
6
5
1
b
Figure 6.8.
3
2
89
90
6. PROBLEM S-6
b
P
A
B
4
6
1
3
c
5
2
Figure 6.9.
b
P
A
d
6
c
5
1
2
3
Figure 6.10.
B
4
6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE
P
A
c
B
6
1
4
5
3
2
b
Figure 6.11.
B
d
5
A
P
3
c
4
6
b
Figure 6.12.
1
2
91
92
6. PROBLEM S-6
b
P
B
A
5
4
c
6
1
3
2
Figure 6.13.
b
B
A
5
1
Figure 6.14.
6
4
2
c
P
3
6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE
b
B
5
4
6
2
P
c
A
3
1
Figure 6.15.
b
P
B
A
3
5
6
1
2
Figure 6.16.
c
4
93
6. PROBLEM S-6
6
B
d
94
4
P
A
c
5
3
1
2
b
Figure 6.17.
b
P
B
5
4
3
c
6
1
Figure 6.18.
A
2
6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE
P
B
c
A
4
5
2
6
3
1
b
Figure 6.19.
P
b
A
B
c
5
1
Figure 6.20.
6
4
3
2
95
96
6. PROBLEM S-6
d
6
B
4
5
c
3
A
2
P
1
b
Figure 6.21.
b
B
4
1
2
Figure 6.22.
3
6
c
P
5
6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE
A
P
c
d
6
5
4
1
B
3
2
b
Figure 6.23.
P
B
3
5
1
6
b
Figure 6.24.
2
4
c
A
97
98
6. PROBLEM S-6
5
b
B
A
P
c
4
2
6
1
Figure 6.25.
d
6
P
B
A
5
1
2
b
Figure 6.26.
4
c
3
3
6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE
b
P
A
B
5
c
6
1
3
4
2
Figure 6.27.
B
A
4
P
3
5
2
1
b
Figure 6.28.
c
d
6
99
100
6. PROBLEM S-6
b
2
A
3
5
c
B
P
4
1
6
Figure 6.29.
b
B
P
A
c
5
6
3
1
2
4
Figure 6.30.
6.2
SAMPLE PROBLEM
A construction is schematically shown in Figure 6.31. Determine the reaction forces in the supporting rods 1, 2, 3,
4, 5, and 6 if G = 6 kN; P = 9 kN; a = 5 m; b = 5 m; c = 4 m; and d = 2 m.
d
6.3 SOLUTION 101
6
B
P
c
A
5
4
3
2
1
b
Figure 6.31.
6.3
SOLUTION
Let’s consider the system of the forces in equilibrium applied to the rectangular plate (Figure 6.32). The given forces
are the gravity force G applied in the center of the gravity of the plate, and the force P. Assuming that all rods are
in extension, their reaction forces S1, S2, S3, S4, S5, and S6 are directed from the joints. We select the Cartesian coordinate system Oxyz, so its axes will be crossing as much rods as possible. In this case the equilibrium equations will
be simple. We draw right parallelepiped and introduce the angles η, ψ, θ, and φ. The force S6 is resolved into three
components along the axes (Figure 6.32):
S6′ = S6 sin θ sin η; S6″ = S6 sin θ cos η; S6‴ = S6 cos θ.
6. PROBLEM S-6
θ
z
d
b
z1
B
A
η
P
S6
φ
S4
S6’’’
G
S2
S6’
S6 ”
S5
D
S3
x
S1
c
102
ψ
Figure 6.32.
Next, we write six equations of equilibrium:
b
∑ Mix = 0; G ∙ 2 + S4 ∙ b + S5 sin φ ∙ b = 0,
(6.1)
∑ Miz = 0; ‒P ∙ b + S5 cos φ ∙ b ‒ S6″ ∙ a = 0,
(6.3)
∑ Yi = 0; ‒S3 cos ψ ‒ S6″ = 0,
(6.5)
∑ Miy = 0; G ∙ a2 + S6‴ ∙ a = 0,
∑ Xi = 0; ‒P + S1 cos φ + S5 cos φ ‒ S6′ = 0,
∑ Zi =0; ‒G ‒ S1 sin ϕ ‒ S2 ‒ S3 sin ψ ‒ S4 ‒ S5 sin φ + S6‴ = 0.
