Academic Year: 017-018 1701GEX04 ENGINEERING MECHANICS Programme: Question Bank B.E Mechanical Year / Semester: I / II Course Coordinator: Mr.S.K.Krishnakumar/ Mr.V.Manathunainathan Course Objectives To familiarise on various methods of adding and resolving various force systems in a real world environment. To provide knowledge on understanding the effects of forces on a point and at a distance and to arrive at equivalent systems from the given force system. To provide knowledge on various support conditions of a rigid body and deciding a support system for given condition. To expose students with impact of geometries of load bearing systems and make them calculate moment of inertia of various cross sections. To make students understand concepts of friction under various applications and make them calculate frictional forces induced. Course Outcomes: On the successful completion of the course, students will be able to CO1: Draw a free body diagram from the given real world system and add or subtract or resolve the forces involved in the system. CO:. Calculate the moment created by the applied force with reference to any reference in a three dimensional space. CO3: Determine the appropriate support system for the given real world system by calculating the reactions generated. CO4: Suggest suitable cross section or geometry for a load bearing support to prevent it from collapsing due to bending CO5: Calculate the frictional force involved in various real world systems. S.No Questions Mar k COs BT L UNIT I 1 Define Engineering mechanics. The branch of physical science that deals with the study of bodies under the state of rest or the state of motion subjected to external mechanical disturbances such as forces, moments etc., The mechanics of the rigid bodies dealing with the bodies at rest is termed as Statics and that dealing with bodies in motion is called Dynamics. CO1 K1 Define force? Force is a physical quantity that changes or tires to change the state of rest or of uniform motion of an object.. 3 Explain concurrent forces. Two or more forces are said to be concurrent at a point if their lines of action intersect at that point. CO1 K1 CO1 K1
4 State the principle of Transmissibility? CO1 K1 According to this law the state of rest or motion of the rigid body is unaltered if a force acting on the body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of the replaced force. Let F be the force acting on a rigid body at point A as shown in Fig. According to the law of transmissibility of force, this force has the same effect on the state of body as the force F applied at point B. 5 Define Coplanar and Non-coplanar forces. Coplanar forces: All the forces act on the same plane Non-coplanar forces : The forces act on different planes. 6 Explain parallelogram law. This law states that if two forces acting simultaneously on a body at a point are presented 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. CO1 K1 CO1 K1 7 State triangle law of forces. What is the use of this law? If the two forces acting simultaneously on a body are represented by the sides of a triangle taken in order, then their resultant is represented by the closing side of the triangle in the opposite order. This law is used to find the resultant of two forces. 8 What are the different laws of mechanics? First law: A body does not change its state of motion unless acted upon by a force or Every object in a state of uniform motion tends to remain in that state motion unless an external force is applied to it. This law is also called law of inertia. Second law: It gives the force in terms of a quantity called the mass and the acceleration of a particle. It says that a force of magnitude F applied on a particle gives it an acceleration a proportional to the force. CO1 K1 CO1 K1
Third Law: Newton's third law states that if a body A applies a force F on body B, then B also applies an equal and opposite force on A. (Forces do not cancel such other as they are acting on two different objects) Law of gravitational attraction : Two particles will be attracted towards each other along their connecting line with a force whose magnitude is directly proportional to the product of the masses and inversely proportional to the distance squared between the particles. where G is called the universal gravitational constant. 9 What is rigid body? The bodies which will not deform or the body in which deformation can be neglected in the analysis are called as Rigid bodies. 10 Define Kinematics and Kinetics. The dynamics dealing with the problems without referring to the forces causing the motion of the body is termed as Kinematics and if it deals with the forces causing motion also, is called Kinetics. 11 Define particle. A particle may be defined as an object which has only mass and no size. Such a body cannot exist theoretically. However in dealing with problems involving distances considerably larger compared to the size of the body, the body may be treated as particle, without sacrificing accuracy. 1 State Lami's theorem. If three forces coplanar and concurrent forces acting on a particle keep it in equilibrium, then each force is proportional to the sine of the angle between the other two and the constant of proportionality is the same CO1 K1 CO1 K1 CO1 K1 CO1 K1 For the system shown in figure, 13 Define Unit Vector. Unit vector is a vector having a unit magnitude or unit length. CO1 K1
14 Differenciate between particles and rigid body? Particle is a body which has mass but no dimension where as rigid body as both mass and dimensions.particle can have only translational motion where as rigid body can have translational as well as rotational motion. 15 Defferentiate between collinear and concurrent forces? UNIT II Collinear forces act along the same line where as concurrent forces have lines of action intersecting at one point. 1 State the Varignon's theorem. It states that the moment of the resultant of a number of forces about any point is equal to the algebraic sum of the moments of all the forces of the system about the same point Differentiate Resultant and equilibrant. Resultant is the single force which is equivalent to the given system of forces on a body. Equilibrant is the single force which brings the body to equilibrium. It has same magnitude as that of resultant and in opposite direction to it. CO1 K1 CO1 K1 CO CO K1 K1 3 Distinguish between a particle and a rigid body. Particle A body of infinitely small volume and is considered to be concentrated at a point. Rigid Body Rigid body is the one which retains its shape and size, if subjected to some external forces. CO K1 Here mass is negligible. Here mass is considered 4 What are the conditions of equilibrium of a two dimensional rigid body? a. Fx = 0 b. Fy = 0 M = 0 CO 5 State the analytical conditions for equilibrium of coplanar forces in a plane. The two conditions for equilibrium of coplanar forces are: The algebraic sum of all the forces of a force system is equal to zero. F = 0 The algebraic sum of the moments of all the forces is equal to zero. M = 0 CO K1 K1 6 Define moment. It state that product of force in to perpendicular distance. M=Force X Distance N-m 7 Define couple? Two non collinear parallel forces having same magnitude but opposite direction from a couple. CO CO K1 K1
8 Distinguish between couple and moment? Moment represents the turning effect of a force where as couple consists of two equal and opposite forces separate by some distance.moment of force various from point to point but moment of a couple is same about any point in the plane. CO K1 9 What is meant by force-couple system? A system of coplanar non concurrent force system acting in a rigid body can be replaced by a single resultant force and couple moment at a point known as force couple system. CO K1 10 Can a coplanar non concurrent system with zero resultant force necessarily be in equilibrium? A coplanar non concurrent system with zero resultant force is not necessarily In equilibrium as it can have a non zero resultant moment. 11 When is moment of force zero about a point? The moment of force about a point is zero its line of action passes through that point. 1 When is moment of force maximum about a point? Moment of force is maximum about a point when, i)its applied at maximum result from the point and, ii)it is applied perpendicular to the line joining the point to the point of application of force. CO CO CO K1 K1 K1 13 When is moment of force zero about a line? Moment of force about a line is zero when, i)force is parallel to that line or, ii)line of action of force intersects that line. CO K1 UNIT III 1 What are the types of supports? The beams usually have three different types of support: Hinged or pinned support Roller support Fixed support List the different types of beams? Cantilever beam Simply Supported beam Fixed beam Overhanging beam Continuous beam CO3 K1 CO3 K1 3 Explain Hinged or pinned support: The hinged support is capable of resisting force acting in any direction of the plane. Hence, in general the reaction at such a support may have two components, one in horizontal and another in vertical direction. To determine these two components two equations of statics must be used. Usually, at hinged CO3 K1
end the beam is free to rotate but translational displacement is not possible. (fig. A and B) 4 Explain Roller support: The roller support is capable of resisting a force in only one specific line or action. The roller can resist only a vertical force or a force normal to the plane on which roller moves. A reaction on this type of supports corresponds to a single unknow. 5 What is Fixed Support The fixed support resists displacement in both horizontal and vertical directions and also resists rotational movement. It has three reaction components (i). Horizontal reaction; (ii). Vertical reaction; (iii). Rotational reaction. CO3 K1 CO3 K1 6 Mention the types of loads: 1. Point load. Uniformly distributed load 3. Uniformly varying load CO3 K1 7 What is point load? A load, acting at appoint on a beam is known as a point load. CO3 K1 8 What is uniformly distributed load? A load which is spread over a beam, in such a manner that each unit length of the beam carries same intensity of the load, is called uniformly distributed load. CO3 K1 9 What is uniformly varying load? Consider a triangular (right angled triangle) shaped wall, weight of the wall on one end is zero, and it is gradually increased to maximum at other end. (i.e, rate of loading on each unit length of the beam varies uniformly). The total load on the beam is equal to the area of the load diagram and it acts at the centre of gravity of the load diagram. (i.e, at one-third of length from maximum load end). CO3 K1 10 Write the equations of static equilibrium condition. 11 How to convert UDL load into a point load? Consider a beam of length l m, subjected to a UDL load of W N/M over the whole length of beam. Total load = (W * l) N and it acts at centre of gravity of the given beam, i.e, at the centre of beam = (l/) m. CO3 K1 CO3 K1 1 How to convert UVL load into a point load? Consider a triangular shaped wall, length of base of the wall is l m. Base is subjected to a Uniformly Varying load of W N/M. Total load = Area of the wall = ((1/) * W * l) N and it acts at a distance of one- third of the l m from the maximum load end or Two-third of the total l m from no load end. CO3 K1 13 What is beam? It is a horizontal structural member which carries a load, transverse (perpendicular) to its axis and transfers the load through support reactions to supporting columns or walls. CO3 K1
14 What is frame? A structure made up of several members, riveted or welded together is known as frame. 15 What is support reaction? The force of resistance exerted by the support on the beam is called as support. Support reactions depends upon the type of loading and the type of support. UNIT IV PROJECTION OF SECTIONED SOLIDS AND DEVELOPMENT OF SURFACES 1 State parallel axis theorem. It states that the moment of inertia of a lamina about any axis in the plane of lamina is equal to the sum of moment of inertia about a parallel centroidal axis in the plane of lamina and the product of area of the lamina and square of the distance between the two axis IAB = IXX + Ah State perpendicular axis theorem. It states that if IOX and IOY be the moment of inertia of a lamina about two mutually perpendicular axes OX and OY in the plane of the lamina and IOZ be the moment of inertia of the lamina about an axis normal to the lamina and passing through the point of intersection of the axes OX and OY, then IOZ = IOX + IOY 3 Find the polar moment of inertia of a hollow circular section of external diameter D and internal diameter d. Moment of inertia about XX axis IXX = ( /4)(D -d ) Moment of inertia about YY axis IYY = ( /4)(D -d ) Polar moment of inertia IP = IXX + IXY IP = ( /)(D -d ) 4 Write down the equations of motion of a particle under gravitation. a. V = u + gt b. h = ut +(gt /) c. V = u + gh Where, V - Final velocity (m/s) U - Initial velocity (m/s) G - acceleration due to gravity (9.81 m/s ) t - time (sec) 5 Distinguish between kinetics and kinematics. Kinetics The study of relation existing between the forces acting on the body, the mass of the body and motion of the body. Kinematics It is the study of the geometry of motion, which is used to relate displacement, velocity, acceleration and time, without reference to the force causing the motion. CO3 K1 CO3 K1 CO4 K1 CO4 K1 CO4 K1 CO4 K1 CO4 K1 6 State the law of conservation of momentum. It states that the total momentum of two bodies remains constant before and CO4 K1
after impact or any other mutual action i.e. Initial momentum = Final momentum m1u1 + mu = m1v1 + mv 7 Define inertia force. Inertia force of a body can be defined as the resistance to change in the condition of rest or of uniform motion of body. 8 Define centroid of gravity. Centroid is the geometrical center of the body whereas center of gravity is the point through which weight of the body acts. 9 State pappus-guldinus theorem forfinding surface area. The area of surface of revolution is equal to the product of the length of the generating curve and the distance travelled by the centroid of the generating curve while generating that surface. 10 Define principal axes and principal moment of inertia. The axes about which moments or inertia is maximum and minimum are known as principal axes.when these two axes are passing through centroid of aera it is known a centroidal principal axis.now the maximum and minimum moments of inertia are called principal moments of inertia. 11 Distinguish between centroid and center of gravity. Center of gravity is defined as the point through which the entire weight of the body acts.the term center of gravity is applied to the solids and centroid refers to the areas. 1 Define polar moment of inertia. It is defined as the moment of inertia of the lamina or plane about the axis perpendicular to the plane of the section. It is denoted by IP or J 13 Differentiate between Mass moment of inertia and Area moment of inertia. Mass moment of inertia Mass moment of inertia about an axis is the product of elemental mass and the square of the distance between the mass centre of the elemental mass and the axis. Mass moment of inertia = X dm Area moment of inertia Area moment of inertia about an axis is the product of elemental area and the square of the distance between the centroid of the elemental area and the axis. Area moment of inertia = X da 14 Define radius of gyration. Radius of gyration about an axis is defined as the distance from that axis at which all the elemental parts of the lamina would have to be placed (i.e., the entire area of the figure is assumed to be concentrated) such that the moment of inertia about the axis is same. rxx = (Ixx/A) ryy = (Iyy/A) 15 Define Co-efficient of restitution. It is defined as the ratio of the relative velocity of their separation after collision to the relative velocity of their approach before collision. CO4 K1 CO4 K1 CO4 K1 CO4 K1 CO4 K1 CO4 K1 CO4 K1 CO4 K1 CO4 K1
UNIT V e = (v-v1) / (u1-u) For perfectly elastic bodies, e = 1 For perfectly inelastic or plastic bodies, e = 0 1 Give mathematical definitions of velocity and acceleration. Velocity is defined as the rate of change of displacement, v = (dx/dt) Acceleration is defined as the rate of change of velocity, a = (dv/dt) Define friction and classify its types. When a body moves over another body, it experiences an opposing force at the contact surfaces. The opposing force is called friction. It is always opposite to the direction of motion. There are two types of friction. They are : a. Dry friction (or) Solid friction (or) Coulomb friction Fluid friction 3 Explain dynamic friction. If one surface starts moving on the other which is at rest, then the force experienced by the moving surface is called dynamic friction. It is also called as kinetic friction. 4 Define Limiting friction. When the external force, which tends to move the body exceeds the limit, then the body will start to move. This maximum resistance offered by the body is called the Limiting friction. 5 Define coefficient of friction and express its relationship with angle of friction. It is defined as the ratio of the limiting friction to the normal reaction between the two bodies and is generally denoted by µ. µ = (F/R) Relationship between coefficient of friction and angle of friction F = µ cosф CO5 K1 CO5 K1 CO5 K1 CO5 K1 CO5 K1 6 Stat coulomb s laws of dry friction. I. Laws of static friction a. The force of friction always acts in the direction, opposite to that in which the body tends to move. b. The magnitude of force of friction, is always equal to the applied force. c. The magnitude of statistical frictional force bears a constant ratio to the normal reaction between the body and the surface. (Static frictional force / Normal reaction) = a constant d. The force of friction depends upon the roughness of the surface. e. The force of friction is independent of the area of contact between the bodies. II. Laws of dynamic friction a. The force of friction always acts in the direction, opposite to that in which the body is moving. b. The magnitude of dynamic frictional force bears a constant ratio to CO5 K1
the normal reaction between the two contact surfaces. (Dynamic frictional force / Normal reaction) = a constant c. For moderate speeds, the force of friction remains constant. But it decreases at a slow rate with the increase of speed. d. The magnitude of dynamic friction is slightly less than that of static friction. e. The force of friction is independent of the velocity of sliding. f. The coefficient of kinetic friction is slightly less than the coefficient of static friction. 7 Define rolling resistance. It is defined as the friction that occurs because of the deformation of the surface under a rolling load. 8 State the principle of work and energy in a rigid body. It states that K.E.1 + W = K.E. Where, K.E.1 - The body s initial translational and rotational kinetic energy K.E. - The body s final translational and rotational kinetic energy W - Work done by the external forces. 9 What us general plane motion? It is the combination of the translation and rotation. If a body moves in such a way that a particle on the body translate through some distance and also displace through a certain angle, such a motion is termed as general plane motion. 10 Define impending motion. The state of motion of a body which is just about to move or slide is called Impending motion. When the maximum frictional force (i.e, Limiting friction ) is attained and if the applied force exceeds the limiting friction, the body starts sliding or rolling. The state is called impending motion. 11 Define angle of repose. It is defined as the angle which an inclined plane makes with the horizontal when a body placed on it is just on the point of moving down is called angle of repose. (ἀ) 1 Define cone of friction. The cone with the point of contact as the vertex and the normal at the point of contact as the axis, having semi-vertex angle (Ф) is known as the cone of friction. CO5 K1 CO5 K1 CO5 K1 CO5 K1 CO5 K1 CO5 K1 13 Define the Ladder friction.ii) Wedge iii) Belt drive. Ladder: The ladder is a device for climbing or scaling on the roofs or walls. It is made by wood, iron or rope connected by a number of cross pieces called ladder. CO5 K1 14 Define the Wedge friction iii) Belt drive. Wedge: A wedge is a piece of wood or metal in the shape of a prism whose CO5 K1
cross section is in the shape of triangle or trapezoid. It is used for lifting heavy loads and tightening fits. 15 Define the Belt drive friction Belt Drive: A belt drive is a power transmission device with belt and pulley arrangement in which the belt friction is used in the application of brakes to stop friction. Note : 15 Questions with answer key must be prepared in each unit CO5 K1 S.N o UNIT I Questions 1 Three coplanar concurrent forces are acting at a point as shown in fig1. Determine the resultant in magnitude and direction Mar k CO s BT L. Step 1. Algebraic sum of horizontal forces H = 00 cos 45 0 400 cos 30 0 + 600 cos 60 0 = 95.01N Step. Algebraic sum of Vertical forces V = 00 sin 45 0 400 sin 30 0 + 600 sin 60 0 = - 178.N Step 3. Magnitude and resultant forces R = (( H ) + ( V )) = 01.95N Step 3. Direction and resultant forces H is positive, draw towards right, V is negative, draw downwards. tan = 1 1 K = 61.93 0 Result and mark split-up H = 95.01N (3) V = -178.N (3) R and = 61.93 0 (3) Direction of resultant force (3) The forces 10N, 0N, 30N and 40N are acting on one of the vertices of a regular pentagon, towards the other four vertices taken in order. Find the magnitude and direction of the resultant forces R. We know, Sum of the interior angles of any regular polygon is (n 4) * 90 0 Sum of the interior angles = ( * 5 4) * 90 = 540 0 1 1 K
each included angle = = 108 0 θ = =36 0 Θ 1 = 0 0, Θ = 36 0, Θ 3 = 7 0 and Θ 4 = 180 - (3*36) = 7 0 Resolving of forces H = 10 + 0 cos 36 0 + 30 cos 7 0-40 cos 7 0 = 3.09N V = 0 + 0 sin 36 0 + 30 sin 7 0 + 40 sin 7 0 = -78.33N R = (( H ) + ( V )) = 81.67N tan = = 73.57 0 Result and mark split-up H = 3.09N (3) V = 78.33N (3) R and =81.67N and 73.57 0 (3) Direction of resultant force (3) 3 Five forces are acing on a particle. The magnitude of the forces are 300N, 600N, 700N, 900N and P and their respective angle with the horizontal are 0 0, 60 0, 135 0, 10 0 and 70 0. If the vertical component of all the forces is -1000N, find the value of P. also calculate the magnitude and the direction of the resultant, assuming that the first forces acts towards the point, while all the reaming forces act away from the point. V =-1000N 1 Or -1000 = 700 sin 45 + 600 sin 60 + 900 sin 30 P Solving, P = 1564.58N H = 700 cos 45 900 cos 30 + 600 cos 60 300 = -174.39N R = (( H ) + ( V )) = 1619.89N tan = 1 K = 38.1 0 Result and mark split-up H = - 174.39N (3) P = 1564.58N (3)
R and =1619.89N and 38.1 0 (3) Direction of resultant force (3) 4 Two rollers, each of weight 50N and of radius 10cm rest in a horizontal channel of width 36cm as shown in fig. find the reaction on the point of contacts A, B and C. To find θ Let the centre of rollers (1) and () are M and N. Draw the horizontal line through M and vertical line through N, to intersect at P, as shown below MN = twice the radius = 0cm MP = base twice the radius = 16 cm θ = cos -1 (16 / 0) = 36.87 0 considering Free body diagram of roller () Apply H = 0 ( + ) R D cos 36.87 R C = 0 R D cos 36.87 = R C ---------- (i) Apply V = 0 ( + ) R D sin 36.87 50 = 0 R D = 83.33N Substituting R D value in eqn (i) We get, R C = 66.66N considering Free body diagram of roller (1) Apply H = 0 R A R D cos 36.87 = 0 R A = 66.66N Apply V = 0 R B - 50 - R D sin 36.87 = 0 R D = 100N Substituting R D value in eqn (i) We get, R C = 66.66N 1 1 K 5 A roller of radius 30cm weighing KN is to be pulled over a rectangular obstruction of height 15cm as shown in figure by a force P applied tangentially at its crest C, through
a string wound around the circumference of the roller. Find the force P required just to turn the wheel over the corner of the obstruction. Also determine the magnitude and direction of the reaction at A and B. Surfaces may be taken as smooth. To find θ In the right angle triangle OBD, OB = radius = 30cm OD = Radius 15cm = 15cm DB = (OB OD ) = 5.98cm 1 Now, In the right angle triangle BCD, CD = CO + OD =45cm Tanθ = CD/DB θ= 60 0 Reaction at A and B When the roller is about to turn over the corner of the obstruction B, the roller lifts at the point A; hence there will be no contact between the roller and the point A. therefore, reaction at A, R A will become zero. Applying H = 0 at C P - R B cos 60 = 0 P = R B cos 60 ------ (i) 1 K Applying V = 0 at C R B sin 60 - = 0 R B =.309 KN Sub R B value in eqn (i) P = 1.154 KN ( ) Reaction at A = zero and Reaction at B =.309KN 6 A cylindrical roller has a weight of 10KN and it is being pulled by a force which is inclined at 30 0 with the horizontal as shown in fig. while moving it comes across an obstacle 10cm high. Calculate the forces required to cross this obstacle, if the diameter of the roller is 1m.
R = reaction offered by the obstacle to the roller, which is normal to the tangent drawn at B. hence, it must pass through the centre of roller P = the force required to pull the roller over the obstacle of 10cm height. In right angle triangle OBC OB = radius = 50cm OC = OA CA =40cm = cos -1 (OC/OB)= 36.87 H = 0 P cos 30 R cos (90 - ) = 0 P cos 30 R cos (90 36.87) = 0 P = 0.693R V = 0 P sin 30 + R sin 53.13 10 = 0 Sub P = 0.693R R = 8.7KN P = 6.04KN 1 1 K UNIT II 1 ABCD is a weightless rod under the action of four forces P, Q, S and T as shown in fig. If P = 10N, Q = 4N, S = 8N, and T = 1N, calculate the resultant and mark the same in direction with respect to the end A of the rod.
1 Algebraic sum of Horizontal components, H = - 10 cos 45 4 cos 30 + 8 cos 30 + 1 cos 60=.393N Algebraic sum of Vertical components, V = 10 sin 45 4 sin 30-8 sin 30 + 1 sin 60 = 11.463N H and V are drqn on co-ordinate axes Magnitude of resultant force, R = ( H + V ) = 11.71N Direction of resultant force, = tan -1 ) = 78. 0 Location of resultant force Let us locate the resultant force with reference to the point A. let Z is the perpendicular distance of resultant force from A. From Varignon s theorem, Algebric sum of moments of all the forces about A is equal to the moment of resultant force about A. i.e., M A = R * Z now, algebraic sum of moments about A, M A = (4 sin 30 * 1) + (8 sin 30 * ) (1 sin 60 * 3) = -1.76Nm (moments of all other forces about the point are zero, as they are passing through it) (-) sign shows that the net moment at A is anticlockwise. Hence, resultant force should also produce anticlockwise moment about A. Hence, R should be taken on the right side of A as shown in fig. M A = R * Z 1.176 = 11.71 * Z Z = 1.808m Note: sometimes, the horizontal distance on the rod between A and the point, at which resultant force cuts the rod will be required To find x. AMN = 90; ANM = 78. 0 Sin 78. = AM/AN = Z/x x = 1.847m K
Four forces of magnitude and direction acting on a square ABCD of side m are shown in fig. calculate the resultant in magnitude and direction and also locate its point of application with respect to the side AB and AD. 1 Algebraic sum of Horizontal components, H = 1 cos 45 + 10 cos 30-6 cos 60-4 cos 30 = 10.681KN Algebraic sum of Vertical components, V = 1 sin 45 10 sin 30 + 6 sin 60-4 sin 30 = 6.681KN H and V are drqn on co-ordinate axes Magnitude of resultant force, R = ( H + V ) = 1.598KN Direction of resultant force, = tan -1 ) = 3 0 Location of resultant force In the problem, location of resultant force with respect to the sides AB and AD is required. Hence, find the algebraic sum of moments of all the given forces about the point A. M A = (4 cos30 * ) + (10 sin30 * ) (10 cos30 * ) (1 sin45 * ) = -17.36KNm (moment of all other forces about A are zero) Hence, to have anticlockwise moment by the resultant force, R is to be taken on the right hand of A as shown in fig. which cut the sides AB and AD. Now, we will find the point of application of resultant force with respect to the sides AB and AD at M and N, let the resultant force cuts at distances z and y K
from A on AB and AD respectively. Location of resultant force w.r.t AB Resolve the resultant force into two components H and V at M as shown in fig. Now M A = sum of the moments of H & V Or M A = V * x (as H moment about A is zero) x =.598m But the side AB is m only. So, the resultant force cuts the side AB extended at a distance of.598m from A. Location of resultant force w.r.t AD Resolve the resultant force into two components H and V at N as shown in fig. Now M A = sum of the moments of H & V Or M A = H * y (as V moment about A is zero) y = 1.65m Resultant force cuts the side AD at N, at a distance of 1.65m from A. The resultant force is shown in fig. 3 A 300 X 150 mm plate is subjected to four forces as shown in fig. find the resultant of the four forces and the two points at which the line of action of the resultant intersects the edge of the plate.
1 Algebraic sum of Horizontal components, H = 50 cos 45-50 cos 45 00 Algebraic sum of Vertical components, V = 50 sin 45-50 sin 45 + 100= 100N Magnitude of resultant force, R = ( H + V ) = 3.6N Direction of resultant force, = - 00N = tan -1 ) = 6.56 0 Algebraic sum of Moments of all forces, about A, M A = (50 cos 45 * 0.15) + (50 sin 45 * 0.15) (50 sin 45 * 0.15) (100 * 3) = -3.48Nm Knowing the direction of resultant force 6.56 0 and to have anticlockwise moment about A. resultant force should be taken on the right side of A. let the perpendicular distance of resultant force from A be x. Sum of moments about A = moment of resultant abut A i.e., M A = R * x x = 0.015m interception of resultant force with the edges let the line of action of resultant force interception the edges AC and FD at M and N respectively as shown in fig. let AM = a and FN = b To find a Resolve the resultant force into two components H and V as shown in fig We know, M A = sum of the moments H and V about A -3.48 = ( H * 0) ( V * a) a = 0.035m or 35mm To find b Resolve the resultant force into two components H and V as shown in fig We know, M A = sum of the moments H and V about A -3.48 = ( H * 0.15) ( V * b) b = 0.335m or 335mm But FD is only 300mm. so, resultant force dose not intercepts the edge FD. It intercepts the line FD extended as shown in fig. K
To check. Draw a perpendicular line through M to FN, to intercept at P. In right angle triangle MPN PN = FN FP = FN AM = 300mm And PM = AF = 150mm Angle MPN = tan -1 (MP/PN) = 6.56 0 Angle of resultant force with horizontal Hence O.K. 4 A 4.8m beam is subjected to the forces shown in fig. Reduce the given system of forces to a) a single force b) an equivalent force couple system at A c) force couple system at B. a) A single force (or resultant force) Hence, magnitude of resultant force, R = 150 600 100 50 = -600 N Direction of resultant force:- vertically downward Location of resultant force: Let, the resultant force acts at a distance of x m from A. Algebraic sum of moment of all forces about A, M A = (150 * 0) + (600 * 1.6) 100 (1.6 + 1.) + 50 (1.6+1.+) = 1880Nm Hence resultant force should also produce clockwise moment at A. Hence it will be on the right side of A. from varginon s theorem, i.e., 1880 = R * x x = 3.13m resultant force is shown in dotted line 1 K
(b) force couple system at A The resultant force is shown in fig. two equal and opposite collinear forces applied at A is shown in fig (b). fig (c) shows the force couple system at A. (c) force couple system at B The resultant force is again shown in fig (a). Distance of resultant force is measured from B. MB = 1.67m. fig(b) shows, equal and opposite collinear forces applied at B, parallel to the resultant force. Moment of couple is -100Nm (anticlockwise). 5 1. The three forces and a couple of magnitude, M = 18Nm are applied bracket as shown in fig. (a) Find the resultant of this system of forces Locate the point where the line of action of the resultant intersects line AB and line BC.
The given force system is again drawn in fig with the components of the inclined force, 15N. (i) Resultant force Algebraic sum of Horizontal forces, H = 15 cos 60 00 = -137.5N Algebraic sum of vertical forces, V = 15 sin 60 50 = 58.5N Magnitude of resultant force, R = ( H + V ) = 149.3N Direction of resultant force, = tan -1 ) =.95 0 Location of resultant force: M A = (00 * 0.) (15 sin 60 * 0.3) 18 = -10.475Nm Let the perpendicular distance of resultant force from A be x m Applying varignon s theorem, i.e., M A = R * x x = 0.07m =70mm (ii) Intersection of resultant force with lines AB and BC To find AM In right angle triangle APM APM = 90 0 ; sin.95 = AP/AM AM = 179.5mm To find BN In right angle triangle BMN BM = AB AM tan.95 = BN/BM BN = 51.01mm PMA = =.95 0 ; AP = 70mm = 10.48mm 1 K
6 For the system of forces shown in fig, determine the magnitudes of P and Q such that the resultant of the system passes through A and B. It is given that, the resultant force passes through A and B, either from A to B (or) from B to A. but in both the cases, angle of inclination of resultant force ( ) with horizontal is same. i.e., = tan -1 (1/) = 6.56 0 hence, from M A = 0 and M B = 0 the two unknown forces P and Q can be determined. We know, moment of resultant forces about A = sum of moment of all forces about A M A = (00 cos 30 * 1) + (Q cos 60 * 1) + 00 (P sin 60 * 4) (P cos 60 * 1) ( M A = 0) 3.464 P 0.5Q = 373. -----(i) Sililarly M B = (00 cos 30 * ) + (00 sin 30 *) + 00 (Q sin 60 *) (P sin60 * ) ( M B = 0) 1.73 P + 1.73Q = 346.41 -----(ii) Solving the equation (i) and (ii) P = 119.37N and Q = 80.63N 1 K 1 Determine the support reactions of a beam shown in fig. UNIT III 3 K
Figure shows a cantilever beam fixed at A and free at D. fixed end has three reaction components, vertical, horizontal and rotational. These reactions and the resolved components of inclined forces are shown in fig. Applying 10 cos 60 1 cos 30 - H A = 0 1 H = 0 H A = 15.39 KN (H A is positive. Hence assumed direction is correct ) Applying V = 0 V A - 10 sin 60 1 sin 30-8 = 0 V A =.66 KN (V A is positive. Hence assumed direction is correct ) Applying M A = 0 (10 sin 60 * 4) + (1 sin 30 * 6) + (8 * 10) - M A = 0 (M A takes negative measure because, because it is initially assumed as anticlockwise) M A = 150.64 KNm (we get M A is positive. Hence assumed direction is correct) Result: HA = 15.39KN ( ) VA =.66 KN ) MA = 150.64KNm (anticlockwise direction) A simply supported overhanging beam 0m long carries a system of loads and a couple as shown in fig. determine the reaction at supports A and B. Applying, H= 0 As three is no horizontal force on the beam, H A = 0 Applying, V= 0 V A + V B 8 (*5) = 0 1 3 K
V A + V B = 18 -----(i) Applying M A = 0 ( * 5 *1.5) (8 * 5) + 0 (V B * 10) = 0 ----- (ii) Solving we get VB = 10.5 KN, Sub V B value in eqn (i) V A = 7.5 KN Result HA = 0, VA = 7.5KN ( ), VB = 10.5 KN ) 3 A beam AB, is simply supported and carries loads as shown in fig. calculate the reaction at A and B. 1 3 K The assumed direction of reaction component and the resolved inclined forces are shown in fig Applying H = 0 ( +) 10 cos 60 6 cos 45 H A = 0 H A = 0.76 KN Applying V = 0 ( +) V A + V B 10 sin 60 9 6 sin 45 = 0 V A + V B = 1.9 KN ----- (i) Applying M A = 0 (9 * 1.5) + (6 sin 45 * 3.5) + (10 cos 60 * 1.5) (10 sin 60 * 1) (V B * 5) = 0 VB = 5.438 KN Sub VB value in eqn i VA = 16.46 KN 4 1. A beam AB of length 6m and of mass 50kg is connected to a wall y means of a pin A, while the order end of the beam B is supported by a cube BC, such that the angle ABC = 5 0. If there is a vertical load of 100N acting on the beam at a distance of 4m to the right of A, calculate the tension in the cable and the total
reaction at A. The beam, connected by cable, and the loads are shown in fig. The beam is in equilibrium under the action of following forces 1. Reaction components at A, H A and V A. Tension in cable BC, 3. Self weight of beam 50kg acting at mid span of AB. 4. Load 100N at m from B. Now applying H = 0 H A T BC cos 5 = 0 H A 0.906T BC = 0 ------ (i) Applying V = 0 V A T BC sin 5 490 100 = 0 V A + 0.43T BC = 590 ------ (ii) Applying M A = 0 (490 * 3) + (100 * 4) ( T BC sin 5 * 6) = 0 T BC = 736.80N Sub T BC value in eqn (i) H A = 667.54N Sub T BC value in eqn (ii) V A = 78.33N Total reaction at A R A = (H A + V A ) = 73.N And θ = tan -1 (V A /H A ) =.6 0 1 3 K 5 A beam AB is simply supported a load as shown in fig. calculate the reaction at A and B.
