6.5 Cables: Concentrated Loads - Civil Engineering · 6.5 Cables: Concentrated Loads Example 1,...

6.5 Cables: Concentrated Loads

6.5 Cables: Concentrated Loads Example 1, page 1 of 3

1.4 kN

D

B

C

A

AA

x

1.4 kN

T BC

A y

B

2 kN

4 m3 m

2 m

2 m

1. For the cable system shown, determine the

reactions at support A and the distance yC.

1 Strategy: Note that both

horizontal and vertical

distances between A and B are

known. Thus summing

moments about B, for a

free-body AB, would give an

equation involving Ax and Ay

only.

2 Free-body diagram of AB

+Equilibrium equation

MB = 0: Ax(2.5 m) Ay(2 m) = 0 (1)

3

2 m

2.5 m yC

2.5 m


1.4 kN

D

BC

A

2 kN

D x

D yA

y

A x

2 m 3 m 4 m

Equilibrium equation

MD = 0: (1.4 kN)(3 m + 4 m)

+ (2 kN)(4 m)

y(2 m + 3 m + 4 m) = 0 (2)

Solving gives

y = 1.978 kN Ans.

Using this result in Eq. 1 gives

Ax(2.5) y(2) = 0 ( Eq. 1 repeated)

1.978 kN

Solving gives

Ax = 1.582 kN Ans.

6

We can get another equation involving the reaction at

A as the only unknown by summing moments about

D for a free body of the entire cable.

Free-body diagram of entire cable

4

5

+


1.4 kN

BC

A

2 kN

A y 1.978 kN

A x 1.582 kN

T CD

2 m 3 m

Now that the values of Ax and Ay are known,

summing moments about C, for a free body ABC,

will give an equation with a single unknown, yc.

Free-body diagram of ABC

+


MC = 0: (1.4 kN)(3 m)

(1.978 kN)(2 m + 3 m)

+ (1.582 kN) yc) = 0

Solving gives

yc = 3.60 m Ans.

9

7

8

yC


15 m

3.5 kN

B

2 kN

P A

C

4 m

2. The horizontal force P is applied to end A of the

cable as shown. Determine the value of P and the

distance d required to keep the cable system in the

configuration shown. Also determine the total

length of the cable.

Strategy: Note that we know the

horizontal distance between points A

and B, and we can compute the vertical

distance between these points. Thus

summing moments about B for a free

body AB, will give an equation from

which P can be found.

1

d

12 m


Free-body diagram of ABC5

Equilibrium equation for ABC

MC = 0: (2.667 kN)(15 m) + (2 kN)(4 m + d)

+ (3.5 kN)(d) = 0

Solving gives

d = 5.819 m Ans.

Free-body diagram of AB

6

+

+

15 m 12 m = 3 m

C

15 m

d

3.5 kN

B

2 kN

P = 2.667 kNA

4 m

C y

C x

P

2 kN

T BC

3.5 kNA

B

4 m

2

Equilibrium equation for AB

MB = 0: (2 kN)(4 m) P(3 m) = 0

Solving gives

P = 2.667 kN Ans.

3

Now that the value of P is known, we can sum

moments about C, for a free-body of the whole

cable, to obtain an equation for d.

4


B

A

C

4 m

12 m

d 5.819 m

15 m 12 m 3 m

The total length of the cable can be

found by applying the Pythagorean

Theorem to segment AB and to BC

7

Geometry8

LTotal = LAB + LBC

= (4 m)2 + (3 m)2 + (5.819 m)2 + (12 m)2

= 5 m + 13.34 m

= 18.34 m Ans.


xC

2 m

4 m

2 m

C500 N

150 NB

6 m

3 m

150 NB

A

A y

A x

T BC

2

A

D

3. The cable supports the 150 N and 500 N

loads shown. Determine the distance xC and

the tension in each segment of the cable.

1 Strategy: Note that both the horizontal

and vertical distances between A and B

are known. Thus summing moments

about B, for a free-body AB would give

an equation involving the reaction

components at A (Ax and Ay), only; no

other unknowns are present.



MB = 0: Ay (4 m) x (2 m) = 0 (1)

3

+

4 m


C500 N

150 N B

A

D

A y

A x

6 m

D y

D

x

2 m

Now summing moments about D, for a free body of the entire

cable, will give another equation for Ax and Ay.

4

Free-body diagram of entire cable5Equilibrium equation

MD = 0: x (2 m + 6 m + 3 m) (150 N)(6 m + 3 m) (500 N)(3 m) = 0

Solving gives

Ax = 259.09 N (2)

Substituting this in Eq. 1 gives

4 Ay 2 Ax = 0 (Eq. 1 repeated)

259.09 N

Solving gives

Ay = 129.55 N (3)

6

+

3 m


9

We can get the tension in AB by considering a

free body of support A.

