Showing posts with label multiple questions of Engineering Mechanics. Show all posts
Showing posts with label multiple questions of Engineering Mechanics. Show all posts

Wednesday 11 November 2009

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
















Monday 9 November 2009


ENGINEERING. MECHANICS:  

Most Common Theoretical Questions

EME - 102; EME - 201


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: Fx = 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, ΣFi = 0;
(ii) Net moment on the body must be zero ie, ΣMi = 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, ΣFi = 0; and we can resolve it along X axis and along Y axis, ie.  (i) ΣFx = 0; and  (ii) ΣFy = 0.

for coplanar non concurrent force system, the equilibrium conditions are
(i) ΣFx = 0; and  (ii) ΣFy = 0.  (iii)  ΣMi = 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: