6.6 cables: uniform loads - civil engineering cables: uniform loads example 1, page 1 of 4 a b 400...

6.6 Cables: Uniform Loads

6.6 Cables: Uniform Loads Example 1, page 1 of 4

A B

400 ft

3800 ft

1. Each cable of the center span of the suspension

bridge supports a uniform load of 10 kip/ft along

the horizontal. If the span is 3,800 ft and the sag

400 ft, determine the maximum and minimum

tensions in the cable.


The maximum tension occurs where the slope is

greatest points A and B.

The minimum tension occurs where the slope is least the

low point of the cable.

2

1

3800 ft

400 ft

BA


C

B

T By

T Bx

400 ft

T o

(1900 ft)/2 950 ft

(3800 ft)/2 = 1900 ft

(Because of symmetry, we know

that the low point occurs at the

middle of the 3800-ft span.)

C(Resultant acts through middle of

1900 ft span)

Resultant load

(10 kip/ft)(1900 ft)

= 19,000 kip

Tension at low point

(minimum tension)

Tension components at B5

7

8

64

3 Free-body diagram of CB

3800 ft

400 ft

BA


T max

T By

19,000 kip

B

Equilibrium equation

Fx = 0: T + TBx = 0

Fy = 0: TBy

19,000 kip = 0

MB = 0: (19,000 kip)(950 ft) T (400 ft) = 0

Solving gives

T = 45,125 kip (minimum tension in cable) Ans.

TBx = 45,125 kip

TBy

= 19,000 kip

9

The maximum tension is the resultant of TBx and TBy

:

TBx = 45,125 kip

+

++

T max = (45,125 kip)2 + (19,000 kip)2

= 49,000 kip Ans.


A

B

C

8 ft

h

50 ft30 ft

2. A length of oil pipeline weighing 3200 lb is supported by a system

of cables as shown. Determine a) the distance h to the lowest point C

on the cable and b) the maximum tension in the cable.


B x

B y

A y

A x

A

B

C

3200 lb

(80 ft)/2 40 ft

(Weight of pipeline acts through

midpoint of 30 ft + 50 ft 80-ft span,

not through low point, C)

8 ft

2

40 ft

The maximum tension occurs at B, where the

slope is largest, so let's draw a free-body

diagram of the entire system, including the

tension components at B:

1 Equilibrium equation

MA = 0: By(40 ft + 40 ft)

Bx(8 ft) (3,200 lb)(40 ft) = 0 (1)

We could write two additional equilibrium

equation, but they would introduce two

additional unknowns, Ax and Ay, so nothing

would be gained. An additional free-body

diagram involving Bx and By is needed.

3

4

+


50 ft

8

7 Free-body diagram of BCPass a vertical section through the low point, C, and

draw a free-body diagram of the portion of the

system to the right of C.

5

The 3200-lb weight spread over an 80-ft length

of pipe is equivalent to a uniformly distributed

load of 3200 lb/80 ft = 40 lb/ft.

6+

++

9 Equilibrium equation for free-body BC

Fx = 0: T + Bx = 0 (2)

Fy = 0: By 2000 lb = 0 (3)

MB = 0: (2000 lb)(25 ft) T (h + 8 ft) = 0 (4)

A

B

C

80 ft

B x

B y

A y

A x

50 ft

B xB

C

B y

h + 8 ft

T o

(50 ft)/2

25 ft

Weight distributed load length of pipe

40 lb/ft 50 ft

2000 lb

(T o is horizontal

because C is

the minimum

point on the

curve.)

h

8 ft


BB

x 4000 lb

B y 2000 lb T

max

Eq. 4 is nonlinear, but Eqs. 1-3 are linear and can be easily

solved to give

Bx = 4000 lb

By = 2000 lb

T = 4000 lb

Using the value of T = 4000 lb in Eq. 4 gives

(2000 lb)(25 ft) T (h + 8 ft) = 0 (Eq. 4 repeated)

Solving gives

h = 4.5 ft Ans.

The maximum tension is the resultant of Bx and By:11

10

Tmax = (4000 lb)2 + (2000 lb)2

= 4470 lb Ans.


A B

The towel near the middle of the clothesline span will touch the

ground if the sag is 75 in. 72 in. = 3 in. We can now work the

problem either of two ways: 1) assume the sag = 3 in., compute

the corresponding maximum tension, and compare it to the

280-lb breaking strength; or 2) assume the maximum tension is

280 lb, compute the corresponding sag, and compare it to the

maximum permissible sag of 3 in.

3. An 18-lb washer load of wet beach-towels is hung on a clothesline

to dry. Each beach towel is 72-in. long, and the clothesline is tied to

supports located 75 in. above the ground. If the breaking strength of

the clothesline is 280 lb, determine if the clothesline can be tightened

enough to keep the middle towel from touching the ground.

1

75 in75 in

28 ft


(28 ft)/4 7 ft(28 ft)/4 7 ft

B x 252 lbB

T max

B y 9 lb

B y

B x

T o

3 in

(18 lb)/2

9 lb

Let's arbitrarily choose the first approach assume a sag

= 3 in. Then a free-body diagram of the right half of the

clothesline and hanging towels would appear like this:

Equilibrium equations

Fx = 0: T + Bx = 0 (1)

Fy = 0: 9 lb + By = 0 (2)

MB = 0: 9 lb(7 ft)

T (3 in. 1 ft/ 12 in.) = 0 (3)

Solving Eq. 1-3 gives

Bx = 252 lb

By = 9 lb

T = 252 lb

The maximum tension occurs where the slope is

greatest at B (or A):

= (252 lb)2 + (9 lb)2 = 252.2 lb

2

3

4

+

++

Since the maximum tension, 252.2 lb, is less than

the breaking strength, 280 lb, the clothesline can

be tightened enough to keep the towel from

touching the ground. Note, however, that the

poles supporting the clothesline must be

well-anchored in the ground since they must

resist a horizontal force of Bx = 252 lb acting at

the point where the clothesline is attached.

B


A

0.5 kg/m0.5 kg/m

3 m s

CB

80 m 60 m

4. Chains AB and BC are attached to a roller support at B

as shown. The chains support beams that have mass per

length of 0.5 kg/m. Determine the maximum tension in

each chain and the sag s in chain BC.

1 The maximum tension in chain AB occurs where the

slope is greatest, points A and B.


We know the location of the low point of chain AB, so

let's choose a free-body diagram of the right half of

the chain because then we will know the distances

appearing in the moment equation.

2 Equilibrium equation

Fx = 0: T + Bx = 0 (1)

Fy = 0: By 196.2 N = 0 (2)

MB = 0: T (3 m)

+ (196.2 N)(20 m) = 0 (3)

5

Solving Eqs. 1-3 gives

T = 1308 N

Bx = 1308 N

By = 196.2 N

6

+

++

B

3 m

B y

B x

T o

(40 m)/2 20 m

Resultant force

(0.5 kg/m)(9.81 m/s2 )(40 m)

196.2 N

By symmetry, we know the low point occurs at mid span:

(80 m)/2 40 m

4

3 Sag distance is given:


s

(60 m)/2 30 m

Resultant weight

(0.5 kg/m)(9.81 m/s2 )(30 m)

147.15 N

(30 m)/2 15 m

T'o

B'x

B'y

T max

B x 1308 N

B y 196.2 N

B

Maximum tension in AB

Next consider a free-body diagram of the

left half of chain BC:


Fx = 0: Bx + T = 0 (4)

Fy = 0: By 147.15 N = 0 (5)

MB = 0: T (s) (147.15 N)(15 m) = 0 (6)

+

++

= (1308 N)2 + (196.2 N)2 = 1323 N Ans.

7

8 9

B


B

147.15 N196.2 N

T o 1308 N T'

o

F B, vertical force

acting on roller support

Eqs. 4-6 are three equations in four unknowns, Bx ,

By , T and s. An additional free-body is needed.

The free-body diagram below shows that no

horizontal force acts on the base of the roller support at B.

10

11

T max

B'x 1308 N

B'y 147.15 N


Fx = 0: 1308 N + T = 0 (7)

Solving Eqs. 4-7 simultaneously gives

T = 1308 N

s = 1.688 m Ans.

Bx = 1308 N

By = 147.15 N

12

+

Maximum tension in BC13

Tmax = (1308 N)2 + (147.15 N)2

= 1316 N Ans.


40°

A

0.2 kg/m

B

3 m

2 m

5. Cable AB supports a uniformly distributed mass of

0.2 kg/m. The slope of the cable at B is known to be

40°. Determine the maximum tension in the cable and

the length of the cable.

The maximum tension occurs at B, where the slope of

the cable is the greatest.

1


40°

T max cos 40°

T max sin 40°

2 mA

y

A x

A Weight = (0.2 kg/m)(9.81 m/s2 )(3 m)

= 5.886 N

(3 m)/2 1.5 m

B

3 m

T max

Free-body diagram2


MA = 0: (Tmax sin 40°)(3 m) (Tmax cos 40°)(2 m)

(5.886 N)(1.5 m) = 0 (1)

Solving gives

Tmax = 22.28 N Ans.

+

To compute the length of the cable, we need

to use the equation of the cable curve,

y = (2) wx2

2T

3

4


Here, w is the distributed load per horizontal meter,

w = (0.2 kg/m)(9.81 m/s2)

= 1.962 N/m (3)

The quantity T is the horizontal component of the cable tension.

Since T is the same at all points, we can evaluate it at support B:

T = Tmax cos 40°

= (22.28 N) cos 40° (4)

Substituting Eqs. 3 and 4 in Eq. 2 gives

y = (Eq. 2 repeated)

or,

y = 0.057478x2 (5)

2T

wx2

1.962 N/m by Eq. 3

(22.28 N) cos 40° by Eq. 4

We will also need the slope,

= 2(0.057478x)

= 0.114956x (6)

dx

dy

5


A(x A, y

A)

B(x B, y

B)

y

x

40°

dy40°

dx

dy

dx

ds

3 m

y 0.05784 x2

The length of the cable is

sAB = ds

= (dx)2 + (dy)2

= 1 + ( )2 dx

A

B

dy

dx

B

A

0.114956x by Eq. 6

(Change the variable of

integration from s to x)6

dy

dx

Also at B, the slope is known:

= tan 40° (9)

xBxA

xA

xB

Thus the length of the cable can be expressed as

sAB = 1 + (0.114956x)2 dx (7)

To evaluate this integral, we have to find the values of xA

and xB. From the figure, we see that

xA = xB 3 m (8)

7


Using Eq. 6 to evaluate the left-hand side of Eq. 9 gives

( )B = tan 40° (Eq. 9 repeated)

0.114956xB, by Eq. 6

Solving for xB gives

xB = (tan 40°)/0.114956

= 7.299311 m

and using this result in Eq. 8 gives

xA = xB 3 m

= 7.299311 3

= 4.299311 m

dy

dx

This integral is best evaluated numerically with the

integral function of a calculator.

The result is

sAB = 3.61 m Ans.

8

9

7.299311

4.299311

By Eq. 7, then, the length of the cable is

sAB = 1+ (0.114956x)2 dx


C

4 ftB

A

48 ft

w lb/ft

6 ft

6. Determine the largest uniform load, w lb/ft, that the cable

can support if it will fail at a tension of 3,000 lb. Also

determine the location of the low point C of the cable.


+

2 Moment equilibrium equation

MB = 0: (48w)(24 ft) (3000 lb)(sin )(48 ft)

+ (3000 lb)(cos )(4 ft) = 0 (1)

C

B

A

B x

B y

(3000 lb) sin

(3000 lb) cos

Resultant load

w lb/ft 48 ft

(48w) lb

48 ft

24 ft(48 ft)/2 24 ft

4 ft

6 ft

The maximum tension in the cable occurs where the

slope is greatest, point A. The cable will fail if the

tension there exceeds 3,000 lb. A free-body diagram of

the entire system, with a maximum tension of 3,000 lb at

A, would appear as below:

1

The line of action of the weight of the cable does

not pass through the low point C, because the ends

of the cable are at different elevations.

As part of our solution

to the problem, we will

calculate .


C

B

A

B x

B y

(3000 lb) sin

(3000 lb) cos 4 ft

6 ft

d

(Unknown distance)

We could write two additional equilibrium equation, but they

would introduce two additional unknowns, Bx and By, so there

would be no advantage gained. Instead, we need another

free-body diagram. To get it, consider a free body of the entire

cable, and pass a vertical section through the low point C.

