CONCEPTS OF BASIC THERMODYNAMICS

¤ Introduction:

The most of general sense of thermodynamics is the study of energy and its relationship to the properties of matter. All activities in nature involve some interaction between energy and matter. Thermodynamics is a science that governs the following:

• (i) Energy and its transformation
• (ii) Feasibility of a process involving transformation of energy
• (iii) Feasibility of a process involving transfer of energy
• (iv) Equilibrium processes

More specifically, thermodynamics deals with energy conversion, energy exchange and the direction of exchange.

¤ Areas of Application of Thermodynamics:

All natural processes are governed by the principles of thermodynamics. However, the following engineering devices are typically designed based on the principles of thermodynamics.

Automotive engines, Turbines, Compressors, Pumps, Fossil and Nuclear Power Plants, Propulsion systems for the Aircrafts, Separation and Liquefaction Plant, Refrigeration, Air-conditioning and Heating Devices.

The principles of thermodynamics are summarized in the form of a set of axioms. These axioms are known as four thermodynamic laws:

• Zeroth law of thermodynamics,
• First law of thermodynamics,
• Second law of thermodynamics, and
• Third law of thermodynamics.

The Zeroth Law deals with thermal equilibrium and provides a means for measuring temperatures.

The First Law deals with the conservation of energy and introduces the concept of internal energy.

The Second Law of thermodynamics provides with the guidelines on the conversion of internal energy of matter into work. It also introduces the concept of entropy.

The Third Law of thermodynamics defines the absolute zero of entropy. The entropy of a pure crystalline substance at absolute zero temperature is zero.


¤ Different Approaches in the Study of Thermodynamics:

There are two ways through which the subject of thermodynamics can be studied

• Macroscopic Approach
• Microscopic Approach

¤ Macroscopic Approach:

Consider a certain amount of gas in a cylindrical container. The volume (V) can be measured by measuring the diameter and the height of the cylinder. The pressure (P) of the gas can be measured by a pressure gauge. The temperature (T) of the gas can be measured using a thermometer. The state of the gas can be specified by the measured P, V and T . The values of these variables are space averaged characteristics of the properties of the gas under consideration. In classical thermodynamics, we often use this macroscopic approach. The macroscopic approach has the following features.

• The structure of the matter is not considered.
• A few variables are used to describe the state of the matter under consideration. The values of these variables are measurable following the available techniques of experimental physics.

¤ Microscopic Approach:

On the other hand, the gas can be considered as assemblage of a large number of particles each of which moves randomly with independent velocity. The state of each particle can be specified in terms of position coordinates ( x_i , y_i , z_i ) and the momentum components ( p_xi , p_yi , p_zi ). If we consider a gas occupying a volume of 1 cm³ at ambient temperature and pressure, the number of particles present in it is of the order of 10²⁰. The same number of position coordinates and momentum components are needed to specify the state of the gas. The microscopic approach can be summarized as:

• A knowledge of the molecular structure of matter under consideration is essential.
• A large number of variables are needed for a complete specification of the state of the matter.

¤ Zeroth Law of Thermodynamics: 

This is one of the most fundamental laws of thermodynamics. It is the basis of temperature and heat transfer between two systems. Suppose we take three thermodynamic system named System A, System B and System C. Now let that system A is in thermal equilibrium with system B. By thermal equilibrium we mean that there is no heat transfer between system A and system B when they are brought in contact with each other. Now, suppose system A is in thermal equilibrium with system C too and there is no contact between system B and system C. It implies that although system B and C are isolated from each other, they will remain at thermal equilibrium to each other. It means that there will be no heat transfer between system B and C, when they are brought in contact with each other. This is called the Zeroth Law of thermodynamics.


¤ Basis of Temperature: 

When two bodies are kept at contact with each other and if there is no heat transfer between them we say that their body temperatures are same. It means that temperature is the property of a system which decides whether there will be any heat transfer between two different bodies. Heat transfer always occur from a higher temperature body to a lower temperature body. Further whenever there is any heat inflow to a body, it raises its temperature and conversely, if heat outflow occurs from a system it lowers its temperature.

Suppose we take two bodies one of which is at higher temperature than the other. Now when we bring the bodies at contact, heat will be transformed from a higher temperature body to that of lower temperature. Then what will be its effect, we may ask as a result of this heat transfer? Is this heat transfer a perpetual process? Our common life experiences tell us that it will not be the case. Although, at first heat transfer will take place, but its amount will be gradually decreased and after some time, a situation will come when there will be no heat transfer between the bodies or the bodies will come to a state of thermal equilibrium with each other. So, what is the reason for that? Can we justify the situation?

