Cables and Arches
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Cable Subjected to Concentrated Loads
Cable Subjected to Uniform Distributed
Loads
Arches
Three-Hinged Arch
1
Cable Subjected to Concentrated Loads
A
Ay
Ax
θ
θ
D
yC
yD
B
L2
C
P1
P2
L1
yD
B
C
P1
D
yC
L3
P2
L1
L
TCD
L2
L3
L
+ Σ MA = 0:
Obtain TCD
y
y
TBA
TCD
TCB
C
B
x
P2
+ ΣF = 0:
x
+ ΣFy = 0:
x
TBC
P1
2
Example 5-1
Determine the tension in each segment of the cable shown in the figure below.
Also, what is the dimension h ?
A
h
D
B
2m
C
3 kN
2m
2m
8 kN
2m
1.5 m
3
Ay
SOLUTION
Ax
A
TCD
D5 4
3
h
B
2m
C
3 kN
2m
8 kN
2m
1.5 m
+ ΣMA = 0:
TCD(3/5)(2 m) + TCD(4/5)(5.5 m) - 3kN(2 m) - 8 kN(4 m) = 0
TCD = 6.79 kN
4
y
Joint C
TCD = 6.79 kN
C 5 4
θBC 3
8 kN
TCB
x
+ ΣF = 0: 6.79(3/5) - T cos θ
x
CB
BC
+
TCB = 4.82 kN
Joint B
TBA
+ ΣF = 0: - T cos θ + 4.82 cos 32.3o = 0
BA
BA
x
B 32.3o
θBA
ΣFy = 0: 6.79(4/5) - 8 + TCB sin θCB = 0
θBC = 32.3o
y
=0
x
TBC = 4.82 kN
+
ΣFy = 0: TBA sin θBA - 4.82 sin 32.3o - 3 = 0
3 kN
θBA = 53.8o
A
h
TBA = 6.90 kN
D
B
3 kN
C
8 kN
h = 2 tanθBA = 2 tan53.8o = 2.74 m
5
Cable Subjected to Distributed Load
Concepts & Conclusion:
T
y
θ
T
θ
To
x
W
x=L
W
To
T cos θ = To = FH = Constant
T sin θ = W
dy
W
= tan θ =
dx
To
6
Parabolic Cable: Subjected to Linear Uniform distributed Load
T
wo = force / horizontal distance
θx
y
To
B
A
x
L
x
y
wox
wo x
x
2
wo x
dx
To
2
0
wo x
+ C1
2To
wo x 2
To =
2y
T
x
x
y=∫
y=
θx
To
dy
wx
= tan θ = o
dx
To
at x = L , T = TB = Tmax
2
Tmax = To + ( wo L)
2
Tmax
woL
θΒ
To
7
Derivation:
wo(∆x)
y
∆x
2
T + ∆T
wo
O
h
x
x
θ
T
∆s
∆x
θ+ ∆θ
∆y
∆x
L
+ ΣF = 0:
x
-T cosθ + (T + ∆T) cos (θ + ∆θ) = 0
ΣFy = 0:
-Tsinθ + wo(∆x) + (T + ∆T) sin(θ + ∆θ) = 0
+ ΣMO = 0:
wo(∆x)(∆x/2) - T cos θ(∆y) - T sinθ(∆x) = 0
+
8
Dividing each of these equations by ∆x and taking the limit as ∆x
∆y 0, ∆θ 0, and ∆T 0, we obtain
T
θ
To
wox
d (T cos θ )
=0
dx
----------(5-1)
d (T sin θ )
= wo
dx
----------(5-2)
dy
= tan θ
dx
----------(5-3)
0, and hence
Integrating Eq. 5-1, where T = FH at x = 0, we have:
T cosθ = FH
----------(5-4)
Integrating Eq. 5-2, where T sin θ = 0 at x = 0, gives
T sin θ = wo x
----------(5-5)
Dividing Eq. 5-5 by Eq. 5-4 eliminates T. Then using Eq. 5-3, we can obtain the slope
at any point,
tan θ =
dy wo x
=
dx FH
----------(5-6)
9
Performing a second integration with y = 0 at x = 0 yields
y=
wo 2
x
2 FH
----------(5-7)
y
This is the equation of a parabola. The constant FH
may be obtained by using the boundary condition y =
h at x = L. Thus,
wo L2
FH =
----------(5-8)
2h
h 2
x
L2
h
x
L
Finally, substituting into Eq. 5-7 yeilds
y=
wo
----------(5-9)
From Eq. 5-4, the maximum tension in the cable occurs when θ is maximum; i.e., at
x = L. Hence, from Eqs. 5-4 and 5-5,
Tmax = FH 2 + ( wo L) 2
----------(5-10)
Tmax
woL
θΒ
To
10
Example 5-2
The cable shown supports a girder which weighs 12kN/m. Determine the tension
in the cable at points A, B, and C.
30 m
A
C
12 m
B
6m
11
SOLUTION
TA
θA
30 m
A
y
TC
C
12 m
6m
B
30 - L´
x2
θC
x
wo = 12 kN/m
L´
x1
12
y
TC
θC
C
To
6m
B
x
wo = 12 kN/m
12 L´
L´
12 x1
dy1
= tan θ =
dx1
To
y1 = ∫
12 x1
dx1
To
0
12 x
y1 =
+ C1
2To
2
1
12 L'2
6=
2To
x1
To = L'2
Tx1
----------(1)
12x1
θ
To
13
TA
θA
y
A
12 x2
dy2
= tan θ =
dx2
To
12 m
B
To
wo = 12 kN/m
12 (30 - L´)
30 - L´
x2
x
0
12 x2
12 x22
y2 = ∫
dx2 =
+ C2
To
2To
12 x2
y2=
2To
2
12(30 − L' ) 2
12 =
2To
(30 − L' ) 2
1=
2To
----------(2)
Tx2
12 x2
θ
To
14
To = L'2
----------(1)
(30 − L' ) 2
1=
2To
From (1) and (2),
----------(2)
L´ = 12.43 m, To = 154.5 kN
TB = To = 154.5 kN
TC
12 L´
TA
12 (30 - L´ )
θA
θC
To
To
2
TC = To + (12 L' ) 2
= (154.50) 2 + (12 ×12.43) 2
= 214.8 kN
2
TA = To + [12(30 − L' )]2
= (154.50) 2 + [12(30 − 12.43)]2
= 261.4 kN
15
Example 5-3
The suspension bridge in the figure below is constructed using the two stiffening
trusses that are pin connected at their ends C and supported by a pin at A and a
rocker at B. Determine the maximum tension in the cable IH. The cable has a
parabolic shape and the bridge is subjected to the single load of 50 kN.
I
H
8m
D
A
Pin
E
B
F
G
4 @ 3 m = 12 m
C
6m
rocker
50 kN
4 @ 3 m = 12 m
16
SOLUTION
Hy
Iy
To
8m
H
I
8m
D
A
F
Ay
To
E
6m
Ax
To
G
12 m
To
6m
C
Cx
Cy
+ ΣMA = 0:
Cx
C
Cy
50 kN
3m
B
By
9m
+ ΣMB = 0:
− 12C y + 50(9) − 8To = 0
− 12C y + 8To = 0
To = 1.5C y
----------(1)
From (1) and (2),
Cy = 18.75 kN,
To = −1.5C y +56.25
To = 28.125 kN
----------(2)
17
From (1) and (2),
TI
y
θI
Cy = 18.75 kN,
I
8m
To = 28.12 kN
Tx
wox
To = 28.12 kN
x
wo
wox
12 m
θ
28.12 kN
dy
wx
= tan θ = o
dx
28.12
y=∫
x
y=
wo x
dx
28.12
2
wo x
+ C1
28.12
0
wo (12) 2
8=
2(28.12)
wo = 3.125 kN/m
18
TH
TI
θΙ
H
I
8m
8m
To = 28.12 kN
To = 28.12 kN
12wo = 37.5 kN
12wo = 37.5 kN
12 m
37.5 kN
θH
12 m
TI
θI
TI = (37.5) 2 + ( 28.12) 2
= 46.88 kN
28.12 kN
Tmax = TI = TH = 46.88 kN
Tmin= To = 28.12 kN
19
T
D
0
Ax
T
A
F
T
T
T
T
T = wo × 3 = 3.125 × 3
= 9.375 kN
E
G
Ay
4 @ 3 m = 12 m
+ ΣMA = 0:
T
C
50 kN
B
By
4 @ 3 m = 12 m
9.375(3 + 6 + 9 + 12 + 15 + 18 + 21) − 50(15) + B y ( 24) = 0
By = -1.56 kN,
+
ΣFy = 0:
A y +7(9.375) − 50 − 1.56 = 0
Ay = -14.07 kN,
20
Example 5-4
For the structure shown:
(a) Determine the maximum tension of the cable
(b) Draw quantitative shear & bending-moment diagrams of the beam.
