Transcript ENGR-36_Lec-20_Cables

Engineering 36
Chp 7:
Flex Cables
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Recall Chp10 Introduction

Examine in Detail Two Important
Types Of Engineering Structures:
1. BEAMS - usually long, straight,
prismatic members designed to support
loads applied at various points along
the member
2. CABLES - flexible members capable
of withstanding only tension, designed
to support concentrated or
distributed loads
Engineering-36: Engineering Mechanics - Statics
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Load-Bearing Cables
Göteborg, Sweden
Straight
Curved
WHY the Difference?
http://www.nmt.edu/~armiller/bridgefu.htm
 Cables are applied as structural elements in
suspension bridges, transmission lines, aerial
tramways, guy wires for high towers, etc.
Engineering-36: Engineering Mechanics - Statics
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Concentrated Loads on Cables

To Determine the Cable
SHAPE, Assume:
a) Concentrated vertical loads
on given vertical lines
b) Weight of cable is negligible
c) Cable is flexible, i.e.,
resistance to bending
is small
d) Portions of cable between successive loads may be
treated as TWO FORCE MEMBERS
•
Internal Forces at any point reduce to
TENSION Directed ALONG the Cable Axis
Engineering-36: Engineering Mechanics - Statics
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Concentrated Loads (2)
 Consider entire cable
as a free-body
 Slopes of cable at A
and B are NOT known
 FOUR unknowns
(i.e., Ax, Ay, Bx, By)
are involved and the
equations of equilibrium
are NOT sufficient to
determine the
reactions.
Engineering-36: Engineering Mechanics - Statics
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Concentrated Loads (3)
 To Obtain an additional equation
• Consider equilibrium of
cable-section AD
• Assume that CoOrdinates of SOME
point D, (x,y), on the cable have been
Determined (e.g., by measurement)
 Then the added Eqn:  M D  0
 With pt-D info, the FOUR Equilibrium Eqns
1`  Fx  0  Ax  Bx
2  Fy  0  Ay  By  P1  P2  P1
3  M A  0  Bxd  By L  P1x1  P2 x2  P3 x3
4  M D  0  P1x  x1   Ax y  Ay x
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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Concentrated Loads (4)
 The 4 Eqns Yield Ax & Ay
 Can Now Work our Way Around
the Cable to Find VERTICAL
DISTANCE (y-CoOrd) For
ANY OTHER point
known
 Example  Consider Pt C2
 M  0 yields y
 F  0,  F  0 yields T ,T
C2
x
UNknown
2
y
x
Tx  T cos  Ax  constant
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known
known
y
known
known
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known
Example  Concentrated Loads

 The cable AE
supports three
vertical loads from
the points indicated.
Point C is 5 ft below
the left support
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For the Given
Loading &
Geometry,
Determine:
a) The elevation of
points B and D
b) The maximum slope
and maximum
tension in the cable.
Bruce Mayer, PE
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Example  Concentrated Loads
Known
CoOrds
•
 Solution Plan
• Determine reaction
force components at
pt-A from solution of
two equations
•
formed from taking
entire cable as a
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free-body and summing
moments about E, and from
taking cable portion ABC
as a free-body and
summing moments about C.
Calculate elevation of B by
considering AB as a
free-body and summing
moments B. Similarly,
calculate elevation of D
using ABCD as a free-body.
Evaluate maximum slope
and maximum tension
which occur Bruce
in DE.
Mayer, PE
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Example  Concentrated Loads
entire cable as a
free-body and
summing moments
about E:
M
E
 0:
20Ax  60Ay  406  3012  154  0
20Ax  60Ay  660  0
 Determine two
OR
reaction force
components at A
Ax  3 Ay  33
from solution of two
equations formed
from taking the
Engineering-36: Engineering Mechanics - Statics
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Example  Concentrated Loads
M
C
 0:
 5 Ax  30Ay  106  0
OR

 Next take Cable
Section ABC as a
Free-Body, and
Sum the Moments
about Point C
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Ax  6 Ay  12
Recall from ΣME
Ax  3Ay  33

Solving 2-Eqns in
2-Unknowns for
Ax & Ay
Ax  18 kips
Ay  5 kips
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Example  Concentrated Loads

Determine elevation of B
by considering AB as a
free-body and summing
moments about B.
y B 18  520  0
MB  0:
y B  5.56 ft