Yes, we can justify it as the hotter body releases heat to the colder body, the temperature of the hotter body decreases where as the temperature of the colder body increases and after sufficient time both the bodies will have equal temperature and a state of thermal equilibrium will be achieved.


¤ Temperature Measurement: 

We know the temperature of a body can be measured with a thermometer. How can we actually calculate the temperature of a body with the help of thermodynamics?


¤ Thermometer:

A thermometer is a temperature measuring instrument. It is made of a thin capillary glass tube, one end is closed and the other end is fitted with metallic bulb full of mercury. The mercury is in thermal equilibrium with the metallic bulb. Therefore, the temperature of the mercury is equal to the temperature of the metallic bulb. 
Mercury has a good coefficient of volume expansion and it means that as the temperature of the mercury increases, its volume increases too and as a result mercury column inside the capillary rises up. 

The capillary tube has been graduated with the help of calibrating with standard temperature sources. Therefore, the temperature of the mercury can be measured from the height of mercury column as the tube is finely graduated. 

Whenever we want to measure the temperature of a body, we kept the body in contact with the metallic bulb of the thermometer. When thermal equilibrium is established between the body and the metallic bulb of the thermoneter, the temperature of both the body will be equal again the metallic bulb is in thermal equilibrium with mercury then the temperature of the mercury will be equal to the temperature of the metallic bulb and the temperature of the object.


As we can measure the temperature of the mercury from the column height, hence we can also determine the temperature of the object as they are equal to each other.

DISCUSSION:
Microscopic basis of temperature and pressure:
Here we shall try to discuss the basis of temperature and pressure only qualitatively, without any mathematical expression. 






.....................contact me at email: subhankarkarma@gmail.com for more notes

SOLUTION OF EME-102; EQUILIBRIUM OF FORCES

EQUILIBRIUM OF FORCES IN 2D

A light string ABCDE whose extremity A is fixed, has weights W₁ & W₂ attached to it at B & C. It passes round a small smooth pulley at D carrying a weight of 300 N at the free end E as shown in figure. If in the equilibrium position, BC is horizontal and AB & CD make 150° and 120° with BC, find (i) Tensions in the strings and (ii) magnitudes of W₁ & W₂

Although ABCDE is a single string/rope but still the tensions in the string/rope will be different at different segments like in the segment AB the tensions will be T₁ , but in BC segment it will be different as the weight is attached at a fixed point (point B) on the string, hence it will be T₂ here and in CD it will be T₃  there. As at point D the string is not attached rather passes over a smooth pulley hence the tension in DE and CD will be same ie. T₃  again.

. 

To solve for equilibrium of forces in 2D, follow these steps:

1. Draw a free body diagram: Draw a diagram of the object in question and identify all the forces acting on it. This will help you visualize the problem and identify any unknown forces or angles.

2. Break forces into components: Resolve all forces into their x- and y-components. This is done by using trigonometry to determine the horizontal and vertical components of each force.

3. Apply Newton's second law: For the object to be in equilibrium, the sum of the forces in the x-direction and the sum of the forces in the y-direction must be equal to zero. This gives you two equations to solve simultaneously.

4. Solve the equations: Solve the equations for the unknown forces or angles by using algebraic methods. This will give you the values of the unknown forces or angles required for the object to be in equilibrium.

5. Check for consistency: Check that the forces and angles you have calculated are consistent with the problem and the free body diagram. For example, make sure that the direction of the forces makes sense and that the angles are reasonable.

6. Interpret the results: Interpret the results and explain what they mean in the context of the problem. This might involve calculating the tension in a rope, the force required to lift an object, or the angle required for an object to remain stationary.

QUESTION BANKS: Analyse the following Trusses:

 Analyse the following Trusses:

(1)      A cantilever truss has been as shown in the figure. Find the value of W which will produces a force of magnitudes 15 kN  in the member AB.

(2)        A cantilever truss is loaded as shown in the figure. Find the nature and magnitudes of the forces in each link.

(3)      A cantilever truss has been as shown in the figure. Find the value of W which will produces a force of magnitudes 15 kN  in the member AB.

(4)      A truss has been loaded as shown in the figure. Find the nature and the magnitudes of the forces in the links BC, CH and GH by the methods of sections.

(5) A truss has been loaded as shown in the figure. Find the internal forces in each of the beam.

(6)      A truss has been loaded as shown in the figure. Find the forces in each member and tabulate them by any methods.

compiled by Subhankar Karmakar

more content: click the following links for more questions.

THEORETICAL QUESTIONS ON SIMPLE TRUSSES part-3

QUESTION BANK : ENGINEERING MECHANICS PART-2

QUESTION BANK : ENGINEERING MECHANICS

 

THEORETICAL QUESTIONS ON SIMPLE TRUSSES

SHORT QUESTIONS: TOPIC - TRUSS ANALYSIS

1)      What is a truss? Classify them with proper diagrams.

2)      State the differences between a perfect truss and an imperfect truss.

3)      Distinguish between a deficient truss and a redundant truss.

4)      Write the Maxwell’s Truss Equation.

5)      What are the assumptions made, while finding out the forces in the various members of a truss?

6)      What are the differences between a simply supported truss and a cantilever truss? Discuss the method of finding out reactions in both the cases.

Analyse the following Trusses:

 

(1)      Analyse the Truss by the method of Joints.

(2) Find the internal forces on the links 1, 2 and 3 by the method of Sections.

(3) Determine the magnitude and the nature of the forces in the members BC, GC and GF of the given truss.

(4)   A truss of span 10 m is loaded as shown in the figure. Find the forces in all the links by any method.






compiled by Subhankar Karmakar 

QUESTION BANK : ENGINEERING MECHANICS PART-2

TOPICS: NUMERICALS ON FORCE SYSTEM- UNIT-1

5) A bar of AB 12 m long rests in horizontal position on two smooth planes as shown in the figure. Find the distance X at which 100 kN is to be placed to keep the bar in equilibrium.

 

6) A light string ABCDE whose extremity A is fixed, has weights W₁ & W₂ attached to it at B & C. It passes round a small smooth pulley at D carrying a weight of 300 N at the free end E as shown in figure. If in the equilibrium position, BC is horizontal and AB & CD make 150° and 120° with BC, find (i) Tensions in the strings and (ii) magnitudes of W₁ & W₂  

 

7) Find reactions at all the contact points if weight of P is 200 N & diameter is 100 mm, where as weight of Q is 500 N and diameter is 180 mm.

 

8) Determine the force P required to begin rolling the uniform cylinder of mass (m) over the obstacle of height (h) as shown in the figure.  

 

9) A roller of weight 500 N has a radius of 120 mm and is pulled over a step of height 60 mm by a horizontal force P. Find magnitudes of P to just start the roller over the step.

 

10) Two identical rollers each of weight 100 N are supported by an inclined plane of 30° with horizontal and a vertical wall as shown in the figure. Find all the reactions at each contact point.

 

11) A smooth cylinder of radius 500 mm rests on a horizontal plane and is kept from rolling by a rope OA of 1000 mm length. A bar AB of length 1500 mm and weight 1000 N is hinged at point A and placed against the cylinder of negligible weight. Determine the tension in the rope.

 

12)      A flat belt connects pulley B, which drives a pulley A; attached to an electric motor. μ_s =  0.25 and μ_k = 0.2 between both the pulleys and the belt. If maximum allowable tension in the belt is 600 N, determine the largest torque which can be exerted by belt on pulley B.

      

13)       Two blocks of mass MA & MB are kept at equilibrium as shown in the figure. The friction between the block B & the floor is 0.35 and between the blocks is 0.3, then find the minimum force P to just move the block B.

CENTROID OF A COMPOSITE LINE

In the figure, a composite line A-B-C-D is made of three straight lines AB, BC, CD as shown in the figure. If the coordinate of point A is (5,5), find the centroid of the composite line.


Solution: At first, the composite line is divided into three parts.








Part -1 : The line AB : Let the centroid of the line be G₁(X₁,Y₁)

length, L₁ = 40 mm;                  


X₁ = 4 + (40*cos 60⁰)/2 = 14  
Y₁ = 4 + (40*sin 60⁰)/2 = 21.32








Part -2 : The line BC : Let the centroid of the line be G₂(X₂,Y₂)


length, L₂ = 15 mm; 


X₂ = 4 + (40*cos 60⁰) + 15/2 = 31.5 
Y₂ = 4 + (40*sin 60⁰) = 38.64




Part -3 : The line CD : Let the centroid of the line be G₃(X₃,Y₃)


length, L₃ = 20 mm; 

X₃ = 4 + (40*cos 60⁰) + 15 = 39 
Y₃ = 4 + (40*sin 60⁰) - 20/2 = 28.64



If the centroid of the composite line be G  (X_g,Y_g)

X_g = (∑L_iX_i)/(∑L_i) 

    = (L₁X₁ + L₂X₂ + L₃X₃)/(L₁ + L₂ + L₃)
    = (40 x 14 + 15 x 31.5 + 20 x 39)/(40 + 15 + 20) 
    = 24.17
     

Y_g = (∑L_iY_i)/(∑L_i) 

    = (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)
    = (40 x 21.32 + 15 x 38.64 + 20 x 28.64)/(40 + 15 + 20) 
    = 26.74

CENTROID OF COMPLEX GEOMETRIC FIGURES:

So in this articles, we are going to discuss the concepts of centroid for one dimensional as well as two dimensional objects. 

