Chapter 2
Engineering Mechanics
CHAPTER 2 STATICS Introduction Statics deals with system of forces that keeps a body in equilibrium. In other words the resultant of force systems on the body are zero. Force: A force is completely defined only when the following three characters are specified. Magnitude Point of application Line of action/Direction Scalar and Vector: A quantity is said to be scalar if it is completely defined by its magnitude alone. e.g. length, energy, work etc. A quantity is said to be vector if it is completely defined only when its magnitude and direction is specified. e.g. force, acceleration.
Equivalent Force System Coplanar force system: If all the forces in the system lie in a single plane, it is called coplanar force system. Concurrent force system: If line of action of all the forces in a system passes through a single point it is called concurrent force system. Collinear force system: In a system, all the forces parallel to each other, if line of action of all forces lie along a single line then it is called a collinear force system. Force system Coplanar like parallel force is straight.
Example Weight of stationary train on rail off the track
Coplanar concurrent Coplanar non- concurrent force
Forces on a rod resting against wall. Forces on a ladder resting against a wall when a person stands on a rung which is not at its center of gravity. The weight of benches in class room A tripod carrying camera Forces acting on moving bus
Non- coplanar parallel Non- coplanar concurrent force Non- coplanar non-concurrent force
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Chapter 2
Engineering Mechanics
Newton’s law of motion First Law: Everybody continues in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by force acting on it. Second law: The rate of change of momentum of a body is directly proportional to the applied force & it takes place in the direction in which the force acts. F (m
dv ) dt
Third law: For every action, there is an equal and opposite reaction. Principle of transmissibility of force: The state of rest or motion of rigid body is unaltered if a force action on a body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of applied force. P A
B P Parallelogram law of forces: If two forces acting simultaneously on a body at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through the point of intersection of the two sides representing the forces.
Equilibrium and Free Body Diagrams Coplanar Concurrent Forces Triangle law of forces: If two forces acting simultaneously on a body are represented by the sides of triangle taken in order, their resultant is represented by the closing side of the triangle taken in the opposite order. Polygon law of forces P3
P2 P4
E
D P3
R2 R
R1
P1 P4
A
P1
C P2 B
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Chapter 2
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If a number of forces acting at a point be represented in magnitude and direction by the sides of a polygon in order, then the resultant of all these forces may be represented in magnitude and direction by the closing side of the polygon taken in opposite order
P2
E
D
P1 A
C
B
Resultant (R) = √ tan
(
)
= angle between two forces,
= inclination of resultant with force P1
When forces acting on a body are collinear, their resultant is equal to the algebraic sum of the forces. Lami’s theorem: (only three coplanar concurrent forces) If a body is in equilibrium under the action of three forces, then each force is proportional to the sine of the angle between the other two forces. c P2
P1
P2
b
P3 P3
P sin
P sin
P1 a
P sin
Free body diagram: A free body diagram is a pictorial representation used to analyze the forces acting on a free body. Once we decide which body or combination of bodies to analyze, we then treat this body or combination as a single body isolated from all our surrounding bodies A free body diagram shows all contact and non-contact forces acting on the body.
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Chapter 2
Engineering Mechanics
Sample Free body diagrams 600N W SMOOTH
600N
P
R1
G
P
SMOOTH R2
A ladder resting on smooth wall F2
V
F1
V
V
V
F3
F
y M
W=m g
x
A cantilever beam ̂ ̂ mg
m
Free body diagram of just the block
A block on a ramp In a free body diagram all the contacts/supports are replaced by reaction forces which will exert on the structure. A mechanical system comprises of different types of contacts/supports. Types of contacts/supports Following types of mechanical contacts can be found in various structures: Flexible cable, belt, chain or rope
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Chapter 2
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Weight of cable negligible
θ T
θ Weight of cable not negligible
θ T
θ
Force exerted by the cable is always a tension away from the body in the direction of the cable.
Smooth surfaces
N
Contact force is compressive and is normal to the surface.
Rough surfaces F R
N
Rough surfaces are capable of supporting a tangential component F (frictional force as well as a normal component N of the resultant R.
Roller support
N
N
Roller, rocker or ball support transmits a compressive force normal to supporting surface.
Freely sliding guide
N
N
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Chapter 2
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Collar or slider support force normal to guide only. There is no tangential force as surfaces are considered to be smooth.
Pin connection
M
Rx
Rx
Ry
R y
A freely hinged pin supports a force in any direction in the plane normal to the axis; usually shown as two components Rx and Ry. A pin not free to turn also supports a couple M.
Built in or fixed end
A
M
A O r
Weld
A
F V
A built-in or fixed end supports an axial force F, a transverse force V, and a bending moment M.
Coplanar Non-Concurrent Forces Varignon’s theorem: The algebraic sum of the moments of a system of coplanar forces about a momentum center in their plane is equal to the moment of their resultant forces about the same moment center. B R
A
R.d = P1.d1 +P2.d2 Effect of couple is unchanged if Couple is rotated through any angle. Couple is shifted to any position. The couple is replaced by another pair of forces whose rotated effect is the same. Couple is free vector THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 2
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Condition for body in Equilibrium The algebraic sum of the components of the forces along each of the three mutually perpendicular direction is zero. The algebraic sum of the components of the moments acting on the body about each of the three mutually perpendicular axis is zero. When a body is in equilibrium, the resultant of all forces acting on it is zero. Thus, the resultant force R and the resultant couple M are both zero and we have the equilibrium equations
R F 0
&
M= M=0
For collinear force system ∑F
∑F
∑F
For non-collinear force system ∑
∑
∑
These requirements are both necessary and sufficient conditions for equilibrium. Two forces can be in equilibrium only if they are equal in magnitude, opposite in direction, and collinear in action. If a system is in equilibrium under the action of three forces, those three forces must be concurrent. Wrench: When the direction of resultant couple ‘ ’ and resultant force ‘F’ are parallel then it is called ‘wrench’ When direction of resultant couple & direction of resultant force is same then it is called ‘Positive wrench’ and when the direction opposite to each other it is called negative wrench. Example of wrench is screw driver Types of Equilibrium There are three types of equilibrium as defined below: Stable Equilibrium: A body is in stable equilibrium if it returns to its equilibrium position after it has been displaced slightly. Unstable Equilibrium: A body is in unstable equilibrium if it does not return to its equilibrium position and does not remain in the displaced position after it has been displaced slightly. Neutral Equilibrium: A body is in neutral equilibrium if it stays in the displaced position after if has been displaced slightly.
Stable Equilibrium
Unstable Equilibrium
Neutral Equilibrium
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Friction Types of Friction (i). Dry friction: Friction between the contact surface (ii). Fluid Friction: Friction between the layers of fluid element. (iii). Internal friction: When cyclic load applied on the solid then, friction between the elements. When applied force is, F And static friction coefficient μ Fmax μN FFmax F μKN= Friction force μK= Kinetic friction co-efficient.
Virtual Work Work: When a force acts on a body and moves it through some distance in its own direction, then work is said to be done. Thus, work may be defined as the product of the force and the distance moved in the direction of the force. Mathematically, we can write that Work = Force × distance U =F × S When the distance moved by the body is not in the direction of the force then to determine the work done, the component of the force in the direction of the distance moved may be multiplied with the distance moved For example if the force F is acting at an angle θ with the direction of the distance S moved, then work done is given by U F cos θ × S Virtual Displacement: It may be defined as the infinitesimally small imaginary (or hypothetical or virtual) displacement given to a body or to a system of bodies in equilibrium, consistent with the constraints. The virtual displacement may be either rectilinear or angular. Virtual Work: The product of the force F and the virtual displacement δs in the direction of the force is called virtual work. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 2
δU
Engineering Mechanics
F δs
Principle of virtual Work It states that if a system of forces acting on a body or a system of bodies is in equilibrium and if the system is supposed to undergo a small virtual displacement consistent with its geometrical constraints, the algebraic sum of the virtual work done by the system of forces is zero.
Trusses and Frames Trusses are commonly used for construction of roofs of workshop factories and bridges. The trusses are subjected to mainly three types of loads, viz, dead load, live load and wind load. The dead load is self weight of truss live, load is the load which is applied to the truss e.g. the load acting on a bridge truss due to the passing of a train, load acting on a workshop truss due to an electric overhead travelling and the wind load due to the high velocities of wind blowing in a particular region. When the number of members in a truss satisfies the condition, m = 2j – 3 where j is the number of joints, then the truss is known as a perfect truss, otherwise imperfect. The truss is called deficient or redundant, if m (2j – 3), respectively. A pin jointed frame which has just sufficient number of members to resist the loads without undergoing deformation in space is called perfect frame. If number of member in frame is less than that that required for a perfect frame then it is called deficient frame. If number of members in frame is more than that required for perfect frame then it is called redundant frame. A redundant frame is indeterminate. The following assumptions are made in solving trusses: 1. 2. 3. 4. 5.
The members of truss are connected at the joints by friction less joints. The members of truss lie in a common plane (plane truss). The loads are applied only on the pins connecting the members and that the lines of action of the loads lie in the plane of the truss. The weight of members is negligible as compared to the applied loads. The truss is rigid and that it does not deform or change its shape upon the application of the loads.
The member of a truss may be in tension or compression. A member in tension is called a tie and a member in compression a strut. Methods of Solution: Two methods are generally used for determining the forces in various members of a truss. These methods are 1.
Analytical methods (a) Method of joints (concurrent force system). (b) Method of sections (non-concurrent force system).
2.
Graphical method
This method of solution is used when THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Large truss in which only few forces are required Situation where method of joints fail.
While determining the reactions at the supports, the following points should be remembered (i) At simply supported (i.e., pinned or roller support) support there can be only a vertical reaction. (ii) At fixed support, the reaction can take an arbitrary direction. A frame in which all the member lies in a single plane is called plane frame. While a frame in which all the members do not lie in a single plane is called space frame. For perfect frame, m = (2j -3) 4
2
4
3
1
5 1
3 2
For deficient frame, m < (2j -3)
If there is only one force acting at joint, then for the equilibrium, this force should be equal to zero.
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Chapter 2
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If there are two forces acting at a joint then, for the equilibrium, forces should act along the same straight line. The two forces should be equal and opposite. If (only) two forces acting at a joint are not along the same straight line, then for the equilibrium of the joint each force should be equal to zero. If three forces act at a joint and two of them are along the same straight line then, for the equilibrium of the joint, the third force should be equal to zero.
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Solved examples Example 1 ABCD is a string suspended from points A and D and carries a weight of 5 N at B and a weight of W N at C. The inclination to the vertical of AB and CD are and respectively and angle ABC is . Find W and thetensions in the different parts of the string. Solution Let ,
and
be the tensions in the parts AB, BC and CD respectively, as shown in Fig. A T
45
D T
1
120
3
B C
5N W For the equilibrium of point B, we have =
=
=5
=
=5
=
×
= 16.73 N
×
= 13.66 N
For the equilibrium of point C, we have = =
= = 13.66 × =
×
× = 23.66 N
= 27.32 N
Example 2 A fine string ABCDE whose extremity A is fixed has weights and attached to it at B and C and passes over a smooth pulley at D carrying a weight of 20 N at the free end E. If in the position of equilibrium, BC is horizontal and AB, CD makes angles and respectively with the vertical, find (A) Tensions in the portions AB, BC, CD and DE (B) The value of the weights, and , and (C) The pressure on the pulley axis. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 2
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Solution Since the string passes over a smooth pulley at D, the tension in CD portion of string is 20 N. Let the tension in AB and BC be and respectively, as shown in Fig. For the equilibrium of point B, we have =
=
and for the equilibrium of point C, =
= 20 N
D
A 3 60 20 N C
B
W2
W1
Hence
= 20 ×
= 20 × Thus
=
=
×
E 20 N
= 20 ×
= 17.32 N
= 20 × = 10 N ×
×
=
= 10 × ×
= 11.55 N = 5.77 N
Pressure on the pulley F=√ = 20 √
× ×
= 20 √
×
cos
= 38.6 N
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Example 3 A beam AB hinged at A and is supported at B by a vertical chord which passes over two frictionless pulleys C and D. If the pulley D carries a vertical load W, find the position x of the load P if the beam is to remain in equilibrium in the horizontal position.