Now applying H = 0 -H A + 10 cos 60 - R B cos 60 = 0 H A + 0.5 R B = 0 ------ (i) Applying V = 0 V A 8 1-10 sin 60 +R B sin 60 = 0 V A + 0.866 R B = 8.66 ------ (ii) Applying M A = 0 (8 * ) + (1 * 4) (R B sin 60 * 6) + (1.5 * 10 cos 60) (10 sin 60) = 0 V B = RB sin 60 = 10.41KN R B = 1.09KN Sub R B Value in eqn (i), We get H A = -1.046KN Sub R B Value in eqn (ii), We get, V A = 18.19KN H B = 6.045KN 1 3 K 6 Find the reaction at the supported A and B of the beam as shown in fig. Now applying H = 0 40 cos 60 50 cos 60 H A H B = 0 ( H B = R B cos 60) H A + 0.5R B = -5 ----- (i) Applying V = 0 V A V B - 40 sin 60 80-50 sin 60 (0 * ) = 0 ( V B = R B sin 60) V A + 0.866 R B = 197.94 ----- (ii) Applying M A = 0 (40 sin 60 * ) + (80 * 4) +(50 sin 60 * 6) + (0 * * 3) (V B * 8) = 0 V B = 96.135KN But, V B = R B sin 60 R B = 111KN Sub R B value in eqn (i) H A = - 60.5KN 1 3 K
(-) sign indicates that, assumed direction of H A is wrong. It acts in the opposite direction. Sub R B value in eqn (ii) V A = 101.81KN Total reaction at A. R A = (V A + H A ) = 118.43KN And direction of R A, θ = tan -1 (VA/HA) = 59.7 0 UNIT IV 1.a. Locate the centroid of the L-Section shown in below fig. 1 4 K Portion 1: Area, a1 = 6 * = 1 cm Horizontal distance of the centroid from OY axis, x1 = 6/ = 3 cm Horizontal distance of the centroid from OX axis, y1 = / = 1 cm Portion : Area, a1 = 8 * = 16 cm Horizontal distance of the centroid from OY axis, x = / = 1 cm Horizontal distance of the centroid from OX axis, y = + (8/) = 6 cm Using the relation, X = (a1 * x1 + a * x) / (a1 + a) = ((1*3) + (16*1)) / (1+16) = 1.857 cm Using the relation, Y = (a1 * y1 + a * y) / (a1 + a)
1.b. = ((1*1) + (16*6)) / (1+16) = 3.857 cm Locate the centroid of the T-Section shown in below fig. The T-section is symmetrical about y axis, hence the yy axis can be drawn directly X = 100/ = 50mm 1 4 K To find Y Portion 1: (Flange 100 mm * 30 mm) Area a1 = 100 * 30 = 3000 mm ; y1 = 80 + (30/) = 95 mm Portion : (web 0 mm * 80 mm) Area a =0 * 80 = 1600 mm ; y = 80/ = 40 mm using the relation, Y = (a1 * y1 + a * y) / (a1 + a) = ((3000*95) + (1600*40)) / (3000+1600) = 75.87 mm Locate the centroid and moment of inertia about XX for the below shown fig.(c- section) The channel is symmetrical about XX axis. Hence, 1 4 K
Y = (1 + 40 + 1) / = 8 mm Therefore, horizontal centroidal axis XX is at a height of 8 mm from the horizontal reference axis OX To find X, sub divide the channel into three portions, Portion 1: (Top flange: 110mm * 1 mm) Area, a1 = 110 mm * 1 mm = 130 mm x1 = (110/) = 55 mm Portion : (web: 1mm * 140 mm) Area, a = 1 mm * 140 mm = 1680 mm x = (1/) = 6 mm Portion 3: (Bottom flange: 110mm * 1 mm) Area, a3 = 110 mm * 1 mm = 130 mm x3 = (110/) = 55 mm Using the relation, X = (a1 * x1 + a * x + a3 * x3) / (a1 + a+a3) = [(130 * 55) + (1680 * 6) + (130 * 55)] / (130 + 1680 + 130) = 35.94 mm Therefore, the vertical centroidal axis YY axis is at a distance of 35.94 mm from the vertical reference axis OY Moment of Inertia about XX axis Ixx = I1 + I + I3 Where I1, I, I3 are moment of inertia of component 1, and 3 respectively about XX axis of I- section I 1 = I G1 + (A 1 * h 1 ) h 1 = y 1 Y = 158 8 = 76 mm = [(110 * 1 3 )/1] + [(130 * 76 )] = 7.640 * 10 6 mm 4 I = I G + (A * h ) = [(1 * 140 3 )/1] + [(1680 * 0 )] =.744 * 10 6 mm 4 I 3 = I G3 + (A 3 * h 3 ) = [(110 * 1 3 )/1] + [(130 * 76 )] = 7.640 * 10 6 mm 4 h = y Y = 8 8 = 0 mm h 3 = y 3 Y = 6 8 = -76 mm Therefore, I xx = (7.640 * 10 6 ) + (.744 * 10 6 ) + (7.640 * 10 6 ) = 1.80 * 10 7 mm 4
3 Locate the centroid and moment of inertia about YY for the below shown fig.(z- section) To find X, sub divide the channel into three portions, 1 4 K Portion 1: (Top flange: 1 cm * 3 cm) Area, a 1 = 1 cm * 3 cm = 36 cm x 1 = (1/) = 6 cm Portion : (web: cm * 6 cm) Area, a = cm * 6 cm = 1 cm x = 10 + (/) = 11 cm Portion 3: (Bottom flange: 1 cm * 3 cm) Area, a 3 = 1 cm * 3 cm = 36 cm x 3 = 10 + (1/) = 16 cm Using the relation, X = (a 1 * x 1 + a * x + a 3 * x 3 ) / (a 1 + a +a 3 ) = [(36* 6) + (1 * 11) + (36 * 16)] / (36 + 1 + 36) = 11 cm Therefore, the vertical centroidal axis YY axis is at a distance of 11 mm from the vertical reference axis OY To find Y, sub divide the channel into three portions, Portion 1: (Top flange: 1 cm * 3 cm) Area, a 1 = 1 cm * 3 cm = 36 cm y 1 = 3 + 6 + (3/) = 10.5 cm Portion : (web: cm * 6 cm) Area, a = cm * 6 cm = 1 cm y = 3 + (6/) = 6 cm Portion 3: (Bottom flange: 1 cm * 3 cm) Area, a 3 = 1 cm * 3 cm = 36 cm y 3 = (3/) = 1.5 cm Using the relation,
Y = (a 1 * y 1 + a * y + a 3 * y 3 ) / (a 1 + a +a 3 ) = [(36*10.5) + (1 * 6) + (36 * 1.5)] / (36 + 1 + 36) = 6 cm Therefore, horizontal centroidal axis XX is at a height of 6 mm from the horizontal reference axis OX Moment of Inertia about YY axis I yy = I 1 + I + I 3 Where I 1, I, I 3 are moment of inertia of component 1, and 3 respectively about YY axis of Z-section I 1 = I G1 + (A 1 * h 1 ) h 1 = x 1 X = 6 11 = -5 cm = [(3 * 1 3 )/1] + [(36 * 5 )] = 133 cm 4 I = I G + (A * h ) h = x X = 6 6 = 0 cm = [(6* 3 )/1] + [(1 * 0 )] = 4 cm 4 I 3 = I G3 + (A 3 * h 3 ) = [(3 * 1 3 )/1] + [(36 * 5 )] = 133 cm 4 Therefore, I yy = (133 + 4 + 133) = 668 cm 4 h 3 = x 3 X = 6 11 = 5 cm 4 Locate the centroid and find moment of inertia about XX axis of an symmetrical I section as shown in below fig. All dimensions are in mm. Due to symmetry about X-axis, X = 00/ = 100 mm Therefore, vertical centroidal axis YY is at a distance of 100 mm from the vertical reference axis OY. Due to symmetry about Y-axis, Y = 150/ = 75 mm Therefore, horizontal centroidal axis XX is at a distance of 75 mm from the vertical reference axis OX. 1 4 K Portion 1: (Top flange: 00 mm * 30 mm) Area, a 1 = 00 mm * 30 mm = 6000 cm y 1 = 30 + 90 + (30/) = 135 mm Portion : (web: 40 mm * 90 mm) Area, a = 40 mm * 90 mm = 3600 mm y =30 + (90/) = 75 cm