3


C

A

(3000 lb) sin

(3000 lb) cos

Resultant

w d

4 ft + 6 ft 10 ft

T o

d/2 d/2

d

Free-body diagram of portion to left of section4

5

++

Equation of equilibrium

Fy = 0: (3000 lb)(sin ) wd = 0 (2)

MC = 0: (3000 lb)(cos )(10 ft)

(3000 lb)(sin )d

+ (wd)(d/2) = 0 (3)

We could write a third equilibrium equation, but it

would introduce an additional unknown, T so no

advantage would be gained.

Eqs. 1-3 are three nonlinear equations in three unknowns,

, w, and d. These equations are best solved with a

calculator capable of solving simultaneous nonlinear

equations. Alternatively, proceed as follows. First note

that Eq. 2,

3000 sin wd = 0 (Eq. 2 repeated)

can be solved for d:

d = (3000 sin )/w (4)

This equation can be used to eliminate d from Eq. 3:

3000 cos (10) (3000 sin )d

+ wd2/2 = 0 (Eq. 3 repeated)

(3000 sin )/w 2 (3000 sin )/w

Multiplying through by w and combining terms gives

(30,000 cos )w (3,000 sin )2/2 = 0 (5)

6


Next note that Eq.1,

(48)(24)w (3,000 sin )(48)

+ (3,000 cos )(4) = 0 (Eq. 1 repeated)

can be solved for w:

w = 125 sin 10.4167 cos (6)

and this equation can be used to eliminate w from Eq. 5:

(30,000 cos ) w (3,000 sin )2/2 = 0 (Eq. 5 repeated)

125 sin 10.4167 cos

Carrying out the multiplication gives

(3.75 106) cos sin (0.312501 106) cos2

(4.5 106) sin2 = 0 (7)

Dividing both sides by 106 gives

(3.75) cos sin (0.312501) cos2

(4.5) sin2 = 0 (8)

Solving this equation by trial and error gives

= 36.48°

Using this value for in Eq. 6 then gives w:

w = 125 sin 10.4167 cos (Eq. 6 repeated)

= 125 sin 36.48° 10.4167 cos 36.48°

= 65.94 lb/ft Ans.

Eq. 4 then gives distance d:

d = (Eq. 4 repeated)

=

= 27.0 ft Ans.

w3000 sin

3000 sin 36.48°

65.94

7 8


The 100-kg mass produces a cable

tension of 100 kg 9.81 m/s2 = 981 N

at point A.

1

7. The cable system shown supports a uniformly

distributed mass of 5 kg/m along the horizontal.

Determine the tension at B and the length of portion AB

of the cable. Assume that the pulleys are frictionless.

4 m

20 m

B

A

5 kg/m

100 kg


B B x

B y

(981 N) sin

(981 N) cos

Resultant weight of cable between A and B

(20 m)(9.81 m/s2 )(5 kg/m)

981 N

20 m

10 m

A

981 N

4 m

2 Free-body diagram


Fx = 0: (981 N) cos + Bx = 0 (1)

Fy = 0: (981 N) sin + By 981 N = 0 (2)

MB = 0: (981 N) cos (4 m)

(981 N) sin (20 m)

+ (981 N)(10 m) = 0 (3)

+

++

Eqs. 1-3 are best solved with a calculator that solves

simultaneous nonlinear equations. Alternatively, Eq. 3

involves only one unknown, , and can be solved by

trial-and-error to yield

= 40.67° (4)

Using this value in Eq. 1 and 2 then leads to

Bx = 744.1 N (5)

By = 341.7 N (6)

3

4


The tension at B is then

Bx2 + By

2 = (744.1 N)2 + (341.7 N)2

= 818.8 N Ans.