C
A
B
D
P W
Solution From pulley D 2T
W W
T
A P Taking moments about A W
×
Wl
x
Wl P
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Example 4 The wire passing round a telephone pole is horizontal and the two portions attached to the pole are inclined at an angle of to each other. The pole is supported by another wire attached to the middle point of the pole and inclined at to the horizontal. Show that the tension in this wire is 4 times that of the telephone wire. Solution Let the tension in the two portions of the telephone wire be wire be , as shown in Fig.
each and the tension in another
A
C
D
B
Then
T=2
cos
Let AC = BC = l Taking moments about B, we get T× cos × =2 × ×2=4 =4 . Example 5 Two halves of a round homogeneous cylinder is held together by a thread wrapped round the cylinder with two equal weights, P attached to its ends, as shown in Fig. The complete cylinder weighs W N. The plane of contact of both of its halves is vertical. Determine the minimum value of P for which both halve of the cylinder will be in equilibrium on a horizontal plane. P r G P
A
P W/2
W (a)
(b)
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Chapter 2
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Solution Given the problem as shown below. We draw the Free Body Diagram as follows. Note the following salient points
Figure 1: Question Description
Figure 2: Free Body Diagram In the FBD.
As the question is to find the minimum value of force F on rope for which the two halves just remain in contact, we see that the limiting case is that the two halves are just about to touch. In this case, the two halves rotate about point of contact C. So, the point of contact as acts as revolute joint about which the two semi-circular cylinders rotate and hence has two normal reactions Nx & Ny as shown in FBD.
In case of a half turn rope (rope which goes around the cylinder just half a turn around the top half once), to split the two halves as we have to cut the rope once. On cutting the rope, the rope tension force P is exposed once on the top tip each half, which is why it is marked on top.
The left two force, are gravity of each half which is acting on Center of Gravity of each semi cylindrical half. To calculate the CG location, we know that ∫ ∫ Now, employing polar coordinates and taking a infinitesimal element as shown in figure, we get as follows. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Chapter 2 ∫ ∫
∫ ∫
Engineering Mechanics ∫
∫ ∫
∫
∫ ∫
= (which is
∫ ∫
clear from Symmetry of the body about X axis)
Figure 3: Semi-Cylinder CoG Calculation Diagram Now, as the body is to be in static equilibrium in the limiting condition, we get by zero moment sum about point P as follows. ( )( ) Modification/Extension for Multiple Turns: When the number of turns on the cylinder increases, by physical intuition, clearly, the minimum value of P required to just hold the two halves together must be lesser, right? Let’s check if the solution gives this analytically Assume that the rope turns n full turns around the cylinder. In this case, when we try to draw FBD of two halves separately, we have to cut the ropes n+1 times on top edge of cylinder which means a force of (n+1)P acts on top edge. In addition, we have to cut rope n times at bottom edge ie at point C, which means force acting at bottom point is nP. With these modifications, we get the new FBD as follows. Now,
Figure 4: FBD for the general n rope turn case
Again taking moment about point C, we get ( ) ( ) clearly, as number of turns the halves together
(
)
The minimum force to hold
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Chapter 2
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Example 6 Given the pulley rotating at constant angular velocity 𝜔 as shown below, which is using a rope of thickness d meters to lift a mass M. What is the tension in the rope?
Figure:5 Pulley Question Solution This is a deceptively simple question. The important point to be noted here is the fact that the rope is of finite non-zero thickness. So, as time proceeds, more and more rope winds in the pulley and hence, radius of pulley increases and hence by the relation, velocity of mass V = no, mass starts accelerating by 𝜔 Now for every πr rad. rotation of pulley, radius changes by rope thickness d
for every 𝜔
rad/s rotation, radius changes by Now, on drawing the FBD for the mass we get as follows. From the FBD, we see that by Newton’s second law,
Figure 6: Free Body Diagram
(
𝜔
)
Hence, solved. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
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Example 7 A smooth circular cylinder of radius 2 m is lying in a triangular groove, one side of which makes an angle of 10 and the other an angle of 30 with the horizontal, as shown in Fig. Find the reactions at thesurface of contact if there is no friction and the weight of the cylinder is 150 N.
40 140
O
170
10 30 W
30
B
10
A
Solution Let
and
be the reaction of the 10 and 30 planes respectively.
Using Lami’s theorem we get =
=
=W × =W
W× = 0.778 W = 116.6 N 150 ×
= 40.52 N
Example 8 Two smooth spheres of weight W and radius r each are in equilibrium in a horizontal channel of width(b