Portion 3: (Bottom flange: 00 mm * 30 mm) Area, a 3 = 00 mm * 30 mm = 6000 mm y 3 = (30/) = 15 cm Moment of Inertia about XX axis I xx = I 1 + I + I 3 Where I 1, I, I 3 are moment of inertia of component 1, and 3 respectively about XX axis of I-section I 1 = I G1 + (A 1 * h 1 ) h 1 = y 1 Y = 135 75 = 60 mm = [(00 * 30 3 )/1] + [(6000 * 60 )] =.05 * 10 6 mm 4 I = I G + (A * h ) = [(40 * 90 3 )/1] + [(3600 * 0 )] =.43 * 10 6 mm 4 I 3 = I G3 + (A 3 * h 3 ) = [(00 * 30 3 )/1] + [(6000 * 60 )] =.05 * 10 6 mm 4 h = y Y = 75 75 = 0 mm h 3 = y 3 Y = 15 75 = -60 mm Therefore, I xx = (.05 * 10 6 ) + (.43 * 10 6 ) + (.05 * 10 6 ) = 46.53 * 10 6 mm 4 5 Locate the centroid and find moment of inertia about horizontal axis for figure shown below. The moment of inertia (M.I) of the section about the horizontal centroidal axis is required; hence location of the centroid is necessary to draw the centroidal axes xx and yy. Divide the given section into three portions, a rectangle, triangle and semi- circle. The net are of the section is determined by subtracting the area of semi- 1 4 K
circle from the sum of the areas of rectangle and triangle. Portion 1: Rectangle Area, a 1 = 0 * 18 = 360 cm x 1 = 0/ = 10 cm; y 1 = 18/ = 9 cm Portion : Triangle Area, a = (1/) * 0 * 6 = 60 cm x = 0/ = 10 cm; y = 18 + (6/3) = 0 cm Portion 3: Semi-circle Area, a 3 = (1/) * /4) * 18 = 39.7 cm x 3 = 10 + (10/) = 15 cm; y 3 = = (4 * =.1 cm Using the relation, X = (a 1 * x 1 + a * x - a 3 * x 3 ) / (a 1 + a -a 3 ) = [(360 * 10) + (60 * 10) (39.7 * 15)] / (360 + 60-39.7) = 9.484 cm Using the relation, Y = (a 1 * y 1 + a * y - a 3 * y 3 ) / (a 1 + a -a 3 ) = [(360 * 9) + (60 * 0) (39.7 *.1)] / (360 + 60-39.7) = 11.443 cm Moment of Inertia about XX axis I xx = I 1 + I - I 3 Where I 1, I, I 3 are moment of inertia of portions 1, and 3 respectively about XX axis of composite section I 1 = I G1 + (A 1 * h 1 ) h 1 = y 1 Y = 9 11.443 = -.443 mm = [(0 * 18 3 )/1] + [(360 *.443 )] = 1.186 * 10 4 cm 4 I = I G + (A * h ) h = y Y = 0 11.443 = 8.56 mm = [(0 * 6 3 )/36] + [(60 * 8.56 )] = 4516.41 cm 4 I 3 = I G3 + (A 3 * h 3 ) h 3 = y 3 Y =.1 11.443 = -9.318 mm = (0.11 * r 4 ) + [(39.7 * 9.318 )] = 3478.36 cm 4 Therefore, I xx = (1.186 * 10 4 ) + (4516.41) - (3478.36 ) = 1898 cm 4 6 Locate the centroid and find moment of inertia about vertical axis for figure shown below.
The section is not symmetrical about any axis, hence we have to find both X and Y. Divide the given section into three portions, a rectangle, triangle and semi- circle. The net are of the section is determined by subtracting the area of semi- circle from the sum of the areas of rectangle and triangle. 1 4 K Portion 1: Rectangle Area, a 1 = 8 * 6 = 48 cm x 1 = 8/ = 4 cm; y 1 = 6/ = 3 cm Portion : Triangle Area, a = (1/) * 3 * 1.5 =.5 cm x = 8 (3/) = 6.5 cm; y = 6 - (1.5/3) = 5.5 cm Portion 3: Semi-circle Area, a 3 = (1/) * ) * 4 = 5.13 cm x 3 = 4 cm; y 3 = = (4 * = 1.698 cm Using the relation, X = (a 1 * x 1 - a * x - a 3 * x 3 ) / (a 1 - a -a 3 ) = [(48 * 4) - (.5 * 6.5) (5.13 * 4)] / (48 -.5 5.13) = 3.73 cm Using the relation, Y = (a 1 * y 1 - a * y - a 3 * y 3 ) / (a 1 - a -a 3 ) = [(48 * 3) - (.5 * 5.5) (5.13 * 1.698)] / (48.5 5.13) = 4.31 cm Moment of Inertia about YY axis I yy = I 1 - I - I 3 Where I 1, I, I 3 are moment of inertia of portions 1, and 3 respectively about YY axis of composite section I 1 = I G1 + (A 1 * h 1 ) h 1 = x 1 X = 4 3.73 = 0.7 cm = [(3 * 8 3 )/1] + [(48 * 0.7 )]
= 59.50 cm 4 I = I G + (A * h ) = [(1.5 * 3 3 )/48] + [(.5 *.77 )] = 18.107 cm 4 h = x X = 6.5 3.73 =.77 cm I 3 = I G3 + (A 3 * h 3 ) = [(1/) * d 4 /64)] + [(5.13 * 0.7 )] = 10.36 cm 4 Therefore, I yy = (59.50-18.107-10.36 ) = 139.03 cm 4 h 3 = x 3 X = 4 3.73 = 0.7 cm UNIT V 1. a. A body of weight is placed on a rough horizontal plane, and pushed by a force of as shown in fig., to just cause sliding over the horizontal plane. Determine the co-efficient of friction. 1 5 K Horizontal component of the force, will tend to move the body towards right, hence is in the left hand side direction; and normal reaction acts upwards. m = 0 R m = 0 m = = = 0 R = Co-efficient of friction, = m / R = 1. b. A man can pull horizontally with a force of. A mass of is resting on a horizontal surface for which the co-efficient of friction is. The vertical cable of a
crane is attached to the top of the block as shown in fig. what will be the tension in the cable if the man is just able to start the block to the right? 1 5 K Given: ; mass Weight of the block External force, towards right The block tends to move towards right; hence, the frictional force will be developed towards left. The free body diagram of the block is shown in fig. Let T be the tension in the cable Solving, => m ; therefore, => R ; therefore, => m R /( ) Tension in the cable is. Block () rests on block (1) and is attached by a horizontal rope to the wall as shown in fig.. What R m force is necessary to cause motion of block (1) to impend?. The co-efficient of friction between the blocks is and between the floor and block (1) is. Mass of blocks (1) and () are and respectively.
1 5 K Given, weight of the block (1), weight of the block (), 1 ; 1 Block () is restrained by the cable Let, be the tension in the cable. The forces acting on the blocks are drawn in fig. With the application of the force, block (1) tends to move towards left. The movement of block () along with block (1) is prevented by the tension in the cable. Hence, the frictional force at the contact of two blocks is acting towards left. These forces are shown in fig. Consider F BD of block () = 0 => N 88. = 0; therefore, N = 88. N (0.5 * N ) = 0 => T.N ) = 0; => T.N )= 0; therefore, T = Therefore, F =.N = (1/4) * 88. =.05 N Consider F BD of block () with values F and N
The force P tends to move the block (1) towards left; hence frictional force F 1 is towards right. Applying, = 0 ( + ).05 + ((1/3) * N 1 ) P cos 45 = 0 P cos 45 =.05 + (0.333 * N 1 ) Applying, = 0 ( + ).(i) N 1 + P sin 45 88. 137. = 0 P sin 45 = 88. + 137. - N 1 = 5.4 - N 1.(ii) (i)/(ii) => tan 45 = 1 = Solving, N 1 = 15.55 N Substituting N 1 = 15.55 N in equation (i) or (ii), P = 103 N 3 What should be the value of the angle so that motion of the block impends down the plane? (fig.). The co-efficient of friction for all surfaces is. The free body diagrams of the blocks are shown in fig. 1 5 K Let T be the tension in the cable.
