CHAPTER 3:
CABLES
AND
ARCHES
Dr. ZaynabTarhini +
Dr.Hanadi Elkhansa
ARCHES
o Definition
*An arch acts as inverted cable so it
receives its load mainly in
compression.
*Because of its rigidity, it must also
resist some bending and shear
depending upon how it is loaded &
shaped.
Arches
o Definition
*If the arch has a parabolic shape and it is subjected to a uniform horizontally
distributed vertical load
→From the analysis of cables,
Only compressive forces will be resisted by the arch
(arch shape: funicular → because no bending or shear forces
occur within the arch)
*Depending upon the application, several types of arches can be selected to
support a loading:
4
Arches
o Definition
“Fixed arch”
(statically indeterminate, 3𝑟𝑑 degree)
Made from reinforced concrete
“Two hinged arch”
(statically indeterminate, 1𝑠𝑡 degree)
Made from metal or timber
5
Arches
o Definition
“Three hinged arch” **
(statically determinate)
Made from metal or timber
“Tied arch”
Allows the structure to behave as a rigid unit
Tie rod carries the horizontal component of
thrust at the support 6
Arches
o Three hinged arch
*The third hinge is located at the crown &
the supports are located at different elevations.
The arch is pinned at its top
7
Arches
o Three hinged arch
*To determine the reactions at the supports → the arch is disassembled
Solution:
M A = 0 equation(C x , C y )
M B = 0 equation(C x , C y )
We get (C x , C y )
Then, reaction supports ( R Ax , R Ay ) are
determined from +Fx = 0
6 unknowns
+Fy = 0
→ we need 6 equations 8
Similar for ( RBx , RBy )
Arches
o Three hinged arch
*Once the reactions are obtained → the internal normal force, shear and
moment loadings at any point along the arch can be found using the
method of section (section ⊥ to the axis of the arch at the point considered)
Arches
Exercise 5
The three-hinged arch bridge has
a parabolic shape and supports
the uniform load.
Show that the parabolic arch is
subjected only to axial
compression at any intermediate
point such as point D.
Assume the load is uniformly
transmitted to the arch ribs.
10
Arches
1st step: Determine the reactions at supports and the force at the internal
hinge B
1*Entire arch:
MA = 0 +
− R Cy ( 40) + 320( 20) = 0
R Cy = 160kN
Fy = 0 RAx RCx
R Ay + R Cy = 320 R Ay = 160kN
RAy RCy
Fx = 0
R Ax = R Cx (1) 11
Arches
2* Local study: Consider the arch segment BC (or AB):
MB = 0
160
160(10m) − R Cy ( 20) + R Cx (10) = 0
R Cx = 160kN
(1) R Ax = 160kN →
+ Fx = 0
Bx − RCx = 0 Bx = 160kN →
RCx
+ Fy = 0
B y − 160 + RCy = 0 B y = 0 RCy 12
160
Arches
2nd step: A section of the arch taken through point D: Consider part BD
− 10 2
Equation: y = 2
x
(20)
At point D: x= 10m
− 10
y= 2
10 2
= −2.5m
(20)
The slope of the segment at D is :
ϴ
dy − 2(10)
tan = = 2
x = −0.5 ϴ
dx (20) x =10 m
= −26.6
13
Arches
Applying the equations of equilibrium
+ Fx = 0
160 + N D cos 26.6 o − VD sin 26.6 o = 0
+ Fy = 0
− 80 − N D sin 26.6 o − VD cos 26.6 o = 0
N D = −178.9kN VD = 0
MD = 0
Then, point D is subjected only
M D + 80(5) − 160( 2.5) = 0 M D = 0 to a compressive axial force 14
N D 0;VD = 0; M D = 0
Arches
Exercise 6
Consider the three-hinged arch
Determine the forces at A, B
and C
15
Arches
1*Entire arch:
MA = 0 +
2(3) + 3( 4) + 4(5) + 4(10) + 5(13) − R Cy (15) = 0
R Cy = 9.53kN
Fy = 0
R Ay + R Cy = 18 R Ay = 8.47 kN
RAx
Fx = 0
RCx
R Ax = R Cx (1) RAy RCy
16
Arches
2*Local study: Consider the arch segment BC (or AB)
MB = 0 +
4( 2) + 5(5) − R Cy (7) + R Cx (5) = 0
R Cx = 6.742kN
Bx
Fy = 0
B y + R Cy = 9 B y = −0.53 = 0.53kN By
RCx
Fx = 0
B x − R Cx = 0 B x = 6.742kN → RCy
(1) R Ax = 6.742kN → 17
Arches
The forces at A, B and C:
F A= R Ax + R Ay = 10.8kN
2 2
F B= R Bx + R By = 6.72kN
2 2
F C= R Cx + R Cy = 11.7 kN
2 2
18
Arches
Exercise 7
19
20
21
| END OF CHAPTER 3
