Cables and Suspension Bridges

CABLES AND SUSPENSION BRIDGES Cables are used to support loads over long spans such as suspension bridges, roof of  large open buildings etc. The only only force in the cable is direct tension. Since the cables are flexible they carry zero B.M.

Analysis of cables Analysis of cable involves determination of reactions at the support and tension over  different parts of the cable. To determine the reactions at the support and tension equilibrium conditions are used. In addition to that BM about any point of the cable can be equated to z ero. P1: Determine the reactions components and tension in different parts of the cable shown in figure. Also find the sag at D and E.

HA = 220 kN

A

20 m

20 m

20 m

φ1 10 m

T1 VA = 55 kN

C

yD = 14.55

φ2

T3 T2

30 kN

20 m

B

HB = 110 kN

yE = 11.82 T4 E

VB = 65 kN

D 40 kN

50 kN

∑F =0 x

−H +H = 0 A

=H

HA

B

------ (1) B

∑F =0 y

+ VB =120kN

VA

------ (2)

∑M =0 A

30 x 20 + 40 x 40 + 50 x 60 − VB x80 = 0 VB = 65 kN VA

Mc =0

= 55 kN

(About the point where sag is given)

− H A x 10 + 55 x 20 = 0 HA

=110kN = H B

1

 Point A:

110 kN

φ = 26.56 55 kN

∑F =0 x

T1 Cos 26.56 − 110 = 0 T1 = 123 kN

tan

φ = 1

φ

1

10

20 = 26.56

To find Y E :We have B.M E

=0 − 110 x YE

+ 65 x 20 = 0

YE = 11.82m To Find Y D :-

B.M D

=0 − 50 x 20 + 65 x 40 − H

B

x YD

=0

YD = 14.55m

2

T1

 Point A:

110 kN

φ = 26.56 55 kN

∑F =0 x

T1 Cos 26.56 − 110 = 0 T1 = 123 kN

tan

φ = 1

φ

1

10

20 = 26.56

To find Y E :We have B.M E

=0 − 110 x YE

+ 65 x 20 = 0

YE = 11.82m To Find Y D :-

B.M D

=0 − 50 x 20 + 65 x 40 − H

B

x YD

=0

YD = 14.55m

2

T1

 Point D:

T2

T3

12.81 = φ

φ3 = 7.77

110 kN 40 kN

∑F =0 x

T3 Cos 7.77 − T2 Cos12.81 = 0 T3 = T2 0.984

tan

φ =

14.55 − 10

2

φ = 12.81

20

1

tan φ 3

=

14.55 − 11.82

φ = 7.77

20

3

∑F =0 Y

T1 Sin 7.77 + T2 Sin 12.81

− 40 = 0 T2 0.984 Sin 7.77 + T2 Sin 12.81 − 40 = 0 T2 = 112.75 kN T3 = 110.95kN

3

 Part B:

30.18

110 kN 65 kN

T4

∑F =0 x

110 − T4 Cos 30.58 = 0 T4 =127.77 kN tan

φ =

11.82

4

20 φ4 = 30.58

Note: It can be observed that more the inclination or the slope more is the tension. Near the supports slope will be more and hence tension will also be more.

2) A Chord Supported at its ends 40 m apart carries loads of 200 kN, 100 kN and 120 kN at distances 10 m, 20 m and 30 m from the left end. If the point on the chord where 100 kN load is supported is 13 m below the level of the end supports. Determine (a) Reactions at the support. (b) Tension in different part. (c) Length of the chord. HA = 200

A

10 m

10 m

10 m

10 m

E

HB = 200

φ1 yB

T1 VA = 230 B

13 m φ3 φ3 T3

φ2 T2

200 kN

C 100 kN

∑F =0 x

4

yD T4 V = 190 E D

120 kN

−H +H = 0 A

E

=H

HA

------ (1) E

∑F =0 y

VA

+ V = 420kN

------ (2)

E

∑M =0 A

200 x 10 + 100 x 20 + 120 x 30 − VE x 40 = 0 VE = 190 kN VA

Mc

= 230 kN

= 0 (Always take BM about the point for which sag is given) −H

A

x 13 − 200 x 10 + 230 x 20 = 0

HA

=200kN = H

E

5

 Point A:

200 kN

49° 230 kN

T1

B.M B = 0 230 x 10 − 200 x YB = 0

= 11.5m 11.5  φ = tan −       10   YB

1

1

φ = 49 1

∑F =0 x

T1 Cos 49 − 200 = 0 T1 = 304.85kN

 Point B:

204.85 kN 49° φ2 200 kN T2

∑F =0 x

6

T2 Cos 8.53 − 304.8 Cos 49 = 0 T2

tan φ 2 =

13 11.5 10

= 202.22kN φ 2 = 8.53

 Point C:

202.22 kN T3 8.53

19.3

100 kN

∑ B.M = 0 D

− 200 x YD

YD + 190 x 10 = 0

= 9.5m

∑F =0 x

T3 Cos19.3 − 202.22 Cos 8.53 = 0

tan φ 3 =

T3 = 211.90kN

13 − 9.5 10

φ =19.3 3

7

 Point E:

φ4 T4

200 kN 190 kN

∑F =0 x

200 − T4 Cos 43.53 = 0

tan φ 4 =

T4 = 275.85kN

9.5 10

φ 4 = 43.53 Length of the Chord = Cos

10

φ= 1

AB =

AB + BC + CD + DE

AB 10

Cos 49

AB = 15.22m Cos

φ = 2

BC =

10 BC 10

Cos 8.53

= 10.12m

BC

Cos φ 3

=

10 CD 10

CD

=

CD

= 10.6m

Cos

Cos19.3

φ = 4

10 DE

8

10

DE

=

DE

= 13.8m

Cos 43.53

∴ Total length of Chord = 49.76m 3) Determine reactions at supports and tension indifferent parts of the cable shown in figure. HA

A

φ1

25 m

25 m

25 m

25 m

7.5 m determined  8.75 φ2 C

T1 VA

T2 30 kN

10 m (given) 26.25 φ3 D 8.95 T4 T3 50 kN E

15 m

2.5

VB

40 kN

∑F =0 x

− H +H = 0 A

HA

=H

B

------ (1) B

∑F =0 y

VA + VB =120kN

------ (2)

∑M =0 A

30 x 25 + 50 x 50 + 40 x 75 − VB x 100 − H B x15= 0

15H B + 100 VB

= 6250

H B + 6.67VB = 416.67

------ (3)

B.M D = 0 − H x 2.5 + V x50 − 40 x 25 = 0 B

B

− 50V = −1000 H − 20V = − 400 2.5H B B

B

------ (4)

B

9

(3) – (4) gives

26.67VB = 416.67 + 400 VB = 30.62kN H B = 212.42kN = H A VA = 89.38kN

 Point A:

∑F =0 x

T1 Cos 19.29

− 212.42 = 0

T1 = 225.05kN

212.42 kN

φ1 = 19.29° 89.38 kN

T1

8.75  φ = tan −       25   φ = 19.29 1

1

1

10

 Point C:

2.5  φ = tan −       25  φ = 5.71 8.75  φ = tan −       25   φ = 19.29 11.25  φ = tan −     25     φ = 24.22 1

2

225.05 kN

2

1

3

19.29 φ2 = 5.71

3

30 kN

T2

1

4

4

4) A cable is used to support loads 40 kN and 60 kN across a span of 45 m as shown in figure. The length of the cable is 46.5 m. Determine tension in various segments. HA = 159.2 kN A

15 m

15 m yB = 4.4 m

T1 VA = 46.67 = 0

15 m

HD

yC = 5.025 T3 V = 53.33 D

B 40 kN T2

D

D 60 kN

∑F =0 x

−H +H = 0 A

D

=H

HA

------ (1) D

∑F =0 y

VA + VD = 100kN

------ (2)

