Statics - Philadelphia University · 2 Equilibrium condition ... Coplanar Forces. 11/7/2015 10...
Transcript of Statics - Philadelphia University · 2 Equilibrium condition ... Coplanar Forces. 11/7/2015 10...
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Statics
Equilibrium
Moves at
constant
velocity
V =C
At rest
V = 0
Constant Velocity
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Equilibrium condition
Satisfying Newtown’s 1st Law of Motion: ∑ F = 0
This condition implies that ma = 0. here m could not
equal zero and so a = 0 (i.e. not accelerating or
decelerating)
Free Body Diagram (FBD)
Definition:
Is a diagram for a particle or a body isolated
from its surrounding.
It shows all the forces acting on the particle
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Procedures to draw FBD
Drawing outlined shapes
Show all the forces
Identify each force
Example[1]
F1
F2
Mg
M
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Example[2]
AB C
D
T1
M1g
B
T2
C
M2g
T2
T3
M1
M2
Connection types: linear springs
LoLo
L
F
δ
Applying Load
deflection (δ) = L – Lo
F = K δ where K is thespring stiffness
increasing K makesthe spring stiffer.
Stiffer springs needsmore force to deflect it
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Forces in Linear Springs
As the load affects the spring, an internal resistance load creates.
The relation between the external load and the internal force of thespring is proportional and the proportional factor is the stiffness (K)
Linear springs could be:
tension spring compression springs
Equivalent of Linear Springs (parallel)
K1
K2
K3
Kn
Force Force
Keq = K1 + K2 + …. + Kn
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Equivalent of Linear Springs (series)
Force
Force
K1 K2 Kn
n21eq K
1...
K
1
K
1
K
1
Connection types: cables and pulleys
it has always tension
force in the direction of
the cable.
In most of engineering
problems, cables are
assumed massless and
unable to stretch.
T
T
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Example[3]
T
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Equilibrium conditions:
∑F = 0
∑Fx i + ∑Fy j = 0
∑Fx = 0
∑Fy = 0
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Procedures for analysis FBD
Coordinate system (x and y axes)
Force Labeling (magnitude and direction)
Assuming the sense of the unknown forces
Equilibrium equations
Positive and negative assigning
Define the direction of the solution
∑ Fx = 0 and ∑ Fy = 0
Coplanar Forces
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Coplanar Forces
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Coplanar Forces
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Equilibrium conditions:
∑F = 0
∑Fx i + ∑Fy j + ∑Fz k= 0
∑Fx = 0
∑Fy = 0
∑Fz = 0
S
t
a
t
i
c
s
.
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Procedures for analysis FBD
Coordinate system (x, y and z axes)
Force Labeling (magnitude and direction)
Assuming the sense of the unknown forces
Equilibrium equations
Positive and negative assigning
Define the direction of the solution
∑ Fx = 0, ∑ Fy = 0 and ∑ Fz = 0
S
t
a
t
i
c
s
.
Three dimensional Forces
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Three dimensional Forces
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Three dimensional Forces