BEAMS, JOINTS AND SUPPORTS

CONCEPT OF FRICTION

CONCEPT OF FRICTION

FRICTION IS ALL PERVADING, FRICTION IS OMNIPRESENT
FRICTION IS AN UNIVERSAL PHENOMENA

FRICTION: Friction may be defined as the resistive force acting in opposite direction in which the body tends to move or it moves. Frictional force always acts tangentially at points of contact.


Friction may be classified into two categories.


• Static Friction and
• Kinetic Friction


Static Friction is the friction experienced by a body when it is at rest under the action of external friction.


Kinetic Friction is the friction experienced by a body when it moves. Kinetic friction may be classified as Sliding Friction and Rolling Friction.


As a body can move in two ways, one is sliding and the other is rolling, there are two types of kinetic friction


• Sliding Friction and
• Rolling Friction


Sliding Friction: When a body slides over a surface without having any rotational tendency about a horizontal axis, it experiences a sliding friction. Sliding frictional force always try to dampen the movement as quickly as possible.


Rolling Friction: It is the friction experienced by a body when it rolls over the surface with an angular velocity as well as linear velocity. Rolling is a combination of translation as well as rotation about a horizontal axis passing through the centre of gravity.

Suppose we have a block of weight (W) lying on the ground as shown in the figure. The block will be at equilibrium as the weight will be neutralised due to the normal reaction provided by the ground.


Now let a horizontal (P) force is gradually applied to the block. Initially, when the force P is small, the block will not move, but if we increase the magnitude of P, then one moment will come when the block will start to move along the direction of applied force.


But from Newton's 2nd law, we know that whether the magnitude of force P is small or large, in all the cases the block should start to move with an acceleration F/m. But, practically the block starts to move when the magnitude of the applied forces reaches a definite value. Why does the block behave so? What does it mean?


Well, it means that there must exist an opposite resistance force that acts opposite to the applied force, having the same magnitude as that of applied force. But, this resistance force has a limit. When the resistance force reaches a maximum value, then the further increase in applied force can not be neutralized and as a result the body starts to move. When the value of the resistance force becomes maximum, any further increase of applied force set the movement in the body. The condition just before a body starts to move is called as "Limiting Conditions." The resistance force is called Frictional force and it becomes maximum, when the body attains the limiting condition. At this position the maximum frictional force is called "Limiting Friction."


Now just look at the experiment again, if we applied an infinitesimal force, the block doesn't move, which means the frictional force thus generated must be equal to the infinitesimal applied, else the body will experience a net force and the body will start to move. Now, if we increase force by some amount, the body will still be static, until the applied force reached the value of limiting friction. So, what does this indicate?

It indicates that frictional forces are variable. As we increase applied load, the frictional force also increases from zero to the highest value of limiting friction.


When the horizontal applied force is zero, the frictional force must be zero else the body will experience a net force. It means that when there is no applied force, frictional force just vanishes. Hence, it is known as pseudo force as its existence is dependent upon the applied force. (Remember there is another force centrifugal force which depends upon the centripetal force in rotational motion, hence, it is also a pseudo force.)


So, as the applied force starts to increase, the frictional force also increases thus maintaining the equilibrium conditions. But, the frictional force can not neutralize the applied force P, when its magnitude crosses the maximum value frictional force that can be generated due to friction on the contact surfaces.

Therefore when P > (fs)max, where (fs)max is the maximum magnitude of frictional force, the body will be in motion. When P = (fs)max, we call it as the case of impending motion or limiting condition. 

So, on what factor does the maximum frictional force on a surface depend upon? Certainly, it doesn't depend upon the applied force, although frictional force depends upon the applied force, but how much will be the maximum value of friction entirely depends upon the surface and the geometry of that surface and the Normal reaction the surface produces to counter balance the weight of the body. Here, one should remember normal reaction depends upon the mass of the body and the inclination of the surface with horizontal. 

COULOMB'S LAW OF DRY FRICTION: 

So, Coulomb's Law of dry friction states that, when there is a body resting on a surface is subjected to an applied force P, the maximum frictional force that would be generated directly depends upon the value of Normal reaction experienced by the body. 

(f_s)_max ∝ N 

N = mg cos θ, where mg is the weight of the body and θ is the inclination angle of the plane with horizonal. For a flat plane N = mg.

(f_s)_max = µN 

where µ is the proportionality constant and N is the normal reaction the body experiences from the surface. 

COEFFICIENT OF FRICTION: 

Here the constant (µ) plays a vital role. On what factor does the constant mu depend upon? It has been observed that the value of (µ) is greatly affected by the roughness of the surface upon which the body rests. It's value is a combined property of the contact surface as well as the surface roughness of the body itself. If we replace the body with another body of same mass but different material the value of (µ) changes. Also, if we place the body upon a different surface then also the value of mu changes. So, the value of mu is such a property that defines the characteristics of friction between the body and the contact surface. Hence, it is aptly named as the coefficient of friction. 

There are basically two types of Co-efficient of Friction. 

• Co-efficient of Static Friction 
• Co-efficient of Kinetic Friction 

ANGLE OF FRICTION (φ) 

When a block of mass is at rest on a surface and a horizontal force P is applied on the body to move it, a frictional force will be there to oppose any movement of the body. This force will act on the contact surface. Normal reaction is also acting upward on the contact surface. So total force on the contact surface will be resultant of normal reaction and frictional force. The angle made by this resultant force with normal reaction is called the angle of friction. 

FRICTION ON AN INCLINED PLANE: 

The direction of a frictional force depends upon the tendency of movement. 

Suppose we get two identical block of weight W in identical planes at angle α with horizontal as shown in the figure. 

Due to the force component W.sin α acting downwards along the plane, the body will have a tendency to move downwards along the plane. 

As the body would try to move downwards, a frictional force will be generated at the contact surface which would try to oppose the tendency to move downwards of the body, i.e., it would try to resist the downwards movement of the body. So, it will act upwards along the plane. 

Normal reaction produced by the inclined surface at the contact point or area. The normal reaction will be equal and opposite the force component of the weight of the body at a perpendicular direction to the inclined plane hence, N = W cos α, where N is the normal reaction. 

Now suppose we plane adjust the inclination of the plane, it means we can either increase or decrease the inclination of the plane. When the inclination is very small, the downward force component W sin α will be small and an equal magnitude frictional force will be produced and neutralize the downward force. Hence, the body will be at rest. 

Now if we increase the inclination of the plane, the downward force component W sin α will increases too, and frictional force will also be increased. Gradually, a condition will arrive when the downward weight component becomes equal to the maximum frictional force generated on the contact surface. This is limiting condition and also known as impending motion. If we increase the inclination angle α by a small amount, the body will start to move downwards. The angle of the plane when the body is at limiting condition is known as angle of repose. 

ANGLE OF REPOSE (α) 

We can define angle of repose as the angle of the inclination of a plane when a body on the plane is at limiting condition of impending motion due to its self weight component along the inclined plane. 

It is numerically equal to the angle of friction. It is denoted by (α). 

CONE OF FRICTION: 

It is an imaginary cone generated by revolving resultant reaction R about the normal reaction N. R is the resultant of the frictional force and normal reaction. 

Properties of Cone of Friction: 

• The radius of this cone represents the frictional force (fs)max. 
• The semi apex angle of the cone represents the angle of friction. 
• For co-planar forces, in order for motion not to occur the reaction R must be within the cone of friction. 

QUESTIONS BANK 5 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

QUESTIONS BANK 4 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTIONS BANK 3 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

QUESTIONS BANK 2: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

1)      Explain the principle of Super-position.

Ans: The principle of superposition states that “The effect of a force on a body does not change and remains same if we add or subtract any system which is in equilibrium.”

In the fig 4 a, a force P is applied at point A in a beam, where as in the fig 4 b, force P is applied at point A and a force system in equilibrium which is added at point B. Principle of super position says that both will produce the same effect.