Through the given sizes we can determine the trigonometric functions of the angles:
sin φ =
c
4
=
= 4 ;
√a2 + c2 √52 + 42 √41
cos φ =
a = 5 ;
√a2 + c2 √41
sin ψ =
c
4
=
= 4 ;
√b2 + c2 √52 + 42 √41
(6.2)
(6.4)
(6.6)
6.3 SOLUTION 103
b = 5 ;
cos ψ =
√b2 + c2 √41
sin η =
cos η =
sin θ =
cos θ =
a
5
1
=
=
;
√a2 + b2 √52 + 52 √2
b
√a2 + b2
=
1
;
√2
√a2 + b2
√a2 + b2 + d 2
=
√52 + 52
√52 + 52 + 22
=
5√2
5
=
;
√54 3√3
d
2
=
.
√a2 + b2 + d 2 √54
From Equation (6.2):
S6″ = 12 ∙ G = 12 ∙ 6 = 3 kN.
As:
S6″ = S6 cos θ, From Equation (6.3):
S5 =
then:
S6 =
S6″
3√54
=
= 11.03 kN.
cos θ
2
P ∙ b + S6″ ∙ a P
S ∙ a sin θ cos η 9√41 1.5 ∙ √54 ∙ 5 ∙
=
+ 6
=
+
5 ·5
b cos φ
cos φ
b cos φ
5
5√2 1
54 ∙ √2 ∙
√41
= 3.3 ∙ √41 = 21.1 kN.
From Equation (6.1):
4
S4 = ‒ 12 ∙ G ‒ S5 sin φ = ‒ 12 ∙ 6 ‒ 3.3∙√41 ∙
= ‒3 ‒ 13.2 = ‒16.2 kN.
√41
From Equation (6.5):
‒S ″
S sin θ sin η ‒1.5 ∙ √54 ∙ 5√2
∙ 1 ∙ √41
√54 √2
S3 = cos6ψ = ‒ 6 cos ψ
=
= ‒1.5 ∙ √41 = ‒9.62 kN.
5
From Equation (6.4):
1
P
S′
S sin θ sin η 9√41
P
1.5 ∙ √54 ∙ 5√2
√54 ∙ √2 ∙√41 = (1.8 ‒ 3.3 +
S1 = cos φ ‒ S5 + cos6 φ = cos φ ‒ S5 + 6 cos φ
=
‒ 3.3√41 +
5
5
1.5) ∙ √41 = 0.
From Equation (6.6):
S2 = ‒G ‒ S1 sin φ ‒ S3 sin ψ ‒ S4 ‒ S5 sin φ + S 6‴ = ‒G ‒ (S1 + S5) sin φ ‒ S3 sin ψ ‒ S4 + S6 cos θ = ‒6
4
4
2
‒ (0 + 3.3 ∙ √41) ∙
‒ (‒1.5 ∙ √41) ∙
‒ (‒16.2) + 1.5 ∙ √54 ·
= 6 kN.
√41
√41
√54
The results of simulations are shown in the following table.
Rod number
Reaction forces, kN
1
0
2
6
3
-9.62
4
-16.2
5
21.1
6
11.03
As seen from the table, the rods 2, 5, and 6 are in extension, the rods 3 and 4 are in compression, and the rod
1 has zero reaction force.
104
6. PROBLEM S-6
To verify the obtained results we will write an equilibrium equation, for example, a moment equation about
axis z1:
∑ Miz1 = ‒S1 cos φ ∙ b + S3 cos ψ ∙ b + S6′ ∙ b = ‒S1 cos φ ∙ b + S3 cos ψ ∙ b + S6 sin θ sin η ∙ b = 0 + (‒1.5√41)
5
5√2 1
∙
∙ 5 + 1.5 ∙ √54 ∙
∙
∙ 5 = ‒37.5 + 37.5 = 0.
