by Dr. ANIL PATIL Associate Professor M.E.D., D.I.T., Dehradun 1
2
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
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
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
6
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
Force system Coplanar force system Concurrent Parallel Non concurrent Non Coplanar force system Concurrent Parallel Non concurrent 8
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
Resolution of Force 10
Composition of Forces When two forces act at a point (Law of Parallelogram of Forces) R = P 2 + Q 2 + 2 PQ cos θ tan α = [ Q sin θ / P+ Q cos θ ] 11
When θ=90 o Q 90 R P α ------------------ R = P² + Q² tan α = Q / P 12
When more than two forces act at a point θ3 θ2 θ1 θ4 13
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
200 sin 65 150 sin 25 100 sin 60 250 sin 45 Y-axis 100 N 250 N 100 cos 60 60 200 cos 65 45 25 250 cos 45 150 cos 25 X-axis 65 25 150 N 200 N F x = + 250 cos 45 + 150 cos 25-100 cos 60-200 cos 65 = 178.2 N F y = + 250 sin 45 + 100 sin 60-200 sin 65-150 sin 25 = 18.73 N 15
F x = 142.84 N F y = 18.73 N Magnitude of the Resultant, R = F x ² + F y ² = 178.2 ² + 18.73² R = 179.2 N Direction of Resultant, F y R α F x tan α = F y / F x = 18.73 / 178.2 = 0.1311 α = 6º 16
Graphical Method (Law of Polygon of forces) θ4 θ3 θ θ2 θ1 17
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
Free Body Diagram Isolated view of body (under consideration) showing all the external forces acting on it 19
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
γ α β 21
From Lami s thoerem T1 Sin 150 = T2 Sin 120 = 1000 Sin 90 Answer T1 = 500 N T2 = 866.03 N 22
T CA T CB 75 135 150 15 N From Lami s theorem Answer TCB Sin 150 = TCA Sin 135 = 15 Sin 75 TCB = 7.76 N TCA = 10.98 N 23
60 30 105º 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 60 183.01 sin 30 200 = 0 T3 = 336.6 N Fx = 0 336.6 cos 60 + 183.01 cos 30 T4 = 0 T1 = 224.14 N T2 = 183.01 N T4 = 326.79 24
125 mm 125 mm 110 Cos α = 250 α = 63º 53 46 α 110 25
Consider F B D of cylinder o1-100 N -100 N -R A 63º 53 46 -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 = 111.36 N 26
Consider F B D of cylinder o2 -R B -R B -R D 63º 53 46 -R D 100 N 100 N Fx = 0 -R C Fy = 0 -R C R D 111.36 cos 63º53 46 = 0 R D = 49 N R c 100 111.36 sin 63º53 46 = 0 R c = 200 N 27
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
Consider F B D of cylinder 1 200 N R B R A 200 N R B 30º 60º R A 29
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 90 200 N RA = 173.2 N RB = 100 N 30
R D Consider F B D of cylinder 2 200 N R B R D 200 N 173.2 N R B 30 60 R C R C Resolving forces Horizontally F x = 0 X or Y First R D 173.2 cos 30 R C cos 60 = 0 R D R C cos 60 = 150 R D = 330.94 cos 60 + 150 Resolving forces vertically F y = 0 Rc sin 60-173.2 sin 30 200 = 0 Rc = 330.94 N R D = 315.47 N 31
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
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
Find Resultant of the following force system y 34
120 KN 1120 KN 420 KN 5000 KN F x = 5000 KN F y = +120 1120-420 = - 1420 KN Magnitude of Resultant ------------------------- R = F x ² + F y ² ------------------------- R = 5000 ² + 1420 ² R = 5197.7 kn 1420 R 5000 α tan α = 1420 5000 α = 15º 51 16 ( Fourth Quadrant ) 35
α Applying Varignon s Theorem d Taking Moments about o M o = R x d R=5197.7N + 5000 x 4 + 1120 x 2 120 x 4 + 420 x 5 = 5197.7 x d d = 23860 5197.7 = 4.59 m 36
20 kn 25 kn 50 kn 35 kn F x = 25 20 = 5 kn F y = - 50 35 = - 85 kn ------------------ R = ( 5 ² + 85 ² ) ----------- = 7250 R = 85.15 kn 85 tan α = 5 α = 86 º 38 1 Resultant 85.15 kn acts in Fourth Quadrant at angle of 86 º 38 1 α R 37
d α Applying Varignon s Theorem Taking Moments about o M o = R x d R = 85.15 kn + 25 x 3 + 35 x 4 = 85.15 x d d = 2.53 m 38
Find the reactions at supports 39
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
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