Part A Qn. No SCHEME OF BE 100 ENGINEERING MECHANICS DEC 201 Module No BE100 ENGINEERING MECHANICS Answer ALL Questions 1 1 Theorem of three forces states that three non-parallel forces can be in equilibrium only when they lie in one plane, intersect in one point, and their free vectors build a closed triangle. (give 3 marks) Law of superposition states that, the action of a given system of forces on a rigid body will in no way be changed if we add to or subtract from them another system of forces in equilibrium.(give 2 marks) 2 2 A free body diagram is a graphic, dematerialized, symbolic representation of the body (structure, element or segment of an element) in which all connecting "pieces" have been removed. (give 2 marks) Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. (give 1 marks) Marks (give 2 marks) 3 3 Parallel axis theorem states that moment of inertia of a plane area with respect to any axis in its plane is equal to the moment of inertia with respect to a parallel centroidal axis plus the product of the area and the square of the distance between the two axes. (give 2 marks) Proof: In Fig. 1 OXY axes is set parallel to the centroidal Oxy axes. Now the moment of inertia of the given area A expressed in terms of the elemental area da shown about OX axis is I XX = Y 2 da = (b + y) 2 da = y 2 da + 2b yda + b 2 da Similarly about OY axis is I YY = X 2 da = (a + x) 2 da = x 2 da + 2a xda + a 2 da Here yda and xda represents the first moment of the Al- Ameen Engg. College 1
area da with respect to the centroidal axes. Since Oxy is the centroidal axes, the above in the above terms must be zero. The first terms in the above equations represent the moment of inertia about the centroidal axes and they are denoted by ICx and I Cy. I XX = I Cx + A b 2 I = + A a 2 YY I Cy Adding the above equations gives Polar moment of inertia J O = I XX + I YY = I Cx + I Cy + A (a 2 + b 2 ) J O = J C + Ad 2 (give 3 marks) 4 4 Weight of the body = 00N Angle of friction, = 20 0 We know, Coefficient of friction, µ= tan = tan20 0 = 0.36 Resolving forces normal to the plane, R= 00cos2 0 = 43.1N We know, F=µR = 0.36 x 43.1 = 163.134N Resolving forces along the plane, P = 00sin2 0 + F = 374.44N = Maximum value of P Therefore Minimum value of P = 00sin2 0 F = 48.18N Al- Ameen Engg. College 2
Law of conservation of energy states that the total energy of an isolated system remains constant it is said to be conserved over time. ( 2 marks) Example with explanation of any system. Eg Thermodynamics, Mass, etc - ( 3 marks) 6 Linear Velocity Angular Velocity The rate of change of The rate of change of angular displacement between an displacement is known as object and a fixed point. angular velocity Linear velocity is a vector Angular velocity is vector quantity. quantity Linear velocity is measured in The unit of angular velocity is meters per second. radians per second Denoted by v This is usually denoted by ω. Example Motion of car in a straight line Example - Motions like blades of a rotating fan or a running wheel 7 6 i) In mechanics, the degree of freedom (DOF) of a mechanical system is the number of independent parameters that define its configuration. Example - The position of a single car (engine) moving along a track has one degree of freedom because the position of the car is defined by the distance along the track. The position and orientation of a rigid body in space is defined by three components of translation and three components of rotation, which means that it has six degrees of freedom. ii) the reaction due to the inertia of an accelerated body is equal and opposite to the force causing the acceleration and results in a condition of kinetic equilibrium Example A reaction of baseball is equal to the blow of the bat upon the baseball iii) The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. F = ma 8 6 i) Amplitude is the maximum displacement of points on a wave, which you can think of as the degree or intensity of change. This maximum displacement is measured from the equilibrium position. For a simple harmonic wave, shown below, A o is the amplitude Al- Ameen Engg. College 3
