Lecture 18 and 19-SFA-DFA
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Transcript of Lecture 18 and 19-SFA-DFA
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BITS PilaniPilani Campus
Dynamic Force Analysis
Lecture-18 and 19 J S Rathore
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BITS Pilani, Pilani Campus
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Static Force analysis:
cgamF
cgJM
0F
0 M
Dynamic analysis has to be carried out
1) Situations, where inertial forces are comparable
with magnitudes of external forces
2) Operating speeds are high
Two or three force members FBD of links Pin reactions Stress analysis and dimensions of links
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Coupled Wheel Locomotion: Double crank mechanism
X
X
1. What would be maximum stress in
coupler AB?
2. Safe x-sectional dimension (X-X) ?
3. OR for given dimension and material
properties of AB, what would be
maximum possible speed of
crank/locomotive?
Locomotive is running at 120 Km/hr:
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Dynamic Force analysis:
To complete analysis
1) Variation in velocity and acceleration
2) DAlembert principle
FBD Two or three force members Pin reactions Stress analysis and dimensions of links
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F2
Fn
F1
5
Motion of a rigid body subjected to
a system of forces:
cgamRF
JM
Newtons equation
R
e
CG
cga
Resultant force
Resultant torque
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Motion of a rigid body subjected to
a system of forces:
0)( cgamF
0)( cgJM
D Alemberts Principle
Newtons equation
0 F
0 M
F2
Fn
F1CG
cgam
J
cga
cgamRF
JM
Resultant force
Resultant torque
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Analytical method:
G3
aG3
3
3GamF
3
JM
0)( 3 GamF
0)( 3
JM
0 F
0 M
Newtons eq D Alemberts eq
W3
PA
PB
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D Alembert Principle:
Dynamic problem is converted into static problem by having
Fictitious force (-macg) Inertia Force
Fictitious torque (-J) Inertia Torque
W3
PA
PB
aG3
3
- maG3
- J3
G3
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Four bar mechanism used in coupled wheel locomotion is as shown in Fig. The
crank and follower, O2A and O4B, are of equal length of 600 mm but of negligible
mass. The coupler AB is a rigid uniform circular rod of length 1000 mm, its total
mass being 12 kg. A torque M acts on the crank O2A, causing coupler AB to move
with a linear acceleration aG3 = 25 45o m/s2 and an angular acceleration AB = 60
rad/sec2 (CCW). Find:
1. The FBD of link AB, O4B & O2A.
2. Pin reactions PA and PB.
3. The minimum diameter of the
coupler required, if the allowable
bending stress of the material of the
coupler is 35 MPa.
4. Shifted inertia force in coupler AB.
Ex 1:
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Soln: FBD
aG3
3
A
Torque M acts on the crank O2A, causing coupler AB to move with a linear
acceleration aG3 = 25 45o m/s2 and an angular acceleration AB = 60 rad/sec
2
(CCW).
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Soln: Pin reactions at A and B
A
From FBD of coupler AB
0xF
0AM Gives PBY = 224.93 N in positive y-direction
So PBX = 224.93 N in positive X-direction
So PAX = 12.8 N in negative x-direction
PBY
PBXPAX
PAY
300 N
0BM
So PAY = 104.925 N
05.83tan 1
AX
AYA
P
P
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Soln: FBD
A
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For minimum diameter of coupler AB, draw BMD of AB
3
32
d
Mbending
mm 32 d
Soln: Diameter of coupler AB
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Soln: Shifted inertia force in AB
A
Shifted inertia force 300 N force at a distance
45cos3G
AB
ma
Je
me 2828.045cos300
60 Right from the centre of mass
e
300 N
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Objectives:
Dynamic force analysis of 4-bar mechanism
aG3
3
Crank and follower with negligible mass External force/moment is acting only on crank
DAlembert principle: Dynamic problem is converted into static problem by having
Pseudo force (-macg) Pseudo torque (-J)
F2M2
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Dynamic Force Analysis:
F3M3
F4F2M2
M4
aG22
DAlembert principle: Dynamic problem is converted into static problem by having
Pseudo force (-macg) Pseudo torque (-J)
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FBD
Fj,x = External force in the x-direction applied at CGj
Fj,y = External force in the y-direction applied at CGj
Mj = External Moment on the jth link applied at CGj
Lj = dj + fj
aG22
aG33
aG4
4
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Link 2
222
2
2
22
2
222222222
2
2
cosf sinf- cosd- sind
1 0 1 0
0 1 0 1
k
a
a
m
P
P
P
P
M
F
F
yg
xg
Ay
Ax
yo
xo
y
x
aG2
2
Fj,x = External force in the x-direction applied at CGj
Fj,y = External force in the y-direction applied at CGj
Mj = External Moment on the jth link applied at CGj
Lj = dj + fj
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Link 3
233
3
3
3
333333333
3
3
cosf- sinf cosd- sind
1 0 1 0
0 1 0 1
k
a
a
m
P
P
P
P
M
F
F
yg
xg
By
Bx
Ay
Ax
y
x
aG33
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Link 4
244
4
4
4
4
4444444444
4
4
cosf- sinf cosd- sind
1 0 1 0
0 1 0 1
K
a
a
m
P
P
P
P
M
F
F
yg
xg
yo
xo
By
Bx
y
x
aG4
4
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Assuming gravitational and frictional effects are negligible, Determine all constraint forces
and driving torque (on link 2) required to produce the velocity and acceleration conditions
specified.
m3 = 1.5 kg, m4 = 5 kg
JG2= 0.025 kg m2, JG3
= 0.012 kg m2, JG4= 0.054 kg m2
2 = 0, 3 = 119k rad/s2, 4 = 625k rad/s
2
AG3= [email protected] m/s2, AG4
= 104@233 m/s2
FC = 0.8 j kN.
RAO2= 60 mm
RO4O2= 100 mm
RBA = 220 mm
RBO4= 150 mm
RCO4= RCB =120 mm
RG3A= 90 mm
RG4O4= 90 mm
Ex 1: Problem 14.4 (Uicker-Shigley) p 536
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Four bar mechanism used in coupled wheel locomotion is shown in
Fig. The crank and follower, O2A and O4B, are of equal length of
600 mm but of negligible mass. The coupler AB is a rigid uniform
circular rod of length 1000 mm and its total mass being 15 kg. If
the locomotive is running at constant speed of 120 km/hr then
find:
1. The linear acceleration (aG3) and angular acceleration (AB) ofcoupler AB.
2. The inertial loading on coupler AB.
Ex 2: Home assignment