Download ENGR-36_Lec-22_Wedge-n-Belt_Friction

Engineering 36
Ch08: Wedge &
Belt Friction
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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Outline - Friction
 The Laws of Dry Friction
• Coefficient of Static Friction
• Coefficient of Kinetic (Dynamic) Friction
 Angles of Friction
• Angle of static friction
• Angle of kinetic friction
• Angle of Repose
 Wedge & Belt Friction
• Self-Locking & Contact-Angle
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Basic Friction - Review
 The Static Friction Force Is The force that Resists
Lateral Motion. It reaches a Maximum Value Just Prior
to movement. It is Directly Proportional to Normal Force:
Fm   s N
 After Motion Commences The Friction Force Drops
to Its “Kinetic” Value
F  N
k
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k
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Wedge Friction
 Consider the
System Below
 The Wedge is of
negligible Weight
 Then the FBD of the
Two Blocks using
Newton’s 3rd Law
 Find the Minimum
Push, P, to move-in
the Wedge
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[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Friction
 For Equilibrium of the Heavy Block
F
y
 0  W  FA,n cos   s FA,n sin 
W
 Solve F 
A, n
cos    s sin 
for FA,n
 For Equilibrium of the Wt-Less Wedge
F
F
x
 0  P   s FC ,n   s FA,n cos   FA,n sin 
y
 0  FC ,n   s FA,n sin   FA,n cos 
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Bruce Mayer, PE
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Wedge Friction
 In the last 2-Eqns Sub Out FA,n




W
W



 Fx  0  P   s FC ,n   s  cos    sin   cos    cos    sin   sin 
s
s








W
W
 Fy  0  FC ,n   s  cos    sin   sin    cos    sin   cos 
s
s




 Eliminating FC,n from the 2-Eqns yields
an Expression for Pmin:
Pmin
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

W
2

1  µs sin   2 s cos 
cos    s sin 
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx

Wedge Friction
 MATLAB Plots for P when W = 100 lbs
W = 100 lbs,  = 10°
W = 100 lbs, µ = 0.2
85
90
80
80
75
70
70
65
P (lbs)
P (lbs)
60
60
50
40
55
30
50
20
45
40
2
0
4
6
8
10
 (°)
12
14
16
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18
20
10
0
5
10
15
µ (%)
20
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MATLAB Code
% Bruce Mayer, PE
% ENGR36 * 22Jul12
% ENGR36_Wedge_Friction_1207.m
%
u = 0.2
W = 100
a = linspace(0,20);
P = W*((1-u*u)*sind(a) +2*u*cosd(a))./(cosd(a)-u*sind(a))
plot(a,P, 'LineWidth',3), grid, xlabel('\alpha (°)'), ylabel('P (lbs)'),
title('W = 100 lbs, µ = 0.2')
disp('showing 1st plot - Hit Any Key to Continue')
pause
%
a = 10;
u = linspace(0,0.3);
P = W*((1-u.*u)*sind(a) +2*u*cosd(a))./(cosd(a)-u*sind(a));
plot(100*u,P, 'LineWidth',3), grid, xlabel('µ (%)'), ylabel('P (lbs)'),
title('W = 100 lbs, \alpha = 10°')
disp('showing 2nd plot - Hit Any Key to Continue')
pause
%
u = linspace(0, .50);
aSL =atand (2*u./(1-u.^2));
plot(100*u,aSL, 'LineWidth',3), grid, xlabel('µ (%)'), ylabel('\alpha
(°)'), title('Self-Locking Wedge Angle')
disp('showing LAST plot')
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Friction
 Now What Happens
upon Removing P
 Then the FBD When
P is Removed
• Note that the
Direction of the
Friction forces are
REVERSED
 The Wedge can
• Be PUSHED OUT
• STAY in Place
– SelfLocking condition
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Friction
 For Equilibrium of the Heavy Block
F
y
 0  W  FA,n cos   s FA,n sin 
W
 Solve for FA,n 
K
cos    s sin 
FA,n
 For Equilibrium of the Wt-Less Wedge
 Fx  0   s FC ,n   s FA,n cos   FA,n sin 
F
y
 0  FC ,n   s FA,n sin   FA,n cos 
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Friction
 To Save Writing sub K for FA,n
 s FC ,n   s K cos   K sin   0
FC ,n  K cos    s K sin   0
 Eliminate
FC,n
 s FC ,n   s K cos   K sin   0
FC ,n  K cos    s K sin   0  s 
0  2  s K cos   K sin  µs2  1  0
 Now Divide Last Eqn by Kcosα
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Friction
 Dividing
2
2  s K cos   K sin  µs  1  0
by Kcosα
K cos 
sin  2