ii) The frequency, f, of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz (Hz), since one hertz is equal to one wave per second. B Set 1 Answer ANY 2 Questions 9a 1 Algebraic sum of horizontal components is given by, H = 20cos30 0 30cos4 0-3cos40 0 = -30.70N Algebraic sum of vertical components is given by, V = 20sin30 0 +2+30sin4 0-3sin40 0 = 33.73N Magnitude of resultant, R= [ (HH)] 2 + [ (VV)] 2 = 4.6N The direction of resultant is given by, tanθ = VV HH θ = -47.68 0 Here H = -ve and V = +ve, therefore angle between 90 0 to 180 0 Ie, actual angle, θ = 180 0 47.68 0 = 132.32 0 Al- Ameen Engg. College 4
9b 1 Al- Ameen Engg. College
10a 1 10b 2 Use F y = 0 (give 1 marks) Use M a = 0 (give 1 marks) R A = 80 KN (give 1. marks) R B = 0 KN (give 1. marks) 11a 2 Here given the mass of crate is 163kg, Therefore weight = mg=163 9.81=199.03N From the figure the sum of tensions are equal to the weight of the crate. T AB +T AC +T AD = 199.03j eq(1) T AB = T AB ( 360i+600j 270k) 360 2 +600 2 +270 2 AB (-.48i+.8j-.36k) Similarly, T AC = T AC (.8j+.47k) T AD =T AD (.2i+.78j-.3k) Ie, T AB (-.48i+.8j-.36k) + T AC (.8j+.47k) + T AD (.2i+.78j-.3k) = 199.03j Equate the coefficients of i,j&k on both sides, T AB = 71.3N T AC = 830.3N T AD = 27.4N 11b 2 Free body diagram Al- Ameen Engg. College 6
Use F y = 0 Use F x = 0 (give 1 marks) R A = 1333.33 KN (give 1 marks) R B = 2000 KN (give 1 marks) R c = 1333.33 KN (give 1 marks) R d = 1666.66 KN (give 1 marks) Set 2 Answer ANY 2 Questions 12a 3 Pappus-Guldinus first theorem states that the area of the surface of revolution is equal to the length of the generating curve times the distance travelled by the centroid of the curve while the surface is being generated. (give 2 mark) Pappus-Guldinus second theorem states that the volume of a body of revolution is equal to the generating area times the distance travelled by the centroid of the area while the body is being generated. (give 2 mark) 12 b 3 The area shown in Fig. 7 is formed by adding the area elements 1 and 2 and then removing the element 3. Centroid of the given plane area about OXY axis can be calculated as 4 6 Al- Ameen Engg. College 7
13a 3 A 1 = 72 cm 2 X 1 = -6 cm A 2 = 36 cm 2 X 2 = -4 cm A 3 = 18π cm 2 X 3 = -8/π cm Centroid, x= -2.62 cm 13b 4 26.89+0.11T+69.42 =.94T T=116.04N Case II (Block B):- Weight of block B = 10N Coefficient of friction, µ = 0.32 Resolving forces normal to the plane, R 2 = Tsin20 0 + W= 39.69+10= 189.69N We know, F 2 =µr 2 = 0.32 x 189.69 = 61.6N Resolving forces along the plane, P = F 2 + Tcos20 0 =61.6+109 = 170.6N 14a 4 8 Al- Ameen Engg. College 8
R 2 = 00 0. P Resolving forces along the plane, F 2 + T = Pcos30 0 We know, F 2 = µr 2 ie, µr 2 + T = Pcos30 0 0.2 00 0. P + 144.9 = 0.87PP P = 22.47N 14b 4 Any 2 uses of virtual work- Explanation 2 Set 3 Answer ANY 2 Questions 1a Difference between each with explanations (2 marks) 3 Examples for each case (1 mark each) 1b Free Body Diagram (2 mark) 7 Acceleration a = 0.98 m/s 2 (2 mark) Tension T = 80 N (3 mark) 16a Free Body Diagram (1 mark) Force in upward direction = 61.16N (1 mark) Pressure exerted by man = Tension in cable = 661.16 N (2 marks) 16b 6 Angular Acceleration = 2ππ = 3.93 rad/s (1 mark) ωω Velocity = ωr sin (ωt) = 2.9 m/s 2 (2 marks) Acceleration = -ω2r cos (ωt) = 3.02 m/s 2 (2 marks) 17a 6 Free vibration occurs when a mechanical system is set off with an initial input and then allowed to vibrate freely. Examples - pulling a child back on a swing and then letting go or 3 hitting a tuning fork and letting it ring. (1. marks) Forced vibrations is when a time-varying disturbance (load, displacement or velocity) is applied to a mechanical system. The disturbance can be a periodic, steady-state input, a transient input, or a random input. The periodic input can be a harmonic or a non-harmonic disturbance. Examples - shaking washing machine due to an imbalance, transportation vibration (caused by truck engine, springs, road, etc.), or the vibration of a building during an earthquake. (1. marks) 17b 6 Stress in spindle = 396.12 kn/m 2 (2 mark) Increase in length = 6.932 x 10-6 m (2 marks) Frequency 189.39 (3 marks) 7 Al- Ameen Engg. College 9
1 Subject BE 100 ENGINEERING MECHANICS 2 Name of Faculty SHAFEEK.K 3 Designation ASSISTANT PROFESSOR 4 Department MECHANICAL ENGINEERING e-mail shafeekup@gmail.com 6 Mobile No. 97476898 7 Name of College Al Ameen Engineering College, Kulappully Al- Ameen Engg. College 10