2)      What is “Force-Couple system?”

Ans: When a force is required to transfer from a point A to point B, we can transfer the force directly without changing its magnitude and direction but along with the moment of force about point B.

As a result of parallel transfer a system is obtained which is always a combination of a force and a moment or couple. This system consists of a force and a couple at a point is known as Force-Couple system.

      In fig 5 a, a force P acts on a bar at point A, now at point B we introduce a system of forces  in equilibrium (fig 5 b), hence according to principle of superposition there is no change in effect of the original system. Now we can reduce the downward force P at point A and upward force P at point B as a couple of magnitude Pxd at point B (fig 5 c).

3) What do you understand by Equivalent force systems?

Ans: Two different force systems will be equivalent if they can be reduced to the same force-couple system at a given point. So, we can say that two force systems acting on the same rigid body will be equivalent if the sums of forces or resultant and sums of the moments about a point are equal.

4)      What is orthogonal or perpendicular resolution of a force?

Ans: The resolution of a force into two components which are mutually perpendicular to each other along X-axis and Y-axis is called orthogonal resolution of a force.

If a force F acts on an object at an angle θ with the positive X-axis, then its component along X-axis is F_x = Fcosθ, and that along Y-axis is F_y = Fsinθ

5) What is oblique or non-perpendicular resolution of a force?

Ans: When a force is required to be resolved in to two directions which are not perpendiculars to each other the resolution is called oblique or Non-perpendicular resolution of a force.

   
       F_OA = (P sin β)/ sin (α +β)

 F_OB = (P sin α)/ sin (α +β)

QUESTION BANK 1: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTION BANK: ENGINEERING MECHANICS

by Er. Subhankar Karmakar

Unit: 1 (Force System)

VERY SHORT QUESTIONS (2 marks):

1)      What is force and force system?

Ans: A force is a physical quantity having magnitude as well as direction. Therefore, it is a   vector quantity. It is defined as an "external agency" which produces or tends to produce or destroys or tends to destroy the motion when applied on a body.

Its unit is Newton (N) in S.I. systems and dyne in C.G.S. system.

When two or more forces act on a body or particle, it is called force system. Therefore, a force system is a collection of two or more forces.

2)      What is static equilibrium? What are the different types of static equilibrium?

Ans: A body is said to be in static equilibrium when there is no change in position as well as no rotation exist on the body. So to be in equilibrium process, there must not be any kind of motions ie there must not be any kind of translational motion as well as rotational motion.

We also know that to have a linear translational motion we need a net force acting on the object towards the direction of motion, again to induce an any kind of rotational motion, a net moment must exists acting on the body. Further it can be said that any kind of complex motion can be resolved into a translational motion coupled with a rotating motion.

“Therefore a body subjected to a force system would be at rest if and only if the net force as well as the net moment on the body is zero.”

There are three types of Static Equilibrium

1.      Stable Equilibrium

2.      Unstable Equilibrium

3.      Neutral Equilibrium

3)      What are the characteristics of a force?

Ans: A force has four (4) basic characteristics.

·         Magnitude: It is the value of the force. It is represented by the length of the arrow that we use to represent a force.

·         Direction: A force always acts along a line, which is called as the “line of action”. The arrow head we used to represent a force is the direction of that force.

·         Nature or Sense: The arrow head also represent the nature of a force. A force may be a pull or a push. If a force acts towards a particle it will be a push and if the force acts away from a point it is pull.

·         Point of Application: It is the original location of a point on a body where the force is acting. 

4)      What are the effects of a force acting on a body?

Whenever a force acts on a body or particle, it may produce some external as well as internal effects or changes.

·         A force may change the state or position of a body by inducing motion of the body. (External effect)

·         A force may change the size or shape of an object when applied on it. It may deform the body thus inducing internal effects on the body.

·         A force may induce rotational motion into a body when applied at a point other than its center of gravity.

·         A force can make a moving body into an equilibrium state at rest.

5)      What is composition and resolution of forces?

Ans: Composition of forces: Composition or compounding is the procedure to find out single resultant force of a force system

Resolution of forces: Resolution is the procedure of splitting up a single force into number of components without changing the effect of the same.

6)      What is Resultant and Equilibrant?

Ans: Resultant: The resultant of a force system is the Force which produces same effect as the combined forces of the force system would do. So if we replace all components of the force by the resultant force, then there will be no change in effect.

The Resultant of a force system is a vector addition of all the components of the force system. The magnitude as well as direction of a resultant can be measured through analytical method.

Equilibrant: Any concurrent set of forces, not in equilibrium, can be put into a state of equilibrium by a single force. This force is called the Equilibrant. It is equal in magnitude, opposite in sense and co-linear with the resultant. When this force is added to the force system, the sum of all of the forces is equal to zero.

7)      Explain the principle of Transmissibility?

Ans: The principle of transmissibility states “the point of application of a force can be transmitted anywhere along the line of action, but within the body.”

The fig 3 a shows a force F acting at a point of application A and fig 3 b, the same force F acts along the same line of action but at a different point of action at B and both are equivalent to each other.

NEW SYLLABUS FOR ENGINEERING MECHANICS: FIRST YEAR OF MTU FOR 2012-13

ENGINEERING MECHANICS

L T P

3 1 2

UNIT I

Two Dimensional Concurrent Force Systems: Basic concepts, Units, Force systems, Laws of motion, Moment and Couple, Vectors - Vector representation of forces and moments - Vector operations. Principle of Transmissibility of forces, Resultant of a force system, Equilibrium and Equations of equilibrium, Equilibrium conditions, Free body diagrams, Determination of reaction, Resultant of two dimensional concurrent forces, Applications of concurrent forces.                                                                     8

UNIT II
Two Dimensional Non-Concurrent Force Systems: Basic Concept, Varignon’s theorem, Transfer of a Force to Parallel Position, Distributed force system, Types of Supports and their Reactions, Converting force into couple and vise versa.                                                                                                                   3
Friction: Introduction, Laws of Coulomb Friction, Equilibrium of bodies involving dry-friction, Belt friction, Ladder Friction, Screw jack                                                                                                         3
Structure: Plane truss, Perfect and Imperfect Truss, Assumption in the Truss Analysis, Analysis of Perfect Plane Trusses by the Method of Joints, Method of Section.                                                           4


UNIT III
Centroid and Moment of Inertia: Centroid of plane, curve, area, volume and composite bodies,
Moment of inertia of plane area, Parallel Axes Theorem, Perpendicular axes theorems, Mass Moment of Inertia of Circular Ring, Disc, Cylinder, Sphere and Cone about their, Axis of Symmetry. Pappus-theorems, Polar moment of inertia.                                                                                                               8

UNIT IV
Kinematics of Rigid Body: Introduction, Plane Rectilinear Motion of Rigid Body, Plane Curvilinear Motion of Rigid Body, Velocity and Acceleration under Translation and Rotational Motion, Relative Velocity. 8


UNIT (V)
Kinetics of Rigid Body: Introduction, Force, Mass and Acceleration, Work and Energy, Impulse and Momentum, D’Alembert’s Principles and Dynamic Equilibrium, Friction in moving bodies.              8

Text books:
1. Engineering Mechanics Statics , J.L Meriam , Wiley
2. Engineering Mechanics Dynamics , J.L Meriam , Wiley
3. Engineering Mechanics – Statics & Dynamics by A Nelson, McGraw Hill
4. Engineering Mechanics : Statics and Dynamics, R. C. Hibbler
5. Mechanics of Solids by Abdul Mubeen, Pearson Education Asia.
6. Mechanics of Materials by E.P.Popov, Prentice Hall of India Private Limited.



