CIVL3310 STRUCTURAL ANALYSISProfessor 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 tensions1 geometrical unknown h8 unknowns8 equilibrium conditions

Solution

mmhFinally

kNT

o

BAo

BA

7428532

906 and 853

Bat mequilibriuJoint Similarly,

..tan)(,

..

BCBA

kNTmkNmkNmTmT

M

CD

CDCD

A

796048235554253

0

.)()().)(/())(/(

kNT

TkNkNTkN

BCo

BC

BCBC

BCBC

824 and 332

0854796053796

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()(cos0

0

xTyTxxwM

TTxwT

F

TTTF

o

o

y

x

3eqn tandxdy

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 tandxdy

2eqn wdx

)sinT(d

1eqn 0dx

)cosT(d

o

FH

6eqn tanH

o

Fxw

dxdy

FH

TTsin

Cable subjected to a uniform distributed load

Performing an integration with y = 0 at x = 0 yields

7eqn 2

2xFwyH

o

6eqn tanH

o

Fxw

dxdy

FH

Cable profile:parabola

8eqn 2

2

hLwF o

H y = h at x = L

9eqn 22 xLhy

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

hLwF o

H

Tmax

max

22 )( xwFT oH

Cable subjected to a uniform distributed load

FH

cos HFT

sin xwT o

tanH

o

Fxw

dxdy

T

2

2

hLwF o

H

22 xLhy

)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

axcoshay

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

SolutionThe 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) 62

12kN/m2

222 xF

xF

xFw

yHHH

o

(2) '0.1'66 22 xFxF HH

FH

mxxx

xx

xFH

43.12'0900'60'

)]'30(['0.1

612

)]'30([612

2

22

2

FH

kN.4154

203890 x.

Solution

FH=154.4kN

203890 x.y

x.dxdytan 07770

kN.cosFT

.

.dxdytan

A

HA

oA

.xA

4261

7953

36615717

12.43m17.57m

A

C

CT

AT

kN..cos.

cosFT

.

.dxdytan

oC

HC

oC

.xC

6214044

4154044

96604312

cos HFT

Example 5.3 Determine the max tension in the cable

IHAssume 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

hLwF o

H 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 5-30Determine reactions at A and C and the

cable force3 global Eqs1 hinge condition

Ay

Ax

Cy

Ay

Ax

By

Bx

T

Example 5.4The 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

22010 xy

x

y

o10

2

626

5020

20

.

.x)(dx

dytan

D

mxD

00

9178

D

D

D

MV

kN.N

8 kN/m

Reflection: What Have You Learnt?