To compute the length of the cable, we need to use the

equation of the cable curve,

2T

wx2

y = (7)


w = (5 kg/m)(9.81 m/s2)

= 49.05 N/m (8)

The quantity T is the horizontal component of the cable

tension. Since T is the same at all points, we can evaluate it

at support B:

T = Bx

744.1 N by Eq. 5 (9)

5

6

Substituting Eqs. 8 and 9 in Eq. 7 gives,


or,

y = 0.03296 x2 (10)

We will need the equation for the slope:

= 2(0.3296 x) = 0.06592 x (11)

wx2

2T

49.05 N/m

744.1 N

dx

dy

The length of the cable is

sAB = ds

= (dx)2 + (dy)2

or

sAB = 1 + ( )2 dx (12)

A

B

(0.06592 x)2, by Eq. 11

ds

dxdy

(Change the

variable of

integration

from s to x)

dx

dy

7

8

B

A

xA

xB


x

y

B(x B, y

B)

A(x A, y

A)

20 m

y 0.03296x2

40.67°

dx

dy

40.67°

To evaluate the integral for sAB, we have to find the values

of xA and xB.

From the figure, we see that

xB = xA + 20 m (13)

Also at A, the slope is known, so

= tan 40.67°

Using Eq. 11 to evaluate the left hand-side of this equation gives

9

10

dy

dx

Solving for xA gives

xA = 13.03436 m (14)

Using this result in Eq. 13 gives

xB = xA + 20 m (Eq. 13 repeated)

= 13.03436 m + 20 m

= 6.96564 m (15)

By Eq. 12, then, the length of cable is

sAB = 1 + (0.06592 x)2 dx

This integral is best evaluated numerically with the

integral function of a calculator. The result is

sAB = 21.7 m Ans.

6.96564

-13.03436

dy

dx( )A = tan 40.67°

0.06592xA, by Eq. 11

11


22 kg/m

16 m

A

B

Cx

y

d A

d B

30 m

8. The chain AB supports a horizontal, uniform beam of

mass per length 22 kg/m. If the maximum allowable

tension in the chain is 7 kN, determine distances dA and

dB of the supports above the low point C of the chain.

Also determine the length of the chain.

The maximum tension occurs at support A,

where the slope of the chain is greatest.

1


A

B B x

B y

A x

A y

d A d

B

(46 m)/2 23 m

30 m + 16 m 46 m

Resultant weight

(22 kg/m)(9.81 m/s2 )(46 m)

9928 N

9.928 kN

3

C

Free-body diagram of entire system.


MB = 0: Ax(dA dB) Ay(46 m) + (9.928 kN)(23 m) = 0 (1)

We could write two additional equations of

equilibrium, but they would contain two additional

unknowns, Bx and By, so nothing would be gained.

4

2

+


22 kg/m

A

B

C

B x

B y

A y

A x

C

A

A y

T o

d A

30 m

(30 m)/2 15 m

Resultant weight

(22 kg/m)(9.81 m/s2 )(30 m)

6475 N

6.475 kN

d A

30 m

To obtain an additional free-body diagram, pass a

vertical section through the low point of the chain.

A x

5 Free-body diagram of portion of chain to left of section.6



Fx = 0: Ax + T = 0 (2)

Fy = 0: Ay 6.475 kN = 0 (3)

MC = 0: Ax(dA) Ay(30 m) + 6.475 kN)(15 m) = 0 (4)

Equation 3 gives

Ay = 6.475 kN (5)

Since we know that the maximum tension of 7 kN occurs

at A and is the resultant of Ax and Ay, we have

Ax2 + Ay

2 = 7 kN

Substituting Ay = 6.475 kN and solving for Ax gives

Ax = 2.660 kN (6)

Substituting for Ax and Ay in Eqs. 2 and 4 gives

Ax + T = 0 (Eq. 2 repeated)

7

+

++

2.660 kN

Ax dA Ay(30) + (6.475)(15) = 0 (Eq. 4 repeated)

2.660 kN 6.475 kN

Solving gives

T = 2.660 kN (7)

dA = 36.52 m (8) Ans.

Distance dB can now be found by substitution in Eq. 1:

Ax (dA dB) Ay(46) + (9.928)(23) = 0

2.660 kN 36.52 m 6.475 kN

by Eq. 6 by Eq. 8 by Eq. 5

Solving gives

dB = 10.4 m Ans.