Let's first discuss about 1D and 2D objects, one by one, an 1D object is a line, practically a metallic rod will be considered as a linear, 1D object. Where as any thin plate of negligible thickness can be considered a 2D body. Suppose we have a thin metallic rectangular plate. If it is considered as a 2D rectangular area of b X h.

The concept of centroid has been developed on the basis of resultant of several areas. We know that an area can be represented as the cross product of two vectors, hence it is also an vector. Suppose we have an area A, in a cartesian 2D coordinate system. We just divide the area into n parts, and represent them as a₁, a₂, a₃, ..... a_n.

Let the elemental areas are at a distance x₁, x₂, x₃, ..... x_n, from Y axis and y₁, y₂, y₃, ...y_n from X axis.

The total moments produced about Y axis will be equal to the summation of all the individual moments produced by n elemental areas. 

Now moment is a vector quantity and we know vectors of same kind can be added together, therefore, all the n moment vectors can be added to get a single value of Resultant Moment. 

We also know this resultant moment's position vector. Let the resultant moment passes through a point G. The point through which resultant moment passes through is called Center of the Area or Centroid.

How can we find out the point G, whose coordinates are (X_g,Y_g)?

As moment of an area also obeys VARIGNON'S THEOREM OF MOMENT, then sum of all the moments produced by individual elemental areas will exactly be equal to the moment produced by the total area, i.e. the resultant of all those elemental areas. Now if all the areas are added to have the resultant area which will pass through the centroid G such that it produces a moment of X_gA about Y axis and Y_gA about X axis.

But Varignon's theorem states us that, for a vector system, resultant vector produces the moment about a point, is exactly equal to the sum of all the moments produced by all elemental areas about the same point and in the same plane. Hence, we can write now that,

Sum(a₁x₁+ a₂x₂+ + +a_nx_n) = AX_g
we can use summation sign ∑ to represent these equations,
∑a_ix_i = (∑a_i)X_g
=> X_g = (∑a_ix_i)/((∑a_i)

Sum(a₁y₁+ a₂y₂+ + +a_ny_n) = AY_g
∑a_iy_i = (∑a_i)Y_g
=> Y_g = (∑a_iy_i)/((∑a_i)

Algorithm to find out the Centroid G(X_g, Y_g) of a Complex Geometric Figure.

Step1:
Take a complex 2D figure like an Area or Lamina.

Step2:
Try to identify the basic figures whose algebraic combination produces our problem figure, whose centroid we shall find out.

Step3:
Choose a coordinate system, and make it as our frame of reference. All the distances and coordinate must be define with respect to our frame of reference.

Step4:
Compute the area (a_i), coordinates of their own centroid G_i (x_i, y_i) for each and every elemental areas. While measuring the centroids, all the measurements will be based on according to our chosen Axes.

Step5:
If any particular area has to subtracted to get the complex figure, the area will be negative, where as any area addition will be positive area.

Step6:
If the Centroid of the complex figure be G(X_g,Y_g)then,

=> X_g = (∑a_ix_i)/((∑a_i)

=> Y_g = (∑a_iy_i)/((∑a_i)

Here G₁ is the centroid of the part one where G₂ is the centroid of the circular area that has to be removed where as G₃ is the centroid of the triangular area that has to be removed also.

If we are asked to find moment of inertia of an area, which is nothing but the "second moment of area" then we shall have to find the centroidal moment of inertia first. Then we shall transfer the Moment of Inertia to another axis ie we shall apply parallel axis theorem to transfer moment of inertia from one axis (here centroidal axis) to another parallel axis.

SECOND SESSIONAL TEST (odd SEMESTER 2009-10) B.Tech…first Semester Sub Name: Engineering Mechanics

SECOND SESSIONAL TEST (odd SEMESTER 2009-10)

B.Tech…first Semester

Sub Name: Engineering Mechanics                                         Max. Marks: 30

Sub Code: EME-201                                                          Max. Time: 2: 00 Hr

Group A

Q.1 Choose the correct answer of the following questions 1x6=6

(i) If two forces of equal magnitudes have a resultant force of the same magnitude then the angle between them is

(a) 0⁰          (b) 90⁰         (c) 120⁰          (d) 135⁰

Ans: ( c)

Explanation: P² = P² + P² +2P.P.cos θ

ð      P²= P² (1+2cos θ)

ð      cos θ = -½

ð      θ = 120°

(ii) If a ladder is kept at rest on a vertical wall making an angle θ with horizontal. If co-efficient of the friction in all the surfaces be µ, then the tangent of the angle θ will be equal to

(a) (1-µ²)/2µ                                  (b) (1-µ)² /2µ

(c) (1-µ)/2µ                                   (d) none of the above

Explanation: (a) (1-µ²)/2µ

(iv) Varignon’s theorem is related with _____________ .

Answer: moment

(v) If two forces of equal magnitudes P having an angle (90°- θ) between them, then their resultant force will be equal to ________.

Answer: √2P (1+sin θ)

(vi) A fixed joint produces

(a) 1             (b) 2             (c) 3               (d) 4         reactions

Answer: (c ) 3

(vii) The equilibrium conditions of concurrent force system is_________.

Answer: ∑Fx =0; ∑Fy=0.

Group B                                                                                                                  8x3=24

Attempt any three questions

Q.2. State and explain Varignon’s theorem of moment. Three forces of magnitudes 3 KN, 4 KN and 2KN act along the three side of an equilateral triangle ∆ABC in order. Find the position, direction and magnitude of the resultant force.         4+4

Answer: resultant force: √3 KN

Q.3  (a) Two cylindrical rollers are kept at equilibrium inside a jar or channel as shown in the figure. The channel width is 1000 mm, where as the rollers have diameters 600 mm and 800 mm respectively. The weights are 2 kN and 5 kN respectively. Find all the reactions at contacts.

                                                                                            4+4

(b) What is pure bending? If a stone is thrown with a velocity 400 m/s then find the maximum height that the stone can reach.

Q:4)  a) Classify different types of joints in beam with proper explanations.

(b) Find the reactions at the support for the beam as shown in the figure.                                           4+4

ENGINEERING. MECHANICS:

FORCE AND FORCE SYSTEM

Topic: FORCE SYSTEM

1) What is a FORCE SYSTEM? Classify them with examples and diagrams.

Ans: A force system may be defined as a system where more than one force act on the body. It means that whenever multiple forces act on a body, we term the forces as a force system. We can further classify force system into different sub-categories depending upon the nature of forces and the point of application of the forces.

Different types of force system:


(i) COPLANAR FORCES:

If two or more forces rest on a plane, then they are called coplanar forces. There are many ways in which forces can be manipulated. It is often easier to work with a large, complicated system of forces by reducing it an ever decreasing number of smaller problems. This is called the "resolution" of forces or force systems. This is one way to simplify what may otherwise seem to be an impossible system of forces acting on a body. Certain systems of forces are easier to resolve than others. Coplanar force systems have all the forces acting in in one plane. They may be concurrent, parallel, non-concurrent or non-parallel. All of these systems can be resolved by using graphic statics or algebra.


(ii) CONCURRENT FORCES:

A concurrent coplanar force system is a system of two or more forces whose lines of action ALL intersect at a common point. However, all of the individual vectors might not actually be in contact with the common point. These are the most simple force systems to resolve with any one of many graphical or algebraic options. If the line of actions of two or more forces passes through a certain point simultaneously then they are called concurrent forces. concurrent forces may or may not be coplanar.

(iii) LIKE FORCES:

A parallel coplanar force system consists of two or more forces whose lines of action are ALL parallel. This is commonly the situation when simple beams are analyzed under gravity loads. These can be solved graphically, but are combined most easily using algebraic methods. If the lines of action of two or more forces are parallel to each other, they are called parallel forces and if their directions are same, then they are called LIKE FORCES.

(iv) UNLIKE FORCES: If the parallel forces are such that their directions are opposite to each other, then they are termed as "UNLIKE FORCE".


(v) NON COPLANAR FORCES:
The last illustration is of a "non-concurrent and non-parallel system". This consists of a number of vectors that do not meet at a single point and none of them are parallel. These systems are essentially a jumble of forces and take considerable care to resolve.

