chapter 4: equilibrium of rigid bodies equilibrium of 2-d ... · for a rigid body in static...
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CHAPTER 4: Equilibrium of Rigid Bodies
CHAPTER 4: Equilibrium of Rigid Bodies
Equilibrium of 2-D BodiesEquilibrium of 2-D Bodies
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IntroductionIntroduction
For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.
The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces is equivalent to zero,
( )∑ ∑ =∑ ×== 00 FrMF Orrrr
∑ =∑ =∑ =∑ =∑ =∑ =
000000
zyx
zyxMMMFFF
In rectangular components,
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Reactions at Supports & ConnectionsReactions at Supports & Connections
A smooth roller resists motion perpendicular to the surface only.
Normal force
Reactions equivalent to a force with known line of action.
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Reactions at Supports & ConnectionsReactions at Supports & Connections
Reactions equivalent to a force with known line of action.
A cable can resist a pull only directed along its length.
Tensileforce
Guess the reaction of a short link and frictionless pin in slot:
axialforce
Normalforce
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Reactions at Supports & ConnectionsReactions at Supports & Connections
Reactions equivalent to a force of unknown direction and magnitude.
A smooth pin resists horizontal and vertical translation but not rotation
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Reactions at Supports & ConnectionsReactions at Supports & Connections
Reactions equivalent to a force of unknown direction and magnitude and a couple of unknown magnitude
Guess for a fixed or cantilever support
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An illustration of the object showing ALL forces acting on it.
The Free Body Diagram (FBD)The Free Body Diagram (FBD)
Indicate all forcesacting on the object and dimensions of the object.
IMPORTANT:
Clearly identify the object.
FBD of the STATIONARY PICK-UP
x b
da
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FBD ExampleFBD ExampleDetermine the FBD of the beam.The mass of the beam is 100 kg.
1200N2m
6m
A
1200N2m
A3m
G
981N
Ay
Ax
MA
Due to applied force
Due to beam weight
Due to fixed support
FBD of beam
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FBD ExampleFBD Example
FBD of the pedal
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FBD ExampleFBD ExampleTwo smooth pipes, each having a mass of 300kg, are
supported by the forks of the tractor. Draw the FBD for:
(a) each pipe and(b) both pipes together.
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FBD ExampleFBD Example3rd law of motion:“Forces come in pair of action and reaction.”
FBD of pipe A
FBD of pipe B
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FBD ExampleFBD ExampleFBD of pipes A and B together
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Some simple cases of rigid body equilibriumSome simple cases of rigid body equilibrium
Two force body – a body held in equilibrium by forces acting at only two points in the body. Two force body
(LOAs coincide)
NOTE:LOAs of a two force body is known.
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Some simple cases of rigid body equilibriumSome simple cases of rigid body equilibrium
Three force body – a body held in equilibrium by three concurrent forces.
Three force body(concurrent LOAs)
The forces in a three force body form a closed triangle.
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Equilibrium Conditions on a PlaneEquilibrium Conditions on a PlaneF x 0
F y 0
Mo 0
3 available equations3 unknown quantities (max.)
When the equilibrium equations are enough to solve the reactions, the reactions are “determinate.”
Reactions become indeterminate when there are more unknown quantities than equations.
The 3 equations can not be augmented with additional equations, but they can be replaced
∑ ∑ ∑ === 000 BAx MMF
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Determinate vs. IndeterminateDeterminate vs. Indeterminate
=
• More unknowns than equations
• 3 eqns & 3 unknowns• Fewer unknowns but
partially constrained
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STEPS IN SOLVING PROBLEMS INVOLVING EQUILIBRIUMSTEPS IN SOLVING PROBLEMS INVOLVING EQUILIBRIUM
1. Identify the object, its surrounding and boundary.
2. Draw the necessary FBD.3. Identify the required quantities.4. Set-up the necessary equilibrium
equations based on the FBD.5. Solve for the unknown quantities
based on the equations.6. Verify when in doubt.
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Example 33Example 33Draw the FBD of the platform and solve the unknown reactions on it.The weight of the platform is 1967N.
G70º
1.4m0.8m
1.0mA
B T = tensionW = weight of platfromAx = horizontal reaction at AAy = vertical reaction at A
G70º
1.4m0.8m
1.0m
A
B
Ax
Ay
T
W
FBD of platform
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Solution (scalar approach)Solution (scalar approach)
o
o
oo
r
r
r
70sin19670
0)(
70cos0
0)(
)70cos1()70sin2.2()19674.1(0
0)(
TNA
F
TA
F
mTmTNm
Mccw
y
y
x
x
A
+−=
=↑+
+=
=→+
⋅+⋅+⋅−=
=+
∑
∑
∑
G70º
1.4m0.8m
1.0m
A
B
Ax
Ay
T
W
FBD of platform
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Solution (scalar approach)Solution (scalar approach)
NTmNmT
TmNm
1.1143)(4.28.2753
4.28.27530
+=
=
+−=
NA
NNA
y
y
8.892)(
70sin)1.1143(19670
+=
++−= o
G70º
1.4m0.8m
1.0m
A
B
Ax
Ay
T
W
FBD of platform
NANA
x
x
391)(70cos)1.1143(0
+=++= o
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Solution (scalar approach)Solution (scalar approach)
G70º
1.4m0.8m
1.0m
A
B
Ax
Ay
T
W
FBD of platform
→=
↑=
=
NA
NANT
Summary
x
y
391
8.8921.1143:
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Solution (vector approach)Solution (vector approach)
)70sin70cos(
)1967()(00
0
jTiT
NjjAiATWA
F
yxrr
rrr
rrr
r
oo +−+
−++=
++=
=∑
G70º
1.4m0.8m
1.0m
A
B
Ax
Ay
T
W
FBD of platform
070sin70cos012.2)19674.1(0
)()(0
0
oo
rrr
rr
rrrr
r
TTmm
kjiNjmi
TrWr
M
Tw
A
−+−×=
×+×=
=∑
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Solution (vector approach)Solution (vector approach)
G70º
1.4m0.8m
1.0m
A
B
Ax
Ay
T
W
FBD of platformN
mNmT
TmNmmkTmkTNmk
1.11434.28.2753
4.28.2753070cos70sin2.2)19674.1(0
==
+−=++−⋅=
rrroo
NA
NNA
y
y
8.892)(
70sin)1.1143(19670
+=
++−= o
NANA
x
x
391)(70cos)1.1143(0
+=++= o
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Example 34Example 34The link shown is pin connected at A and rests against a smooth support at B. Compute the vertical and horizontal reaction at the pin A.
FBD of link
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Solution (vector approach)Solution (vector approach)
NmkNmkkNm
NmkNjrNr
M
NjjAiAjNiN
NjAN
F
B
BB
A
yxBB
B
rrr
rrrrr
r
rrrrr
rrr
r
oo
9060)75.0(0
)90()60()(0
0
60)30cos30sin(0
)60(0
0
1
−−⋅=
−+−×+×=
=
−++−−=
−++=
=
∑
∑
FBD of link
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Solution (vector approach)Solution (vector approach)
NA
NNA
NjjAjN
NANA
iAiN
NjjAiAjNiN
NNNmkNmkkNm
y
y
yB
x
x
xB
yxBB
NB
B
2.233)(
6030cos)200(
6030cos0
100)(30sin)200(
30sin0
60)30cos30sin(0
2009060)75.0(0
75.0150
+=
+=
−+−=
+==
+−=
−++−−=
==
−−⋅=
o
o
o
o
oo
rrr
rr
rrrrr
rrr
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Solution (vector approach)Solution (vector approach)
NjA
NiA
NNSummary
y
x
B
r
r
2.233
100
200:
=
=
=FBD of link
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Example 35Example 35The lever ABC is pin-supported at A and connected to a short link BD as shown. The weight of the members are negligible. Determine the reaction of the pin to the lever at A.
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FBD of lever(3-force body)
FBD of link(two-force body)
SolutionSolution
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30FBD of lever(3-force body)
( ) o26.60tan 4.02.05.01 == +−θ
FA
F
400
45º
o26.60o74.119
15.26º
45º
NF
F
A
A
FA
6.1074
40026.15sin
45sin
40026.15sin45sin
=
=
=
o
o
oo
SolutionSolution