10

Free-body diagram of A for calculating tension TAB

Summing horizontal and vertical forces gives

TABx = 259.09 N (4)

TABy

= 129.55 N (5)

and the magnitude of the tension in AB is

TAB = (TABx)2 + (TABy

)2

= (259.09 N)2 + (129.55)2

= 290 N Ans.

12

x C

+

C500 N

150 NB

A

11

A y 129.55 N

A x 259.09 N

T ABy

T ABx

T CD

A y 129.55 N

A x 259.09 N

6 m

T AB

2 m

Now that Ax and Ay are known, we can calculate xC by

summing moments about C for a free-body ABC.

Free-body diagram of ABC

7

8


MC = 0: (150 N)(6 m) (259.09 N)(2 m + 6 m)

+ (129.54 N)(xC) = 0

Solving gives

xC = 9.05 m Ans.


B

TABy

129.55 N

T BCy

TABx 259.09 N

T BC

T BCx

14

150 N

Free-bodies of B and C will yield the

values of the tension in BC and CD

13

Free-body diagram of B for

calculating tension TBC

The magnitude of the tension in BC is

TBC = (TBCx)2 + (TBCy

)2

= (109.09 N)2 + (129.55 N)2

= 169.4 N Ans.

+

Summing vertical forces gives

TBCy

= 129.55 N

Summing horizontal forces gives

Fx = 0: 150 N TBCx + 259.09 N = 0

Solving gives

TBCx = 109.09 N

15

16


C

500 N

17

T BCy

129.55 N

T CDy

T CD

TBCx 109.09 NT

CDx

Free-body diagram of C for

calculating tension TCD


TCDy

= 129.55 N


Fx = 0: 500 N TCDx + 109.09 N = 0

Solving gives

TCDx = 390.91 N

The magnitude of the tension in CD is

TCD = (TCDx)2 + (TCDy

)2

= (390.91 N)2 + (129.55)2

= 412 N Ans.

+

C500 N

150 NB

6 m

3 m

A

D

4 m

xC

2 m


2.2 kN

E

B C

A

1.8 kN

1.2 kN

2 m

3 m

D

y D

0.5 m0.25 m

4. For the cable system shown, determine the distance yC for

which segment BC will be horizontal. Also determine yD.

Strategy: We have to make use of the fact

that segment BC is horizontal. One way to

do this is to pass a section through BC and

then consider the portion of the cable to the

left of the section.

1

2 m

yC


B

A

2.2 kN

A y

A x

2

T BC (horizontal)


Equilibrium equation for AB (Because BC is

horizontal, we use the sum of vertical forces

so that the unknown tension TBC will not

appear in the equation.):

Fx = 0: Ay 2.2 kN = 0

Solving gives

Ay = 2.2 kN (1)

3

+2.2 kN

E

B C

A

1.8 kN

1.2 kN

2 m

3 m

D

y D

0.5 m0.25 m 2 m

yC


2.2 kN

E

B C

A

1.8 kN

D

1.2 kN

2 m 2 m

y C

A y 2.2 kN

A x

E x

E y

3 m

0.5 m

Now that Ay is known, we can solve for

Ax by summing moments about E for a

free-body diagram of the entire cable.

4

Free-body diagram of entire cable5

+

6 Equilibrium equation for entire cable

ME = 0: (1.2 kN)(0.5 m)

+ (1.8 kN)(0.5 m + 2 m)

+ (2.2 kN)(0.5 m + 2 m + 2 m)

(2.2 kN)(0.5 m + 2 m + 2 m + 0.25 m)

x(3 m) = 0

Solving gives

Ax = 1.517 kN (2)

0.25 m


yB

A y 2.2 kN

2.2 kN

0.25 m

B

A

T BC

A x 1.517 kN y

B y C (because BC is horizontal)

Summing moments about B for a free body AB

will now give us the value of yC.

7

Free-body diagram of AB8


MB = 0: (1.517 kN)(yC) (2.2 kN)(0.25 m) = 0

Solving gives

yC = 0.36 m Ans.

9

+

2.2 kN

E

B C

A

1.8 kN

1.2 kN

2 m

3 m

D

y D

0.5 m0.25 m 2 m

yC


2.2 kN

B C

A

1.8 kN

D

1.2 kN

2 m 2 m0.25 m

A y 2.2 kN

A x 1.517 kN

T DE

y D

Finally, summing moments about D

for a free body ABCD will give an

equation for yD.