_________________________________________________________________________________
N.B. Almost any system of known forces can be resolved into a single force called a resultant force or simply a Resultant. The resultant is a representative force which has the same effect on the body as the group of forces it replaces. (A couple is an exception to this) It, as one single force, can represent any number of forces and is very useful when resolving multiple groups of forces. One can progressively resolve pairs or small groups of forces into resultants. Then another resultant of the resultants can be found and so on until all of the forces have been combined into one force. This is one way to save time with the tedious "bookkeeping" involved with a large number of individual forces. Resultants can be determined both graphically and algebraically.The Parallelogram Method and the Triangle Method. It is important to note that for any given system of forces, there is only one resultant.


It is often convenient to decompose a single force into two distinct forces. These forces, when acting together, have the same external effect on a body as the original force. They are known as components. Finding the components of a force can be viewed as the converse of finding a resultant. There are an infinite number of components to any single force. And, the correct choice of the pair to represent a force depends upon the most convenient geometry. For simplicity, the most convenient is often the coordinate axis of a structure.


A force can be represented as a pair of components that correspond with the X and Y axis. These are known as the rectangular components of a force. Rectangular components can be thought of as the two sides of a right angle which are at ninety degrees to each other. The resultant of these components ...


is the hypotenuse of the triangle. The rectangular components for any force can be found with trigonometrical relationships: F_x = Fcosθ, Fy = Fsinθ. There are a few geometric relationships that seem to common in general building practice in North America. These relationships relate to roof pitches, stair pitches, and common slopes or relationships between truss members. Some of these are triangles with sides of ratios of 3-4-5, 1-2-sqrt3, 1-1-sqrt2, 5-12-13 or 8-15-17. Committing the first three to memory will simplify the determination of vector magnitudes when resolving more difficult problems.


When forces are being represented as vectors, it is important to should show a clear distinction between a resultant and its components. The resultant could be shown with color or as a dashed line and the components as solid lines, or vice versa. NEVER represent the resultant in the same graphic way as its components.


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. A non-concurrent or a parallel force system can actually be in equilibrium with respect to all of the forces, but not be in equilibrium with respect to moments.
__________________________________________________________________________________


2) What is STATIC EQUILIBRIUM? 
    What are the conditions of static equilibrium for
            (i) concurrent force system
            (ii) coplanar non concurrent force system.

Ans: A body is said to be in 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 be zero. Therefore the general condition of any system to be in static equilibrium we have to satisfy two conditions

(i) Net force on the body must be zero ie, ΣF_i = 0;
(ii) Net moment on the body must be zero ie, ΣM_i = 0.

Now we can apply these general conditions to different types of Force System.

For concurrent force system total moment about the concurrent point is always zero as all the forces pass through the point, and we know the moment of a force passing through the point about which we shall take moment is always zero. Hence, the conditions of equilibrium for concurrent forces will be  

Net force on the body must be zero ie, ΣF_i = 0; and we can resolve it along X axis and along Y axis, ie.  (i) ΣF_x = 0; and  (ii) ΣF_y = 0.

for coplanar non concurrent force system, the equilibrium conditions are

(i) ΣF_x = 0; and  (ii) ΣF_y = 0.  (iii)  ΣM_i = 0.

 Moment on a plane:

For a force system the total resultant moment about any arbitrary point due to the individual forces are equal to the moment produced by the resultant about the same point. Now if the system is at equilibrium condition, then the resultant force would be zero. Hence, the moment produced by the resultant about any arbitrary point is zero. In case of coplanar & concurrent force system, as the forces are concurrent ie. each of the force passes through a common point. Hence, about that common point total moment of all the forces will be zero.

3) What are different types of joint? discuss them in details.

Answer: The Concepts of Joints. In Engineering terminology any force carrying linear member is called as links. Links can be attached to each other by the fasteners or joints. Hence, we can say to prevent the relative motion between two links completely or partially we use fasteners or joints.



Basically there are three types of joints which we shall discuss and they are named as,
(i) pin/ hinged joints, 
(ii) roller joints and 
(iii) fixed joints.


PIN JOINTS:

They are classified according to the degrees of freedom of the links they would allow. Like a pin or hinge joint is consisted of two links joined by the insertion of a pin at the pivot hole. A pin joint doesn't allow a vertical or horizontal relative velocities between the two links.

For better understanding of the mechanism of pin joint we would like to make a simplest type of pin joints. Suppose we would take two links and make holes at one of the ends of each link. Now if we insert a bolt through the holes of both the links, then what we get is an example of pin/hinge joints.

A pin joint although restricts any kind of horizontal or vertical displacement but they can not restrict rotation about an axis passing through the hole, in clockwise or anti clockwise direction. Hence it provides two reactions one vertical and one horizontal to restrict any kind of movement along that direction.

