Engineering mechanics - جامعة ذي قارeng.utq.edu.iq/images/lectures/Engineeringmechanicwasan.pdf · 4 F1 F2 F3 F4 Fig. (a) b. Parallel ... U is the process of replacing a

Engineering mechanics Lecturer: Wasan Q. Fayyadh

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UChapter 1

U1.1 Mechanics:-

Is that branch of physical science which concerned with the state of rest or motion of bodies that are subjected to the action of forces. In engineering mechanics attention is directed primarily to the external effects of a system of forces acting on a rigid body.

U1.2 Forces:-

Is the action of one body on another. A force tends to move a body in the direction of its action. The action of a force is characterized by:

- Magnitude. - Direction - Point of application.

Thus force is a vector quantity. U1.3 Scalars and vectors: Most of the physical quantities in mechanics can be expressed mathematically by means of scalars and vectors. UScalar: A quantity characterized by a positive or negative number is called a scalar. For example, mass, volume, and length are scalar quantities. UVector: A vector is quantity that has both a magnitude and direction. A vector is represented graphically by an arrow, which is used to define its magnitude, direction, and sense. The magnitude of the vector is the length of the arrow, the direction is defined by the angle between reference axis and the sense is indicated by the arrowhead. For


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example, the vector A shown in the Fig (1) has a magnitude of 4 units, a direction which is 20° measured counterclockwise from the horizontal axis, and sense which is upward and to the right.

head

Line of action

20

o Fig (1)

U1.4 Types of vectors: There are three types of vector: *fixed vectors Fixed vectors, in addition of their magnitude and direction, also have a point of application (see Figure 1) Example: a force on a deformable body.

y

x

z

A

(Figure 1) fixed vector with point of application A

*sliding vectors Sometimes the location of the point of application is of no importance and may be moved in the direction of the vector. This is called a sliding vector. Sliding vectors do not have a fixed point of action, but have only a line of action (see Figure 2). Example: the force on a rigid body.


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y

x

z

L

(Figure 2) sliding vector with line of action L

*free vectors When the place of the line of action of a vector is not important either, one refers to a free vector. Example: the translation of a rigid body. All points of the body are subject to the same displacement. The free vector stands for the entire collection of displacement vectors (see Figure 3).

(Figure 3) Free vector

U1.5 force system: Are a number of forces acting in a given situation and can be classified according to the arrangement of the lines of action of the forces of the system. Types of force system:-

1. Coplanar force system: - system consisting of several forces the lines of action of which all lie in one plane.

a. A concurrent coplanar force system: - is shown in Fig. (a). this system consists of several forces the lines of action of which all intersect at a.


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F1F2

F3

F4

Fig. (a)

b. Parallel coplanar force system:- consists of several forces the lines of action of which are all parallel, as shown in Fig. (b).

F1 F2 F3

Fig. (b)

c. General coplanar force system: - consists of several forces the lines of action of which are in various directions and do not intersect in a common point, as shown in Fig. (c).

F1F2

F3

F6

F5

F4

Fig. (c)

d. Couple: - consists of two equal and opposite parallel forces that do not have a common line of action, as shown in Fig. (d).


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a

F

F

Fig. (d)

2. Non coplanar force system: - systems consisting of several forces the lines of action of which lie in two planes, as shown in Fig. (E).

F1

F2

F3y

x

z

Fig. (E)

U1.6 composition & resolution of forces:-

A force is a vector quantity since it has a specified magnitude, direction, and sense and it adds according to the parallelogram law. Two common problems in statics involve either finding the resultant force, knowing its components, or resolving a known force into two components.

UComposition:- U is the process of replacing a force system by its resultant.

UResolution:- U is the process of replacing a force by its components.

1. Resolving a force in two given directions analytically

In a plane, we can resolve a force F into components with given lines of action.


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a. Resolving a force along rectangular components. i. In plane force (two dimensional)

For most purpose, rectangular components of a force are more useful.

θ

Fcosθ

Fsinθ

A

B

O X

Y

F

Fx=OA=FcosƟ

Fy=OB=FsinƟ

UNote:- U the two components of a force must intersect on the action line of the force. In other words OA & OB are noting the components of (F).

Ex1:

Resolve the force (F=1000 N) with α=30° into components in the x & y direction.

α=30°

F=1000 N

X

Y


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Ex2:-

A force F perpendicular to a sloping roof has to be resolved into the components Fx & Fy, parallel to the x & y axis. Respectively.

Determine these components.

30°

F=500 N

X

Y

ii. For a force in space(3 dimensional)

The resolution is as following

Fx=OA=FcosƟx

Fy=OB=FcosƟy

Fz=OC=FcosƟz

- The angles Ɵx, Ɵy, Ɵz are the angles between the components force & the positive coordinate axis.

- The cosines of the angles Ɵx, Ɵy, Ɵz are called direction cosines.


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- Note that the three components always intersect at a point on the line of action of the resultant force.

θx

θy

θz x

y

z

F

Fx

Fy

Fz

A

B

C

O

Ex1:

Determine a set of three rectangular components of the 170 KN force of the figure below.

170 KN

X

Y

8 m

9 m

12 mZ

0

b- Resolving a force along non-rectangular components.

ii. Using law of sine.

Procedure for analysis:-


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In a plane, we can resolve a force F into two components with given lines of action by using the parallelogram law.

UParallelogram law:-

- Make a sketch showing the vector addition using the parallelogram law.

- To resolve a force into components along two axes directed from the tail of the force, then start at the head of the force & construct lines parallel to the axes, thereby forming the parallelogram. The sides of the parallelogram represent the components.

- Label all the known and unknown force magnitudes & the angles on the sketch & identify the two unknowns.

- The magnitude of two force components is determined from the law of sine.

A B

C

ab

c

Sine law:

A/sin a = B/sin b=C/sin c


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Ex1:- A console is subject to a force F=1000 N. In addition α=β=30°. Resolve these forces into components in respectively:-

a- The a & b direction.

b- The c & d direction.

c- The a & c direction.

d- The b & c direction.

α

βa

b

c

d

F=1000 N


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EX:-2

Resolve the 170kN. Force of the Fig. below into two components, one along AB & the other parallel to CD.

12

5

815

43

170 KN

A

B

C

D

ii. Resolving a force in two given directions graphically

A force F can be resolved graphically into two components Fa&Fb with given lines of action a & b using the parallelogram rule in figure (a) or the force polygon in figure (b). The graphical approach has the advantage that you can at once see in which directions the components Fa&Fb .


