engineering mechanice lecture 03
TRANSCRIPT
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Lecture 03BY
Engr Muhammad UsmanMechanical Engineering
DepartmentCECOS University
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COURSE OUTLINE
Fundamental concept and principle of mechanics, important vector quantityForce system i.e concurrentNon concurrent and parallel force system
• Resultant of forces (Remaining)• Moment and couple (Start)• Equilibrium of forces (law & type) concept of
free body diagram
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MOMENT
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Moment of A Force The rotation or turning effect of a force is called
MOMENT. The moment M of a force F about a fixed point A
is defined as the product of the magnitude of force F and the perpendicular distance d from point A to the line of action of force F.
MA = F x d Where force F is in newtons, NAnd distance d is in meters, mThus moment MA is in newton-meter, Nm.
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A
d F
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Moment Sign Convention
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Anti-clockwise : + VE(Counter-clockwise)
Clockwise : - VE
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Principle of TransmissibilityIf we move (TRANSMIT) the Force P from point A to point B which lies on the Line of Action of Force P, the same effect would be expected.
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Principle of transmissibility states that a force acting on a rigid body at different points along the force’s line of action will produce the same effect on the body.
Point of Application
Line of action
P
A
BP
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Sample Problem # 01• Calculate the moment of the 500 N
force about the point A as shown in the diagram.
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A1.5 m
500 N
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Solution
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A 1.5 m
500 N
Since the perpendicular distance from the force to the axis point A is 1.5 m, from
MA = F x d
MA = - 500 x 1.5 = - 750 Nm
= 750 Nm
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Sample Problem # 02 • Calculate the moment about point A
caused by the 500 N force as shown in the diagram.
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A 1.5 m
60o
500 N
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Solution:
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MA = F x d = -500
x AB
= -500 x 1.5 sin 60
= -
649.5 Nm= 649.5 Nm
1.5 m
500 N
Line of action
600
600
d
B
A
A 1.5 m
60o
500 N
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Sample Problem # 03 Determine the resulting moment about point A of the system of forces on bar ABC as shown in the diagram.
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A B C
60 0
500 N 800 N
1.5 m 0.5m
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Solution:
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60 0
AD = (1.5 + 0.5) sin 60= 2.0 sin 60= 1.732 m
MA = (-500 x 1.5) + (-800 x 1.732)
= - 750 – 1385.6 = - 2135.6 Nm
= 2135.6 Nm
A B C
500 800 N1.5 m 0.5 m
A1.5 m 0.5 m
Line of Action
600
D
C
800 N
500 N
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Varignon’s Theorem • Varignon’s Theorem states that “ the
moment of a force about any point is equal to the sum of the moments of its components about the same point”.
• To calculate the moment of any force with a
slope or at an angle to the x or y-axis, resolve the force into the Fx and the Fy components, and calculate the sum of the moment of these two force components about the same point.
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Sample Problem # 04 Employing Varignon’s Theorem.
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A B C
60 0
500 N 800 N
1.5 m 0.5m
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Solution
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A
MA
= (-500 x 1.5) + (- 800 sin 60 x 2)
= -2135.6 Nm = 2135.6 Nm
B C
60 0
500 N 800 N
1.5 m 0.5m
1.5 m 0.5m600
A
800 N
B C
Fx
Fy = 800 sin 600
500 N
600
Fx
Fy
= 800 cos 600
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Sample Problem # 05Determine the resulting moment about point A for the system of forces acting on the plate ABCD as shown in the diagram.
15 cm
8cm
10.38 cm 45 N
20 N
35 N
80o
45o
30o
A B
C
D
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Solution
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45 N force: Fx = 45 cos 80 = 7.814 N Fy = 45 sin 80 = 44.32 N
15 cm
8cm
10.38 cm 45 N
20 N35 N
80o
45o
30o
A B
C
D
35 N force: Fx = 35 cos 60 = 17.5 N Fy = 35 sin 60 = 30.31 N
FxFy
45 N
800
Fx
Fy
35 N300
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20 N force: Fx = 20 cos 45 = 14.14 N Fy = 20 sin 45 = 14.14 N
0.15 m
0.08m
0.1038 m 45 N
20 N
35 N80o
45o60o
A B
C
D
MA = (- 45 sin 80 x 0.1038) + (- 35 cos 60 x 0.08) + (20 cos 45 x 0.08) + (20 sin 45 x 0.15)
= - 4.591 – 1.4 + 1.131 + 2.121 = - 2.74 Nm = 2.74 Nm
FyFx
20 N
450
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Problem 2/29
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Problem 2/30
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Problem 2/31