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UNIT - 2Analysis of Perfect Frame Q1) Explain simple and compound machine?A1) Simple Machine A machine a device by which heavy load can be lifted by applying less effort as compared to the load. Eg. Heavy load of car can be lifted with the help of simple screw jack by applying small force.Compound Machine: Compound machine is a device which may consists of number of simple machines. A compound machine may also be defined as a machine which has multiple mechanisms for the same purpose. Compound machines do heavy work with less efforts and greater speed. e.g. In a crane, one mechanism (gears) are used to drive the rope drum and other mechanism (pulleys) are used to lift the load. Thus, a crane consists of two simple machines or mechanisms i.e. gears and pulleys. Hence, it is a compound machine.Q2) What is efforts and load?A2) Effort: It may be defined as, the force which is applied so as to overcome the resistance or to lift the load. It is denoted by „P‟. Magnitude of effort (P) is small as compared to the load (W). Load: The weight to be lifted or the resistive force to be overcome with the help of a machine is called as load (W). Q3) Explain reversible and irreversible machine?A3) Reversible Machine: When a machine is capable of doing some work in the reverse direction even on removal of effort, it is called as reversible machine. Condition for Reversible Machine: The efficiency of the machine should be more than 50%. Irreversible Machine / Non-reversible Machine / Self Locking Machine: When a machine is not capable of doing some work in the reverse direction even on removal of effort, it is called as irreversible machine or non-reversible machine or self-locking machine. Condition for Irreversible Machine: The efficiency of the machine should be less than 50%. Q4) Explain static and kinetic friction?A4) A) Static Friction. In order to set in motion a body lying on an even surface in aState of rest, a critical force, the force of static friction Fs, must be overcome. ThisForce is roughly proportional to the normal force FN 1:Fs = μs N F.The coefficient μs is called the coefficient of static friction. It is dependent on thePairing of the contacting materials, however, shows almost no dependence on contact area or roughness. B. Kinetic Friction FR is the resisting force which acts on a body after the Force of static friction has been overcome. Coulomb experimentally determined the following properties of kinetic friction:Kinetic friction is proportional to the normal force FN:FR = μk N F It shows no considerable dependence on the contact area or roughness of theSurface.The coefficient of kinetic friction is approximately equal to the coefficient ofStatic friction:μk s ≈ μ This proportionality is known as Amontons’ law.Q5) Determine the forces in each member of the plane truss as shown in fig. in terms pf the external loading and state if the members are in tension or compression. Use 0+ 30 deg, L = 2 m and p =100N.Diagram
A5)Consider FBD of Truss,For equilibrium, 
Fx =0RHA+100 =0RHA =-100 KN
Fy =0 RVA = RD – 100 ….. 1MA = 0 -------Taking moment @ A- (Roxz) + (100*1) + (100*1.732) =0- - 2 Rd + 100 + 1.7320 =0- -2 Rd + 100 + 173.2 =0- RD =136.66 N ()From eqn (1)RVA = 100 – 136.6RVA= 36.6NRVA = 36.6 NConsider Joint D, for equilibrium,Fx =0-FAD + FCD cos 60 =0 – (11)Fy = 0
136.6 +FCD sin 60 +0 FCD = -157.73 N © From eqn (11), Fad =- 78. 87 N (c) Consider point A, for the equilibrium of point A,
Fx = 0- 100 + FAD + FAc cos + 30 FAb cos60 = 0-100 + (-78.87) +FAC cos 30 +FAb cos60 =0FAc cos 30 + FAB cos60 + 178. 87 (3)
fx =0-36.6 + FAC sin 30 + FAB sin 60 =0FAC sin 30 + FAb sin 60 = 36.6Solving eqn (3) and (4)FAC = 273.21 N (T)FAB = -115.47 NFAB= 115.47 N ©Consider point B, for the equilibrium of point,
Fx =0-FAB cos60 + FBC =0-[9-115.47) cos60]+ FBC =0
FBC = -57.73 NFBC =57.73N
Q6) Determine the axial forces in each member of the plane truss as shown in figure.