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p

FFb

Fa

a

b

A

B

F// a

// b

Fa

Fb

// b// a

A

Fb

F

FaB

(a) (b)

Resolving a force in two given directions graphically using (a) the parallelogram rule & (b) a force polygon.

Ex.1

A force F=30 KN acts on the trestle in the figure below. This has to be resolved into the components Fa&Fb with lines of action a & b.

b

a

30 kN

15

60

80 40

Solution

In the figure below the force has been resolved using a force polygon. The force scale is 1 cm= 5 KN. By measuring, you find (in the force polygon)Fa&Fb have lengths 5.7 cm & 6.3 cm respectively, so that

60

55

65

Fa

15

30KN

Fb


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Fig(3) the forces drawn to scale in a force polygon

Fa≈5.7x5 KN=28.5 KN

Fb≈6.3x5 KN=31.5 KN

In figure below forces Fa&Fb are shown as they act on the trestle. It is clear that force Fa is pulling on the beam while force Fb is pushing against it. Together, they exert the same load as the single force F.

≈28.5 KN

≈31.5 KN

Check:

The magnitude of Fa&Fb can also be calculated from the force triangle by using the sine rule (see figure 3)

60

55

65

Fa

15

30KN

Fb

Fa\sin 55=Fb\sin 65=F\sin60

So that

Fa=F(sin55\sin60)=28.4 KN

Fb=F(sin65\sin60)=31.4 KN

The values measured in figure (3) correspond closely to the calculated values.


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1.7 Umoment of a force:

A force applied to a body tends to cause the body to translate in the direction of the force. Depending on the point of application of the force on the body. The force may also tend to cause the body to rotate. This tendency to produce rotation is called the moment of the force.

- The magnitude of moment is equal to the product of the magnitude of the force & the perpendicular distance from the line of action of the force to the point consideration.

MoMo=F*r

O90

r

F

- The moment M of a force F about a point O is thus simply Mo=F*r, where r is the perpendicular distance from the line of action F to the point O. R is often called the moment arm of the force.

- A moment has the units of force times distance (N.m) or (KN.m).

- If the force does not lie in a plane perpendicular to the moment axis, thus the force may be resolved into two components one parallel to the axis & the other is a plane perpendicular to the axis.

- The moment of (F) about the point A may be found by resolving it into two components.


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The moment of the components (F1) about point A =F1*d1.

The moment of the components (F2) about point A=0(F2//AB).

- A force // to an axis has no moment with respect to the axis.

I

E

J

AB

CG

dF

F1

F2

- Moment has magnitude as well as direction, as can be seen by considering two cases of a wheel where the same force F is applied to the rime at a distance d as shown in the Fig(a)&(b). in the first case the wheel rotates in the clockwise direction while in the second case it rotates in the anticlockwise direction. In other words the moment M not only has a magnitude of F.d but also has a direction. In other words anticlockwise rotation is positive.

ddo o

FF

(a) (b)


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- UPrinciple of moments of forces:- - Moments due to multiple forces: the total rotational effect

produced by several forces about the same point or line is merally the algebraic sum of their in dvidual moments about that point or line. Thus

Mo

F1

F2

d1

d2o

Mo=(F1*d1)+(F2*d2)+……+(Fn*dn)

- UPrinciple of transmissibility:-

Force F has the properties of a sliding vector & can therefore act at any point along its line of action & still create the same moment about point o.

F

A

B

C

O

MO

X

Y

Z

rA

rBrC


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UEX1:-

For each case illustrated in the figure below, determine the moment of the force about point o.

100N

2m

O

50N2m

O0.75m

60N

O

1 sin45 m451m

3040N

2m

4m 2 cos30 m

O


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2m

70N

1m

4m

O

UEX2:-

A 200N force acts on the bracket shown in the figure, determine the moment of the force about A.

F=200N

45

A

B100mm

100mm 100mm


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UEX3:- Determine the moment of the forces about O.

50 N

60 N

20 N

40 N

30

2 m2 m

3 m

O

UEX4:- Determine the moment of the force with respect to point A for the system shown in the figure below. The (100 N) force is vertical & passes through the center of the rectangle.

100 N

43

A

6 m

3 m


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1.8 UCouples:-

Couples are special cases of force system which have zero resultant force but anon-zero moment. A couple is defined as a pair of equal & opposite parallel forces at a distance of d. figure (a) shows such a system. An important property of a couple is that a part from the fact that the resultant force is zero, the magnitude of the moment of the couple about any axis is a constant & is equal to Fd.

F

F

d/2

d/2B

a

A

MB=F(d/2)+ F(d/2)=F.d

MA=F.a-F(d+a)=-F.d

UNote:-

The magnitude of the couple is independent of the distance which locates the forces with respect to the moment center o. It follows that the moment of a couple has the same value for all moment center.


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The properties of a couple are:-

- The magnitude of the moment of the couple. - The sense of rotation of the couple. - The moment of the couple is the algebraic sum of the moments of

its forces about any axis perpendicular to the plane of the couple.

UNote:-U the moment of the couple is the same for all axes perpendicular to the plane.

- A couple is avector quantity having both magnitude (of moment) & direction (aspect of plane & sense of rotation).

UTransformation of a couple:-

Transformations of a couple are operations on the couple that do not change any of its characteristics.

- The properties of a couple are not changed if:- 1- The couple is rotated in its plane.

x

y

x

y

F1

F2

F1F2

d

d


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2- The couple is moved to a parallel position in its plane.

x

y

x

y

A AB BC C

F1 F1F2 F2

d d

3- The couple is moved to a parallel plane.

10

10d

10

10

4- If the distance between the forces of the couple changed the magnitude of the forces are changed, provided the moment remains the same.

x

y

x

y

A AB C

C

510

10

105

5

M=F.d F=M/d=50/10=5


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EX1:-

If the couple moment has a magnitude of 220 N, determine the magnitude F of the couple forces.

5m

3m

1m

4m

Y

X

A

B

F

-F

0

34

34

EX2:-

By using the transformations of a couple, replace the four couples of the figure below by one couple with the forces acting vertically at A & B.

2.5

2

32

424

4

2A B

50 N

30 N

70 N

60 N60 N

50 N

70 N

30 N


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EX3:-

Determine the moment of the couple with respect to

a) Point A

b) Point B

c) Point C

150 N

150 N

6m

3m

4m

A

B

C

EX4:-

Replace the 80 N force by a force through C & a couple whose forces act horizontally at A & B.

80 N

5m

5m

6m

4m

3m 3m

A

B

C


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UResolution of A force into A force & A couple.

A force may be replaced by an equal parallel force through any other point of the body & a couple. This can be done by adding two equal collinear forces of opposite sense to a force system on a rigid body.

UEx1

Replace the 60 N force shown in the fig. below by a force through (D) & a couple where forces act vertically through A & B.

5m

6m4m

5m

60 N

A

D

UEX 2U

By using the transformations of a couple replace the three couples of the fig. below by one couple with the forces acting horizontally at A & B.

100 N

200 N

50 N

100 N

200 N

50 N

3m

2m

4mA

B


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UChapter two

URESULTANT:-

A resultant of a force system is defined as the simplest force system which can replace the original system without changing its external effect on rigid body. When the resultant is zero, then the body is in equilibrium & the original force system in this case called a balanced.

2.1 Uresultant of a concurrent, coplanar force system:

The resultant of a concurrent force system is a single force passing through the point of concurrence.

Procedure for solution:-

To find the resultant of a force system shown in the figure below, we must follow the following procedure:

F1y

F3y

F2y

F1x F2x F3x

F1F2

F3

x

y

1- Each force in the force system is resolved into a pair of rectangular components.

2- Find the components of the resultant in (x)& (y) direction (Rx&Ry) which are equal to the algebraic sum of the x&y components of the forces, respectively Rx=∑Fx & Ry=∑Fy


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Rx&Ry: the components of the resultant in the x&y direction respectively.

∑Fx & ∑Fy: the algebraic sum of the x&y component of the force system.

3- Find the magnitude of the resultant :

4- Find the slope of the resultant

Where is the angle between the resultant & the x-axis, & the sense can be determined from the components Rx&Ry.

UEXAMPLES

UEx.1

A number of coplanar cables are attached to a console. They exert the forces as shown in the figure. Determine the magnitude & direction of the resultant force on the console.

1500 N800 N

1000 N

1100 N

X

y

20º

30º15º


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UEX.2

Determine the magnitude of the resultant force & its direction, measured counterclockwise from the negative x-axis.

150 N

200 N

X

y

45º

UEX.3

A boat is kept in the middle of a river by means of two cables. The direction in which the boat pulls on both cables is shown by the dashed arrow. Calculate FR2 RIF F1=1000 N

60º

F1

F2

30º


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2.2 UResultant of a non concurrent coplanar force system:

The resultant of this force is either a single force or a couple.

F2

F1F4

F3

F1X

F1yF4y

F4X

F3X

F3y


To determine the resultant analytically:-

Each force is resolved into rectangular components then:-

a- If the algebraic sum of the components in either x & y direction for both is different from zero, the resultant is a force.

- The magnitude of the resultant force is obtained as following:

Rx=∑Fx , Ry=∑Fy , R=

- The angle between the x-axis & the resultant is obtained from the rotation

- The distance between the resultant & a specified point is determined by the principle of moment.


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d

R

R

O

F1

F2F3

F4

R.q=∑mo Where q is perpendicular distance from the moment axis through O to the resultant (R). UNOTE:- The direction of distance q is determine from the sense of R & (∑Mo) b- If the algebraic sum of the components of the forces is zero in two

different directions, the resultant is a couple in the plane of the forces.

- The magnitude & the sense of rotation of the couple may be obtained as the algebraic sum of the moments of the forces with respect to any point in the plane.

UNote:-

If Rx=0 & Ry=0

The resultant is a couple.

If Rx≠0 & Ry=0


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The resultant is a force & it is // to x-axis

If Rx=0 & Ry≠0

The resultant is a force & it is // to y-axis

Examples

UEx- 1

Determine the resulting of the force system.

X

Y

F2=40 N

F1=50 N

F3=80 N

F5=90 N

F4=20 N

F6=40 N

F7=170 N

UEx- 2

Determine the magnitude & position of the resultant of the four forces which act on the plate shown.

X

Y

80 N

50 N60 N

40 N 0

45°

30°

2m

2m

2m 5m

1m

A


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U2.3 Resultant of a concurrent, non coplanar force system

The resultant of any set of concurrent force must be a force passing through the point of concurrent of the forces of the system.


The components of the resultant in three perpendicular directions are obtained as following:

1. Resolve each force into rectangular components.

Rx=∑Fx Ry=∑Fy Rz=∑Fz

The magnitude of the resultant is

1. To specifying the action line of (R) in space is to give its direction cosines.


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UEx- 1

IF Rx=150 N , Ry=25 N, Rz=75 N. Determine the resultant of the force system & cosine direction.

150 N

25 N

75 N

X

Y

Z

U2.4 resultant of a parallel, non coplanar force system:

The resultant of parallel, non-coplanar force system is either a single force or a couple.

- If the resultant is a single force, its magnitude is equal to the algebraic sum of the forces of the system, & its position in space can be determined by the coordinates of the intersection of its action line with a plane perpendicular to the forces of the system.

As an example in fig.( ) . the resultant of forces is parallel to the y-axis & it is completely determined by the equations.

R=∑Fy

R

R


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R

F1

F2

F3X'

Z'

FIG.( )

- If the sum of forces is zero & the sum of the moments about one or both of two rectangular axes in a plane perpendicular to the forces is not zero, the resultant is a couple.

(for forces // to y-axis, the resultant is a couple ie. ∑Fy=0 & ∑Mx≠0 or ∑Mz≠0 or ∑Mx≠0 & ∑Mz≠0.

The resultant couple can be shown an a sketch by placing an upward force on one moment axis & an equal downward force on the other moment axis, so spaced as to produce the proper resultant moments.

EXAMPLE


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UEX-1

The slab in the figure below is subjected to four parallel forces. Determine the magnitude & direction of a resultant force equivalent to the given force system & locate its point of application on the slab.

400 N500 N

600 N 100 N

X

Y

Z

O D

B

A C

8 m

2 m

5m5m

2.5 RESULTANT of a system of couples in space:

The resultant of any force system composed of couples is a couple.

a- The resultant of couples in same plane or in parallel planes is equal to algebraic sum of the moments of the original couples.

b- The resultant of couples in non-parallel planes is also a couple.


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b

aQ

Q

P

P

p q

c- The resultant of couples parallel, non-coplanar force.

R=0

The resultant is a couple, in a plane // to ab.

b

a

Q

Q

P

P

p q


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UEX-1

Determine the resultant of the two couples shown in fig. below.

50 N

50 N

3 m

100N

100N2 m

UEX-2

Determine the resultant of the two couples shown in Fig. below.

75N

75N

4m

60N

60N

3m

X

Y

Z


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2.6 Resultant of a non-concurrent, non-parallel, non-coplanar force system.

The resultant of this force system can be a single force or a couple, but in general it is a force & a couple.

The resultant of non-concurrent, non-parallel & non-coplanar force system can be obtained by resolving each force into a parallel force through some common point origin & a couple.

X

Y

Z

F1

F4F2

F3

Y

X

Z

F1

F4

F2

F3

cy

cx

xz

UNote:-

FR1R is in x-y plane & passing through point o.

FR4R is in x-y plane.

FR2R is in y-z plane.

FR3R is in z-x plane.

The resultant force is:-


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R=ⱱ(RRxRP

2P+RRyRP

2P+RRzRP

2P) when RRxR=ΣFRxR, RRyR= ΣFRyR & RRzR=ΣFRz

& the resultant couple is:-

RRcR=ⱱ(cxP

2P+cyP

2P+czP

2P)

Where:-

Cx=ΣMx Cy=ΣMy Cz=ΣMz

Y

X

Z

R

Rc

examples

UEX1:-

Determine the resultant of the force system shown in fig.

X

Y

Z

F2=50 N

F1 =600 N

F3=100 N

4m

3m

6m

4m


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UChapter three

U3.1 Equilibrium

The body is in equilibrium provided it is at rest when a system of forces acting on it has no resultant, (equal to zero).

∑F=0, ∑M=0

To apply the equation of equilibrium we must account for all the known & unknown forces which act on the body. The best way to do this is to draw the free-body diagram. This diagram is simply a sketch which shows the body (free) from its surroundings with all the forces that act on it.

3.2 Free Body Diagram (F.B.D):-

To apply the equation of equilibrium, we must account for all the known & unknown forces (∑F) which act on the particle. The best way to do this is to draw the particle’s free-body diagram.

This diagram is simply a sketch which shows the particles (free) from its surroundings with all the forces that act on it.

Procedure for drawing a free-body diagram:-

1. Draw outlined shape:- Imagine the particle to be isolated or cut (free) from its surrounding by drawing its outlined shape.

2. Show all forces:- Indicate on this sketch all the forces that act on the particle. These forces can be active forces, which tend to set the particle in motion or they can be reactive forces which are the result of the constraints or supports that tend to prevent motion.


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To account for all these forces it may help to trace around the particle’s boundary, carefully noting each force acting on it.

3. Identify each force:- the forces that are known should be labeled with their proper magnitudes & directions. Letters are used to represent the magnitude & directions of forces that are unknown.

Note:-

The sense of unknown forces may be assumed & corrected later if it is incorrect.

Modeling the action of force:-

No. Name of body to be removed

Sketch of reaction bodies Action of body removed

Description

1. Earth body

earth

G

W

Always a vertical force equal to weight & passing thought the center of gravity of the body.

2. Flexible cord, rope, cable (weight neglected)

BodyƟ

cord

Ɵ

Always a single force (tension) along the cord.


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3. Smooth surface Ɵ

Body

Ɵ N

Always a single force perpendicular to the smooth surface.

4. Roller or ball

body body

QQ

Always a force perpendicular to the surface on which roller or ball can move.

5. Smooth pin

Body

RyRx

A force through the pin at an unknown angle, usually shown as two independent components.

6. Knife edge

Ry

Rx

A force through the point of contact at an unknown angle, usually shown as two independent components. If the knife edge is specified as smooth, the components tangent to the body surface is


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zero.

7. Smooth pin with additional force on pin.

80 N

50 N

A

B

C

Bx1

By1

Ax

Ay

Bx2

By2

Cx

Cy

Bx1

By2

Bx2

By1

80 N50 N

A force throught the pin at an unknown angle, or two independent components as shown here. Notic equal opposite action of members on pin with applied loads acting on pin.

8. Support for a beam or post fixed at the end.

Body

Pp

Pn M

A force at the cut section at an unknown angle, usually shown as two independent components.

9. Smooth bearing on a shaft

Body

Rx

Ry

A force normal to the shaft at an unknown angle usually shown as two independent components.


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10. Pin or runner in a smooth guide or slot. Ɵ

body

Ɵ

R

A force normal to the guide or slot.

11. Ball & socket

Body Ry

Rx

Rz

A force at the ball at an unknown angle, usually shown as three independent components.

Examples

EX1-

Draw the free body diagram (FBD) of the body is shown in the figures below:-

ƟA

Bcable

m

(1)

Mass m

F1F2F3


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EX2-

Draw the F.B.D for the two balls, ball A weight 300 N, ball B weight 500N.

A

B 40°

3.3 General procedure for the solution of problems in equilibrium:-

1- Determine the given data or what the unknown’s results are required.

2- Draw F.B.D for the member on which the unknown forces are acting.

3- Determine the type of force system acting on the F.B.D. 4- Compare the no. of unknowns on the F.B.D with the no. of

independent equations of equilibrium & :- a. If the no. of equation = the no. of unknowns, then start the

solution. b. If the no. of unknowns> the no. of independent equations, then

draw F.B.D for another body & repeat step 3&4. 5- If the no. of unknowns in the second F.B.D = the no. of equations

then solve the problem.

If it is not then repeat step (4-6).


46

6- If there are still too many unknowns after drawing F.B.D for all bodies, then the problems statically indeterminate.

3.4 categories of equilibrium:-

1. Category (1):- equilibrium of collinear forces clearly requires only the one force equation in the direction of the forces (x-direction).

2. Category (2):- equilibrium of forces which lie in a plane (x-y) plane & are concurrent at a point o, requires the two force equations only, since the moment sum about o, that is about a z-axis through o, is necessarily.

3. Category (3):- equilibrium of parallel forces in a plane requires the one force equation in the direction of the forces (x-direction) & one moment equation about an axis (z-axis) normal to the plane of the forces.

4. Category (4):- equilibrium of a general system of forces in a plane (x-y), requires the two force equations in the plane & one moment equation about an axis (z-axis) normal to the plane.


47

Force system Free body diagram (FBD) Independent equations

1. collinear x

F1

F2

F3

ΣFRXR=0

2. Concurrent at point

x

F1F2

F3F4

y

0

ΣFRXR=0

ΣFRyR=0

3. parallel

xF3

F1F2

F4

y

ΣFRXR=0

ΣMRZR=0

4. general

x

F3

F1

F2

F4

y

M

ΣFRXR=0

ΣFRyR=0

ΣMRZR=0


48

EX1:-

Determine the magnitude of FR1R & FR2R so that particle P is in equilibrium.

F1

F2

30°

60° 34

5

400 N

X

Y

EX2:-

The uniform beam has a mass of 50 kg per meter of length. Compute the reactions at the support O. The force loads shown lie in a vertical plane.

oA

B

C3 KN

1.4 KN

0.6m

0.6m

0.6m1.8m

1m

Ɵ

4 KN.m


49

EX3:-

Determine the force F of the fig. which must be applied to ring A in order to keep the 100 N cylinder B in equilibrium.

