Dr. ANIL PATIL Associate Professor M.E.D., D.I.T., Dehradun

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1 by Dr. ANIL PATIL Associate Professor M.E.D., D.I.T., Dehradun 1

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3 Mechanics The study of forces and their effect upon body under consideration Statics Deals with the forces which are acting on a body and keep it at rest Dynamics Deals with the forces which are acting on a body and keep it in motion 3

4 Kinetics Deals with the forces acting on moving bodies Kinematics Deals with the motion of bodies without considering the forces responsible for the motion 4

5 Rigid Body A body is said to be rigid if it does not undergo deformation under the action of forces Elastic Body A body is said to be elastic if it undergoes deformation under the action of forces 5

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7 Force Agent which changes or tends to change the state of rest or uniform motion of a body Important about force Magnitude Point of application direction 7

8 Force system Coplanar force system Concurrent Parallel Non concurrent Non Coplanar force system Concurrent Parallel Non concurrent 8

9 PRINCIPLE OF TRANSMISSIBILITY OF FORCES The condition of rest or motion of a rigid body is unaffected if a force, (F) acting on a point (A) is moved to act at a new point, (B) provided that the point (B) lies on the same line of action of that force 9

10 Resolution of Force 10

11 Composition of Forces When two forces act at a point (Law of Parallelogram of Forces) R = P 2 + Q PQ cos θ tan α = [ Q sin θ / P+ Q cos θ ] 11

12 When θ=90 o Q 90 R P α R = P² + Q² tan α = Q / P 12

13 When more than two forces act at a point θ3 θ2 θ1 θ4 13

14 Magnitude of the Resultant, R = F x 2 + F y 2 Direction of Resultant, tan θ = F y / F x F x = F1cos θ1 +F2cos θ2 + F3cos θ3 +F4cos θ4 F y = F1sin θ1 +F2 sin θ2 + F3 sin θ3 +F4 sin θ4 (where θ is taken from the + X axis) 14

15 200 sin sin sin sin 45 Y-axis 100 N 250 N 100 cos cos cos cos 25 X-axis N 200 N F x = cos cos cos cos 65 = N F y = sin sin sin sin 25 = N 15

16 F x = N F y = N Magnitude of the Resultant, R = F x ² + F y ² = ² ² R = N Direction of Resultant, F y R α F x tan α = F y / F x = / = α = 6º 16

17 Graphical Method (Law of Polygon of forces) θ4 θ3 θ θ2 θ1 17

18 Conditions of Equilibrium Body is said to be in equilibrium if it remain in the state of rest after application of forces For equilibrium of coplanar concurrent force system: F x = 0 & F y = 0 For equilibrium of non concurrent force system: F x = 0, F y = 0 & M = 0 18

19 Free Body Diagram Isolated view of body (under consideration) showing all the external forces acting on it 19

20 Lami s Theorem When three concurrent forces are acting on a body simultaneously, be in equilibrium then each force is directly proportional to the SINE of angle between the other two forces F1 sinα = F2 sinβ = F3 sinγ 20

21 γ α β 21

22 From Lami s thoerem T1 Sin 150 = T2 Sin 120 = 1000 Sin 90 Answer T1 = 500 N T2 = N 22

23 T CA T CB N From Lami s theorem Answer TCB Sin 150 = TCA Sin 135 = 15 Sin 75 TCB = 7.76 N TCA = N 23

24 º 120º 135º At Joint B At Joint D From Lami s theorem T1 = T2 Sin 120 Sin 135 = 250 Sin 105 Fy = 0 T3 sin sin = 0 T3 = N Fx = cos cos 30 T4 = 0 T1 = N T2 = N T4 =

25 125 mm 125 mm 110 Cos α = 250 α = 63º α

26 Consider F B D of cylinder o1-100 N -100 N -R A 63º R A -R B -R B Lami s theorem -R A -R B -100 = = Sin 153º53 46 Sin 90 Sin 116º6 14 RA = 49 N RB = N 26

27 Consider F B D of cylinder o2 -R B -R B -R D 63º R D 100 N 100 N Fx = 0 -R C Fy = 0 -R C R D cos 63º53 46 = 0 R D = 49 N R c sin 63º53 46 = 0 R c = 200 N 27

28 Find reactions at all points of contacts RA, RB, RC, and RD Both cylinders weigh 200 N each R B R D R B R A R C 28

29 Consider F B D of cylinder N R B R A 200 N R B 30º 60º R A 29

30 Consider F B D of cylinder 1 R A 90º R B Lami s theorem 150º 120º R A = Sin 120 RB 200 Sin 150 = Sin N RA = N RB = 100 N 30

31 R D Consider F B D of cylinder N R B R D 200 N N R B R C R C Resolving forces Horizontally F x = 0 X or Y First R D cos 30 R C cos 60 = 0 R D R C cos 60 = 150 R D = cos Resolving forces vertically F y = 0 Rc sin sin = 0 Rc = N R D = N 31

32 Moment of a Force about a point The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis 32

33 Varignon s Theorem Algebric Summation of moments of all forces about a point is equal to the moment due to RESULTANT about the same point R.d = F1.d1+F2.d2 33

34 Find Resultant of the following force system y 34

35 120 KN 1120 KN 420 KN 5000 KN F x = 5000 KN F y = = KN Magnitude of Resultant R = F x ² + F y ² R = 5000 ² ² R = kn 1420 R 5000 α tan α = α = 15º ( Fourth Quadrant ) 35

36 α Applying Varignon s Theorem d Taking Moments about o M o = R x d R=5197.7N x x x x 5 = x d d = = 4.59 m 36

37 20 kn 25 kn 50 kn 35 kn F x = = 5 kn F y = = - 85 kn R = ( 5 ² + 85 ² ) = 7250 R = kn 85 tan α = 5 α = 86 º 38 1 Resultant kn acts in Fourth Quadrant at angle of 86 º 38 1 α R 37

38 d α Applying Varignon s Theorem Taking Moments about o M o = R x d R = kn + 25 x x 4 = x d d = 2.53 m 38

39 Find the reactions at supports 39

40 Couple Two equal and unlike forces acting on a body separated by a distance constitute a couple Moment of couple = P a (Anticlockwise) (Torque) 40

41 Force replaced by force and couple F A d B = A d B = B M= F x d F F F Moment of the pair of forces = F x d F 41

Unit 1. (a) tan α = (b) tan α = (c) tan α = (d) tan α =

Unit 1. (a) tan α = (b) tan α = (c) tan α = (d) tan α = Unit 1 1. The subjects Engineering Mechanics deals with (a) Static (b) kinematics (c) Kinetics (d) All of the above 2. If the resultant of two forces P and Q is acting at an angle α with P, then (a) tan