√41
√54 √2
105
CHAPTER
7
Topic S-7
7.1
DETERMINATION OF CENTER OF GRAVITY
Determine the center of gravity of the construction consisting of the homogeneous rods of the same weights per
unit of their lengths (Figures 7.1‒7.6), the plane objects (Figures 7.7‒7.18 and 7.24‒7.30) or the 3D objects (Figures
7.19‒7.23) (Figures 7.1‒7.30). In Figures 7.1-7.6 the sizes are in meters, in Figures 7.7‒7.30 in centimeters.
2.5
2.5
y
x
O
4
Figure 7.1.
106
7. PROBLEM S-7
2.5
5
y
45°
x
O
Figure 7.2.
y
30° 45°
15°
30°
O
Figure 7.3.
x
7.1 DETERMINATION OF CENTER OF GRAVITY 107
y
2
1.5
45°
2
45°
x
O
Figure 7.4.
y
90°
60°
30°
O
3.5
Figure 7.5.
3.5
x
108
7. PROBLEM S-7
y
90°
60°
60°
60°
O
2
Figure 7.6.
2
x
7.1 DETERMINATION OF CENTER OF GRAVITY 109
4
y
x
O
60
5
Figure 7.7.
x
O
30
Figure 7.8.
110
7. PROBLEM S-7
y
10
4
2
8
2
2
2
2
x
O
Figure 7.9.
12
8
4
15
y
O
8
2
20
Figure 7.10.
x
7.1 DETERMINATION OF CENTER OF GRAVITY 111
y
60°
60°
x
O
30
Figure 7.11.
y
18
R2
2
2
9
x
16
Figure 7.12.
112
7. PROBLEM S-7
y
4
x
4
O
2 2
16
Figure 7.13.
y
7.5
7.5
7
20
2
7
O
7
Figure 7.14.
x
7.1 DETERMINATION OF CENTER OF GRAVITY 113
y
10
7
10
O
30
Figure 7.15.
x
114
7. PROBLEM S-7
y
6
4
3
6
20
O
x
10
Figure 7.16.
45°
Figure 7.17.
O
20
10
5
y
x
7.1 DETERMINATION OF CENTER OF GRAVITY 115
y
15
x
O
Figure 7.18.
25
z
x
Figure 7.19.
10
15
O
25
y
10
116
7. PROBLEM S-7
17
z
15
O
10
y
40
x
Figure 7.20.
z
10
10
10
25
O
20
30
x
Figure 7.21.
y
7.1 DETERMINATION OF CENTER OF GRAVITY 117
z
10
15
O
5
13
15
y
25
x
Figure 7.22.
10
20
z
O
y
15
x
35
10
Figure 7.23.
15
118
7. PROBLEM S-7
40
y
x
O
15
40
20
Figure 7.24.
y
15
15
25
10
x
O
15
Figure 7.25.
40
7.1 DETERMINATION OF CENTER OF GRAVITY 119
40
R15
y
x
O
65
Figure 7.26.
y
40
20
x
O
70
Figure 7.27.
120
7. PROBLEM S-7
y
7
15
25
35
x
O
40
60
Figure 7.28.
y
5
Figure 7.29.
40
O
30
5
4 4
x
7.2 SAMPLE PROBLEM 121
y
10
30
10
20
x
O
50
Figure 7.30.
7.2
SAMPLE PROBLEM
Determine the coordinates of the gravity center of the plane object shown in Figure 7.31.
y
30
R20
80
Figure 7.31.
x
122
7.3
7. PROBLEM S-7
SOLUTION
The coordinates of the gravity center are determined by formulas:
xc =
∑ Fi xi
;
F
yC=
∑ Fi xi
.
F
(7.1)
In order to use these formulas, we need to cut the area into small objects with known centers of gravity. In this
problem these objects are a rectangle, a triangle, and a semicircle (Figure 7.32). The area of the semicircle will be cut
from the rectangle area, so we will consider it with the negative sign.
The area of the rectangle:
F1 = 40 ∙ 30 = 1,200 cm2.
The area of the triangle:
F2 =
40 ∙ 50
= 1,000 cm2.
2
The area of the semicircle:
F3 =
π ∙ 202
= 200π = 628 cm2.
2
y
40
30
C1
C
C2
yC
C3
50
x
xC
Figure 7.32.
The gravity centers of selected elements will have the following coordinates.
1. Rectangle:
x1 = 15 cm; y1 = 20 cm.
7.3 SOLUTION 123
2. Triangle:
50
40
x2 = 30 + 3 = 46.7 cm; y2 = 3 = 13.3 cm.
3. Semicircle:
x3 =
4R 4 ∙ 20
=
= 8.5 cm; y3 =20 cm.
3π
3π
To calculate the coordinates of the gravity center of the entire plane object, we provide the following table.
Element #
1
2
3
Total
Fi, cm2
1,200
1,000
-628
1,572
Xi, cm
15
46.7
8.5
-
Yi, cm
20
13.3
20
-
Siy = Fi xi, cm3 Six=Fi yi, cm3
18,000
24,000
46,700
13,300
-5,338
-12,560
59,362
24,700
The coordinates of the gravity center of the given plane object will be calculated by Equations (7.1):
xC =
59,362
24,700
= 37.8 cm; yC =
= 15.7 cm.
1,572
1,572
The center of gravity is shown in Figure 7.32.
Note: The areas and the gravity centers’ coordinates of some plane objects are given in the following table.
Plane object
Triangle
y
B
1
F= 2 aha
1
xC = 3 (x1 + x2 + x3),
where x1, x2, and x3 are the
coordinates of the vertices 0,
A, and B;
1
yC = 3 h a
F = αR2
xC =
ha
C
yC
Gravity Centers’
Coordinates
Area
A
0
x
xC
a
Circular Sector
R
a
a
xC
C
x
2R sin α R2b
=
3α
3F
124
7. PROBLEM S-7
π
α=2
F=
πR2
2
xC =
4R
3π
π
α= 6
F=
πR2
6
xC =
2R
π
(semicircle)
Circular
Segment
R
a
a
C
xC
x
1
F = 2 R2 ∙ (2α ‒
sin 2α)
xC =
b3
4R sin3 α
=
3(2α ‒ sin 2α) 12F
125
Author Biography
Dr. Sayavur I. Bakhtiyarov is a Professor at New Mexico Institute of Mining and
Technology (Socorro, NM, USA) and a Fulbright Fellow. Dr. Bakhtiyarov obtained his
Ph.D. from the Russian Academy of Sciences in 1978, and in 1992 a D.Sc. from the
Azerbaijan National Academy of Sciences. His areas of expertise are: multiphase flows,
nanotechnology, nonlinear fluid mechanics, tribology, rheology, and self-healing composites. Dr. Bakhtiyarov taught engineering courses over four decades in several countries (Azerbaijan, Russia, China, UK, Turkey, U.S.). Dr. Bakhtiyarov authored 350+
scientific publications in refereed scholarly journals, books, international conferences
and symposia proceedings, and 14 patents. Dr. Bakhtiyarov was elected as a foreign
member of Russian Academy of Natural Sciences and International Ecoenergy Academy. He served as a Program director of US DOE and NASA research projects,
INSRP US DOD coordinator for NASA’s Mars Science Lab mission. Dr. Bakhtiyarov is a lead organizer of the
ASME annual symposia and forums, Editor in Chief of two international journals, Mechanics and Solids (IJM&S)
and Manufacturing Science and Technology (IJMS&T), and an Editorial Board Member of i-manager's Journal on
Engineering and Technology (IJET), Mathematics Applied in Science and Technology (MAST), International Journal of
Applied Engineering Research (IJAER), International Journal of Dynamics of Fluids (IJDF), and Far-East Journal of
Mathematics (FEJM).