B

34

34

12

5A

F

EX4:-

Determine the force P of the Figure below that will maintain equilibrium.

24

13

400 N

P


50

Chapter four

4.1 Trusses

A truss is by definition a structure assembled with straight bars (members), which are connected by hinged joints, and loaded by forces which have their point of application at these joints. To design both the members & the connections of a truss, it is first necessary to determine the force developed in each member when the truss is subjected to a given loading. In this regard, two important assumptions will be made.

1. All the loadings are applied at the joints, in the force analysis the weight of the members is neglected since the forces supported by the members are usually large in comparison with their weight. If the member’s weight is to be included in the analysis, it is generally satisfactory to apply it as a vertical force, half of its magnitude applied at each end of the member.

2. The members are joined together by smooth pins. Because of these two assumptions, each truss member acts as a two-force member, & therefore the forces at the ends of the member must be directed along the axis of the member. If the force tends to elongate the member, it is a tensile force (T), fig.(a); whereas if it tends to shorten the member, it is a compressive force (C) FIG.(b).

C

C

T

TTension

(a)Comression

(b)


51

Pratt

Home

Warren

Baltimore

Commonly used Bridge Trusses


52

Fink

Pratt

Home

Warren

Commonly used Roof Trusses


53

4.2 Determining member forces In a plane truss, let b represent the number of members, r the number of reaction components, n the number of joints, & p the index of static indeterminacy, then p can be determined from the expression

P=b+r-2n Where the following conditions on p define the structure to be Geometrically unstable when p< 0 Statically determinate if not geometrically unstable when =0 statically indeterminate to degree p if not geometrically unstable

When p> 0 Examples

A B

CD

(1)


54

(2) (3)

(4)

(5)


55

4.3 U Truss analysis:-

There are various methods for calculating member forces in statically determinate trusses. We will look at two: • The method of sections; • The method of joints. In the method of sections, one introduces a suitable section across the truss and calculates the member forces from the equilibrium of one of the isolated parts. In the method of joints, we consistently determine the member forces from the equilibrium of the joints. U4.3 The method of joints. Because the truss members are all straight two-force members lying in the same plane, the force system acting at each-joint is coplanar & concurrent. Consequently, rotational or moment equilibrium is automatically satisfied at the joint (or pin), & it is only necessary to satisfy ∑Fx=0 & ∑Fy=0 to ensure equilibrium. UProcedure for analysis: The following procedure provides a typical means for analyzing of truss using the method of joints.

1. Draw the free body diagram of a joint having at least one known force & at most two unknown forces.(if this joint is at one of the supports, then it may be necessary to know the external reactions at the truss support).

2. Orient the x & y axes such that the forces on the free-body diagram can be easily resolved into their x&y components & then apply the two forces equilibrium equation ∑Fx=0 & ∑Fy=0, solve for the two unknown member forces &verify their correct sense.


56

3. Continue to analyze each of the other joints, where again it is necessary to choose a joint having at most two unknown & at least one known force.

4. One the force in a member is found from the analysis of a joint at one of its ends, the result can be used to analyze the forces acting on the joint at its other end.

EX1:- By using the method of joint, determine the forces in all members.

A B

CD E

F G H I

6m6m6m

8m

8m

200 N


57

U4.4 Zero-force members

Truss analysis using the method of joints is greatly simplified if

one is first able to determine those members which support no loading. These zero-force members are used to increase the stability of the truss during construction & to provide support if the applied loading is changed.

The zero-force members of a truss can generally by determined by inspection of each of its joints.

- If only two members form a truss joint & no external load or support reaction is applied to the joint, the members must be zero- force members.

- If three members form a truss joint for which two of the members are collinear, the third member is a zero – force member provided no external force or support reaction is applied to the joint.

Examples

AB

C

D

EF

P

(EX.1)


58

A

B

C

D

EP

(EX.2)


59

U4.5 The method of sections:-

The method of sections is used to determine the loadings acting within a body. It is based on the principle that if a body is in equilibrium that any part of the body is also in equilibrium. In the method of sections, the member forces in a (statically determinate) truss are determined by introducing a section and investigating at the equilibrium of one of the isolated parts. Since there are only three equilibrium equations available, you have to select a section such that there are no more than three unknowns. In general, the support reactions have to be determined previously. The method of sections can be used to “cut” or section the members of an entire truss, if the section passes through the truss & the free body diagram of either of its two parts is drawn, we can then apply the equations of equilibrium to that part to determine the member forces at the “cut section”. Since only three independent equilibrium equations (∑Fx=0, ∑Fy=0, ∑Mo=0) can be applied to the isolated part of the truss. Examples


60

EX.1 Determine the force in members (10,11,12)

120 KN

1

2

3

4 5

6 7

8 9

10

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17 4m

4m4m4m4m

a

a UEX.2 Determine the force in members 6 TO 9 & in member 13.

120 KN

1

23

4 5

6 7

8 9

10

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17 2m

2m2m2m2m

a

a

11

12

13

1418

19

20

21

22

23

24

25

b

b

C

C2m 2m

40 KN


61

UEX.3 Determine the force in members 6 TO 9 & in member 13.

120 KN

1

2

3

4

5

6 7

8

9

10

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

2m

3m3m3m3m

11

12

13

1418

19

20

21

22

23

24

25

3m 3m

1m

1m

1m


62

U4.6 Frames: A structure is called a frame if at least one of its individual

members is a multiforce member. a multiforce member is defined as one with three or more forces acting on it. Frames are structures which are designed to support applied loads & are usually fixed in position. Because frames contain multiforce members, the forces in these members in general will not be in the directions of the members. Therefore, we cannot analyze these structures by the methods developed because these methods apply to simple trusses composed of two-force members where the forces are in the directions of the members. Procedure for analysis:- The joint reactions on frames or machines (structures) composed of multiforce members can be determined using the following procedure. F.B.D: - Draw F.B.D of the entire structure, a portion of the structure, or each of its members. The choice should be made so that it leads to the most direct solution of the problem. - When the F.B.D of a group of members of a structure is drawn. The forces at the connected parts of this group are internal forces & are not shown on the F.B.D of the group. - Forces common to two members who are in contact act with equal magnitude but opposite sense on the respective F.B.D of the members. - Two – force members, regardless to their shape, have equal but opposite collinear forces acting at the ends of the member.


63

- In many cases it is possible to tell by inspection the proper sense of the unknown forces acting on a member, however, if this seems difficult, the sense can be assumed. - A couple moments is a free vector & can act at any point on the F.B.D. Also, a force is a sliding vector & can act at any point a long its line of action. Equations of equilibrium:

- Count the no. of unknowns & compare it to the total no. of equilibrium equations that are available. In two dimensions, there are three equilibrium equations that can be written for each member.

- Sum moments about a point that lies at the intersection of the lines of action of as many unknown forces as possible.

- If the solution of a force or couple moment magnitude is found to be negative, it means the sense of the force is the reverse of that shown on the F.B.D.

EXAMPLS

p

P2

p

P2

M

truss frame

P3


64

UEX.1 Determine the horizontal & vertical components of force which the pin at c exerts on member CB of the frame in the Fig. below.

2000 N

A

B C

2 m2 m

3 m

60°

UEX.2 The compound beam is fixed at A & supported by a rocker at B & C. There are hinges (pins) at D & E. Determine the reactions at the supports.

A B CD E

15 KN

6 m2 m2 m2 m6 m


65

UEX.3 Determine the horizontal & vertical components of force at pins A,B & C & the reactions to the fixed support D of the three – member frame.

D

C

A

B

0.5m0.5m0.5m0.5m

2 KN2 KN2 KN2 KN

2 m

2 m

UEX.4 Determine the horizontal & vertical components of the pin reaction at B of the figure below.

2000 N

A

B

C

D

2 m

4 m

4 m

3 m

3 m3 m2 m3 m


66

Chapter five U5.1 Center of gravity: The center of gravity G is a point which locates the resultant weight of a system

- The resultant weight

X

Y

Z

G

XYZ

nW1W

2W

- To determine the of G →(∑My)

-

- To determine the of G →(∑MX)

-

- To determine the of G , Particles fixed in it & rotated about the x (or y) axis,


67

-

………………….. (1)

Here:-

Represent the of the G of the system particles.

Represent the of each particle in the system. is the resultant sum of the weights of all the particles in the

system. when the rigid body is composed of an infinite number of

particles. Considering the arbitrary particle located at ( ) & having a weight dw,

…………………. (2)


68

UEX.1

Determine the of the center of gravity G of the plane system of four particles which have Weights of 3N , 2N , 4N & 6N & are located at (2,1) , (3,5) , (-3,2) & (0,-6). UEX.2

Determine the of the center of gravity for the system shown. Y

X

Z

8 N2 N

7 N1 N

5 N

6 N

3 N

4 N

8m

4m

6m


69

U5.2 Centroid:-

The centroid C is a point which defines the geometric center of an object.

There are three cases will be considered. U1-volume:-

X

Z

Y

C

dv

y

xz

x~

y~

z~

If an object is subdivided into volume elements. (dv)

U2- Area:-

X

Z

Y

C

dA

y

xz

x~

y~

z~

The centroid for the surface area of an object, such as a plate or shell.


70

U3- Line:-

X

Z

Y

C

y

xz

x~

y~

z~dL

If the geometry of the object, such as a thin rod or wire.

USymmetry:- The centroids of some shapes may be partially or completely specified by using conditions of symmetry.

x~x~

Y

X

C

Symmetry in y-axis, C lies on y-axis.

Symmetry in y & x axis, C lies on the intersection of x & y axis.


71

Z

X

Y

C

UProcedure for analysis:- The center of gravity or centroid of an object or shape can be determined by single integrations using the following procedure:-

• Differential element: - Select an appropriate coordinate system, specify the coordinate

axes, & then choose adifferetial element for integration. - For lines the element (dL) is represented as a differential line

segment. - For areas the element (dA) is generally a rectangle having a finit

length & differential width. - For volumes the element (dV) is either a circular disk having a finit

radius & differential thickness, or a shell having a finit length & radius & differential thickness.

- Locate the element at an arbitrary point (x,y,z) on the curve that defines the shape.

• Size & moment arms:- - Express the length dL, area dA, volume dV of the element in terms

of the coordinates of the curve used to define the geometric shape.

- Determine the coordinates or moment arms for the centriod or center of gravity of the element.

• Integrations:- - Substitute the formulations for & dL, dA or dV into the

appropriate equations (eq. 1-5) & perform the integration.

Symmetry in y, x & z axis, C lies on the intersection of x, y & z axis.


72

- Express the function in the integrand in terms of the same variable as the differential thickness of the element in order to perfume the integration.

- The limits of the integral are defined from the two extreme locations of the element’s differential thickness, so that when the elements are “summed” or the integration performed, the intire region is covered.

UEX.1 Locate the centroid of the circular wire segment shown in the figure below.

R

dƟ

dL=RdƟ

y=RsinƟ

x=RcosƟ

o

(R,Ɵ)

(x,y)¯ ¯

Ɵ

y

x


73

UEX.2 Locate the centroid of the area in the figure below.

(x,y)¯ ¯

(x,y)

x

y

1m

2xy =

dx

y

1m

x

y

UEX.3 Locate the xˉ centroid of the shaded area bounded by the two curves y=x & yxP

2P in the figure below.

(y2-y1)

dx (x,y1)

(x,y2)

Y=x

y=x2

(x,y)

y

x


74

U5.3 Composite bodies:-

A composite body consists of a series of connected (simpler) shaped bodies, such a body can often be sectioned or divided into its composite parts & provided the weight & location of the center of gravity of each of these parts are known. The method for doing this requires treating each composite part like a particle. We have:-

Here:-

:- represent the of the center of gravity G of the composite body.

:- represent the of the center of gravity G of each composite part of the body.

:- is the sum of the weights of all the composite parts of the body, or simply the total weight of the body. UProcedure for analysis:-

1- Composit parts:- - using as a sketch, divide the body or object into a finit number of

composite parts that have simpler shaper. - If a composite part has a hole, or a geometric region having no

material, then consider the composite part without the hole & consider the hole as an additional composite part having negative weight or zero.


75

2- UMoment arms:-

- Establish the on the sketch & determine the of the center of gravity or the centroid of each part.

3- USummations:- - Determine by applying the center of gravity equations. - If an object is symmetrical about an axis, the centroid of the

object lies on this axis.

segment L or A or V

x y z x(L or A or V)

y(L or A or V)

z(L or A or V)

1 2 3 Σ= Σ= Σ= Σ=


76

UEX.1 Locate the centroid of the wire shown in the fig. below.

r=60mm

12

3

40 mm

20mm y

z

x

UEX.2 Locate the centroid of the plate area shown in the fig. below.

y

x

3m2m

1m

2m

1m


77

U5.4 Distributed force or pressure:-

A force is exerted on a body over a larger area. The distributed force or pressure can be considered to be a number of concentrated forces, forces each acting on a small part of the surface. Usually these forces act normal to the surface & for a plane surface they constitute a parallel force system.

If the force is the same on all equal elements of area no matter

how small the elements, the pressure is said to be uniform & the resultant force acts at the centroid of the plane area. If the force is not the same on equal elements of area, the pressure is variable & the line of action of the resultant force must be determined by the principle of moments. UEX.1 Determine the reaction at point A,B,C & D in the figure below.

A BC D

200N/m

800N/m

4m 6m

10m 5m


78

UEx.2 Determine the reaction at point A,B in the figure below.

A B

600N/m

200N.m3m 2m 2m

300N

UEX.3 Determine the reaction at point A,B in the figure below.

A B6m 4m 4m

500N/m 400N/m

200N/m


79

U5.5 Moment of inertia:- Is the angular motion of rigid bodies.

• The moment of inertia of the differential planar area dA about the x & y axes are:-

& So • The second moment of dA about the pole O or z- axis. (Polar

moment of inertia).

Here:- R= is the perpendicular distance from the pole (z-axis) to the element dA.

• The polar moment of inertia of an area is equal to the sum of the rectangular moments of inertia with respect to any two perpendicular axes intersecting the polar axis.

• The second moment of an element of area is always a positive quantity.


80

• The second moment of an area has dimensions of length raised to the fourth power & common units of , , , & so on.

• The moment of inertia of an area is the sum of the second moments of the elements of the area.

Y

X

x

yr

dA

O

∫= dArJO2

222 YXr +=

∫ ∫ ∫ +=+=+= YXO IIdAYdAXdAYXJ 2222 )(


81

U5.6 Radius of Gyration:- Is a measure of the distribution of the area from the axis in question.

C

X

Y

x′y′A

Y

X

A

Kx

Y

X

A

Ky

X

Y

KZ

O

• The square of the radius of the radius of gyration about a polar axis equals to the sum of the squares of the radii of gyration about the two corresponding rectangular axes.


82

UEX.1 Determine the moments of inertia of the rectangular area about the centroidal xRoR & yRoR axes & polar axis zRoR through c.

X

Y yo

xo

h/2

b

Ch/2

UEX.2 Determine the moments of inertia of the triangular area about the centroidal xo & yo axes & polar axis zo through c.

h

bx

y


83

U5.7 Transfer of axes:- The moment of inertia of an area about a non centroidal axis may be easily expressed a parallel centroidal axis.

d

CdA

XO

yO

dy

dx

X

Y

XO

yO

O

• The moment of inertia of the element dA about the x-axis is:-

By integration

(Moment of inertia about the centriodal xRoR-

axis.

These equations are called parallel-axis theorems. UNote:-

1- The axes between which the transfer is must be parallel. 2- One of the axes must pass through the centroid of the area.


84

• If transfer is desired between two parallel axes neither of which passes through the centroid, it is first necessary to transfer from the centroidal axis to the second axis.

The parallel-axis theorems also hold for radii of gyration. With substitution of the definition of k into equations 1 , the transfer relation becomes

Where:-

= is the radius of gyration about a centroidal axis. K= is the radius of gyration about parallel axis. d= the distance between the two axes. UEx.1:- Determine the IRxoR & IRyoR & IRzoR, and then find the IRxR, IRyR & IRzR.

h

b

Y

X


85

UEX.2 Determine the Ix about its base & about parallel axes through its vertex.

b

h

Ex.3 Determine the moment of inertia of the area under the parabola about the x-axis.

3mm

X=4y2 /9

04mm

y

x


86

UEx.4 Determine the rectangular & polar radii of gyration of the shaded area about the axes shown.

0.5

2.5

12

0

y

x

Y=x3 /4

UEx.5 Determine the radius of gyration of the shaded area of the fig. below with respect to the axis. The equation of the curve is (x P

2P=12y).

y

x

a a

6m

1m

2m


87

U5.8 Moment of inertia of composite areas:-

A composite area consists of two or more simple areas, such as rectangles, triangles, & circles. The cross-sectional area of standard structural elements, such as channels, I-beam, & angles, are frequently included in composite area with respect to any axis is equal to the sum of the moments of inertia of its component areas with respect to the same axis. When an area, such as a hole, is removed from a larger area, its moment of inertia of the larger area to obtain the net moment of inertia.

12

3bhIc = 3

3bhI X =c

xb

h

36

3bhIc =c 12

3bhI X =h

b

rc

x4

4rIcπ

=4

5 4rI Xπ

=

cπ3

4r

x40549.0 rIc = 16

4rI xπ

=


88

a

b c4

3abIcπ

=

part Area,A dx dy Adx2 Ady2 Ix¯ Iy¯

sum ∑A ∑Adx2 ∑ Ady2 ∑∑ Ix¯ Iy¯

∑∑ += 2 dyAyIIy

∑∑ += 2 dxAxIIx

Where:- Ix= the moment of inertia for the composite area. Iy= the moment of inertia for the composite area.

Where:- K= the radius of gyration for the composite area. I= is the total moment of inertia. A= is the total area of the composite figure.

Iz=Ix+Iy


89

EX1:- Determine the moment of inertia & radius of gyration about the x-axis, y-axis & the z-axis.

30mm

30mm30mm

40mm40mm

30mm

EX2:- Determine the radius of gyration of the shaded area of the fig. with respect to the x-axis.

26mm

6mm

8mm


90

EX3:- Determine the radius of gyration of the shaded area of the fig. with respect to the x-axis.

3

4 4 m

6 m

12 m


91

Chapter 6 UFriction:- Is a force of resistance acting on a body which prevents or retard slipping of the body relative to a second body or surface with which it is in contact. This force always acts tangent to the surface at points of contact with other bodies & is directed so as to oppose the possible or existing motion of the body relative to these points. Friction:- 1- Fluid friction 2- Dry friction Fluid friction:- Occurs between the contacting surfaces are separated by a film of fluid. Dry friction:- Occurs between the contacting surfaces of bodies in the absence of a lubricating fluid.

p

w

p

w

Rough surfaceFn∆

Nn∆


92

a/2

p

x

N

o F

h

w

a/2

When x = a/2

w

p

h

Fs

RsNs

x

Impending motion

ɸs

Fs= limiting static frictional force

Where:-

= coefficient of static friction.

• When the magnitude of p increased so that it becomes greater

than Fs, the frictional force at the contacting surfaces drops slightly to a smaller value Fk, called the kinetic frictional force.