ROLLER JOINTS:


 

MULTIPLE CHOICE QUESTIONS: sub: engg. mechanics. Sub: Engineering Mechanics, Sub Code: EME-202, Semester: 2nd Sem, Course: B.Tech

Q.1) The example of Statically indeterminate structures are,
a. continuous beam,
b. cantilever beam,
c. over-hanging beam,
d. both cantilever and fixed beam.

Q.2) A redundant truss is defined by the truss satisfying the equation,
a. m = 2j - 3,
b. m < 2j + 3, 

c. m > 2j - 3,
d. m > 2j + 3

Q.3) The property of a material to withstand a sudden impact or shock is called,
a. hardness 
b. ductility, 
c. toughness, 
d. elasticity of the material

Q.4) The stress generated by a dynamic loading is approximately _____ times of the stress developed by the gradually applying the same load.

Q.5) The ratio between the volumetric stress to the volumetric strain is called as
a. young's modulus
b. modulus of elasticity
c. rigidity modulus,
d. bulk modulus

Q.6) In a Cantilever beam, the maximum bending moment is induced at
a. at the free end
b. at the fixed end
c. at the mid span of the beam
d. none of the above

Q.7) The forces which meet at a point are called
a. collinear forces
b. concurrent forces
c. coplanar forces
d. parallel forces

Q.8) The coefficients of friction depends upon
a. nature of the surface
b. shape of the surface
c. area of the contact surface
d. weight of the body

Q.9) The variation of shear force due to a triangular load on simply supported beam is
a. uniform 
b. linear 
c. parabolic 
d. cubic

Q.10) A body is on the point of sliding down an inclined plane under its own weight. If the inclination of the plane is 30 degree, then the coefficient of friction between the planes will be

a. 1/√3

b. √3

c. 1

d. 0

11. A force F of 10 N is applied on a mass of 2 kg. What is the acceleration of the mass?

A. 2 m/s²

B. 5 m/s²

C. 10 m/s²

D. 20 m/s²

Answer: B

12. What is the moment of a force of 50 N applied at a distance of 2 meters from a fixed point?

A. 25 Nm

B. 50 Nm

C. 100 Nm

D. 200 Nm

Answer: C

13. A 2000 kg car traveling at 20 m/s collides with a 500 kg car traveling at 10 m/s in the opposite direction. What is the velocity of the cars after the collision?

A. 6.7 m/s

B. 10 m/s

C. 13.3 m/s

D. 16.7 m/s

Answer: A

14. A 500 N force is applied to a 100 kg object on a flat surface. What is the coefficient of static friction if the object is just about to move?

A. 0.5

B. 0.7

C. 0.8

D. 1.0

Answer: D

15. A beam of length 4 m and moment of inertia of 1000 kg/m² is supported at each end. What is the maximum load that the beam can support if it is uniformly loaded?

A. 500 N

B. 1000 N

C. 2000 N

D. 4000 N

Answer: C

16. A block of mass 2 kg is hanging from a string. What is the tension in the string if the block is stationary?

A. 19.6 N

B. 20 N

C. 29.4 N

D. 30 N

Answer: B

17. A roller coaster car of mass 500 kg is traveling at 20 m/s at the bottom of a  loop-the-loop. What is the minimum radius of the loop required for the car to remain in contact with the track?

A. 40 m

B. 50 m

C. 60 m

D. 70 m

Answer: D

18. A body of mass 10 kg is moving with a velocity of 5 m/s. What is the kinetic energy of the body?

A. 50 J

B. 100 J

C. 125 J

D. 250 J

Answer: B

19. A body of mass 5 kg is placed on an inclined plane which makes an angle of 30° with the horizontal. What is the force acting on the body parallel to the plane?

A. 4.9 N

B. 7.5 N

C. 8.7 N

D. 10 N

Answer: B

20. A force of 100 N is applied on a body of mass 20 kg. What is the work done by the force in moving the body through a distance of 5 meters?

A. 250 J

B. 500 J

C. 1000 J

D. 2000 J

Answer: B

21. What is the principle of moments?

A. The sum of the moments about any point of a system in equilibrium is zero.

B. The sum of the forces acting on a system in equilibrium is zero.

C. The sum of the torques acting on a system in equilibrium is zero.

D. The sum of the accelerations of a system in equilibrium is zero.

Answer: A

22. What is the difference between static and dynamic equilibrium?

A. In static equilibrium, there is no motion, while in dynamic equilibrium, there is motion.

B. In static equilibrium, the forces are balanced, while in dynamic equilibrium, the forces are unbalanced.

C. In static equilibrium, the sum of the forces and moments is zero, while in dynamic equilibrium, the sum of the forces and moments is not zero.

