1! Cable Subjected to Concentrated Loads! Cable Subjected to Uniform Distributed

Loads! Arches! Three-Hinged Arch

Cables and Arches

2Cable Subjected to Concentrated Loads

L

BC

D

L1 L2 L3

P1 P2

yC yD

C

P2

TCDTCB x

y

AyAx

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

3Example 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

42 mh

2 m 2 m 1.5 m

A

BC

D

3 kN8 kN

SOLUTION

+ MA = 0:

AyAx

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

56.79(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

TCBBC

x

y

Joint B

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

3 kN

x

y

TBC = 4.82 kN

TBABA 32.3

o

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

BA = 53.8o TBA = 6.90 kN

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

A

B C

D

3 kN 8 kN

h

6y

x

x = L

To

Cable Subjected to Distributed Load

WTo

WT

T cos = To = FH = Constant

T sin = W

oTW

dxdy

== tan

Concepts & Conclusion:

T

7wo = force / horizontal distance

Tx

To

x

y

x

wo x

2x

To

wox

o

o

Txw

dxdy

== tan

= dxTxwy

o

o

1

2

2C

Txwyo

o +=

at x = L , T = TB = Tmax

22max )( LwTT oo +=

yxwT oo 2

2

=

0

To

woLTmax

x

A

B

T

x

Parabolic Cable: Subjected to Linear Uniform distributed Load

y

xL

8wo

y

x

h

L

xx

wo(x)2x

xs

Oy

T

+ T + T

Fx = 0:+

Fy = 0:+

+ MO = 0:

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

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

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

Derivation:

9Dividing 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( =dx

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

owdxTd

=)sin(

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

tan=dxdy

----------(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 o=sin ----------(5-5)

Dividing Eq. 5-5 by 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)

To

woxT

10

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 FHmay be obtained by using the boundary condition y =h at x = L. Thus,

hLwF oH 2

2

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

Finally, substituting into Eq. 5-7 yeilds

22 xL

hy = ----------(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)

To

woLTmax

wo

y

x

h

L

11

Example 5-2

The cable shown supports a girder which weighs 12kN/m. Determine the tensionin the cable at points A, B, and C.

12 m

30 m

6 m

A

B

C

12

SOLUTION

L30 - L

12 m

30 m

6 m

A

B

C

y

xwo = 12 kN/m

x1x2

TCC

TA

A

13

12x1

TCC

wo = 12 kN/m

L

x1

6 mB

C

y

x

12 L

To

To

Tx1

oTx

dxdy 1

1

1 12tan ==

= 11112 dxT

xyo

0

oTL

2'1262

=

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

1

21

1 212 C

Txyo

+=

14

12 m

A

B

y

x

wo = 12 kN/m

30 - L

x2

To

TA

A

12 (30 - L)

12 x2

To

Tx2

oTx

dxdy 2

2

2 12tan ==

2

22

22

2 21212 C

Txdx

Txy

oo

+== 0

oTL

2)'30(1212

2=

oTxy

212 22

2=

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

2)'30(1

2=

15

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

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

2)'30(1

2=

From (1) and (2), L = 12.43 m, To = 154.5 kN

12 L

To

TC

C

TB = To = 154.5 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( +== 214.8 kN

= 261.4 kN

16

Example 5-3

The suspension bridge in the figure below is constructed using the two stiffeningtrusses that are pin connected at their ends C and supported by a pin at A and arocker at B. Determine the maximum tension in the cable IH. The cable has aparabolic shape and the bridge is subjected to the single load of 50 kN.

I H

A B

D

F G C

8 m

6 m

50 kN

4 @ 3 m = 12 m 4 @ 3 m = 12 m

Pin rocker

E

17

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

SOLUTION

I

A

D

F G C

E

12 m

8 m

6 m

+ MA = 0:

0812 =+ oy TC

yo CT 5.1=

To

Cy

Cx

To

Iy

Ay

Ax

H

BC

8 m

6 m

50 kN

9 m3 m

To

To

Hy

ByCy

Cx

+ MB = 0:

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

08)9(5012 =+ oy TC

25.565.1 += yo CT

From (1) and (2), Cy = 18.75 kN, To = 28.125 kN

18

I

12 m

wo

8 m

x

y

wox

To = 28.12 kN

From (1) and (2), Cy = 18.75 kN, To = 28.12 kN

x

I

TI

28.12 kN

woxTx

12.28tan xw

dxdy o==

= dxxwy o12.28

0

1

2

12.28Cxwy o +=

)12.28(2)12(8

2ow=

wo = 3.125 kN/m

19

8 m

H

8 m

I

12 m 12 m

37.5 kN

12wo = 37.5 kN12wo = 37.5 kN

To = 28.12 kN To = 28.12 kN

THH

22 )12.28()5.37( +=IT

= 46.88 kN

Tmax = TI = TH = 46.88 kN

Tmin= To = 28.12 kN

28.12 kN

TI

I

TI

20

A B

D

F G C50 kN

4 @ 3 m = 12 m 4 @ 3 m = 12 m

E

ByAy

Ax

TTTTTTT 3125.33 == owT

= 9.375 kN0

+ MA = 0: 0)24()15(50)21181512963(375.9 =+++++++ yB

Fy = 0:+ 056.150)375.9(7 =+yA

Ay = -14.07 kN,

By = -1.56 kN,

21

Example 5-4

For the structure shown:(a) Determine the maximum tension of the cable(b) Draw quantitative shear & bending-moment diagrams of the beam.

8 m

8 m

0.5 m

AB

C

5 m 20 m

D

E

1 kN/m

Hinge

22

8 m

0.5 m

A B

5 m

D

1 kN/m

5 kN

To

By

BxAy

Ax

To

Dy

8 m

B C

20 m

E

1 kN/m

20 kN

To

To

Ey

CyBy

Bx

+ MA = 0:

0)5.0()5.2(5)5( =+ oy TB

+ MC = 0:

0)8()10(20)20( =+ oy TB

From (1) and (2), By = 0, To = 25 kN

SOLUTION

23

20wo

To= 25 kN

8 m

20 m

E

x

y

TE = Tmax

22max )20()25( +== ETT

Tmax = 32.02 kN

TE = Tmax

To= 25 kN

20wo = 20 kN

25tan xw

dxdy o==

1

2

)25(2

25

Cxw

dxxwy

o

o

+=

= 0

)25(2)20(8

2ow=

wo = 1 kN/m

Tx

To= 25 kN

wox

24

A BC

1 kN/m

CyAy

Ax

5 m 20 m

10 @ 2.5 m = 25 m

T = wo(2.5 m) = (1kN/m)(2.5 m) = 2.5 kN

2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5

=1.25 kN = 1.25 kN

1.25

-1.25

1.25

-1.25

1.25

-1.25

1.25

-1.25

1.25

-1.25

x (m)

V (kN)

x (m)

M (kNm) 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78

25

Example 5-5

The cable AB is subjected to a uniform loading of 200 N/m. If the weight of thecable is neglected and the slope angles at points A and B are 30o and 60o,respectively, determine the curve that defines the cable shape and the maximumtension developed in the cable.

30o

60o

A

B

15 m200 N/m

x

y

26

SOLUTION

(0.2 kN)(15 m) = 3 kN30o

60o

A

B

15 m

TA

60oTB

3 kN

TA30o

TB

60o

30o

30o 30o

120o

oA

ooB TT

30sin30sin3

120sin==

TB = 5.20 kN

TA = 3 kN

27

30o3 kN

30o

Ax

y

TA = 3 kN

0.2x

x

T

0.2x

3 sin 30o = 1.5

3 cos 30o = 2.6

T

6.25.12.0tan +== x

dxdy

0 += 577.00769.0 xy

577.00769.0 += xdxdy

1

2

577.02

0769.0 Cxxy ++=

y = 0.0385x2 + 0.577x

28

Example 5-6

The three-hinged open-spandrel arch bridge shown in the figure below has aparabolic shape and supports the uniform load . Show that the parabolic arch issubjected only to axial compression at an intermediate point D along its axis.Assume the load is uniformly transmitted to the arch ribs.

7 kN/m

15 m 7.5 m 7.5 m

7.5 m

A C

B2

2)15(5.7 xy =

y

xD

29

210 kN

15 m

B2

2)15(5.7 xy =

15 m

SOLUTION

Ax

Ay Cy

Cx

Entire arch :

+ MA = 0: 0)15(210)30( =yC

Fy = 0:+ 0105210 =+yA

Ay = 105 kN

Cy = 105 kN

30

105 kN

B

7.5 m 7.5 m 105 kN

CxB

Arch segment BC :

+ MB = 0:

Fy = 0:+ 0105105 =+yB

Bx

By

Fx = 0:+

Cx = 105 kN

Bx = 105 kN

By = 0

0)5.7()15(105)5.7(105 =+ xC

31

52.5 kN

B

D

105 kN

0

3.75 m

26.6o

26.6o

A section of the arch taken through point D, x = 7.5 m, y = -7.5(7.52)/(15)2 = -1.875 m,is shown in the figure. The slope of the segment at D is

Fy = 0:+

ND = 117.40 kN, VD = 0, MD = 0 kN

Fx = 0:+ 105 - ND cos 26.6o - VD sin 26.6o = 0

Arch segment BD :

VDND

MD

5.0)15(

15tan5.7

2 =

===x

xdxdy

+ MD = 0: MD + 52.5(3.75) - 105(1.875) = 0

-52.5 + ND sin 26.6o - VD cos 26.6o = 0

, = 26.6o

32

52.5 kN

B

D

105 kN

0

3.75 m

26.6o

26.6o

A section of the arch taken through point D, x = 7.5 m, y = -7.5(7.52)/(15)2 = -1.875 m,is shown in the figure. The slope of the segment at D is

Arch segment BD :

VDND

MD

5.0)15(

15tan 5.72 =

== =xxdxdy , = 26.6o

No= 25 kN

7.5 wo = (7.5)(7)= 52.5 kNND

22max )5.52()105( +== ETT

Tmax = 117.4 kNNotes : Since the arch is a parabola, there are noshear and bending moment, only ND is present

Alternate Method

33

Example 5-7

The three-hinged tied arch is subjected to the loading shown in the figure below.Determine the force in members CH and CB. The dashed member GF of the trussis intended to carry no force.

E

15 kN20 kN

15 kN

A

BC

D

FG

H

3 m 3 m 3 m 3 m

4 m

1 m

34

E

15 kN20 kN

15 kN

AB

CD

FG

H

3 m 3 m 3 m 3 m

4 m

1 m

SOLUTION

+ MA = 0:

Fy = 0:+

Fx = 0:+

025152015 =+yA

0)9(15)6(20)3(15)12( =yE

Ax

Ay Ey

Ax = 0

Ay = 25 kN

Ey = 25 kN

35

15 kN20 kN

AB

C

GH

3 m 3 m

5 m0

25 kN

+ MC = 0:

Fy = 0:+

Fx = 0:+

0201525 =+ yC

0)3(15)6(25)5( =+AEF

0Cx

CyFAE

-Cx + 21= 0Cx = 21.0 kN

Cy = 10 kN

FAE = 21.0 kN

36

Fy = 0:+

Fx = 0:+

020 =GCF

FGC = 20 kN (C)

FHG = 0

20 kN

0

FGC

FHG G

Joint G :

Fy = 0:+

Fx = 0:+

FCH = 4.75 kN (T),

Joint C :

20 kN

21 kN

10 kNFCB

C18.43o18.43o

FCH

-FCH cos18.43 - FCB cos18.43 - 21= 0

FCH sin18.43 - FCB cos18.43 - 20 + 10 = 0

FCB = -26.88 kN (C)

Thus,

37

Archescrownextrados

(or back)

Intrados(or soffit)

huanch

centerline rise

abutment

springline

fixed arch two-hinged arch

three-hinged archtied arch

38

P1

A

C

P2

B

C

Three-Hinged ArchP1

P2

AB

C

Bx

By

Cx

Cy

Ax

Ay

Cx

Cy

D

NDMD

VDAx

Ay

D

39

Example 5-8

The tied three-hinged arch is subjected to the loading shown. Determine thecomponents of reaction at A and C and the tension in the cable.

10 kN

15 kN B

A

C

0.5 m 1 m

2 m 2 m

D

2 m

40

SOLUTION

10 kN

15 kN B

A

C

0.5 m 1 m

2 m 2 m

D

2 m

Cy

Ax

Ay

Entire arch :

+ MA = 0: 0)5.0(15)5.4(10)5.5( =yC

Cy = 9.545 kN

Fy = 0:+ 0545.91015 =+yA

Ay = 15.46 kN

0

41

TD

10 kN

C

1 m

2 m

D

BBx

By

Cy = 9.545 kN

15 kN B

A

0.5 m

2 m

2 m

Bx

By

TA

Ay = 15.46 kN

Member AB :

+ MB = 0: 0)2()5.2(455.15)2(15 =+ AT TA = 4.319 kN

Fy = 0:+ 015455.15 = yB By = 0.455 kN

Fx = 0:+ 0319.4 = xB Bx = 4.319 kNMember AB :

Fx = 0:+ 0319.4 = DT TD = 4.319 kN