F= Frictional force

N= Normal force

∑Mo= 0

w.x = p.h

x = p.h/ w


93

By comparison,

• Characteristics of Dry friction:- - The frictional force acts tangent to the contacting surfaces in a

direction opposite to the relative motion or tendency for motion of one surface against another.

- The maximum static frictional force Fs that can be developed is independent of the area of contact, provided the normal pressure is not very low nor great enough to severely deform or crush the contacting surfaces of the bodies.

- The maximum static frictional force is generally greater than the kinetic frictional force for any two surfaces of contact, if one of the bodies is moving with a very low velocity over the surface of a nother, Fk becomes approximately equal to Fs, .

- When slipping at the surface of contact is about to occur the maximum static frictional force is proportional to the normal force, such that .

- When slipping at the surface of contact is occurring, the kinetic frictional force is proportional to the normal force, .


94

EX1:- Determine the maximum angle θ which the adjustable incline may have with the horizontal before the block of mass m begins to slip. The coefficient of static friction between the block & the inclined surface is μRsR.

θ

mμs

EX2:- Determine the range of values which the mass mo. May have so that the 100 kg block shown in the fig. will neither start moving up the plane nor slip down the plane. The coefficient of static friction for the contact surfaces is 0.30.

mo

100 k

g

20°


95

EX3:- The three flat blocks are positioned on the 30° incline as shown & a force parallel to the incline is applied to the middle block. The upper block is prevented from moving by a wire which attaches it to the fixed support. The coefficient of static friction for each of the three pairs of mating surfaces is shown. Determine the maximum value which P may have before any slipping takes place.

40kg50kg30kg

P

μ=0.30μ=0.40

μ=0.45


96

Chapter 7 Rectilinear motion:- The rectilinear motion of a particle is defined as motion along a straight line. The instantaneous position of a particle moving in a plane is completely determined when the path of the particle is known & the distance S, measured along the path from some fixed point on the path to the particle, is specified as a function of time.

- Linear displacement :- Is defined as the change of position of the particle during the time interval.

- q (األزاحة)= المسافة بين النقطة األولى والنقطة األخيرة . بغض النظر عن النقاط الوسطية .

- Q. (المسافة) = مجموع المسافات بين النقاط التي مر بها الجسم جميعها - UمالحظةU ـ دائما : Q هي أكبر أو تساوي q واليمكن أن تكون Q أقل من q.

نعتبر اإلتجاه الموجب هو باتجاه اليمين.

A P B CO

S

خالل مدة من الزمن فإن اإلزاحة B إلى النقطة A اذا تحرك من النقطة Pالجسم •

).q = B-A وكقيمة تساوي الفرق بين النقطتين . (B و Aتساوي المسافة بين فهناك B ومن ثم رجع إلى النقطة C إلى النقطة A اذا تحرك من النقطة Pالجسم •

أزاحتان (أزاحة كمية وأزاحة أتجاهية) . ) .q (,) q = B-Aاألزاحة اإلتجاهية ( • .Q = AC + BC)) = المسافة بين النقاط التي مر بها الجسم (Qاإلزاحة الكمية ( •


97

- Linear velocity (v) of a particle is defined as the time rate of change of position of the particle. The average velocity of a particle during an interval of time is :-

S positive → to the right of the origin. V positive → to the right of the origin.

- Linear acceleration (a) of a particle is defined as the time rate of change of the linear velocity of the particle.

(deceleration)

UDrawing method:-

1- (V-t) diagram:-

t

V

t1

dt

V

t2

t


98

- The slope of the (v-t) curve, (dv/dt) for any time represents the magnitude & sense of the linear acceleration of the point or particle at that instant.

- The area under the curve between any two ordinates at t1 & t2 is given by the definite integral of (v dt).

SR2R-SR1R = area under the curve from tR1R to tR2R.

area under the (V-t) curve for this time interval.

- When a point with rectilinear motion has a constant acceleration,

the (v-t) diagram is a straight line, since the slope of the (v-t) diagram is the acceleration.

Vf

Vt

∆t

1a

V

t


99

UEX1:- The position, s, of a particle moving along a horizontal straight line is given by the equation (s=6t P

2P-4), where S is in meter & t is the time in

seconds. The particle is 4m to the right of the origin when t is zero. a) Determine the displacement of the particle during the time

interval from t=2 sec to t=4 sec. b) Determine the velocity & acceleration of the particle when t is 4

sec. UEX2:- The magnitude of the linear acceleration of a point motion along a vertical path is given by the equation (a=6t-24), where a is in meter per second per second (m/sP

2P) & t is in seconds. The acceleration is up ward

when t=5 sec, the point is 4m below the origin when t=0 & 23m above the origin when t=3 sec. Determine:-

a) The velocity when t=3 sec. b) The displacement during the time interval from t=0 to t=4 sec. c) The total distance traveled during the time interval from t=0 to

t=4 sec.


100

UMethod of Force – Mass – Acceleration:- When a force system acts on a particle, the system must be concurrent (particle has no dimensions), & the resultant is a single force acting through the particle, Newton’s second law of motion for a particle can be expressed as :- R = m.a … (1) Where:- R = is the resultant force. m = is the mass of the particle. a = is the acceleration of the particle. The equation resolved into x,y & z components.

…. (2)

…. (3)

The equation (3) is called motion equation.

مالحظة:- باتجاه الحركة.(+a)دائما يكون التعجيل باألتجاه الموجب - . اما السرعة a) فتكون دائما عكس اتجاه اذا كانت هناك قوى احتكاك ( -

.aفتكون باتجاه ) يساويان az , ay) اذا كانت الحركة مثال باتجاه ء فقط فان (az , zy , ax الى (aيحلل -

.axصفر وتكون هناك قيمة فقط لل -


101

EX1:- The small body A weight 1.5 N & swings in a horizontal plane on the end of the cord AB. The line AO has a constant angular velocity of 30 rpm. Determine the tension in the cord & the angle θ.

X

Y

O

B

A

2mωZ

θ

EX2:- Block A weights 10,000 N & block B weights 6.44 N. The blocks are connected by a flexible inextensible cord which passes over a smooth pulley at C. The weights of the cord & pulley can be neglected. The coefficient of friction between the plane & block A is 0.30. the blocks are initially at rest. Determine the tension in cord & the velocity of A after it has traveled 5m.

AB

5 N

C

3

4

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Engineering mechanics - جامعة ذي قارeng.utq.edu.iq/images/lectures/Engineeringmechanicwasan.pdf · 4 F1 F2 F3 F4 Fig. (a) b. Parallel ... U is the process of replacing a