Similarly, Calc elevation at D
using ABCD as a free-body
M
D
 0:
 yD 18  455  256  1512  0
yD  5.83 ft
Engineering-36: Engineering Mechanics - Statics
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Maximum Tension
Analysis
 By the ΣFx = 0
F
x
 0  Ax  Tx OR
Tx   Ax
but
Tx  T cos 
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T

 Solving T   Ax
cos 
for T
 Thus T is Maximized
by Maximum θ
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Find Maximum Segment Angle
 Use the y-data just calculated to find the
cable segment of steepest slope
Pt x (ft) y (ft) Seg ∆x (ft)
A 0
0
AB 20
B 20 -5.56
BC 10
C 30
-5
CD 15
D 45 5.83
DE
15
E 60
20
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∆y (ft) θ by aTan(∆y/∆x) (°)
-5.56
-15.54
0.56
3.21
10.83
35.83
14.17
43.37
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Example  Concentrated Loads

Use yD to Determine
Geometry of tanθ
14.17
tan  


43
.
4

15
 Evaluate maximum  Employ the
Just-Determined θ
slope and maximum
to Find Tmax
tension which occur
18 kips
in the segment with
Tmax 
cos
the STEEPEST
Slope (large θ); DE
Tmax  24.8 kips
in this case
Engineering-36: Engineering Mechanics - Statics
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Distributed Loads on Cables

For a negligibleWeight Cable
carrying a
Distributed Load of
Arbitrary Profile
y
a) The cable hangs in
shape of a CURVE
b) INTERNAL force is
a tension force
directed along the
TANGENT to the
curve
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x
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Distributed Loads (2)

Consider the Free-Body for a
portion of cable extending from
LOWEST point C to given point
D. Forces are T0 (FH in the Text
Book) at Lo-Pt C, and the
tangential force T at D

From the Force Triangle
T cos  T0
T
T02
W
T sin   W
2
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W
tan  
T0
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Distributed Loads (3)

Some Observations based on
Tx  T cos  T0
Ty  T sin   W
T  T W
2
0
•
•
•
2
Horizontal component of T is
uniform over the Cable Length
Vertical component of T is equal to
the magnitude of W
Tension is minimum at lowest point
(min θ), and maximum at A and B
(max θ)
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W
tan 
T0
Bruce Mayer, PE
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Parabolic Cable


Consider a cable supporting a
uniform, horizontally distributed
load, e.g., support cables for a
suspension bridge.
With loading on cable from lowest
point C to a point D given by internal
tension force, T, and Vertical Load
wx
W = wx, find: T  T 2  w2 x 2
tan 
0


Summing moments about D
x
M

0
:
wx
 T0 y  0
 D
2
The shape, y, is PARABOLIC: y
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T0
2
wx

2T0
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T0 for Uniform Vertical Load
 Consider the
uniformly Loaded
Cable
• In this case: w(x) = w
– w is a constant
L
 L is the Suspension
Span: L  xB  xA
wx 2
 From
y
Last Slide
2T0
 Or x   2T0 y w
 Then x B   2T0 yB w
xB & xA x A   2T0 y A w
 Thus L
L  2T0 yB w  2T0 y A w
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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T0 for Uniform Vertical Load
 Factoring Out 2T0/w
L  T0 2 w

yB  y A

 Isolating T0
T0 
2

L2 w
yB  y A

2
 If WE design the
Suspension System,
then we KNOW
• L (Span)
• w (Load)
• yA & yB (Dims)
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 Example
• L= 95 m (312 ft)
• w = 640 N/m
(44 lb/ft)
• yA = 19m
• yB = 37m
 Then
T0 = 26 489 N
(5955 lb)
 Tmax
by
2
Tmax  T02  w2 xmax
Bruce Mayer, PE
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Tmax for Uniform Vertical Load
 Find xmax from
x B   2T0 yB w
 In this case
xmax = 55.3 m
(181 ft)
 And finally
Tmax = 44 228 N
(9 943 lbs)
• Buy Cable rated to
20 kip for Safety
factor of 2.0
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>> L = 95
L =
95
>> w = 640
w =
640
 MATLAB
Calcs
>> yA = 19
yA =
19
>> yB = 37
yB =
37
>> TO =L^2*w/(2*(sqrt(yB)+sqrt(yA))^2)
TO =
2.6489e+04
>> xB = sqrt(2*TO*yB/w)
xB =
55.3420
>> Tmax = sqrt(TO^2 + (w*xB)^2)
Tmax =
4.4228e+04
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
Load-Bearing Cables
Göteborg, Sweden
Vertically
Loaded
End Loads
NO Deck-Support Cables
Deck-Support COLUMNS
http://www.nmt.edu/~armiller/bridgefu.htm
 Tension in the Straight-Section is roughly Equal
to the Parabolic-Tension at the Tower-Top.
• So the Support Tower does Not Bend
Engineering-36: Engineering Mechanics - Statics
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UNloaded Cable → Catenary

