Structure Analysis IChapter 5

Structure Analysis IChapter 5

Cables & Arches

Cables

• Cables are often used in engineering structure to

support and or transmit loads from one member to

another

Structural Analysis IDr. Mohammed Arafa

Cables subjected to concentrated

loads

Example 1Determine the tension in cables and what is the dimension h ?

kNT

TT

M

CD

CDCD

A

79.6

0)4(8)2(35.52

0

54

53

o

BC

BC

BCBCy

BCBCx

T

TF

TF

3.32

82.4

0sin8)(79.60

0cos)(79.60

54

53

o

BA

BA

BABA

o

y

BABA

o

x

T

TF

TF

8.53

90.6

0sin3)3.32(sin82.40

0cos)3.32(cos82.40

mh 74.28.53tan2

Cables subjected to a Uniform

Distributed Load

Cables subjected to a Uniform

Distributed Load

Analysis Procedure

0

0

0 cos cos 0

0 sin sin 0

0 cos sin 02

Dividing each eq. by Δx and taking the limit as Δx 0

and hence y 0, 0 and T 0

x

y

o

F T T T

F T w x T T

xM w x T y T x

Analysis Procedure

Structural Analysis IDr. Mohammed Arafa

0

cos0 (1)

sin(2)

tan (3)

d T

dx

d Tw

dx

dy

dx

Analysis Procedure

H

0

0

at x=0 T=F

Integrate eqs. 1 where T=F at x=0

cos (4)

Integrate eqs. 2 where Tsin =0 at x=0

sin (5)

dividing 4 by 5

tan =

H

H

H

T F

T w x

w xdy

dx F

Analysis Procedure

2

0 0

2

0

2

2

2

,2

H H

H

w x w xdyy

dx F F

w Lat x L y h F

h

hy x

L

Analysis Procedure

max

2 2

max

0

222

max 0 0

cos

cos

is at where is maximum

( )

In our Case at x=L

12

H

H

H

H

T F

FT

T

at x L

T F V

V w L

LT F w L w L

h

Analysis Procedure

2

2

2

0

2 2

22 2 2

max max 0

2

cos

H

HH

H H

hy x

L

w LF

h

FT F V

T F V F w L

Summary

Suspension Bridge

Cable Stayed Bridge

Example 2Determine the tension of the cable at points A, B, C

Assume the girder weight is 850 lb/ft

Structural Analysis IDr. Mohammed Arafa

2 2 2 2

max

500 307500 Ib=7.5k

2 2

7.031 7.5 10.280

A B

A B

WLV V

T T T H V

Example 3Determine the tension of the cable at points A, B

2

500 157031.25

2 8HF

2

0

2H

w LF

h

HF

AVBV

HF

Example 4

The beams AB and BC are supported by the cable that

has a parabolic shape. Determine the tension in the cable

at points D and E.

Example 5

Referring to the FBD of member BC, Fig. b,

Then from the FBD of member AB, Fig. a,

Draw the shear and moment diagrams for members

AB and BC.

Example 5 -continue

100

35.035.020.0

-5.0-5.0

SFD

BMD

Example 6: The cable AB is subjected to a uniform loading of

200kN/m. If the weight of the cable is neglected and the slope angles at

points A and B are 30 and 60, respectively, determine the curve that

defines the cable shape and the maximum tension developed in the

cable.

0 0 1

Integrate the equations

(1)

(2)

cos0 cos

sinsin

H

d TT F

dx

d Tw T w x C

dx

1

0

Substitute at 0 

sin 30 tan 3

30cos cos30

eq.(2) will be:

sin tan

0

(3)30

A

H H

H

A H

F FT T

T

x

C T

F

F

w x

0

2

tan 30tan

10.2 tan 30

60

1tan 60 0.2 15 tan 30

Divide

2.6

1 10.2 tan 30 0.2 2.

(3) by (1)

Substitute

6 0.5772.60

0

at 15 

(Equation of thecablecurv.0385 0.577 e)

H

H

H

H

H

H

H

H

H

w x F

F

dyx F

dx F

FF

F kN

dyx F x

dx F

x x

x

y

max

cos

2.6

cos cos

At 0 30

2.63.0

cos3

Substitute in (1)

0

At 15 60

2.65.2

cos 60

H

H

A

B

T F

FT

x

T kN

x

T kN T

When the weight of a cable becomes important in the force analysis, the

loading function along the cable will be a function of the arc length s rather

than the projected length x.

Cable Subjected to its own Weight

w = w(s)

To perform a direct integration, it is necessary to replace dy/dx by ds/dx.

one obtains the following relationships

2 2

2 2

1 1

sd dx dy

dy ds ds dy

dx dx dx dx

Separating the variables and integrating we obtain

The two constants of integration, say C1 and C2, are found using

the boundary conditions for the curve.

2

1/22

2

1

11

H

ds dy

dx dx

dsw s ds

dx F

1/2

2

2

11

H

dsx

w s dsF

continued on next slide

Example 7

Arches

Arches Types

Example of Fixed Arch

Three Hinge Arches

Example 5

Determine the internal forces at Section D

kNA

kNA

F

kNB

BM

kNB

BM

y

x

x

x

xRightC

y

yA

93

0F

86

0

86

0)20(67)8(60)6(0

67

0)28(60)10(100)40( 0

y

86

93

86 cos20

93 sin20

86 sin20

93 cos20

242

242

112.6

58

Example 6

Determine the internal forces at Section D

y

0 25

F 0

25

0 25 (50) (25) 25 (25) 0

25

0

25

A y

y

B xRight

x

x

x

M C kips

A kips

M C

C kips

F

A kips

yF 0

0

0

25

y

x

x

B

F

B kips

2

2

2

1

2

25

25

50

252

50

25tan 2 25 0.5

50

26.6

x

o

y x

dyx

dx

dy

dx

25cos 12.5sin 27.95

25 12.5cos 0.0

25(6.25) 12.5(12.5) 0

D

D

D

N

V sin

M

26.6o

Example 7

Structural Analysis IDr. Mohammed Arafa

Problem 3Determine the internal forces at B