A6)Consider FBD of Truss,For the equilibrium of Truss, 
fx =0 RHA + 10 =0RHA = -10 KN RHA =10KN (
Resolving forces vertically,
fy =oRVA + RD -15 =0 RVA + RD = 15 …… (1)Taking moment about point A,
Fy =0(10*3) – 3 Rd =0 30+ 3 Rd =0RD= 10 KN 
RVA = 5KN (
)Now Consider joint B, FBD of joint B is shown below.Assuming forces developed in all members to be Termile,For the equilibrium of joint We have 
fx =0
10+FBC =0 FBC = -10 KNFBC = 10KN (c)
fy =0 FAB = -15 KN FAB = 15 KN (c) Now consider joint c,For the equilibrium of joint,
fx =0-FBC – FAC cos 45 =0- (-10) – FAC cos 45=0 10= FAC cos 45 FAC = 10/cos 45
FAc = 14.14 KN (T)
fy =0-Fac sin 45 – FCD=0- 14.14 sin45 = FCDFCD = -10 KN (c)FCD= 10 KN (c) Consider joint D,By observation, FAD=0
Q7) Explain the difference between frame and truss?A7) Difference between Frame and Truss:
Q8) Explain single band brake?A8) Simple Band Brake:Friction can be used as a power-transmitting agent (in belts-pulleys) as well as a power-absorbing agent (in band brakes). The idea is, the band resists the motion of rotating wheel through friction. The braking torque is obtained by the relations
In this case, the values of 
and 
depend on the applied force 
as well. Drawing the free body diagram of lever and pulley is necessary.Q9) What is imperfect truss?A9) Imperfect truss: A truss which does not collapse under the load is called Imperfect or unstable truss.Here n
2j -R Over stable (Redundant Truss)A truss in which n > 2j- R, Then it is over stable truss.
Deficient truss It is a truss in whichN <2 j-R.
Q10) Explain Cantilever Truss?A10) Cantilever Truss: A truss which is fixed on one side & free of other end is called as cantilever truss.Diagram
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Fx =0RHA+100 =0RHA =-100 KNFy =0 RVA = RD – 100 ….. 1MA = 0 -------Taking moment @ A- (Roxz) + (100*1) + (100*1.732) =0- - 2 Rd + 100 + 1.7320 =0- -2 Rd + 100 + 173.2 =0- RD =136.66 N ()From eqn (1)RVA = 100 – 136.6RVA= 36.6NRVA = 36.6 NConsider Joint D, for equilibrium,Fx =0-FAD + FCD cos 60 =0 – (11)Fy = 0
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Fx = 0- 100 + FAD + FAc cos + 30 FAb cos60 = 0-100 + (-78.87) +FAC cos 30 +FAb cos60 =0FAc cos 30 + FAB cos60 + 178. 87 (3)
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fx =0-36.6 + FAC sin 30 + FAB sin 60 =0FAC sin 30 + FAb sin 60 = 36.6Solving eqn (3) and (4)FAC = 273.21 N (T)FAB = -115.47 NFAB= 115.47 N ©Consider point B, for the equilibrium of point,Fx =0-FAB cos60 + FBC =0-[9-115.47) cos60]+ FBC =0
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Member | AB | BC | CD | AD | AC |
Force | 115.47N | 57.73N | 157.73N | 78.87N | 273.21N |
Nature | c | C | c | c | T |
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fx =0 RHA + 10 =0RHA = -10 KN RHA =10KN (Resolving forces vertically,fy =oRVA + RD -15 =0 RVA + RD = 15 …… (1)Taking moment about point A, Fy =0(10*3) – 3 Rd =0 30+ 3 Rd =0RD= 10 KN RVA = 5KN ()Now Consider joint B, FBD of joint B is shown below.Assuming forces developed in all members to be Termile,For the equilibrium of joint We have fx =0
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fy =0 FAB = -15 KN FAB = 15 KN (c) Now consider joint c,For the equilibrium of joint, fx =0-FBC – FAC cos 45 =0- (-10) – FAC cos 45=0 10= FAC cos 45 FAC = 10/cos 45
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fy =0-Fac sin 45 – FCD=0- 14.14 sin45 = FCDFCD = -10 KN (c)FCD= 10 KN (c) Consider joint D,By observation, FAD=0
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Sr .No | Member | Force | Nature |
1 | AB | 15 KN | c |
2 | BC | 10KN | c |
3 | CD | 10KN | c |
4 | DA | 0 | - |
5 | AC | 14.14 | T |
Truss
| Frame
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And
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and depend on the applied force as well. Drawing the free body diagram of lever and pulley is necessary.Q9) What is imperfect truss?A9) Imperfect truss: A truss which does not collapse under the load is called Imperfect or unstable truss.Here n 2j -R Over stable (Redundant Truss)A truss in which n > 2j- R, Then it is over stable truss. Diagram
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