D. In static equilibrium, the sum of the forces and moments is not zero, while in dynamic equilibrium, the sum of the forces and moments is zero.

Answer: C

23. What is the moment of inertia?

A. The resistance of an object to angular acceleration.

B. The force required to rotate an object.

C. The distance between the center of mass and the axis of rotation.

D. The angular velocity of an object.

Answer: A

24.What is the difference between stress and strain?

A. Stress is the deformation per unit length, while strain is the force per unit area.

B. Stress is the force per unit area, while strain is the deformation per unit length.

C. Stress is the force applied to an object, while strain is the resulting deformation.

D. Stress is the resistance of an object to deformation, while strain is the resistance of an object to stress.

Answer: B

25. What is Hooke's Law?

A. The stress applied to an elastic material is proportional to the strain produced.

B. The strain produced in an elastic material is proportional to the stress applied.

C. The deformation produced in an elastic material is proportional to the force applied.

D. The force applied to an elastic material is proportional to the deformation produced.

Answer: A

26.What is the difference between a beam and a truss?

A. A beam is a one-dimensional structure, while a truss is a two-dimensional structure.

B. A beam is made up of several members connected at their ends, while a truss is made up of several members connected at their joints.

C. A beam is used to support loads that are perpendicular to its axis, while a truss is used to support loads that are parallel to its axis.

D. A beam is a rigid structure, while a truss is a flexible structure.

Answer: B

27. What is the difference between a force and a moment?

A. A force is a vector quantity, while a moment is a scalar quantity.

B. A force is a scalar quantity, while a moment is a vector quantity.

C. A force is a push or a pull, while a moment is a twist or a turn.

D. A force is a linear motion, while a moment is a rotational motion.

Answer: C

28. What is the center of mass?

A. The point where the weight of an object is concentrated.

B. The point where the forces acting on an object are balanced.

C. The point where the moments acting on an object are balanced.

D. The point where the acceleration of an object is zero.

Answer: A

29. What is the method used to determine the forces in a truss?

A. Method of joints

B. Method of sections

C. Both A and B

D. None of the above

Answer: C

30. In a truss, which members are in tension and which members are in compression?

A. All members are in tension.

B. All members are in compression.

C. Members with angled force vectors are in tension, and members with vertical force vectors are in compression.

D. Members with vertical force vectors are in tension, and members with angled force vectors are in compression.

Answer: C

31. What is the difference between a simple truss and a compound truss?

A. A simple truss is made up of one triangle, while a compound truss is made up of two or more triangles.

B. A simple truss is made up of straight members only, while a compound truss may have curved members.

C. A simple truss is statically determinate, while a compound truss may be statically indeterminate.

D. A simple truss is used for short spans, while a compound truss is used for long spans.

Answer: A

32.How many unknown forces are there in a simple truss?

A. 2

B. 3

C. 4

D. It depends on the number of joints in the truss.

Answer: B

33. What is the method used to analyze a truss with multiple loadings?

A. Superposition method

B. Substitution method

C. Iterative method

D. None of the above

Answer: A

34. What is the maximum number of reactions that can be present in a truss?

A. 1

B. 2

C. 3

D. 4

Answer: B

35. What is the difference between a statically determinate and a statically indeterminate truss?

A. A statically determinate truss has only one solution for the unknown forces, while a statically indeterminate truss may have more than one solution.

B. A statically determinate truss has more unknown forces than the number of equations available to solve them, while a statically indeterminate truss has fewer unknown forces than the number of equations available to solve them.

C. A statically determinate truss is easier to analyze, while a statically indeterminate truss requires more advanced techniques.

D. A statically determinate truss is always more efficient than a statically indeterminate truss.

Answer: C

36. What is the difference between a pinned support and a roller support?

A. A pinned support allows rotation but not translation, while a roller support allows translation but not rotation.

B. A pinned support allows both rotation and translation, while a roller support allows neither.

C. A pinned support is used for horizontal loads, while a roller support is used for vertical loads.

D. A pinned support is always more stable than a roller support.

Answer: A

37. What is the maximum number of members that can be present in a simple truss?

A. 2n-2, where n is the number of joints

B. 2n-3, where n is the number of joints

C. n-1, where n is the number of joints

D. n+1, where n is the number of joints

Answer: B

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