∑M =0 A

40 x 15 + 60 x 30 − VD x 45 = 0

VD

= 53.33kN

VA

= 46.67 kN

------ (3)

B.M B = 0 46.67 x 15 − H A x YB = 0 H A YB

= 700.05

11

MC = 0 46.67 x 30 − 40 x 15 − H A x YC H A YC H A YB

= 800.10

H A YC

800.10

=

=0

700.05

YB = 0.875 YC We have

AB + BC + CD + DE = 46.5m 15 2 + YB

+

2

15 2 + (YC − YB ) 2

+

15 2 + YC

2

= 46.5m

[15 + (0.875 Y ) ] + [15 + (Y −0.875 Y ) ] + [15 + Y ] = 46.5 2

2

1

/2

2

2

C

2

2

1

C

1

/2

C

/2

C

[15

2

+ 0.766Y

2

C

+ [15 + Y

2

2

C

]

1

/2

] + [15 1

/2

2

+ 0.0156 Y

C

2

]

1

/2

= 46.5

 1 + 0.766Y 2   /  1 + 0.0156Y 2   /   Y 2   /  C  C  C   = 46.5   + 15  + 1 + 2 2 2         15 15 15              2 2  1 0.766 YC 2 1 0.0156 YC 1 YC  15 1 + x +1 + x + 1 + + 2  = 46.5 2 2 2 2 2 15  15 15  1

1

2

1

2

2

[3 + 0.00396 Y ]= 4615.5 2

C

0.00396 YC

2

= 0.1

YC = 5.025m

∴Y = 4.4m B

We have

H A YB = 700.05 4.4  φ = tan −       15  

700.05

HA

=

HA

= 159.2kN = H

1

1

4.4 D

φ1 =16 0.35 12

 Point A:

∑F =0 x

T1 Cos 16.350

− 159.2 kN = 0 T1 =16.5.90 kN

159.2 kN

φ1 46.67 kN

 Point C:

∑F =0 x

T3 Cos 18.52 − T2 Cos 2.385 = 0 T3 = 1.053 T2 5.025 − 4.4  φ = tan −       15   φ = 2.385 5.025  φ = tan −       15   φ = 18.52 1

2

2

1

3

3

13

T1

∑F =0 Y

T3 Sin 18.52 + T2 Sin 2.385 − 60 = 0

1.053 T2 Sin 18.52 + T2 Sin 2.385 − 60 = 0 0.376 T2 = 60 T2 =159.44 kN T3 = 167.89 kN

T2

T3

φ2

φ3

60 kN

General Equation of a cable or Differential Equation. General shape of a cable depends on nature of loading, location of loads, type of  supports etc. The equilibrium of a part of a cable shall be considered to obtain equation for  cable when the cable is subjected to all over UDL. Let us consider the equilibrium of a small length ‘ds’ of the cable shown in figure. Let the cable be subjected to UDL of intensity W over horizontal span figure shows tension and horizontal, vertical reactions in the part of the cable we have.

14

D C W kN/m VD = v + dv ds H = HC = TCos φ C

φ φ

T + d  φ + 2φ HD = H dy

dx

T V = VC = TSin φ H = T Cos φ V = T Sin φ

∴T = V H

=

H

= H Sec φ

Cos φ T Sin φ

T Cos φ

= tan φ

V = H tan φ Let us consider the equilibrium of the part of the cable shown in fig. We have.

∑F =0 Y

(V + dv) – V – W x dx = 0 dv = wdx

d [Htanφ] dx

=W

 dy   dx       =W

Hd 

dx d 2 Y W dx 2

=

H

To derive equations for cable profile and tension in the cable when it is supported at the same level and subjected to horizontal UDL. Let us consider a cable of span L and max sag H subjected to UDL of intensity ‘W’ as shown in fig. From general equation we have

d y

2

dx 2

15

=

W H

y

A

B

L

h x W Integrating with respect to x we have

d y

We have at X = 0 Substituting

dx

=

W H

. x + c1

=0

dx W

0=

d y

. 0 + C1

H

C1 = 0

∴

d y

W

=

dx

.x

H

Integrating with respect to x we get We have at X=0

Y=

W X2 . H 2

Y=0

∴0 = 0 + C

2

C 2 =0

=

W X2 . H 2

At X=  / 2

Y=h

Y

Substituting

H=

h

W

=

H

W

( / 2)

2

.

2

2

8h Y=

We have

Y=

W X2 . w 2 2 8h 4h 2 x 2 L

If A is considered as the origin then

Y=

4hx 2

L

(L − x )

16

+C

2

To fin the tension at any point on the cable: Tension at any point T = H Sec φ

=H

Sec 2 φ

=H

1 + tan 2 φ

 dy  1+   dx 

T=H

2

To derive an expression for cable profile when it is subjected to horizontal UDL and  supports are at different levels: General equation for cable profile is

d y dx

2

= 2

W H y

L2

B

a L1  b O

w

x

L1 Let us consider each part separatel y we have

Y=

Wx 2 2H

At x = -L 1, Y = b

W (− L1 )

2

∴ b =

OR  b =

2H

WL1

2

------ (1)

2H

At X = L 2, Y= a + b

∴a + b =  b a + b

∴

L1 L2

=

=

WL 2

2

------ (2)

2H

L1

2

L2

2

 b

(a + b )

Add (1) on either side

17

L1 L2

L1 + L 2

L2 L2

a + b

=

L2

L

 b

+1 =

+1

 b +

(a + b ) (a + b )

( L 1 + L 2 = L)

 b + a + b

=

a + b L a + b

=

 b + a + b

Substituting in (2)

(a + b) = H=

L2 (a + B)

W

[  b +

2H

(a + b)

]

2

WL2

(

2  b + a + b

)

2

Tension at any point is given by T = H Sec φ T=H

Sec 2 φ

T=H

1 + tan 2 φ

=H

 dy  φ 1+   dx 

2

T=H

Y=

 wx  1+    H  

2

wx 2 2H dy dx

=

2 wx 2H

To derive an expression for length of the parabolic cable profile when the supports are at the same level

18

ds

d y

d x

dx

We have ds =

Ds = dx

2

+ dy

2

 dx   1 +     dy  

2

When the supports are at the same level we have

Y= dy dx

=

4hx 2 L2 8hx

(taking origin at the center)

L2

ds = dx

1 + 64h 2 x 2 L4

Total length of the cable is given by

  64h 2 x 2    ∫ ds = ∫ 1 + 4 L   − / 2   /2

LC =

½

dx



½

  64h 2 x 2    dx = 2 ∫ 1 + 4 L   0   /2   1 64h 2 x 2    dx + .......... .. = 2 ∫ 1 + 4 2 L   0   /2  32h 2 x 3  LC = 2 x + 4  3 0 L   L 32h 2 L3  LC = 2  + 4   2 L 24  8h 2 LC = L + 3L /2





(expanding by binomial theorem)

To derive an expression for length of the cable profile when the supports are at different levels

19

L2

A

B

a

L1

 b c L C = Length of cable Between A & C + Length of cable between C&B LC =

−1

Length of cable of span 2L 1 +

2

But L C = L +

−1

2 8  b 2

1

length of cable of span 2L 2

  8 (a + b )  i.e L = 2L + 1 / 2 2 L . + +    3L 2 3 2L  3 2L   2 b 2 (a + b) =L + +L +

8h 2

C

2

1

2

1

2

2

1

2

2

3 L1

3 2   b 2

(L + L ) +  + 3  L

LC =

1

L2 (a + b) 2  

2

L2

1

  

1) A cable suspends across a gap of 250m and carries UDL of 5kN/m horizontally calculate 1 th the maximum tensions if the maximum sag is / 25 of the span. Also calculate the sag at 50m from left end. A

