in table 1. Тable 1. F irst n u m b er o f cip h er a. (m). F1. (kN). S eco n d n u m b ..... crane which consist of two weightless rods in joints AB and AC, ... Weight of a cable and of a block and dimension of a ..... two supports is in equilibrium under the action of given ..... Ball â and âSocket Joint in point A have three components.
«APPROVED» Scientific Council
Ministry of Education and Science of Ukraine
Institution of Civil Engineering
Protocol №3 (5.12.2018)
Odessa State Academy of Civil Engineering and Architecture Composers:
METHODICAL INSTRUCTIONS AND TASKS
Kozachenko T.O., Ph.D. in Physics and Maths, Associate Professor.
for realization of calculated-graphic works
from course of Theoretical Mechanics Section «STATICS»
Leshchenko D.D., D.Sc. in Physics and Maths, professor; Head of the Department of Theoretical Mechanics;
Reviewers: professor,
Orobey V.F., D.Sc. in Engineering, Head of the Department, Odessa National Polytechnic University; Rachinskaya A.L., Ph.D. in Physics and Maths, Associate Professor, Odessa I.I. Mechnikov National University
Оdesа 2019
Responsible for issue: Leshchenko D.D., D.Sc. in Physics and Maths, professor; Head of the Department of Theoretical Mechanics.
Тheme 4. Equilibrium of plane arbitrary system of forces……………………………………… 28 Тheme 5. Equilibrium of planar system of two bodies………………………………………… 37 Theme 6. Equilibrium of arbitrary system of forces……………………………………………… 45 Theme 7. Determination of position of centre of gravity of planar homogeneous figure……………. 53 References……………………………………………….. 59
2
Тable 1.
1 2 3 4 5 6 7 8 9 0
a F1 (m) (kN) 1,5 2 2,5 3 3,5 4 3,5 3 2,5 2
10 20 30 40 50 60 70 80 90 25
1 2 3 4 5 6 7 8 9 0
b F2 (m) (kN) 2 2,5 3 3,5 4 3,5 3 2,5 2 1,5
10 20 30 40 50 60 70 80 90 25 3
1 2 3 4 5 6 7 8 9 0
c F3 (m) (kN) 1 1,5 2 2,5 3 3,5 4 3,5 3 2,5
10 20 30 40 50 60 70 80 90 25
Number of problem
Тheme 3. Equilibrium of space system of forces whose lines of action intersect at point………………. 20
Third number of cipher
Theme 2. Equilibrium of planar system of forces, whose lines of action intersect at point………………. 12
Second number of cipher
Theme 1. Calculation of projections of forces and moments of forces about coordinate axes……….. 3
For solving problems of this theme it is necessary to study following sections and questions: Fundamental concepts of statics; Elementary operations and equivalent transformations over the vectors; Rules of projection of vector of force on the axis and on the plane; Moment of force about pole and its properties; Moment of force about coordinate axes and its properties; Varignon’s theorem. Problems In problems 0-9 you must find projections of forces F1 , F2 , F3 on the coordinate axes and moments of these forces about coordinate axes. Output data for all problems were cited in table 1. First number of cipher
TABLE OF CONTENTS
Theme 1. Calculation of projections of forces and moments of forces about coordinate axes.
1 2 3 4 5 6 7 8 9 0
Problem 0
Problem 2
Z
F2 0
F3
c
F1
F1
F3
Y
F2
a
b X
Problem 3
Problem 1
F1
F1
F3
4
F2
F2 F3
5
Problem 4
Problem 6
F2
F2
F3
F1
F3
F1
Problem 5
Problem 7
F1
F1
F3 F3
F2
6
F2
7
Problem 8
For defence problems of the theme it is necessary F1 F2
F3
Problem 9
І. Answer the questions: 1. To define aims and problems of statics. 2. To give definitions for basic concepts of statics (rigid body Force System of forces, couple). 3. What elementary operations and simplest equivalent transformation of systems do you know? 4. Give the definition of moment of force about pole and formulate its properties. 5. Formulate the Varignon’s theorem for two converged forces. 6. Give the definition of moment of forces about axes and formulate its properties. ІІ. To know how conduct: 1. Projection of vector of force on the coordinate axis (on the example). 2. Finding of the moment of force about axes (on the example). EXAMPLE. Forces F1 , F2 , F3 act at the rigid body has a shape parallelepiped (Figure 1).