ENGINEERING MECHANICS- LAB

(Any 10 experiments of the following or such experiments suitably designed)

1. Polygon law of Co-planer forces (concurrent)
2. Bell crank lever -Jib crane
3. Support reaction for beam
4. Collision of elastic bodies(Law of conservation of momentum
5. Moment of inertia of fly wheel.
6. Screw fiction by using screw jack
7. To study the slider-crank mechanism etc. of 2-stroke & 4-stroke I.C. Engine models.
8. Friction experiment(s) on inclined plane and/or on screw-jack.
9. Simple & compound gear-train experiment.
10. Worm & worm-wheel experiment for load lifting.
11. Belt-Pulley experiment. .
12. Experiment on Trusses.
13. Statics experiment on equilibrium
14. Dynamics experiment on momentum conservation
15. Dynamics experiment on collision for determining coefficient of restitution.
16. Simple/compound pendulum

SOLUTION OF EME-102; TRUSS ANALYSIS

SOLVE THE TRUSS GIVEN BELLOW WITH THE HELP OF METHODS OF JOINT

________________________________________________________________________________

a)      REPLACE JOINTS WITH REACTIONS at A and at B

              

       

b)      Draw FBD of the TRUSS

 

Applying the conditions of Equilibrium of Coplanar Non-concurrent Force System,

 

∑ F_X = 0;        R_b – R_ah = 0  ------ (i)

(-) ← ● → (+)

∑ F_Y = 0;        R_av – 10 – 5 – 15 = 0 => R_av = 30 kN ----- (ii)

∑ M_A = 0;       10 x 4 + 5 x 4 + 15 x 2 – R_b x 3 = 0  ----- (iii)

                        R_b = 30 kN

                        Hence R_ah = R_b = 30 kN

Calculation of Angle θ

The angle θ = tan^-1(3/2) = 56.3°

All the unknown forces will be taken as Tensile, if their magnitudes is found negative, then they will be treated as compressive forces.

First we shall choose a joint having only two unknown forces, either we shall choose joint D or joint A

Let us choose joint D first.

We shall consider point D first, as it has only two unknown force. FBD of the point D is drawn.

∑ F_X = 0;      F₂ = 0

∑ F_Y = 0;      F₁ – 5 = 0

                      F₁ = 5 kN

 

Our next joint will be point E. FBD of the joint E is drawn. As F₁ = 5 kN, hence unknown forces are two. F₃ and F₄

∑ F_X = 0;     – F₃ – F₄ cos 56.3° = 0

∑ F_Y = 0;     – F₁ – 10 –  F₄ sin 56.3° = 0  [ as F₁ = 5 kN]

                           F₄ = –15/ sin 56.3° = – 18.02 kN

                    F₃ = – F₄ cos 56.3° = 10 kN

 

Our next joint is C

 F₅ and F₉ are unknown where as F₄ = – 18.02 kN

 F₂ = 0

 ∑ F_X = 0;     – F₉ + F₄ cos 56.3° = 0

            F₉ = F₄ cos 56.3° = 10 kN

 ∑ F_Y = 0;     F₅ + F₄ sin 56.3° – 15 = 0

                            F₅ = – F₄ sin 56.3° + 15 = 30 kN

 

F₃ = 10 kN;   F₅ = 30 kN

        ∑ F_X = 0;     F₃ = F₆ + F₇ cos 56.3°

        ∑ F_Y = 0;     – F₅ – F₇ sin 56.3° =0

       F₇ = – F₅/ sin 56.3° = – 36.05 kN

F₆ = F₃ – F₇ cos 56.3° = 10 + 20 = 30 kN

R_av = 30 kN;  R_ah = 30 kN

 ∑ F_X = 0;      F₆  = R_ah = 30 kN

 ∑ F_Y = 0;     F₈ = R_av = 30 kN

Sl no	Link	Force	Magnitude	Nature
01	ED	F1	5 kN	T
02	CD	F2	0
03	FE	F3	10 kN	T
04	CE	F4	18.02 kN	C
05	FC	F5	30 kN	T
06	AF	F6	30 kN	T
07	BF	F7	36.05 kN	C
08	AB	F8	30 kN	T
09	BC	F9	10 kN	T