chapter 6 equilibrium of rigid bodies - drexel universitycac542/l11.pdf · equilibrium of rigid...
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MEM202 Engineering Mechanics - Statics MEM
Chapter 6Equilibrium of Rigid Bodies
0
0
=++=
=++=
∑∑∑∑∑∑
kMjMiMC
kFjFiFR
zyx
zyxrrrr
rrrr
000000======
∑∑∑∑∑∑
zyx
zyx
MMMFFF
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MEM202 Engineering Mechanics - Statics MEM
Equilibrium of Rigid Bodies
m 6m 51
N 300 N 450
m 42
m 4
N/m 100=w
A B
Roller Support
m 6m 51
N 300 N 450
m42
m4
N/m 100=w
A BFree endPin
SupportFixed
Support
Simply-Supported Cantilever
m 4.31=cx
N ,2502=R
N 909=yA N 341,1=yB
N 0=xA
m 4.31=cx
N ,2502=R
N 250,2=yA
N 0=xA
m-N 175,32=AM
Free-Body DiagramSupports and Connections
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MEM202 Engineering Mechanics - Statics MEM
6.2 Free-Body DiagramsIdealized Supports and Connections in 2-D
(Table 6-1, pp. 252-254)
Transmit a tensile force in the direction of the cable.
Transmit a force (tensile or compressive) in the direction of the axis of the link.
Transmit a compressive force perpendicular to the surface at the point of contact.
Transmit a compressive force perpendicular to the surface supporting the ball.
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MEM202 Engineering Mechanics - Statics MEM
6.2 Free-Body DiagramsIdealized Supports and Connections in 2-D
(Table 6-1, pp. 252-254)
Transmit a force, usually expressed in terms of its components: a compressive normal force Rnand a tangential frictional force Rt. The angle θis related to the coefficient of frictional, µ, i.e., µ = tanθ = Rr/Rn
Transmit a force, usually expressed in terms of its components Rxand Ry.
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MEM202 Engineering Mechanics - Statics MEM
6.2 Free-Body DiagramsIdealized Supports and Connections in 2-D
(Table 6-1, pp. 252-254)
Transmit a force (tensile or compressive) perpendicular to the surface of the guide.
Transmit a force (tensile or compressive) perpendicular to the axis of the shaft.
A smooth collar with a fixed connection.Transmit a force, usually expressed in terms of its components Rxand Ry, and a moment.
Transmit a force (tensile or compressive) perpendicular to the axis of the shaft, and a moment.
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MEM202 Engineering Mechanics - Statics MEM
6.2 Free-Body DiagramsIdealized Supports and Connections in 2-D
(Table 6-1, pp. 252-254)
Transmit a force (tensile or compressive) along the axis of the spring.
Transmit and change the direction of a tensile force.For a frictionless pulley, T1 = T2.
More about pulleys
xR
yR
( )θpT
T
( )θpxR
X
Y
yR
MxR
yR
T
T
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MEM202 Engineering Mechanics - Statics MEM
6.2 Free-Body DiagramsIdealized Supports and Connections in 2-D
Flexible cable
Rigid link
Ball, roller, or rocker
Fixed support
Smooth
pin
or hin
ge
Linear
elastic
spring
αLinkF
RollerFPinF
θ
FixY
FixXFixM
CableF
SpringF
LinkX
LinkYθtan=
Link
Link
XY
PinX
PinY
(3)
(1)(1)
(2)
(1)
(1)
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two Dimensions
( ) 0or 000 :2 ====− ∑∑∑∑ MMFFD zyx
000000
:3======
−∑∑∑∑∑∑
zyx
zyx
MMMFFF
D
Three equations, thus, can only solve for three unknownsAlternatively
000or
000or
000
===
===
===
∑∑∑∑∑∑∑∑∑
CBA
BAy
BAx
MMM
MMF
MMF
where A, B, and C are three points on the x-yplane which cannot be chosen arbitrarily
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two Dimensions
022
0
0
41
4321
31
=−=
=−+−=
=+−=
∑∑∑
bFhFM
bFhFbFhFM
FFF
A
O
x
1Fr
2Fr
3Fr
4Fr
h2
b2
x
y
A B
CD
O
0
0
0
4321
42
31
=−+−=
=−=
=+−=
∑∑∑
bFhFbFhFM
FFF
FFF
O
y
x
Example
022
022
0
41
32
31
=−=
=+−=
=+−=
∑∑∑
bFhFM
hFbFM
FFF
A
C
x
( )⎪⎪⎩
⎪⎪⎨
⎧
====
==
hbFFFFhbFFFF
FFFF
2341
4321
42
31
or
:Solution
022
022
0
43
21
4321
=−=
=−=
=−+−=
∑∑∑
bFhFM
bFhFM
bFhFbFhFM
D
B
O
Choice Bad022
0220
21
32
31
⇐⎪⎭
⎪⎬
⎫
=−==+−=
=+−=
∑∑∑
bFhFMhFbFM
FFF
B
C
x
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsThe Two-Force Body (Two-Force Members)
////0 BABA FFFF −=⇒==
⊥⊥
α
//BF
//AF
• The two forces acting on a two-force member must be equal, opposite, and collinear.• The weight of the member must be neglected.• The connections to the member must not support a moment about the connection (i.e., it
must be a pin connection).
