CED, UET, Taxila 1

CablesTheory of Structure - I

CED, UET, Taxila 2

Contents Introduction to Cables

Cables subjected to concentrated loads

Cables subjected to uniformly distributed loads

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Introduction Cables are pure tension members. Used as

Supports to suspension roofs Suspension bridges Trolley wheels

Self weight of cable is neglected in analysis of above structures

When used as guys for antennas or transmission lines, weight is considered.

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Cable Subjected to Concentrated Loads

L

BC

D

L1 L2 L3

P1 P2

yC yD

C

P2

TCD

TCB x

y

Ay

Ax

TCD

B

P1

x

y

TBC

TBA

Fx = 0:+

Fy = 0:+

+ MA= 0: Obtain TCD

L

A

BC

D

L1 L2 L3

P1 P2

yC yD

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Example 5-1

Determine the tension in each segment of the cable shown in the figure below. Also, what is the dimension h ?

2 m

2 m

h

2 m 2 m 1.5 m

A

BC

D

3 kN8 kN

CED, UET, Taxila 6

2 mh

2 m 2 m 1.5 m

A

BC

D

3 kN8 kN

SOLUTION

+ MA = 0:

Ay

Ax

5TCD

34

TCD(3/5)(2 m) + TCD(4/5)(5.5 m) - 3kN(2 m) - 8 kN(4 m) = 0

TCD = 6.79 kN

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6.79(3/5) - TCB cos BC = 0

BC = 32.3o TCB = 4.82 kN

Joint C

Fx = 0:+

Fy = 0:+ 6.79(4/5) - 8 + TCB sin CB = 0C

8 kN

5

34TCD = 6.79 kN

TCB

BC

x

y

Joint B

Fx = 0:+ - TBA cos BA + 4.82cos 32.3o = 0B

3 kN

x

y

TBC = 4.82 kN

TBA

BA32.3o

Fy = 0:+ TBA sin BA - 4.82sin 32.3o -3 = 0

BA = 53.8o TBA = 6.90 kN

h = 2tanBA = 2tan53.8o = 2.74 m

CED, UET, Taxila 8

y

x

x = L

To

Cable Subjected to Distributed Load

T

WTo

WT

T cos = To = FH = Constant

T sin = W

oTW

dxdy

tan

CED, UET, Taxila 9

wo = force / horizontal distance

Tx

To

x

y

x

wo x

2x

To

wox

o

o

Txw

dxdy

tan

dxT

xwyo

o

1

2

2C

Txwyo

o

at x = L , T = TB = Tmax

22max )( LwTT oo

yxwT o

o 2

2

0

To

woLTmax

x

A

B

T

x

Parabolic Cable: Subjected to Linear Uniform distributed Load

y

xL

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wo

y

x

h

L

xx

wo(x)2x

xs

Oy

T

T + T

Fx = 0:+

Fy = 0:+

+ MO = 0:

-Tcos + (T + T) cos ( + ) = 0

-Tsin + wo(x) + (T + T)sin ( + ) = 0

wo(x)(x/2) - T cos y - T sin(x) = 0

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Dividing each of these equations by x and taking the limit as x 0, and hence y 0, 0, and T 0, we obtain

0)cos(

dxTd

----------(5-1)

owdx

Td

)sin( ----------(5-2)

tandxdy

----------(5-3)

Integrating Eq. 5-1, where T = FH at x = 0, we have:

HFT cos ----------(5-4)

Integrating Eq. 5-2, where T sin = 0 at x = 0, gives

xwT osin ----------(5-5)

Dividing Eq. 5-5 Eq. 5-4 eliminates T. Then using Eq. 5-3, we can obtain the slope at any point,

H

o

Fxw

dxdy

tan ----------(5-6)

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Performing a second integration with y = 0 at x = 0 yields

2

2x

Fwy

H

o ----------(5-7)

This is the equation of a parabola. The constant FH may be obtained by using the boundary condition y = h at x = L. Thus,

hLwF o

H 2

2

----------(5-8)

Finally, substituting into Eq. 5-7 yeilds

22 x

Lhy ----------(5-9)

From Eq. 5-4, the maximum tension in the cable occurs when is maximum; i.e., at x = L. Hence, from Eqs. 5-4 and 5-5,

2max )(2 LwFT oH ----------(5-10)

Or, using Eq. 5-8, we can express Tmax in terms of wo, i.e.,

2max )2/(1 hLLwT o ----------(5-11)

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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.

12 m

30 m

6 m

A

B

C

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SOLUTION

L´30 - L´

12 m

30 m

6 m

A

B

C

y

xwo = 12 kN/m

x1x2

TC

C

TA

A

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----------(1)12x1

TC

C

wo = 12 kN/m

L´

x1

6 mB

C

y

x

12 L´

To

To

T

oTx

dxdy 1

1

1 12tan

11

112 dxT

xyo

0

'

01

1126L

o

dxT

x

oTx

2126

21

0

L´

+ C1

oTL

2'1262

2'LTo

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12 m

A

B

y

x

wo = 12 kN/m

30 - L´

x2

To

TA

A

12 (30 - L´)

12 x2

To

T

----------(2)

oTx

dxdy 2

2

2 12tan

22

212 dx

Txyo

0

)'30(

02

21212L

o

dxT

x

oTL

2)'30(1212

2

oTx

21212

22

0

(30 - L´)

+ C2

oTL

2)'30(1

2

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y2

B

y

x

wo = 12 kN/m

x2

x2

To

T

12 x2

12 x2

To

T

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----------(1)2'LTo

----------(2)oTL

2)'30(1

2

From (1) and (2), L´ = 12.43 m, To = 154.50 kN

12 L´

To

TC

C

TB = To = 154.50 kN

12 (30 - L´ )

To

TA

A

22 )'12( LTT oC

22 )43.1212()50.154(

22 )]'30(12[ LTT oA

22 )]43.1230(12[)50.154( = 214.75 kN

= 261.39 kN

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Practice Problems Chapter 5

Structural Analysis by R. C. Hibbeler

Examples and Exercise

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