8


To compute the length of the chain, we need to use the

equation of the chain curve:

y = (9)


w = (22 kg/m)(9.81 m/s2)

= 215.8 N/m

= 0.2158 kN/m (10)

The quantity T has been given in Eq. 7

T = 2.660 kN (Eq. 7 repeated)

Substituting into Eq. 9 gives

0.2158 kN/m by Eq. 10


2.660 kN by Eq. 7

or,

y = 0.040564 x2 (11)

9

2T

wx2

2T

wx2

We will need the equation for the slope,

= 2(0.040564 x) = 0.08113 x (12)dy

dx

(Change the variable of

integration from s to x)

dydx

dsB

A

The length of the chain is

sAB = ds

= (dx)2 + (dy)2

or

sAB = 1 + ( )2 dx (13)dy

dx

10

A

B

xA

xB


x

y 0.040564 x2

B(x B 16 m)

y

A(x A 30 m)

30 m 16 m

Since the location of the low point of the chain is

known, xA and xB are known:

11 The integral for the cable length, Eq. 13, can now be

written as

sAB = 1 + ( )2 dx (Eq. 13 repeated)

This integral is best evaluated numerically with the integral

function of a calculator. The result is

sAB = 69.2 m Ans.

(0.08113x)2, by Eq. 12

-30

16 dy

dx

12


CA B

10 kg

h

d

9. A 40-m length of rope has a uniformly distributed

mass of 0.1 kg/m and has one end fixed and the other

end attached to a cart as shown. Determine the distance

d and the sag h when the cable and cart are in

equilibrium under the force supplied by the 10-kg load.

The total weight of the rope is 40 m 0.1 kg/m 9.81 m/s2 =

39.24 N, and the force of the cart acting horizontally on the

rope is 10 kg 9.81 m/s2 = 98.1 N, or more than twice as much.

Thus it seems reasonable to assume that the sag, h, is small

compared to the span d, h d, and the weight of the rope is

well-approximated as a uniformly distributed load along the

horizontal of

w = (0.1 kg/m) (9.81 m/s2) = 0.981 N/m (1)

At the end of the problem, we can check the reasonableness of

these assumptions.

1


B x

Vertical force from

ground acting on cart

B y

B y

C x

C y

B x T

o

(Horizontal component

of tension)

(10 kg) (9.81 m/s2 )

98.1 N

CB

The equation describing the shape of the rope is

y = (2)

T is the horizontal component of tension in the rope

and can be found by considering free bodies of the

cart and rope.

wx2

2T

2

Free-body diagram of cart3

Free-body diagram of rope5

The equation of the rope is now

0.981 N/m by Eq. 1


98.1 N by Eq. 3

or,

y = 0.005 x2 (4)

We will also need the equation for the slope,

= 2(0.005 x) = 0.01x (5)

2T

wx2

+4

6

Fx = 0: 98.1 N + Bx = 0

Solving gives

Bx = 98.1 N = T (3)

dy

dx


0.005d + 1 + (0.005d)2

0.005d + 1 + (0.005d)2

d/2d/2

CB

hx

C d/2x B d/2

x

dxdy

C

B

B

C

ds

dxdy

The equation for the length of the slope is

sBC = ds

= (dx)2 + (dy)2

= 1 + ( )2 dx (6)

xC and xB can be expressed in terms of the unknown, d:

7

The best way to solve for d in Eq. 7 is to use the solver on a

calculator and to use the calculator integral function to input

Eq. 7 in the solver. Alternatively, use a table of integrals to

evaluate the integral in Eq. 7 to get

40 = 1 + (0.005d)2

+ (50) ln

Solving this equation by trial-and-error gives

d = 39.74 m Ans. (8)

d

2

xB

xC

8

y

d/2

-d/2

-d/2

d/2

dx

dy

Thus Eq. 6 can be written

sBC = 1 + ( )2 dx (Eq. 6 repeated)

Rope length = 40 m (0.01 x) by Eq. 5

or,

40 = 1 + (0.01x)2 dx (7)


hB C

y

x

The sag, h, can now be found from Eq. 4:

y = 0.005x2 (Eq. 4 repeated)

h = by Eq. 839.74

2

The result is

h = 1.974 m Ans.

Note that the sag is much smaller than the span,

h = 1.974 m << 39.74 m = d, as we assumed in

the beginning of the problem.

9

d2

6.6 cables: uniform loads - civil engineering cables: uniform loads example 1, page 1 of 4 a b 400...

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