E
0.5 m
D
8m
1 kN/m
8m
A
B
5m
C
Hinge
20 m
21
Ey
SOLUTION
0.5 m
To
8m
20 kN
Dy
5 kN
D
To
Ax
A
Ay
B
1 kN/m
Bx
By
5m
To
E
8m
To
1 kN/m
Bx
B
C
By
Cy
20 m
+ ΣMA = 0:
+ ΣMC = 0:
B y (5) − 5( 2.5) + To (0.5) = 0
B y (20) + 20(10) − To (8) = 0
From (1) and (2),
By = 0, To = 25 kN
22
y
Tx
wox
θ
To= 25 kN
TE = Tmax
E
θ
8m
x
To= 25 kN
20wo
20 m
TE = Tmax
θ
To= 25 kN
dy
wx
= tan θ = o
dx
25
wx
y = ∫ o dx
25
0
wo x 2
=
+ C1
2(25)
wo (20) 2
8=
2(25)
wo = 1 kN/m
20wo = 20 kN
Tmax = TE = ( 25) 2 + ( 20) 2
Tmax = 32.02 kN
23
T = wo(2.5 m) = (1kN/m)(2.5 m) = 2.5 kN
2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5
1 kN/m
Ax
C
B
A
Cy = 1.25 kN
Ay =1.25 kN
5m
20 m
10 @ 2.5 m = 25 m
V (kN)
1.25
1.25
1.25
1.25
1.25
x (m)
-1.25
-1.25
-1.25
-1.25
-1.25
M (kN•m) 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78
x (m)
24
Example 5-5
The cable AB is subjected to a uniform loading of 200 N/m. If the weight of the
cable is neglected and the slope angles at points A and B are 30o and 60o,
respectively, determine the curve that defines the cable shape and the maximum
tension developed in the cable.
y
B
60o
A
30o
x
200 N/m
15 m
25
SOLUTION
TB
B
60o
60o
A
TA
30o
(0.2 kN)(15 m) = 3 kN
15 m
60o
30o
3 kN
TB
TB = 5.20 kN
120o
30o
30o
TA
TB
3
TA
=
=
sin 120 o sin 30 o sin 30 o
30o
TA = 3 kN
26
y
0.2x
T
θ
TA = 3 kN
A
T
θ
x
30o
3 kN
30o
3 sin 30o = 1.5
3 cos 30o = 2.6
0.2x
x
0.2 x + 1.5
dy
= tan θ =
2.6
dx
dy
= 0.0769 x + 0.577
dx
y = ∫ 0.0769 x + 0.577
y=
2
0
0.0769 x
+ 0.577 x + C1
2
y = 0.0385x2 + 0.577x
27
Example 5-6
The three-hinged open-spandrel arch bridge shown in the figure below has a
parabolic shape and supports the uniform load . Show that the parabolic arch is
subjected only to axial compression at an intermediate point D along its axis.
Assume the load is uniformly transmitted to the arch ribs.
y
7 kN/m
x
B
D
− 7.5 2
y=
x
(15) 2
A
15 m
7.5 m
7.5 m
C
7.5 m
28
SOLUTION
210 kN
B
y=
Ax
Ay
15 m
− 7.5 2
x
(15) 2
15 m
Cx
Cy
Entire arch :
+ ΣMA = 0:
C y (30) − 210(15) = 0
Cy = 105 kN
+ ΣFy = 0:
Ay − 210 + 105 = 0
Ay = 105 kN
29
105 kN
B
Bx
By
Cx
B
7.5 m
7.5 m
105 kN
Arch segment BC :
+ ΣMB = 0:
− 105(7.5) + 105(15) − C x (7.5) = 0
Cx = 105 kN
+ ΣF = 0:
x
+ ΣFy = 0:
Bx = 105 kN
B y − 105 + 105 = 0
By = 0
30
52.5 kN
B
105 kN
0
MD
D
26.6o
ND
VD 26.6o
3.75 m
Arch segment BD :
A section of the arch taken through point D, x = 7.5 m, y = -7.5(7.52)/(15)2 = -1.875 m,
is shown in the figure. The slope of the segment at D is
tan θ =
dy − 15
=
x
= −0.5 ,
2
dx (15) x =7.5
θ = 26.6o
+ ΣF = 0:
x
105 - ND cos 26.6o - VD sin 26.6o = 0
+ ΣFy = 0:
-52.5 + ND sin 26.6o - VD cos 26.6o = 0
+ ΣMD = 0:
MD + 52.5(3.75) - 105(1.875) = 0
ND = 117.40 kN,
VD = 0,
MD = 0 kN
31
52.5 kN
Alternate Method
105 kN
B
0
MD
D
26.6o
ND
VD 26.6o
3.75 m
Arch segment BD :
A section of the arch taken through point D, x = 7.5 m, y = -7.5(7.52)/(15)2 = -1.875 m,
is shown in the figure. The slope of the segment at D is
tan θ =
dy − 15
=
x
2
dx (15)
7.5 wo = (7.5)(7)= 52.5 kN
Tmax = TE = (105) 2 + (52.5) 2
Tmax = 117.4 kN
x = 7.5
= −0.5 ,
θ = 26.6o
ND
θ
No= 25 kN
Notes : Since the arch is a parabola, there are no
shear and bending moment, only ND is present
32
Example 5-7
The three-hinged tied arch is subjected to the loading shown in the figure below.
Determine the force in members CH and CB. The dashed member GF of the truss
is intended to carry no force.
20 kN
15 kN
15 kN
G
H
1m
C
B
4m
F
D
E
A
3m
3m
3m
3m
33
20 kN
SOLUTION
15 kN
15 kN
G
H
1m
C
B
4m
Ax
F
D
E
A
Ay
Ey
3m
+ ΣMA = 0:
3m
3m
3m
E y (12) − 15(3) − 20(6) − 15(9) = 0
Ey = 25 kN
+ ΣF = 0:
x
Ax = 0
+ ΣFy = 0:
Ay − 15 − 20 − 15 + 25 = 0
Ay = 25 kN
34
20 kN
15 kN
G
H
B
5m
0
C
0
Cx
Cy
A
FAE
25 kN
3m
+ ΣMC = 0:
3m
FAE (5) − 25(6) + 15(3) = 0
FAE = 21.0 kN
+ ΣF = 0:
x
-Cx + 21= 0
Cx = 21.0 kN
+ ΣFy = 0:
25 − 15 − 20 + C y = 0
Cy = 10 kN
35
20 kN
FHG
G
20 kN
FCH
18.43o
18.43o
FCB
0
FGC
Joint G :
C
21 kN
10 kN
Joint C :
+ ΣF = 0:
x
FHG = 0
+ ΣFy = 0:
FGC − 20 = 0
FGC = 20 kN (C)
+ ΣF = 0:
x
-FCH cos18.43 - FCB cos18.43 - 21= 0
+ ΣFy = 0:
FCH sin18.43 - FCB cos18.43 - 20 + 10 = 0
Thus,
FCH = 4.75 kN (T),
FCB = -26.88 kN (C)
36
Arches
extrados
(or back)
crown
Intrados
(or soffit)
huanch
springline
centerline rise
abutment
fixed arch
two-hinged arch
three-hinged arch
tied arch
37
Three-Hinged Arch
P1
C
P2
D
B
A
P1
C
Ax
A
Ay
Cy
Cx
Cx
Cy
D
C
B
VD
Ax
P2
Bx
MD N
D
Ay
By
38
Example 5-8
The tied three-hinged arch is subjected to the loading shown. Determine the
components of reaction at A and C and the tension in the cable.
15 kN
B
10 kN
2m
A
D
C
2m
0.5 m
2m
1m
39
SOLUTION
15 kN
B
10 kN
2m
A
Ax
0
D
C
Ay
2m
0.5 m
2m
Cy
1m
Entire arch :
+ ΣMA = 0:
C y (5.5) − 10( 4.5) − 15(0.5) = 0
Cy = 9.545 kN
+ ΣFy = 0:
Ay − 15 − 10 + 9.545 = 0
Ay = 15.46 kN
40
15 kN
B
Bx
By
2m
A
Bx
B
10 kN
By
TA
TD
D
Ay = 15.46 kN
C
2m
Cy = 9.545 kN
2m
0.5 m
1m
Member AB :
+ ΣMB = 0:
15( 2) − 15.455(2.5) + TA (2) = 0
TA = 4.319 kN
+ ΣFy = 0:
15.455 − 15 − B y = 0
By = 0.455 kN
+ ΣF = 0:
x
4.319 − Bx = 0
Bx = 4.319 kN
4.319 − TD = 0
TD = 4.319 kN
Member AB :
+ ΣF = 0:
x
41