Consider a cable uniformly
loaded by the cable itself,
e.g., a cable hanging under
its own weight.
•
With loading on the cable from
lowest point C to a point D given
by W = ws, the Force Triangle
reveals the internal tension force
magnitude at Pt-D:


T  T02  w2 s 2  w2 T02 w2  s 2  w c 2  s 2
– Where
c  T0 w ft or m
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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UNloaded Cable → Catenary (2)

Next, relate horizontal
distance, x, to cable-length s
dx  ds cos

T0
But by Force
cos  
Balance Triangle
T

Also From last slide recall
T  w c2  s2
and
T 0 wc
 Thus
T0
wc
c
dx  ds cos  ds
 ds

ds
T
w c2  s2
c2  s2
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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UNloaded Cable → Catenary (3)

Factoring Out c in DeNom
c
c
dx 
ds 
ds
c2  s2
c c2 c2  s2 c2

Finally the Differential Eqn
dx 

1
1 s c
2
2
Integrate Both Sides using
Dummy Variables of Integration:
•
σ: 0→x
η: 0→s
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ds
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UNloaded Cable → Catenary (4)

Using σ: 0→x
 x


 x

0

0
 s
d  
 0
η: 0→s
1
d
1  2 c2
Now the R.H.S. AntiDerivative is the argSINH
d   
 x
 0
 s

 0
 s

  
d  c  arg sinh 
2
2
 c  0
1  c

1
Noting that
1


argsinh 0  sinh 0  0
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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UNloaded Cable → Catenary (5)

Thus the Solution to the Integral Eqn
 
 s
 x
 0

  
1  s 
 x  0  c  arg sinh   c  sinh    0
 c    0
c

x
s
1  s 
 Then x  c  sinh     sinh  
c
c
c
1

Solving for s in
terms of x by
taking the sinh of
both sides
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s
 x
 sinh 
c
c
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
UNloaded Cable → Catenary (6)

Finally, Eliminate s in favor
of x & y. From the Diagram
dy  dx tan

From the
Force Triangle

And From Before
W  ws and T0  wc

So the Differential Eqn
W
ws
s
dy  dx tan  dx 
dx  dx
T0
wc
c
Engineering-36: Engineering Mechanics - Statics
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W
tan 
T0
Bruce Mayer, PE
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UNloaded Cable → Catenary (7)
s
 x
 sinh 
c
c

Recall the Previous Integration
That Relates x and s

Using s(x) above in the last ODE
W 
s
 ws 
 x


dy  tan  dx   dx  
dx  dx  sinh  dx
c
 wc 
c
 To 

Integrating the ODE with Dummy Variables:
•

 y
 c
Ω: c→y
d  
 y
 c
σ: 0→x

  sinh
 0
c
 x
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


d  c cosh

c

 x


 0
Bruce Mayer, PE
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UNloaded Cable → Catenary (8)

Noting that cosh(0) = 1
 x

  
 y
c

 y  c  c cosh 
 c   0

 x
 c cosh   c
c
Solving for y yields the
Catenary Equation:
y  c coshx c 
•
Where
– c = T0/w
– T0 = the 100% laterally directed force at the ymin point
– w = the lineal unit weight of the cable (lb/ft or N/m)
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Bruce Mayer, PE
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Catenary Comments
y  c coshx c 
c  T0 w

With Hyperbolic-Trig ID: cosh2 – sinh2 = 1
y 2  s 2  c 2 cosh2 x c   c 2 sinh2 x c 


 y 2  s 2  c 2 cosh2 x c   sinh2 x c   c 2

Or: y  c  s

Recall From the Differential Geometry

2
2
and c  y  s
2
2
2
T c, s   w c 2  s 2  w y 2  wy  T  y 
 wx 
 wx 
T0
  T0 cosh 
or T x   wy  w cosh
w
 T0 
 T0 
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2
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Loaded & Unloaded Cable Profiles
550
PARAMETERS
• w/T0 = 0.1 (Parabola)
• c = 50 (Catenary)
Parabola (Loaded)
500
Catenary (Unloaded)
450
Relative Vertical Displacement, y
Loaded and
Unloaded
Cables
Compared
wx 2
y
2T0
400
350
300
250
200
150
y  c coshx c 
100
50
0
-100
-75
file = Catenary-Parabola_0407.xls
-50
-25
25
50
75
Relative Horizontal Displacement, x
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0
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100
y = 0 at Cable Minimum
 Translate the CoOrd
System Vertically
from Previous:
xx
 No Change
yO  y  c 
y  yO  c
 Recall Eqn for y−c
 x
y  c  c cosh   c
c
 Sub y = yO+c
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yO
 x
 yO  c   c  c cosh   c
c
 Thus with Origin at
cable Minimum
 x
yO  c cosh   c
c
Bruce Mayer, PE
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y = 0 at Cable Minimum (2)
 Then

 x 
yO  c cosh   1
c 

yO
L
 Recall c = T0/w
 Thus
 x  
T0 
  1
yO   cosh

w
T
w
 0  
 Next, Change the
Name of the Cable’s
Lineal Specific
Weight (N/m or lb/ft)
w
Engineering-36: Engineering Mechanics - Statics
35
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
y = 0 at Cable Minimum (3)
 With µ replacing w
T0 
x 
yO   cosh  1

T0

 In Summary can
Use either
Formulation based
on Axes Origin:
yO
L
 T0 w
T0
wx
y  cosh
w
T0
Engineering-36: Engineering Mechanics - Statics
36
T0 
x 
yO   cosh  1

T0

Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
Cable Length, S, for Catenary
 Using this Axes Set
yO
L
W = µs
 With Cable macrosegment and
differential-segment
at upper-right
Engineering-36: Engineering Mechanics - Statics
37
 The Force Triangle
for the
MacroW
= µs
Segment
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
Cable Length, S , for Catenary
s
 By Force
tan 
Triangle
T0
 But by Differential
Segment notice
dy
tan  
dx
 By Transitive
Property
dy µs
tan 

dx T0
Engineering-36: Engineering Mechanics - Statics
38
 Now ReCall
T0 
x 
y   cosh
 1

T0

 Then dy/dx
dy d  T0 
x 
x



 1  sinh
  cosh
dx dx   
T0
T0

 Subbing into the
tanθ expression
dy
µx µs
 sinh

dx
T0 T0
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
Cable Length, S , for Catenary
 Solving the last Eqn
for s
T0
µx
s  sinh
µ
T0
 From the CoOrds
 So Finally
T0 
µxB
µx A 

S   sinh
 sinh
µ
T0
T0 
yO
L
S s OB s OA
 s  xB   s  x A 
Engineering-36: Engineering Mechanics - Statics
39
 Now find T(y)
 Recall T = f(x)
 x 
T  T0 cosh 
 T0 
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
T(y) for Catenary
 Also ReCall
T0 
x 
y   cosh
 1

T0

 Solve above for
cosh

 µx 
T  T0 cosh 
 T0 

 µy 
T  T0   1  µy  T0
 T0

x
µ
µy
cosh
 y 1 
1
T0
T0
T0
 Sub cosh into T(x)
Expression
Engineering-36: Engineering Mechanics - Statics
40
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
Catenary Summary
 y(x)
T0 
x 
y   cosh
 1

T0

 µx 
T  T0 cosh 
 T(x)
 T0 
T(y)
T  µy  T0
 S(x)
 Slope at any pt
dy
µx µs
 sinh

dx
T0 T0
 Angle θ at any pt
µx µs
tan  sinh

T0 T0
T0 
µxB
µx A 

S   sinh
 sinh
µ
T0
T0 
Engineering-36: Engineering Mechanics - Statics
41
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
WhiteBoard Work
Let’s Work
These Nice
Problems
Engineering-36: Engineering Mechanics - Statics
42
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
Engineering 36
Appendix
dy
µx µs
 sinh

dx
T0 T0
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
43
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
WhiteBoard Work
Let’s Work
These Nice
Problems
Engineering-36: Engineering Mechanics - Statics
44
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx

T0
w
Engineering-36: Engineering Mechanics - Statics
45
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx
Engineering-36: Engineering Mechanics - Statics
46
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-20_Cables.pptx