F2
z
F1
F3
F2
F2
F3
c
x
8
F1
F2
F3'''
F3
a
b Figure 1. 9
F3
y
Define: 1) projections of forces on the coordinate axes; 2) moments of forces about coordinate axes; DATA: F1 = 5 kN, F2 = 10 kN, F3 = 30 kN, а = 1 m, b = 1 m, с=3m SOLUTION: 1. We define projections of every force on the coordinate axes. А) Force F1 lies on the edge of axes OY. Its projections are equal: X1 = 0 kN ; Y1 = F1 =5 kN Z1 = 0 kN. B) Force F2 lies on the bound of the body and parallel to the plane XOY. Write the angle between the line of action of force F2 and the axes OY as angle . We decompose force F2 on two components F2' and F2' : F2 = F2' + F2'' , for convenience of calculations. Where: F2' F2 cos and F2 F2 sin d1 b 2 a 2 12 12 1.41 (m)
cos b
1 0.71 ; sin a 1 0.71 . 1.41 d1 1.41 Projections of force F2 : X2= F2'' 10 0.71 7.1 kN; d1
Y2= F2' 10 0.71 7.1 kN;
Z2=0 kN. C) Line of action of force F3 coincide with diagonal of parallelepiped. Write the angles between its direction and the axes as , , . We decompose force F3 on three components: F3 = F3' + F3'' + F3''' .
cos a cos b
d2
1
3.32
d2
1
0.31 ;
0.31 ; cos c
Projections of the force F3 equal:
d2
3
3.32
0.91 .
X3 = F3' 30 0.31 9.3 kN; Y3 = F3''' 30 0.31 9.3 kN; Z3 = F3'' 30 0.91 27.3 kN. 2. We find moments of forces about coordinate axes. А) Force F1 : mx ( F1 ) = 0 (force F1 cross axes OX); my ( F1 ) = 0 (force F1 is parallel axis OY); mz ( F1 ) = F1 a= 5 1 5 kNm . B) For calculation of moments of force F2 we use Varignon’s theorem: mx ( F2 ) = mx ( F2' ) + mx ( F2'' )= F2' c 0 7.1 3 21.3 kNm; my ( F2 ) = my ( F2' ) + my ( F2'' ) = 0 + F2'' c = 7.1 3 21.3 kNm; mz ( F2 ) = mz ( F2' ) + mz ( F2'' ) = 0 F2'' b = 7.1 kNm. C) For calculation of moments of force F3 we also use Varignon’s theorem: mx ( F3 ) = F3'' b = 27.3 1 27.3 kNm ; my ( F3 ) = 0 (force F3 cross axes OY); ' mz ( F3 ) = F3 b = 9.3 1 9.3 kNm.
Where: F3' F3 cos ; F3'' F3 cos ; F3''' F3 cos ; d 2 a 2 b 2 c 2 12 12 32 3.32 ;
10
3.32
11
Theme 2. Equilibrium of planar system of forces, whose lines of action intersect at point. It is necessary to study for solving problems of this theme following sections and questions. Basic concepts of statics. Classification of system of forces. Projection of a force on an axis and on a plane. Constraints and its reactions. Analytical conditions of equilibrium of systems of forces.
Problem 0
β
A
C α
1
2
B
a) α=1050 β=450 b) α=1050 β=300
G
5 10 15 20 25 30 35 40 45 50
1 2 3 4 5 6 7 8 9 0 12
α and β
1 2 3 4 5 6 7 8 9 0
a) b) a) b) a) b) a) b) a) b)
Problem 1
Third number of cipher Number of problem
G (kN)
Second number of cipher Angels
First number of cipher
PROBLEMS We must find the stress in rods of freely pivotal construction which is in state of equilibrium under the action of force G . Output data for all problems were cited in table 2. Тable 2.
1 2 3 4 5 6 7 8 9 0
C a) α=1050 β=600 2
b) α=1200 β=300
α
B
β A
1
G
13
Problem 4
Problem 2
A
0
β C
1
2 α
B
a) α=30 β=450 0
a) α=300 β=600
C α
2
b) α=45 β=600
B 1
b) α=600 β=750
β A
G G
Problem 5
Problem 3 C a) α=300 β=600
2 A
b) α=600 β=750
α β
1
2 α
14
G
C
B
β G
a) α=300 β=300
B
1
A
15
b) α=450 β=450
Problem 6
Problem 8 B
B
0
1
α G 2
β A
α
a) α=30 β=300
2
b) α=600 β=150
G
1
b) α=600 β=300
β
C
A
Problem 7
Problem 9
B
A
α
a) α=750 β=450
1
2 G C
C
a) α=300 β=450
β
A
b) α=900 β=450
C
β 2
α
a) α=300 β=1200
1
b) α=600 β=1050
B G
16
17
For defence the problems of the theme it is necessary І. Answer the questions: 1. Whow we can formulate axioms of statics? 2. Whow look constraint Rod and what you can say about its reaction? 3. What elementary operations and simplest equivalent transformations of systems of forces do you know? 4. How we can formulate basic theorem of statics? 5. What equations of equilibrium are used to examination of equilibrium of planar convergent system of forces? ІІ. To be able: 1. To compose equations of equilibrium in solving problem if external loading change? EXAMPLE. Load weighted G = 2 kN (figure 2.1) retain by crane which consist of two weightless rods in joints AB and AC, attached to vertical wall. Angles between the wall and road are α1 = 60° and α2 = 40°. In point A block is suspended. Cable which go till block in point D is throwing over the block. Angle between the wall and a cable is α3 = 60°. Weight of a cable and of a block and dimension of a block are neglected. You may find stresses in the rods. SOLUTION. We consider a load being in equilibrium (figure 2.2). Force of gravity G and force of pull of cable N1 are acted in the load. Therefore the force of pull of a cable directed in the middle of a load and equals weight of a load Ν1 = G by modulus. If for any block (figure 2.3) neglect friction on it axes, forces of pull of a cable are equal, N1 N 2 . We can see it from equations of block.
18
Figure 2.2
Figure 2.1
Figure 2.3
Figure 2.4
Now as object of equilibrium we examine knot in point A. Forces of pull of a cable N1 and N 2 and reactions R1 and R2 of rods AB and АС are acted on a knot (figure 2.4). Reactions of rods directed along these rods. We direct it in the middle of rods. Rods are stretch at the beginning. We compose the equations of equilibrium as equations of proections of forces at the axes x and y (for convergent system of forces). Forces R1 , R2 , N 2 have angles α1, α2, α3 with axes y. YK N1 N 2 cos 3 R2 cos 2 R1 cos 1 0;
X
K
N 2 sin 3 R2 sin 2 R1 sin 1 0;
We consider N1 = N2 = 2 kN and obtain 1.5 R2 cos 2 R1 cos 1 0; 1.73 R2 sin 2 R1 sin 1 0. We solve this system of equations and find R1 = 0.611 kN, R2 = -3.52 kN. Sign «» indicate that rod AC is compressed. Answer: R1 = 0.611 kN, R2 = -3.52 kN.
19
Тheme 3. Equilibrium of space system of forces whose lines of action intersect at point. It is necessary to study following sections and questions for solving problems of this theme. Resultant of system of forces. Moment of system of forces statics. Axioms of statics. Constraints and its reactions. Basic lemma of statics. Basic theorem of statics. Analytical conditions of equibrium of systems of forces.
1 2 3 4 5 6 7 8 9 0
30 60 45 30 45 45 30 60 60 30
60 45 30 30 60 60 75 30 60 45
45 30 60 60 30 45 45 30 75 60
1 2 3 4 5 6 7 8 9 0
30 60 45 60 90 45 90 60 30 90
60 30 60 45 60 45 45 60 90 30 20
90 90 60 60 30 90 45 45 60 60
1 2 3 4 5 6 7 8 9 0
P Q (kN) (kN) 100 120 130 140 150 160 170 180 190 200
60 80 100 120 130 140 150 160 170 180
P Q
Problem 1
Number of problem
Third number of cipher
Second number of cipher
First number of cipher
PROBLEMS We must find the stress in rods of freely pivotal construction which is in the state of equilibrium under the action of forces P and Q . Output data for all problems were cited in table 3. Axes x, y, z are mutually, perpendicular, x'‖x, y'‖ y, z'‖z. Тable 3.
Problem 0
1 2 3 4 5 6 7 8 9 0 21
Problem 2
Problem 4
Q
P
Problem 3
22
Problem 5
23
Problem 6
Problem 8 Y
Z
φ P
X
А
Y
X
О
В
Problem 7 Problem 9
P
Q
VectorP P lies in plane Oyz 24
25
С
DATA: Q = 60 kN, P = 80 kN, = 450 , = 600 , = 600. SOLUTION. Reactions of the rods N1 , N 2 , N 3 , directed along the rods because all rods fixed by joints and their weights neglected. We assume that all rods of frame are strecked. We introduce coordinate system and compose for our space convergent system of forces three equation of equilibrium:
For defence problems of the theme it is necessary I. Answer the questions: 1. Whow we can formulate axioms of statics? 2. Whow look constraint Rod and what you can say about its reaction? 3. What elementary operations and simplest equivalent transfornations of systems of forces do you know? 4. How we can formulate basic lemma of statics? 5. What equations of equilibrium are used to examination of equilibrium of space convergent system of forces? EXAMPLE. We consider that construction being in equilibrium. We neglect the weight of construction. You may find stresses in the rods when
z P
A
Q
N1 N2
1
X Y Z
K
K
K
N 1 cos N 3 cos 0; N 2 cos P 0;
(2)
N 1 sin N 2 sin N 3 sin Q 0.
(3)
We substitute numerical values and we find from the second equation N2 = 200 kN. From the system of equations (1) and (3) we find N1 = 246.67 kN;
3
N3 D
γ
2 β C
α x
y
B Figure 3. 26
(1)
27
N3 = 350.27 kN.
Тheme 4. Equilibrium of plane arbitrary system of forces.
Problem 0
It is necessary to study following sections and questions for solving problems of this theme: Resultant of system of forces. Moment of couple and its properties. Relation between the moments of a force about a point and an axes. Generalized Varignon’s theorem. Principle of release from constraints. Conditions of equilibrium of planar system of forces. PROBLEMS In problems 0-9 we must find constraint forces covered at the construction being in equilibrium. Numerical data were cited in table 4.
a (m)
b (m)
c (m)
Second number of cipher
h (m)
P1 (kH)
P2 (kH)
Third number of cipher
M (kHm)
q (kH /m)
1 2 3 4 5 6 7 8 9 0
4 5 4 6 4 5 4 4 5 5
3 4 3 2 2 3 2 3 3 4
1 2 3 1 2 3 1 2 3 2
1 2 3 4 5 6 7 8 9 0
7 5 7 6 5 7 5 6 5 7
20 40 50 60 70 80 90 80 70 60
90 80 70 60 50 40 30 20 30 40
1 2 3 4 5 6 7 8 9 0
12 14 16 18 20 22 20 18 16 14
2 3 4 5 6 2 3 4 5 6
30 45 60 30 45 60 30 45 60 30
28
Number of problem
First number of cipher
Тable 4.
1 2 3 4 5 6 7 8 9 0
Problem 1
29
Problem 4 Problem 2
P2 P1
Problem 5 Problem 3
P1 P2
P1
P2
30
31
Problem 6 Problem 8
P1 P2
Problem 7
Problem 9
P1
P1
A
q
P1
P2
M B
60o
c a
P2 h
b
P2
α
32
33
For defence problems of the theme it is necessary I. Answer the questions: 1. How formulate principle of release from constraints? 2. How look constraint Cylindrical Pin and its reaction? 3. How look constraint Ball-and Socked Joint and its reaction? 4. How look constraint Stopping up? 5. Where apply resultant of uniformly distributed loading and how define its quantity? 6. How formulate the basic theorem of statics? 7. What equations of equilibrium are used to examination of equilibrium of planar system of forces? II. To be able: 1. To compose equations of equilibrium in solving problem if external loading change? EXAMPLE. Planar rigid weightless construction which have two supports is in equilibrium under the action of given loading. We must define reactions for imposed constraints at the construction. R y B . q α α P α В Q M
F 2) F 3) m 1)
kx
0; X A P sin RB cos 0;
0; Y A P cos RB sin Q 0; z A ( Fk ) 0;
ky
c a
h
YA
DATA: P= 8 kN, М= 5 kNm , q= 3 kN /m, = 300, а= 4 m, b= 3 m, c= 2 m, h= 5 m. SOLUTION. 1. A system of arbitrary situated in space external loadings acts on the beam: concentrated force - P1 , moment of couple – М and uniformly distributed loading of intensity q. 2. Beam has two constraints: freely fixed constraint in point A and rod in point B. According to principle of release from constraints we substitute them by reaction. We substitute freely fixed constraint by reaction which consist from two components X A (horizontal) and YA (vertical). We substitute action of rod by reaction RB which apply in point B and have direction along the rod. 3. We substitute uniformly distributed load of intensity q by resultant of system of forces Q , Q q c 3·2=6 kN, where Q applied in the middle of interval c. 4. We take point A as origin of coordinates and conduct axes x and y (you can see figure 4). 5. We compose three equations of equilibrium for planar arbitrary system of forces
b
P sin h P cos (a b) RB sin a RB cos h M Q
XA
6. We define reaction of rod RB from equation (3): x
А Figure 4. 34
35
c 0; 2
RB
36
PROBLEMS
c (m)
Second number of cipher
P1 (kN)
P2 (kN)
Third number of cipher
M (kN m)
q (kN /m)
In problems 0-9 we must find reactions of external constraints, stress in joint C (internal constraint) for represented systems of two bodies, being in equilibrium. Data were cited in table 5. Тable 5.
b (m)
Then R A X A2 YA2 =24.44 (kN) 7. We conduct control of correctness of definition of reactions. We choose axes z perpendicular to plane of forces in point B and we compose the sum means that conditions of equilibrium are fulfilled and accuracy of solution was verified. c m zB ( Fk ) YA a P cos b M Q(a 2 ) X A h 2 19.29 4 8 3 cos 30 0 5 6(4 ) 15.01 5 0 2 Answer RA=24.44 kN RB=-12.65 kN. Sign «» indicate that RB has direction which opposite to represented on figure.
a (m)
YA P cos RB sin Q 19.29 (kN )
First number of cipher
We define YA from equation (2)
It is necessary to study following sections and questions for solving problems of this theme: Common property of equivalence of two system of forces. Lemma about parallel transfer of a force. Theorem about reduction of a force system to a given centre. Conditions of equilibrium of system of connected bodies. Internal and external forces, constraints and reactions.
1 2 3 4 5 6 7 8 9 0
4 3 2 1 4 3 2 1 2 3
7 5 7 6 5 7 5 6 5 7
1 2 3 1 2 3 1 2 3 2
1 2 3 4 5 6 7 8 9 0
20 40 50 60 70 80 90 80 70 60
90 80 70 60 50 40 30 20 30 40
1 2 3 4 5 6 7 8 9 0
12 14 16 18 20 22 20 18 16 14
2 3 4 5 6 2 3 4 5 6
30 45 60 120 135 150 30 45 60 120
37
Number of problem
1 c P(a b) cos Ph sin M Q (a sin h cos ) 2 12.65 (kN ) We define X A from equation (1) X A P sin RB cos 15.01 (kN )
Тheme 5. Equilibrium of planar system of two bodies
1 2 3 4 5 6 7 8 9 0
Problem 0
Problem 3
P2 P1
P1
Problem 1
P2
P1
q
Problem 4
C
P2
D
E
α
a b
b A
Problem 2
P2
P1
38
P1
c
c
M
Problem 5
39
P2
Problem 8 Problem 6
P1 P2
Problem 9
P2
Problem 7
P1 a b
b
q
C
B c
For defence problems of the theme it is necessary
M c
α
P2
a
A
40
P1
a D
b
I. Answer the questions: 1. How formulate common property of equivalence of two system of forces? 2. How formulate Lemma about parallel transfer of force? 3. How formulate a theorem about reduction of a force system to a given centre (theorem Poinsot)? 4.What reaction form in internal constraint-intermediate joint? 5. How we can conduct static control of correctness of the problem solution? II To be able: 1. To compose equations of equilibrium in solving problem if external loading change? 41
EXAMPLE. Construction consist of two inflexible (rigid) bodies which connected freely to each other in point C. Constraint A -inflexible fix (stopping up). and joint-mobile constraints imposed on construction. Point forces P1 , P2 ; uniformly distributed loading of intensity q , couple with moment M are doing on the construction. We must find reactions of external and internal constraints. a)
A
P2
C
b
P1
M
q
h
B
α c a
Y
C XC
b)
l
YC
P1
h
q
α c) YA A
МА
XA b
P2
C M
RB Q
X С' YС'
42
а = 4 m, b=1 m, c=3 m, h=2 m, l=6 m, 60 0 . SOLUTION. 1. We substitute uniformly distributed loading of intensity q by resultant of system of forces Q , Q q c 12 kN. Force Q applied in the middle of interval c. 2. According to principle of release from constraints we substitute them by reactions. Action of inflexible closing is equivalent to force decomposed into components X A , YA and a couple with moment M A (reactive moment). Action of joint-mobile constraint we substitute reaction RB . Direction of reactions of constraints we choose arbitrarily. For planar system of forces { Q , P1 , P2 , M , X A , YA , M A RB }, under the action of which construction is in equilibrium we can construct only three equations of equilibrium. This is insufficient for finding of fourth unknown external reactions. It is necessary to find also reactions of connective joint. We must divide construction into pasts and we consider equilibrium of it separately. 3. We consider equilibrium of part CB (figure 5b). We construct three equations of equilibrium for system of forces { Q , P1 , X C , YC , RB }:
X Y m
B
c
K
X C P1 sin 60 0 0 ;
(1)
YC RB Q P1 cos 60 0 0 ; (2) c 0 F (3) K R B l Q (l ) P1 sin 60 h 0 . Z (C ) 2 We substitute in equations numerical values of given quantities and we solve system of equations (1)-(3). We define reaction of external constraint RB and reactions X C , YC connective joint C: RВ = 14.77 kN; XС =- 17.32 kN; YС = -7.23 kN. 43 K
X
Figure 5.
DATA: М = 3 kNm, P1 = 20 kN; P2 =10 kN, q = 4 kN/m ,
44
1 2 3 1 2 3 1 2 3 2
1 2 3 4 5 6 7 8 9 0
20 90 40 80 50 70 60 60 70 50 80 40 90 30 80 20 70 30 60 40 45
1 2 3 4 5 6 7 8 9 0
10 15 20 25 30 10 15 20 25 30
30 45 60 30 45 60 30 45 60 45
Number of problem
7 5 7 6 5 7 5 6 5 7
F або Q (kN)
4 3 2 1 4 3 2 1 2 3
Third number of cipher
1 2 3 4 5 6 7 8 9 0
P2 (kN)
P1 (kN)
We substitute in equations numerical values of given quantities and we solve system of equations (4)-(6). We define reaction of external constraints: МА= 35.92 kNm; XА = 17.32 kN; YА = 17.23 kN. Sign «» indicate that component of reaction X С has direction which opposite to reprented on figure. We conduct control of correctness of definition of reactions. We choose axes Z perpendicular to plane of forces in point B and we compose the sum of moments: c m Z ( В ) FK Q 2 P2 (l a b) P1 l cos 60 0 M M A X A h Y A (a l ) 206.94 206.94 0 Obtained sum «0» means that conditions of equilibrium are fulfilled and accuracy of solution was verified.
Second number of cipher
(6)
c (m)
(5)
b (m)
YA YC' P2 0 ; m F Z ( A) K M A YC' a M P2 b 0 . K
It is necessary to study following sections and questions for solving problems of their theme: Constraints and its reactions (Cylindrical Pin, Ball – and – Socket Joint, Step Bearing and so on). Centre of parallel forces. Centre of gravity. Centres of gravity of homogeneous bodies. Centres of gravity of homogeneous symmetric bodies. Basic methods of definition of centre of gravity of a body. Analytical conditions of equilibrium of arbitrary system of forces PROBLEMS Equilibrium of combine construction is considered in problems 0-2. It consists of two rectangular homogeneous plates weighing P1 and weighing P2 . Equilibrium of rectangular homogeneous plate weighing P1 is considered in problems 3-9. We must find reactions of constraints acts on the construction. Тable 6.
a (m)
Y
Theme 6. Equilibrium of arbitrary system of forces.
First number of cipher
4. We consider equilibrium of part AC (figure 5c). We consider that if equal action and reaction X C X C' , YC YC' and X C X C' , YC YC' numerically. For planar system of forces { P2 , M , M A , X A , YA , X C' , YC' } we construct three equations of equilibrium: (4) X K X C' X A 0 ;
1 2 3 4 5 6 7 8 9 0
Problem 2 Problem 0
Problem 1
Problem 3
46
47
Problem 4 Q
Problem 6
Problem 7
Problem 5
Q
48
Problem 8
49
Q
Problem 9
z
Q
D
1
2
3 6
c
4 5
ZA
b
XA
a
YA
A
P
For defence problems of the theme it is necessary I. Answer the questions: 1. How we can define centre of gravity of homogeneous symmetric bodies? 2. What equations of equilibrium we use to solve this problem? 3. How look constraint cylindrical pin? What reaction arise in this case? 4. How look constraint Ball-and-Socket-Joint? What reaction arise in this case? II. To be able: 1. To compose equations of equilibrium in solving problem if external loading change.
EXAMPLE. Homogeneous rectangular plate weighing P is appended to the wall by Ball-and – Socket – Joint in point A and cylindrical pin in point B and retain in horizontal position by weightless supporting rod with joints at the ends in point C. We must find reactions in covered constraints at the construction. 50
В
NC
300
XB
С
x
y a
300
b
ZB
y
Figure 6.
DATA: a 3 m, b 2 m, P 40 kN SOLUTION. 1. We consider equilibrium of construction. We draw coordinate axes and represent given loadings and reactions of covered constraints which act at the construction. We consider that force of weight P apply at the centre of plate. Reaction of Ball – and –Socket Joint in point A have three components X A , YA , Z A . Reaction of point B have two components X B , Z B , which lie in plane perpendicular axes of joint. Reaction N C of supporting rod we direct along the rod.
51
m F Z K
X
B
b P
a m F P 2 N Y
K
C
b N C sin 30 0 b 0; 2
(4)
a sin 30 0;
(5)
m Z FK X B b 0.
(6)
We substitute in equations numerical values of given quantities and we solve system of equations (1)-(6) and define reactions. We conduct control of correctness of definition of reactions. We choose axes Y and we compose the sum of moments: 1 mY FK Z B a Z A a P 2 a a (0 20 20) 0 . Obtained sum «0» means that conditions of equilibrium are fulfilled and accuracy of solution was verified. Answer: X A 17.4 kN, YA 30.28 kN, Z A 20 kN, X B = 0 kN, Z B 0 kN, N C = 40 kN.
52
PROBLEMS We must find coordinate of the centre of gravity of planar homogeneous figures of complex form (problems 0-9). Figures consist of from combination of simple parts – rectangle, rectangular triangle, parabolic segment. Necessary data you must take from tables 7 and 8. Таble 7. Number of problem
(3)
d (cm)
Z A Z B N C sin 30 0 P 0;
Third number of cipher
(2)
c (cm)
Y A N c cos 30 0 cos 30 0 0;
It is necessary to study following sections of this theme: Centre of parallel forces and its radius vector. Coordinates of the centre of parallel forces. Centre of gravity. Centres of gravity of homogeneous bodies: of the volume, of the area, of the line. Centres of gravity of homogeneous symmetric bodies. Methods of determining the coordinates of the centre of gravity of bodies.
b (cm)
K
(1)
Second number of cipher
K
X В X A N C cos 30 0 cos 60 0 0;
h (cm)
K
a (cm)
X Y Z
Theme 7. Determination of position of centre of gravity of planar homogeneous figure.
First number of cipher
2. We compose six equations of equilibrium for arbitrary space system of forces { P , X A , YA , Z A YB , Z B N C }.
1 2 3 4 5 6 7 8 9 0
10 12 14 16 18 20 18 16 14 12
8 10 12 14 12 10 14 8 10 6
1 2 3 4 5 6 7 8 9 0
21 18 15 15 12 15 12 15 18 18
18 15 12 9 6 9 9 15 18 12
1 2 3 4 5 6 7 8 9 0
5 6 7 8 9 5 6 7 8 9
1 2 3 4 5 6 7 8 9 0
53
rectangle
triangle
segment of parabola
Problem 0
Problem 1
Spaces and position of centres of gravity of homogeneous figures (in all figures is accepted: l - length, h – height).
Figure
Таble 8. Position of the centre of gravity х1 х2
Space S
Rectangle 1
Problem 2
l h
1 l 2
1 l 2
1 l h 2
1 l 3
2 l 3
Problem 3
х1 х2 Тriangle 2
х1 х2 Segment of parabola
Problem 4
2 l d 3
3
3 d 5
х1 х2
Problem 5
2 d 5
Problem 6 54
Problem 7 55
superposition of square segment 4 (which have conditionally negative 1 and two triangles 2, 3 (which have positive spaces). y
Problem 8
3
1
Problem 9
b
CC 2
For defence problems of the theme it is necessary I. Answer the questions: 1. How you can define centre of gravity of homogeneous volume body? 2. How you can define centre of gravity of homogeneous thin plate? 3. How you can define centre of gravity of homogeneous thin rod? 4. How you can define centre of gravity of homogeneous symmetric bodies? 5. What methods of definition of centre of gravity are exist? II. To be able: 1. To define centre of gravity of simple figures. 2. To define centre of gravity of complex homogeneous bodies. 3. Definition of centre of gravity of figures conduct with the help of: 1) method of symmetry, 2) method of livision 3) method of supplementation EXAMPLE. You must find coordinates of centre of gravity of figure of complex form with cut. DATA: a 6 сm, b 5 cm, с 3 cm, d 2 cm, h 3 cm . SOLUTION. We use combination of method of division and method of supplementation for solution of a problem. We represent that given complex figure is obtained by method of 56
h d
4
x c
a
c Figure 7.
1. Lower left angle of given figure we select as origin of general system of coordinate. We find spaces S i and
coordinates X i , Yi of centres of gravity Ci of each i – part of figure. А. Rectangle (part 1). Centre of gravity of rectangle point С1 is point of intersection of diagonals. Space of rectangle equals S1 a b 6 5 30 cm2. Coordinates of point С1 in local coordinate system (with origin in lower left angle of rectangle): 1 1 X 1* a 3 cm ; Y1* b 2,5 cm; 2 2 Coordinates of point С1 in general system of coordinate: X 1 c X 1* 3 3 6 cm; Y1 .Y1* 2,5 cm; B. Triangle (part 2). Centre of gravity of triangle – point С2 of intersection of median. Space of triangle equals 1 S 2 c h 4,5 cm2 2 57
Coordinates of point С2 in local system of coordinate (with origin) in lower left angle of triangle: 2 1 X 2* c 2 cm; Y2* h 1 cm; 3 3 Coordinates of point С2 in general system of coordinate: X 2 X 2* 2 cm; Y2 Y2* 1 cm; C. Triangle (part 3). Centre of gravity of triangle – point С3 of intersection of median. Space of triangle equals: 1 S 3 c b 7,5 cm2 2 Coordinates of point С3 in local system of coordinate (with origin in lower left angle of triangle): 1 1 X 3* c 1 cm; Y3* b 1,67 cm; 3 3 Coordinates of point С3 in general system of coordinate: X 3 c a X 3* 10 cm; Y3 Y3* 1,67 cm; D. Segment of parabola (part 4). Centre of gravity of 2 segment – particle С4. Space segment: S 4 a d 8 cm2. 3 Coordinates of particle С4 in local system of coordinate: 1 2 X 4* a 3 cm ; Y4* d 0,8 cm; 2 5 Coordinates of particle С4 in general system of coordinate: X 4 c X 4* 6 cm; Y4 Y4* 0.8 cm; Segment of parabola is a cut in a given figure and space of segment occur with sign minus into formula of general space and coordinates of centre of gravity of complex figure. 2. We find general space of the whole figure:
We find coordinates of centre of gravity of figure by formula:
X S S
X 1 S1 X 2 S 2 X 3 S 3 X 4 S 4 S K 6 30 2 4.5 10 7.5 6 8 6.35 ( cm ); 34 YK S K Y1 S1 Y2 S 2 Y3 S 3 Y4 S 4 YC S SK XC
K
K
2.5 30 1 4.5 1.67 7.5 0.8 8 2.52 (cm). 34
Answer: centre of gravity of given Figure (particle С have coordinates X C 6.35 cm, YC 2.52 cm in selected system of coordinate.
REFERENCES 1. Targ S. Theoretical Mechanics. A Short Course. M: Mir Piblishors, 1976. 2. Pavlovsky M. A. Theoretical Mechanics. – K.: Technika, 2004. 3. Yablonsky A.A., Nikiforova V. A. Course of Theoretical Mechanics. Part 1. – Spb.: Lan, 2001. 4. Bat M. I., Dzhanelidze G. Yu, Kelzon A. S. Theoretical mechanics in examples and problems. Part 1. – Spb: Lan, 2013. 5. Kirilov V. Kh., Leshchenko D. D. Course of Theoretical Mechanics. – Odessa: Astroprint, 2001. 6. Balduk P. G. Statics. Textbook for solving problems. – Odessa, Abzats, 1998.
4
S S K S1 S 2 S 3 S 4 30 4,5 7,5 8 34 cm2 K 1
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