 2 s 
µs  1  0
cos 
 Recognize sinu/cosu = tanu


tan  µ  1  2 s
2
s
 2 s
2 s
2 s
tan   2


2
2
µs  1  µs  1 1  µs

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

 

Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Friction
 After all That Algebra
 2 s 
Find The Maximum  max  arctan 

2
1  
α to Maintain the
s 

Block in the Static Location
 Since Large angles Produce a Large
Push-Out Forces, and
 2 s 
a ZERO Angle Produces

 SL  arctan 
2
NO Push-Out Force,
 1  s 
the Criteria for Self-Locking
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Push-Out
 SMALL PushOut Force
• Likely SelfLocking

 2 s 

 SL  arctan 
2 
 1  s 
 LARGE PushOut Force
• Likely NOT SelfLocking

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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Friction
Self-Locking Wedge Angle
60
50
 (°)
40
30
20
10
0
0
5
10
15
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20
25
µ (%)
30
35
40
45
50
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Belt Friction
 Consider The Belt Wrapped
Around a Drum with Contact
angle .
 The Drum is NOT FreeWheeling, and So Friction
Forces Result in DIFFERENT
Values for T1 and T2
 To Derive the Relationship
Between T1 and T2 Examine a
Differential Element of the Belt
that Subtends an Angle 
• The Diagram At Right Shows
the Free Body Diagram
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Bruce Mayer, PE
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Belt Friction cont
 Write the Equilibrium Eqns for
Belt Element PP’ if T2>T1


 Fx  0  T  T cos 2  T cos 2   s N




 Fy  0  N  T  T sin 2  T sin 2
 Eliminate N from the Equations


 T sin
2
2



 

  Fx  0  T  T  cos
 T cos
  s T  T sin
 T sin
2
2
2
2 






 

0  T cos
 T cos
 T cos
  s T sin
 T sin
 T sin
2
2
2
2
2
2 

 Fy
 N  T  T sin
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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Belt Friction cont.1
 Combining Terms


0  T cos
  s 2T  T sin
2
2
 Divide Both Sides by 
T

sin  2
cos
  s T  T 2

2
 2
 Now Recall From Trig And Calculus
sin 
   d
cos0  1
Lim
1
Lim

 0



0


d
 So in the Above Eqn Let: /2 →0; Which Yields
0
dT
dT
0
  s T  dT 2 
  sT
d
d
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as T  dT 2  T
Bruce Mayer, PE
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Belt Friction cont.2
 The Belt Friction Differential Eqn
dT
  sT
d
1
Sep Vars 
dT   s d
T
 Integrate the Variables-Separated Eqn within Limits
• T( = 0) = T1
• T( = ) = T2
 From Calculus

T2
T1

1
dT   s  d
0
T
 ln T2   ln T1    s   ln T2 T1 
 Now Take EXP{of the above Eqn}
e
lnT2 T1 
e
s 
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 T2 T1  e
s 
Bruce Mayer, PE
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Belt Friction Illustrated
 This is a VERY
POWERFUL
Relationship T2
T1
e
s 
 Condsider the Case at
Right. Assume
• A ship Pulls on the Taut
Side With A force of 4 kip
(2 TONS!)
• The Wrap-Angle = Three
Revolutions, or 6
• µs = 0.3
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 The Tension, T1, Applied by
the Worker
T2
4kip
T1   s   0.36  14lb
e
e
Bruce Mayer, PE
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WhiteBoard Work
Let’s Work
These Nice
Problems
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Bruce Mayer, PE
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Engineering 36
Appendix
dy
µx µs
 sinh

dx
T0 T0
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
WhiteBoard Work
Let’s Work
This Nice
Problem
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx
Wedge Push-Out
 SMALL PushOut Force
• Likely SelfLocking
 LARGE PushOut Force
• Likely NOT SelfLocking
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx