198
CHAPTER 5
CABLES AND ARCHES
EXAMPLE 5.5
The three-hinged tied arch is subjected to the loading shown in
Fig. 5–11a. 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
H
1m
G
1m
20 kN
15 kN
15 kN
15 kN
F
C
B
D
4m
A
5
E
3m
3m
3m
Ax
E
A
3m
3m
3m
3m
3m
Ay
Ey
(b)
(a)
Fig. 5–11
SOLUTION
The support reactions can be obtained from a free-body diagram of
the entire arch, Fig. 5–11b:
20 kN
15 kN
d+ ©MA = 0;
0
C
Cy
5m
FAE
3m
3m
Cx
Ey112 m2 - 15 kN13 m2 - 20 kN16 m2 - 15 kN19 m2 = 0
+ ©F = 0;
:
x
+ c ©Fy = 0;
Ey = 25 kN
Ax = 0
Ay - 15 kN - 20 kN - 15 kN + 25 kN = 0
Ay = 25 kN
The force components acting at joint C can be determined by considering the free-body diagram of the left part of the arch, Fig. 5–11c.
First, we determine the force:
25 kN
d + ©MC = 0;
(c)
FAE15 m2 - 25 kN16 m2 + 15 kN13 m2 = 0
FAE = 21.0 kN
5.5
THREE-HINGED ARCH
199
Then,
+ ©F = 0;
:
x
- Cx + 21.0 kN = 0, Cx = 21.0 kN
+ c ©Fy = 0;
25 kN - 15 kN - 20 kN + Cy = 0, Cy = 10 kN
To obtain the forces in CH and CB, we can use the method of joints
as follows:
20 kN
FHG
G
0
FGC
Joint G; Fig. 5–11d,
(d)
+ c ©Fy = 0;
FGC - 20 kN = 0
20 kN
FGC = 20 kN 1C2
FCH
1
Joint C; Fig. 5–11e,
+ ©F = 0;
:
x
+ c ©Fy = 0;
Thus,
FCB
3
3
1
C
21.0 kN
10 kN
FCB A 1310 B - 21.0 kN - FCH A 1310 B = 0
(e)
FCB A 1110 B + FCH A 1110 B - 20 kN + 10 kN = 0
FCB = 26.9 kN 1C2
FCH = 4.74 kN 1T2
Ans.
Ans.
Note: Tied arches are sometimes used for
bridges. Here the deck is supported by
suspender bars that transmit their load to
the arch. The deck is in tension so that it
supports the actual thrust or horizontal
force at the ends of the arch.
5
200
CHAPTER 5
CABLES AND ARCHES
EXAMPLE 5.6
The three-hinged trussed arch shown in Fig. 5–12a supports the
symmetric loading. Determine the required height h1 of the joints B
and D, so that the arch takes a funicular shape. Member HG is
intended to carry no force.
5k
5k
J
5k
I
5k
5k
H
G
F
C
h1
5
D
15 ft
B
h1
A
E
10 ft
10 ft
10 ft
10 ft
(a)
y
10 ft
10 ft
x
C
SOLUTION
For a symmetric loading, the funicular shape for the arch must be
parabolic as indicated by the dashed line (Fig. 5–12b). Here we must
find the equation which fits this shape. With the x, y axes having an
origin at C, the equation is of the form y = - cx2. To obtain the
constant c, we require
yD
D
-115 ft2 = - c120 ft22
c = 0.0375>ft
15 ft
Therefore,
E
(b)
Fig. 5–12
y ! "cx2
yD = - 10.0375>ft2110 ft22 = - 3.75 ft
So that from Fig. 5–12a,
h1 = 15 ft - 3.75 ft = 11.25 ft
Ans.
Using this value, if the method of joints is now applied to the truss, the
results will show that the top cord and diagonal members will all be
zero-force members, and the symmetric loading will be supported
only by the bottom cord members AB, BC, CD, and DE of the truss.