Consider the free body diagram of upper block (fig.) Resolving the forces along the plane 1 0 or +...(i) (µ=1/3) 1 Resolving the forces normal to the plane 1..(ii) Substitute N 1 in equation (i) + 130 +..(iii) Consider the free body diagram of lower block (fig.) Resolving the forces along the plane or 1 + sin = µ 1 + sin 1 + µ + 1 ) 1 ⅓ cos + N )..(iv) Resolving the forces normal to the plane 1 0 or + 1 + Substitute in equation (iv) ( ) cos ) Note: If the tension in the cable is required. Substitute the value of in eqn.(iii)
4 long ladder rests against a vertical wall, with which it makes an angle of and on a floor. If a man whose weight is one half that of the ladder climbs it, at what distance along the ladder will he be, when the ladder is about to slip? Take co-efficient Of friction between the ladder and the wall is and that between the ladder and the floor is The forces acting on the ladder are shown in fig. Weight of ladder acts at its mid point, from each other end. Let the ladder slips, when the man stands at a distance of from the foot of the ladder. Applying ( ; = 0 A B 1 5 K A B = 0 or N A = N B (i) Applying = 0 ; A + Substitute B + A = B B W/ = 0 or A + B = 0 B = or B = A = = B Taking moments of forces about A, and equating to zero. ie., A = 0 ( +) (W From the geometry of the figure B Substituting these values and F B = B N B B/3) N B Substituting N B =, we get
The ladder tends to slip down, when the climbing man comes at foot of the ladder from the 5 A block overlying wedge on a horizontal floor and learning against a vertical wall and weighing is to be raised by applying a horizontal force to the wedge. Assuming co-efficient of friction between all the surfaces in contact to be, determine the minimum horizontal force to be applied to raise the block. Angle of wedge = Weight of the block, co-eff. Of friction for all surfaces in contact, But, 1 5 K angle of friction, The free body diagram of the block and wedge are shown in fig. and fig. respectively. (identity the directions of reaction components draw the FBD of wedge first and then draw the FBD of block) Consider FBD of block (fig.) Applying 3 3 3 = 0.468 Applying 3
Substitute Solving, 3 3 3 3 Consider FBD of wedge (fig.) Applying 1 1 1 Applying 1 1 (Ans) 6 A concrete block weighing is to be shifted away from the wall with the help of a wedge as shown in fig. calculate the magnitude of the vertical force that has to be applied to the top of the wedge to shift the block. The co-efficient of friction between all the rubbing surfaces is.
Weight of the block, angle of wedge co-eff. Of friction, but, angle of friction, 1 5 K The free body diagram of the body and the wedge are drawn in fig. and fig. respectively (draw FBD of wedge first and then draw FBD of body) As the load is given, the free body of block is solved first. Consider FBD of block (fig.) Applying 3 Applying 3 3 3 3 Substitute 3 3 3
Solving, 3 Consider FBD of wedge (fig.) 3 Applying 1 1 1 Applying 1 1 Substitute 1 Substitute RESULT: The force required to move the concrete block is. Note : 6 Questions with answer key must be prepared in each unit
S.No Questions Mark COs BTL UNIT I PLANE CURVES AND FREE HAND SKETCHING 1. a Two identical rollers, each of weight 50N, are supported by an inclined plane and a vertical wall as shown in fig. find the reactions at the points of supports A, B and C. Assume all the surfaces to be smooth. Free body diagram of roller () The angle of F DE and R A with horizontal are 30 0 and 60 0 respectively. Now, applying the equation of equilibrium at E, H = 0 F DE cos 30 - R A cos 60 = 0 -----(i) V = 0 F DE sin 30 + R A cos 60-50 = 0 -----(ii) Solving the equation (i) and (ii) F DE and R A can be determined. From eqn. (i) F DE cos 30 = R A cos 60 F DE = 0.577R A Sub. F DE value in eqn (ii) F DE sin 30 + R A cos 60-50 = 0 R A = 43.3N. F DE = 5N. Free body diagram of roller (1) Now, applying the equation of equilibrium at D, H = 0 R C 5 cos 30 - R B cos 60 = 0 -----(i) V = 0 R B sin 60 5 sin 30 50 = 0 -----(ii) 10 1 K
Solving the equation (i) and (ii) F DE and R A can be determined. From eqn. (ii) R B sin 60 = 5 sin 30 + 50 = 6.5 R B = 7.17N R C = 57.73N Result Reaction at A = 43.3 N Reaction at B = 7.17 N Reaction at A = 57.73 N b. A circular roller of radius 0cm and of weight 400N rests on a smooth horizontal surface and is held in position by an inclined bar AB of length 60cm as shown in fig.a horizontal forces of 500N is acting at B. Find the tension in the bar AB and the reaction at C. In right angled triangle ABC, Length of bar AB = 60cm Radius BC = 0cm let BAC = θ θ = sin -1 (BC/AB) = 19.47 0 applying equation of equilibrium at B H = 0 500 - T BA cos 19.47 = 0 T BA`= 530.3N V = 0 R C - 400 - T BA sin 19.47 = 0 R C = 576.76N Tension in the bar = 530.3N Reaction at C = 576.76N 10 1 K Three smooth pipes each weighing 0KN and of diameter 60cm are to be placed in a Rectangular channel with horizontal base as shown in fig. calculate the reactions at The points of contact between the pipes and between the channel and the pipes. Take
E. G.S. PILLAY ENGINEERING COLLEGE mous Institution, Affiliated to Anna University, Chennai) gore (An Autono Post, Nagapattinam 61100, Tamilnadu. Na Width of channel as 160cm. In triangle ABC, Side AB = 160 DA BG = 100 cm AC = BC = X radius = 60cm Draw a vertical line through C, perpendicular to AB, to intersect at M. Now, in right angled triangle BMC BM = AB/ = 50cm Cosθ = BM/BC θ= 33.55 0 Free body diagram of pipe (3) Now applying the equation of equilibrium at C, H = 0 F AC cos 33.55 - F BC cos 33.55 = 0 F AC = F BC ------(i) V = 0 F AC sin 33.55 - F BC sin 33.55 0 = 0 In eqn (i) F AC = F BC F BC sin 33.55 = 0 F BC = 18.09KN = F AC (in eqn (i)) Free body diagram of pipe (1) Now applying the equation of equilibrium at A, H = 0 R D F CA cos 33.55 = 0 R D = 15.07KN V = 0 R E 0 - F CA sin 33.55 = 0 R E = 30KN Similarly, the free body diagram of pipe () is analysed for unknown forces 0 1 K
R G and R F But, due to symmetry, we get R G = R D = 15.07KN and RF = RE = 30KN Result Reaction at D = 15.07 KN ( ) Reaction at G = 15.07 KN ( ) Reaction at E = 30 KN ( ) Reaction at F = 30 KN ( ) UNIT II PROJECTION OF POINTS, LINES AND PLANE SURFACES 1 (i) find the resultant of the system shown in fig. (ii) Find the points of intersection of its line of action with AC and CD (iii)the 7N-cm couple applied at C is removed and replaced by a couple of unknown magnitude M. determine the value of M if the resultant force is to pass through C. Algebraic sum of Horizontal forces, H = 8N Algebraic sum of vertical forces, V = 3 9 = -6N Magnitude of resultant force, R = ( H + V ) = 10N Direction of resultant force, = tan -1 ) = 36.86 0 Location of resultant force: M C = (3 * 0) (9 * 1.5) (8 * 7.5) + 7 = -85.5 Ncm We know the direction of resultant force = 36.86 0 Let the perpendicular distance of resultant force from C be x m R * x = M C x = 8.55cm (ii)intersection of resultant force with the lines AC and CD In right angle triangle CMP, sin 36.86 = CP/CM 0 K
CM = 14.5cm Similarly, In right angle triangle CPN, sin 53.14 = CP/CN CN = 10.68cm (iii)magnitude of couple M, if resultant force is to pass through C We know, sum of moments of all the forces about C, M C = moment of resultant force about the same point C M C = (3 * 0) (9 * 1.5) (8 * 7.5) + M M = 11.5Ncm. Reduce the system of forces shown in fig. to a force couple system at A. UNIT III PROJECTION OF SOLIDS 1 Find the reaction at the supported of the overhanging beam shown in fig. 0 3 K The reaction components are shown above. 60N forces is applied through a bracket at C. this force is to be reduced into a force and couple at C. apply two imaginary equal and opposite collinear forces at C, as shown in fig. 0 Now the given downward 60N force and upward 60N force at C from a
clockwise couple of 60Nm, and the downward 60N force at C is unaltered. This substitute load system is placed at C in given beam as below in fig. Applying H = 0 -H A 150 cos 60 = 0 H A = -75N (-) sign indicates that, HA acts in the opposite direction of assumed direction Applying V = 0 V A + V B 60 (0 * ) 150 sin 60 = 0 V A + V B = 9.9 Applying M A = 0 (60 *) + 60 + (0 * * /) + (150 sin 60 * 7) (V B * 5) = 0 V B = 5.86N Sun V B value in eqn (i), we get V A = 4.04N Resultant reaction at A, RA = (V A + H A ) = 75.108KN θ = tan -1 (VA/HA) = 3 0 -----(i) Determine the support reaction of a simply supported beam, subjected to the loads as shown in fig. Determine the support reaction of a simply supported beam, subjected to the loads as shown in fig. The assumed direction of support reaction and the given trapezoidal load split into an equivalent rectangular (P 1 ) and triangular (P ) loads are shown below in fig 0 3 K P 1 = 6KN/m * 6m = 36KN, at 6/ = 3m from D
P = ½ * 6 * (10 6) = 1KN, at 6/3 = m from D Applying H = 0 -H A = 0 H A = 0 Applying V = 0 V A + V B 4 P P 1 = 0 V A + V B = 5KN ----- (i) Applying M A = 0 (4 * ) + (P 1 * 7) + (P * 6) (V B * 10) = 0 V B = 33.KN Sub V B value in eqn (i), we get V A = 18.8KN UNIT IV 1 Locate the centroid and moment of inertia of an un symmetrical I section as shown in below fig. Due to symmetry, X = 100/ = 50 mm Therefore, vertical centroidal axis YY is at a distance of 50 mm from the vertical reference axis OY. To find Y, divide the I-section into three parts Portion 1: (Top flange: 60mm * 0 mm) Area, a 1 = 60 mm * 0 mm = 100 mm y 1 = 0 + 60 + (0/) = 90 mm Portion : (web: 0mm * 60 mm) Area, a = 60 mm * 0 mm = 100 mm y = 0 + (60/) = 50 mm Portion 3: (Bottom flange: 100mm * 0 mm) Area, a 3 = 100 mm * 0 mm = 000 mm y 3 = (0/) = 10 mm Using the relation, Y = (a 1 * y 1 + a * y + a 3 * y 3 ) / (a 1 + a +a 3 ) = [(100 * 90) + (100 * 50) + (000 * 10)] / (100 + 100 + 000) 0 4 K
= 4.7 mm Therefore, horizontal centroidal axis XX is at a height of 4.7 mm from the horizontal reference axis OX. Moment of Inertia about XX axis I xx = I 1 + I + I 3 Where I 1, I, I 3 are moment of inertia of component 1, and 3 respectively about XX axis of I- section I 1 = I G1 + (A 1 * h 1 ) h 1 = y 1 Y = 90 4.7 = 47.8 m = [(60 * 0 3 )/1] + [(100 * 47.8 )] =.7 * 10 6 mm 4 I = I G + (A * h ) h = y Y = 50 4.7 = 7.8 mm = [(0 * 60 3 )/1] + [(100 * 7.8 )] = 4.35 * 10 5 mm 4 I 3 = I G3 + (A 3 * h 3 ) h 3 = y 3 Y = 10 4.7 = -3.7 mm = [(100 * 0 3 )/1] + [(000 * 3.7 )] =.07 * 10 6 mm 4 Therefore, I xx = (.7 * 10 6 ) + (4.35 * 10 5 ) + (.07 * 10 6 ) = 5.35 * 10 6 mm 4 Moment of Inertia about YY axis I yy = I 1 + I + I 3 Where I 1, I, I 3 are moment of inertia of component 1, and 3 respectively about YY axis of I- section I 1 = I G1 + (A 1 * h 1 ) h 1 = x 1 X = 0 mm = [(0 * 60 3 )/1] + [(100 * 0 )] = 3.6 * 10 5 mm 4 I = I G + (A * h ) h = x X = 0 mm = [(60 * 0 3 )/1] + [(100 * 0 )] = 4 * 10 4 mm 4 I 3 = I G3 + (A 3 * h 3 ) h 3 = x 3 X = 0 mm = [(0 * 100 3 )/1] + [(000 * 0 )] = 1.667 * 10 6 mm 4 Therefore, I yy = (3.6 * 10 5 ) + (4 * 10 4 ) + (1.667 * 10 6 ) =.067 * 10 6 mm 4 I xx = 5.35 * 10 6 mm 4 I yy =.067 * 10 6 mm 4 A thin plate 4mm thick is cut and bent as shown in fig. below. If the density of steel is 7850 kg/m 3, determine the mass moment of inertia of the plate with respect to the centroidal axes xx and yy.
Location of centre of gravity Divide the plate into three components Due to symmetry, X = 8 cm Component 1: Rectangular plate Mass, M 1 = * t * b * d = 7580 * 0.004 * 0.016 * 0.00 = 1.005 kg y 1 = 0.01 m Component : Semi-circular plate Mass, M = * t * )/) 0 4 K = 7580 * 0.004 * ) = 0.3157 kg y = 0. + = 0. + = 0.34 m Component 3: Circular hole Mass, M 3 = * t * * r = 7580 * 0.004 * * 0.05 = 0.466 kg y 3 = 0.1 m Using the relation, Y = (M 1 * y 1 + M * y - M 3 * y 3 ) / (M 1 + M - M 3 ) = [(1.005 * 0.1) + (0.3157 * 0.34) - (0.466 * 0.1)] / (1.005 + 0.3157 + 0.0466) = 0.14 m Moment of Inertia about XX axis I xx = I 1 + I + I 3 Where I 1, I, I 3 are moment of inertia of component 1, and 3 respectively about XX axis of I- section
I 1 = I G1 + (M 1 * h 1 ) h 1 = y 1 Y = 0.1 0.14 = -0.04 m = [(1.005 * 0. )/1] + [(1.005 * 0.04 )] = 4.958 * 10-3 kg.m I = I G + (M * h ) h = y Y = 0.34 0.14 = 0.834 m = [(0.176 * 0.3157 * 0.08 )] + [(0.3157 * 0.834 )] = 3.9 * 10-3 kg.m I 3 = I G3 + (M 3 * h 3 ) h 3 = y 3 Y = 0.1 0.14 = -0.04 m = [(0.466 * 0.05 )/4] + [(0.466 * 0.04 )] = 5.486 * 10-3 kg.m Therefore, I xx = (4.958 * 10-3 ) + (3.9 * 10-3 ) - (5.486 * 10-3 ) = 7.6384* 10-3 kg.m Moment of Inertia about YY axis I yy = I 1 + I - I 3 = ((M 1 * b )/1) + ((M * r )/4) - ((M 3 * r )/4) = ((1.005 * 0.16 )/1) + ((0.3157 * 0.08 )/4) - ((0.466 * 0.05 )/4) =.495 * 10-3 kg.m I xx =.495 * 10-3 kg.m I yy =.495 * 10-3 kg.m UNIT V 1 A uniform ladder 3 m long weighs 180 N. It is placed against a wall making an angle 60 with the floor as shown in fig. The co-efficient of friction between the wall and the ladder 0.5 and that between the floor and the ladder is 0.35. The ladder in addition to its own weight, has to support a man weighing 00 N at its top at A. calculate (i). The horizontal force F to be applied to the floor level to prevent slipping. (ii). If the force F is not applied, what should be the minimum inclination of the ladder with the horizontal so that there is no slipping of it with the man at the top?
0 5 K Case (i): Horizontal force at the floor level to prevent slipping The free body diagram is shown in fig. Let F be horizontal force to be the ladder at the floor level to prevent slipping. Resolving the forces horizontally and equating to zero ( ; B A F = 0 F = A B..(i) Resolving the forces vertically and equating to zero = 0 ; B + A B = 380 - F A 00 = 0 or Now applying, = 0 C (00 * 0) + (180 * CD) + (N A * AC) - (N B * CB) = 0 (ii) From fig. CD = CB/ = (3 * cos 60) / = 0.75 m AC = 3 * sin 60 =.598 m CB = 3 * cos 60 = 1.5 m Substitute CD, AC, CB & N B in eqn (ii) (180 * 0.75) + (N A *.598) ((380 F A ) * 1.5) = 0 (180 * 0.75) + (N A *.598) 570 + 1.5 F A = 0 (180 * 0.75) + (N A *.598) 570 + 1.5 * since, F A = A (180 * 0.75) + (N A *.598) 570 + (1.5 * 1.5 * 135 +.598 N A 570 + 0.375 N A = 0.973 * N A = 435 => N A = 146.31 N F A = 0.5 * 146.31 = 36.58 N N B = 380 F A = 380 36.58 = 343.4 N F B =0.35 * 343.4 = 10.19 N F = N A F B = 146.31 10.19 = 6.1 N F = 6.1 N ) = 0 A A ) = 0
Case (ii): Minimum angle, without the force F, to avoid slipping = 0 ; ( ; A F B = 0 B = B + A A or B + F A = 380 N B + ( A) + = 380 00 = 0 ) = 380..(iii) A ( ) + = 380 since, F B = N B & A Solving, N A = 1.3 N Substitute N A in eqn.(iii) N B = 349.4 N Now applying, = 0 C (180 * CD) + (N A * AC) - (N B * CB) = 0 (iv) (180 * 1.5 ) + (N A * ) - (N B * ) = 0 (180 * 1.5 ) + (0.35 * N B * ) - (N B * ) = 0 366.89 + 70 = 1048.6 366.89 = 778.6 => =.11-1 (.11) Find the force P inclined at anangle of 3 to the inclined plane making an angle of 5 with the horizontal plane to slide a block weighing 15 KN. (i) Up the inclined plane (ii) down the inclined plane. Take With the least amount of the force P, the body will starts moving upwards, hence, the equation of equilibrium and F can be applied. Resolving the force along the plane P - F 1 W = 0 0 5 K