B

250 m h=

1 25

x 250

= 10 m 5 kN/m We have T=H

 dy  1+   dx 

2

Take origin as left end 

Y= dy dx

=

4hx 2

[L − x ]

2

[L − 2x ]

L 4h L

Tension is maximum at the supports i.e., at X = 0 from A

20

dy dx dy dx

4 ×10

=

250

2

[250 − 2 × 0]

= 0.16 2

WL

H=

8h

=

5 × 250 8 ×10

∴ T = 3906.25 max

We have

Y=

4hx 2

= 3906.25kN

1 + 0.16

2

= 3955.93kN

(L − x ) [x measured from A]

L 4 ×10 × 50

Y=

2

250 Y = 6.4m

2

(250 − 50)

2) Determine the length of the cable and max tension developed if the cable supports a load  of 2kN/m on a horizontal span of 300m. The maximum sag is 25m A

300 m

B

25 2 kN/m

WL2

H=

8h 2 × 300 2

8 × 25 H = 900kN

T=H

 dy  1+    dx 

2

4hx

[L − x ] [X from A] L2 dy 4h Y= = 2 [L − 2x ] dx L

Y=

Tension is maximum at the supports i.e., at x = 0

dy dx

=

4 × 25 300 2

(300 − 2× 0)

21

dy

= 0.333

dx ∴Tmax = 900 1 + 0.3332 Tmax = 948.68kN 8h 2 Length of cable L C = L + 3L 2 8 × 25 300 + 3 × 300

LC

= 305.55m

3) Determine the maximum span for a mild steel cable between supports at the same level if  1 2 the central dip. is / 10th of the span and permissible stress in steel is 150 N/mm . Steel 3 weighs 78.6 kN/m . Assume the cable to hang in a parabola.

=

Here the weight of the cable itself is acting as UDL on the cable. We have SP weight Wieght Volume A L Weight = Specific Weight x Volume = Specific Weight x Area x Length Weight = Specific Weight x Area Length

W

= 78.6

78.6 A

We have

kN m3

× Am

2

kN m

H=

WL2

8h 78.6A × L2 1 8× × L 10 H = 98.25 LA T =H Y=

 dy  1+    dx 

4hx L2

2

(L − X)

22

dy dx

=

4h 2

L

(L − 2x )

Tension is maximum at X = 0

dy dx

=

4h L

 4× 1 L     10   ∴Tmax = 98.25AL 1 +  L        Tmax = 105.81AL Tmax = 105.81L A Fmax =105.81L Fmax =150

2

 N mm

2

= 150MPa =150×103

kN m

2

150 ×103 = 105.81 L L = 1417.63m

Bridges supported by cables

Anchor  cable Suspension Bridge

23

Fan Type Cable stayed Bridge

Harp type Cable stayed Bridge

Anchoring of cables There are 2 methods by which suspension cable can be anchored: 1) Continuous cable or pulley type anchoring. 2)  Non- Continuous cable or saddle type anchoring.

Continuous cable or pulley type anchoring In this method suspension cable itself passes over roller or guide pulley on the top of  the tower or abutment and then anchored. The tension remains same in the suspension cable and anchor cable at the supports. TA

Pulley Anchor cable

A TA

Suspension cable Abutment TA = H A

+ VA 2  V   α = tan −1  A   H A   2

α TA

B TA

ht

Tower 

24

β is the inclination of the suspension cable with the horizontal. Net horizontal force on tower H T = T A Cos α ~ T A Cos β Where h t is the height of the tower.

2) Saddle type anchoring or Non-continuous cable In this method of anchoring suspension cable are attached to saddles mounted on rollers on the top of the tower as result in suspension cable and anchor cable will be differed. However horizontal components of tension will be equal. Anchor cable

Suspension cable Abutment B Ta

α TA

25

TA = H A

2

+ VA 2

 V   α = tan −1  A   H A   TA Cos α = Ta Cos β∴H T = φ VT = TA Sin α + Ta Sinβ M = HT ×h t M=0 1) A cable of span 150m and dip 15m carries a load of 6 kN/m on horizontal span. Find the maximum tension for the cable at the supports. Find the forces transmitted to the supported   pier if  (a) Cable is passed over smooth rollers or pulleys over the pier. (b) Cable is clamped to saddle with smooth rollers resting on the top of the pier. For each of the above case anchor cable is 30 to horizontal. If the supporting pier is 20m tall. Determine the maximum BM on the pier. A

150 m

B

HA

HB VA

15 m

VB

6 kN/m

HA

=H = B

=

W 2 8h

6 ×150 2 8 ×15

=1125kN VA

= VB =

6 ×150 2

Tmax = TA = H A

2

= 450kN

= VA 2 =1211.66kN

 V   450   α = tan −1  A  = tan −1     H 1125       A   0 = 21 .8

26

Case 1: Cable over smooth pulley 0

α = 21.8 TA = 1211.66 kN

30 = β TA

Tower  H T = 1211.66 Cos 21.8 0 ∼ 1211.66 Cos 30 = 75.73kN 0 V T = 1211.66 Sin 21.8 + 1211.66 Sin 30 V T = 1055.77kN M = HT X h t = 75.73 X 20 = 1513.4kNm Case 2 : Cable clamped to saddle 0

21.8 TA = 1211.66 kN

30 Ta

Here T A Cos

α = Ta Cos β

1211.66 Cos 21.8 = T a Cos 30 T a 1299.05kN In this case H T = 0 ∴M = φ VT = TA Sinα = Ta Sin β

=1211.66 Sin 21.8 +1299.05Sin 30

0

V T = 1099.5 kN 3) A Suspension cable is suspended from 2 pier A and B 200m apart, B being 5m below A the cable carries UDL of 20kN/m and its lowest point is 10m below B. The ends of  the cable are attached to saddles on rollers at the top of the piers and backstays anchor  0 cables. Backstays may be assumed to be straight and inclined at 60 to vertical. Determine maximum tension in the cable, tension in backsta y and thrust on each pier. AA = 8081.6 kN

200 m y 10 m

VA = 2202.05

5m

C 20 kN/m L1 = 110.12 m

L2 = 89.89 m

Let C be the origin 27

x

We have

Y=

Wx 2

2H At x = − L1

Y = 15m

At x = − L 2 w ( − L1 ) 2 15 = 2H w (−L 2 ) 2 10 = 2H 15 L21 10

=

Y = 10m

L2 2 0

15.24 TA = 8376.26 kN

β 30 0 Ta 60

∴

L1 L2

=

15 10

L1 = 1.225 L 2 But

L1 + L2 = 200 1.225 L 2 + L 2 = 200

L 2 = 89.89m L1 = 110.12m

∑F =0 x

− H + H =0 A

HA

=H

B

---- (1) B

∑F =0 y

VA + VB = 4000

---- (2)

∑M =0 A

20 × 200 × 100 − VB × 200 − H B × 5 = 0

5H B

+ 200 V = 400000

---- (3)

B

∴ H + 40 V =80000 B

B

28

MC = 0

− (20× 89.89)

89.89 2

+ VB ×89.89 − H B ×10 = 0

− 10H B + 89.89VB = 80802.12 − H B + 8.989VB = 8080.212

----- (4)

48.989VB = 88080.212

VB =1797.95 kN H B = 8081.6kN H A = 8081.6kN VA

= 2202.05kN

Since V A > V B tension at A is maximum

Tmax = TA = H 2 A + V 2 A

=

2202.05 2 + 8081.6 2

TA = 8376.23kN

 V   α = tan −1  A   H A   2202.05  = tan −1       8081.6  

α =15.24 TA Cos α = Ta Cos β 8376.26Cos15.24 = Ta Cos 30 0 Ta = 9331.94kN

29