BF
AF
BF
AF
BF
AF
00 =⇒=⋅=⊥⊥∑ BBA FLFM
00 =⇒=⋅=⊥⊥∑ AAB FLFM
α
AF
BF
//BF⊥BF
//AF⊥AF L
Examples of two-force members:
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsThe Three-Force Body (Three-Force Members)
(or, Multiple-Force Members)
Concurrent force systems Parallel force systems
AF
BF
CF
BF
CF
AF
aL
0
0
=⇒
=++=
∑∑
MConcurrent
FFFF CBAr
rrrr
0
0
=⋅−⋅=
=−+=
∑∑
aFLFM
FFFF
cBA
CBA
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsStatically Determinate, Statically Indeterminate, Partial Constraints
nu: number of unknowns, ne: number of equations
In general:nu = ne: Statically determinatenu > ne: Statically indeterminatenu < ne: Partial Constraints
In 2-D, ne = 3 (i.e., ΣFx = 0, ΣFy = 0, ΣM = 0), thereforenu = 3: Statically determinatenu > 3: Statically indeterminatenu < 3: Partial Constraints
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsStatically Determinate, Statically Indeterminate, Partial Constraints
Statically Determinate Statically Indeterminate Partial Constraints
nu = 3, ne = 3 nu = 4, ne = 3 nu = 2, ne = 3
nu = 3, ne = 3 nu = 4, ne = 3 nu = 2, ne = 3
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsStatically Determinate, Statically Indeterminate, Partial Constraints
Some Special CasesRigid link
Rigid link
Pin
nu = 3, ne = 3Yet, it is unstable!
(ΣM ≠ 0)
nu = 3, ne = 3Yet, it is unstable!
(No horizontal support)
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsExample 6-11
Pin( ) o87.3643tan 1 == −φ
Tension in cable
( ) ( ) lb 75 03375 =⇒=−=∑ TTMC
(Tension in a cable remains constant)
0sin7575
0cos75
=−−=
=+=
∑∑
φ
φ
yy
xx
CF
CF
↑==
←=−=
lb 120lb 120
lb 60lb 60
y
x
C
CNOTE: As a rule of thumb, you should always assume unknown forces in positive directions.
⇒
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsExample 6-11
Pin Roller
Equal and opposite to forces acting on pulley
( ) ( )
0120
060
0186010
=−+=
=+=
=−=
∑∑∑
BAF
AF
BM
yy
xx
A
↑==
↑==
←=−=
lb 108lb 108
lb 12lb 12
lb 60lb 60
B
A
A
y
x
↑=
←==
lb 120
lb 60lb 75
y
x
C
CT
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsExample 6-12
N 4.8582 == TRA
( )( )N 8.171681.9175
=== mgW
N 4.8582
02
==
=−=∑WT
WTF
C
Cy
C
N 4.858=CT
B
T T
RA
A
N 2.4292
02
==
=−=∑C
Cy
TTTTF
08.17162.4294.8582.4294.858um EquilibriOverall :NOTE
=−−++=−−++ WTTTR CA
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsExample 6-16
Overall F.B.D.
xA
yA
xB
yB
4 unknowns, 3 equations!
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsExample 6-16
A
( ) ( ) ( )
( ) ( ) ( )[ ]01.060044.04006.0 or,
04007.33sin
06007.33cos
01.060016.04007.33sin6.0
=+−=
=+−−=
=−+=
=++−=
∑∑∑∑
yA
yy
xx
B
BM
BAF
BAF
AM
o
o
o
( ) o7.33300200tan 1 == −θ
A
AAC is a two-force member!
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsProblem 6-54
T
W
R
75 kg40 kg
0=−+=∑ WRTFy
If the boy is standing on a scale, what would the reading on the scale be?
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MEM202 Engineering Mechanics - Statics MEM
6.3 Equilibrium in Two DimensionsProblem 6-55, 6-56
2P
23P
WP =29