1

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

Loads! Arches! Three-Hinged Arch

Cables and Arches

2

Cable Subjected to Concentrated Loads

L

BC

D

L1 L2 L3

P1 P2

θ

yC yD

C

P2

TCDTCB x

y

Ay

Ax

TCD

B

P1

x

y

TBC

TBA

ΣFx = 0:+

ΣFy = 0:+

+ Σ MA = 0: Obtain TCD

L

A

BC

D

L1 L2 L3

P1 P2

θ

yC yD

3

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

4

2 mh

2 m 2 m 1.5 m

A

BC

D

3 kN8 kN

SOLUTION

+ ΣMA = 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

5

6.79(3/5) - TCB cos θBC = 0

θBC = 32.3o TCB = 4.82 kN

Joint C

ΣFx = 0:+

ΣFy = 0:+ 6.79(4/5) - 8 + TCB sin θCB = 0C

8 kN

5

34TCD = 6.79 kN

TCB

θBC

x

y

Joint B

ΣFx = 0:+ - TBA cos θBA + 4.82 cos 32.3o = 0B

3 kN

x

y

TBC = 4.82 kN

TBA

θBA32.3o

ΣFy = 0:+ TBA sin θBA - 4.82 sin 32.3o - 3 = 0

θBA = 53.8o TBA = 6.90 kN

h = 2 tanθBA = 2 tan53.8o = 2.74 m

A

B C

D

3 kN 8 kN

h

6

y

x

x = L

To

Cable Subjected to Distributed Load

θ

WTo

WT

θ

T cos θ = To = FH = Constant

T sin θ = W

oTW

dxdy

== θtan

Concepts & Conclusion:

T

7

wo = force / horizontal distance

Tθx

To

x

y

x

wo x

2x

To

wox

o

o

Txw

dxdy

== θtan

∫= 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

θΒ

x

A

B

T

θx

Parabolic Cable: Subjected to Linear Uniform distributed Load

y

xL

8

wo

y

x

h

L

x∆x

wo(∆x)2x∆

∆x∆s

O∆y

θT

θ+ ∆θT + ∆T

ΣFx = 0:+

ΣFy = 0:+

+ ΣMO = 0:

-T cosθ + (T + ∆T) cos (θ + ∆θ) = 0

-Tsinθ + wo(∆x) + (T + ∆T) sin(θ + ∆θ) = 0

wo(∆x)(∆x/2) - T cos θ(∆y) - T sinθ(∆x) = 0

Derivation:

9

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

0)cos(=

dxTd θ

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

owdx

Td=

)sin( θ----------(5-2)

θtan=dxdy

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

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

HFT =θcos ----------(5-4)

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

xwT o=θsin ----------(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

==θtan ----------(5-6)

To

woxT

θ

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 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 θ is maximum; i.e., at x = L. Hence, from Eqs. 5-4 and 5-5,

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

To

woLTmax

θΒ

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

L´30 - L´

12 m

30 m

6 m

A

B

C

y

xwo = 12 kN/m

x1x2

TC

θC

TA

θA

13

12x1

TC

θC

wo = 12 kN/m

L´

x1

6 mB

C

y

x

12 L´

To

To

Tx1

θ

oTx

dxdy 1

1

1 12tan == θ

∫= 11

112 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

θA

12 (30 - L´)

12 x2

To

Tx2

θ

oTx

dxdy 2

2

2 12tan == θ

2

22

22

2 21212 C

Txdx

Txy

oo

+== ∫0

oTL

2)'30(1212

2−=

oTxy

212 2

22=

----------(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

θC

TB = To = 154.5 kN

12 (30 - L´ )

To

TA

θA

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

+ ΣMA = 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

+ ΣMB = 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

θI

TI

28.12 kN

woxTx

θ

12.28tan xw

dxdy o== θ

∫= dxxwy o

12.280

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

TH

θH

22 )12.28()5.37( +=IT

= 46.88 kN

Tmax = TI = TH = 46.88 kN

Tmin= To = 28.12 kN

28.12 kN

TI

θI

TI

θΙ

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

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

ΣFy = 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

Bx

Ay

Ax

To

Dy

8 m

B C

20 m

E

1 kN/m

20 kN

To

To

Ey

CyBy

Bx

+ ΣMA = 0:

0)5.0()5.2(5)5( =+− oy TB

+ ΣMC = 0:

0)8()10(20)20( =−+ oy TB

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

SOLUTION

23

θ

20wo

To= 25 kN

8 m

20 m

E

x

y

TE = Tmax

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

Tmax = 32.02 kN

θ

TE = Tmax

To= 25 kN

20wo = 20 kN

25tan xw

dxdy o== θ

1

2

)25(2

25

Cxw

dxxwy

o

o

+=

= ∫0

)25(2)20(8

2ow

=

wo = 1 kN/m

θ

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 (kN�m) 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

60o

TB

3 kN

TA30o

TB

60o

30o

30o 30o

120o

oA

ooB TT

30sin30sin3

120sin==

TB = 5.20 kN

TA = 3 kN

27

30o

3 kN

30o

Ax

y

TA = 3 kN

0.2x

x

Tθ

0.2x

3 sin 30o = 1.5

3 cos 30o = 2.6

T

θ

6.25.12.0tan +

==x

dxdy θ

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 :

+ ΣMA = 0: 0)15(210)30( =−yC

ΣFy = 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 :

+ ΣMB = 0:

ΣFy = 0:+ 0105105 =+−yB

Bx

By

ΣFx = 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

ΣFy = 0:+

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

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

Arch segment BD :

VD

ND

MD

5.0)15(

15tan5.7

2 −=−

===x

xdxdyθ

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

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

, θ = 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 :

VD

ND

MD

5.0)15(

15tan 5.72 −=−

== =xxdxdyθ , θ = 26.6o

θNo= 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

+ ΣMA = 0:

ΣFy = 0:+

ΣFx = 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

+ ΣMC = 0:

ΣFy = 0:+

ΣFx = 0:+

0201525 =+−− yC

0)3(15)6(25)5( =+−AEF

0Cx

Cy

FAE

-Cx + 21= 0Cx = 21.0 kN

Cy = 10 kN

FAE = 21.0 kN

36

ΣFy = 0:+

ΣFx = 0:+

020 =−GCF

FGC = 20 kN (C)

FHG = 0

20 kN

0

FGC

FHG G

Joint G :

ΣFy = 0:+

ΣFx = 0:+

FCH = 4.75 kN (T),

Joint C :

20 kN

21 kN

10 kNFCB

C18.43o

18.43o

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 :

+ ΣMA = 0: 0)5.0(15)5.4(10)5.5( =−−yC

Cy = 9.545 kN

ΣFy = 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 :

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

ΣFy = 0:+ 015455.15 =−− yB By = 0.455 kN

ΣFx = 0:+ 0319.4 =− xB Bx = 4.319 kN

Member AB :

ΣFx = 0:+ 0319.4 =− DT TD = 4.319 kN