CIVL3310 STRUCTURAL ANALYSIS Professor CC Chang Chapter 5: Cables and Arches.
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Transcript of CIVL3310 STRUCTURAL ANALYSIS Professor CC Chang Chapter 5: Cables and Arches.
CIVL3310 STRUCTURAL ANALYSIS
Professor CC Chang
Chapter 5: Chapter 5: Cables and ArchesCables and Arches
Cables: Assumptions
Cable is perfectly flexible & inextensible No resistance to shear/bending: same as
truss bar The force acting the cable is always
tangent to the cable at points along its length
Only axial force!
Example 5.1 Under Concentrated Forces
Determine the tension in each segment of the cable. Also, what is the dimension h?
4 unknown external reactions (Ax, Ay, Dx and Dy)
3 unknown cable tensions
1 geometrical unknown h
8 unknowns
8 equilibrium conditions
Solution
mmhFinally
kNT
o
BAo
BA
7428532
906 and 853
Bat mequilibriuJoint Similarly,
..tan)(,
..
BCBA
kNT
mkNmkNmTmT
M
CD
CDCD
A
796
048235554253
0
.
)()().)(/())(/(
kNT
TkNkN
TkN
BCo
BC
BCBC
BCBC
824 and 332
0854796
053796
Cat mequilibriuJoint
..
sin)/(.
cos)/(.
Cable subjected to a uniform distributed load
Consider this cable under distributed vertical load wo
The cable force is not a constant.
0sincos)2/)((
0
ve as clockwise-antiWith
0)sin()()(sin
0
0)cos()(cos
0
0
xTyTxxw
M
TTxwT
F
TTT
F
o
o
y
x
3eqn tandx
dy
2eqn wdx
)sinT(d
1eqn 0dx
)cosT(d
o
FH
wo
cos)cos(coscos TTT
Cable subjected to a uniform distributed load
From Eqn 1 and let T = FH at x = 0:
Integrating Eqn 2 realizing that Tsin = 0 at x = 0:
Eqn 5/Eqn 4:
4eqn HFttanconscosT
5eqn sin xwT o
3eqn tandx
dy
2eqn wdx
)sinT(d
1eqn 0dx
)cosT(d
o
FH
6eqn tanH
o
F
xw
dx
dy
FH
TTsin
Cable subjected to a uniform distributed load
Performing an integration with y = 0 at x = 0 yields
7eqn 2
2xF
wy
H
o
6eqn tanH
o
F
xw
dx
dy
FH
Cable profile:parabola
8eqn 2
2
h
LwF oH
y = h at x = L
9eqn 22x
L
hy
Cable subjected to a uniform distributed load
Where and what is the max tension?
T is max when x=L10eqn 22 )Lw(FT oHmax
11eqn 21 2)h/L(LwT omax
FH
4eqn cos HFT
5eqn sin xwT o
8eqn 2
2
h
LwF oH
Tmax
max
22 )( xwFT oH
Cable subjected to a uniform distributed load
FH
cos HFT
sin xwT o
tanH
o
F
xw
dx
dy
T
2
2
h
LwF oH
22x
L
hy
)2/(1 2max hLLwT o
Cable subjected to a uniform distributed load
Neglect the cable weight which is uniform along the length
A cable subjected to its own weight will take the form of a catenary curve
This curve ~ parabolic for small sag-to-span ratio
a
xcoshay
Hangers are close and uniformly spaced
If forces in the hangers are known then the structure can be analyzed
1 degree of indeterminacy
Determinate structure
hinge
Wiki catenary
Example 5.2The cable supports a girder which weighs 12kN/m. Determine the tension in the cable at points A, B & C.
12kN/m
Solution
The origin of the coordinate axes is established at point B, the lowest point on the cable where slope is zero,
Assuming point C is located x’ from B:
From B to A:
(1) 6
2
12kN/m
2222 x
Fx
Fx
F
wy
HHH
o
(2) '0.1'6
6 22 xFxF HH
FH
mxxx
xx
xFH
43.12'0900'60'
)]'30(['0.1
612
)]'30([6
12
2
22
2
FH
kN.4154
203890 x.
Solution
FH=154.4kN
203890 x.y
x.dx
dytan 07770
kN.cos
FT
.
.dx
dytan
A
HA
oA
.xA
4261
7953
36615717
12.43m17.57m
A
C
CT
AT
kN..cos
.
cos
FT
.
.dx
dytan
oC
HC
oC
.xC
6214044
4154
044
96604312
cos HFT
Example 5.3
Determine the max tension in the cable IH
Assume the cable is parabolic
(under uniformly distributed load)
75180 .AIM yyB
HyyC F.AIM 66700
kN.FH 1328
Example 5.3
kN.FH 1328
2
2
h
LwF oH
m/kN.wo 133
kN46.9 21 2 )h/L(LwT omax
Cable and Arch
flip
What if the load direction reverses?
FH
FH
Arches An arch acts as inverted cable so it receives
compression An arch must also resist bending and shear
depending upon how it is loaded & shaped
Arches Types of arches
indeterminate
indeterminate
indeterminatedeterminate
Three-Hinged Arch
Ax
AyBy
Bx
Problem 5-30
Determine reactions at A and C and the cable force
3 global Eqs1 hinge condition
Ay
Ax
Cy
Ay
Ax
By
Bx
T
Example 5.4
The three-hinged arch bridge has a parabolic shape and supports the uniform load. Assume the load is uniformly transmitted to the arch ribs.
Show that the parabolic arch is subjected only to axial compression at an intermediate point such as point D.
Solution
=160 kN=160 kN
=160 kN
=160 kN10 m
=160 kN
=160 kN
160 kN =
0 =
2
220
10xy
x
y
o
102
626
5020
20
.
.x)(dx
dytan
D
mxD
0
0
9178
D
D
D
M
V
kN.N
8 kN/m
Reflection: What Have You Learnt?