Free-body diagram of ABCD

10

11


MD = 0: (1.517 kN)(yD)

+ (2.2 kN)(2 m + 2 m)

+ (1.8 kN)(2 m)

(2.2 kN)(0.25 m + 2 m + 2 m) = 0

Solving gives

yD = 2.01 m Ans.

12

+


2.2 kN

E

B C

A

1.8 kN

D

1.2 kN0.36 m

The minus sign in yD = 2.01 m

indicates point D lies above point A,

not below it, as was assumed in

drawing the free-body diagram.

13

2.01 m

3 m


4 m2 m

P B

B

A

C

200 N

P D

D

E

7 m

5 m

2 m

2 m

3 m

5. For the cable system shown,

determine the value of the forces PB

and PD necessary to maintain the

given configuration.

Strategy: Equilibrium equations for a free

body of the entire cable would involve

six unknowns (the components of the

support reactions: Ax, Ay, Ex, and Ey;

plus PB and PD). Thus using a free body

of the entire cable does not look like a

good place to start.

1

4 m


P B

B

C

200 N

P D

D

5 m

A y

A x

E y

E x

4 m 4 m2 m 3 m

2 m

2 m

D

C

B

C 2 m + 3 m 5 m

2

BC

DC

BC

DC

A better place to start is to observe that we

know the horizontal and vertical distances

between points B, C and D. Thus we can

calculate the angles that segments BC and DC

make with the horizontal, and then we can use

the equilibrium equation for connector C to find

the tension in BC and in CD. Once these

tensions are known, we can use the equilibrium

equations for segments AB and ED to determine

PB and PD.

Geometry3

3 m5 m1

BC = tan ( ) = 59.04° (1)

1 7 m5 mDC = tan ( ) = 54.46° (2)

7 m

5 m

3 m

7 m


4 m

P B

B

C

A y

A x

2 m

C 200 N

T BC

T DC

T BC 177.46 N

A

4

6

BC 59.04°

DC 54.46°

BC 59.04°

Free-body diagram of C

Equilibrium equations for C

Fx = 0: TBC cos 59.04° TDC cos 54.46° + 200 = 0

Fy = 0: TBC sin 59.04° TDC sin 54.46° = 0

Solving gives

TBC = 177.46 N (3)

TDC = 187.02 N (4)

+

+

5



MA = 0: PB(4 m) + (177.46 N)(cos 59.04°)(2 m)

(177.46 N)(sin 59.04°)(4 m) = 0

Solving gives

PB = 106.5 N Ans.

7

+


C

P D

D

E

E y

2 m E x

T DC 187.02 N

DC 54.46°

8

8 m

Free-body diagram of ED

ME = 0: PD(2 m) (187.02 N)(cos 54.46 )(2 m)

+ (187.02 N)(sin 54.46 )(8 m) = 0

Solving gives

PD = 500 N Ans.

9 +


80 lb

50 lbB

C

1.5 ftA

D

6. For the cable system shown, determine

distance yB and the tension in each segment.

Strategy: If we can compute the reactions

at supports A and D, then we can compute

the tensions in AB and CD. Let's start

with support D. Note that both horizontal

and vertical distances between C and D

are known. Thus summing moments

about C, for the free body CD, would give

an equation involving Dx and Dy only.

1

5 ft3.5 ft2 ft

3.5 ft

yB


80 lb

50 lbB

C

A x

A y

D y

D x

2 ft

1.5 ft

50 lb

C

D x

D y

3.5 ft

D

T BC

5 ft

2

5

Free-body diagram of CD


MC = 0: Dy(5 ft) Dx(3.5 ft) = 0 (1)

+

3

4 We can get another equation involving reactions at Dx and

Dy as the only unknowns by summing moments about A

for a free body of the entire cable.


6

+

Equation of equilibrium

MA = 0: Dy(2 ft + 3.5 ft + 5 ft) Dx(1.5 ft)

(50 lb)(2 ft + 3.5 ft) 80 lb(2 ft) = 0 (2)

3.5 ft 5 ft


yB

80 lb

50 lbB

C

D

D y 34.41 lb

D x 49.15 lb

T AB

9

Solving Eqs. 1 and 2 simultaneously gives

Dx = 49.15 lb (3)

Dy = 34.41 lb (4)

Now that the values of Dx and Dy are

known, summing moments about B, for a

free body BCD, will give an equation with

a single unknown, yB.

8

Free-body diagram BCD

MB = 0: (50 lb)(3.5 ft) + (34.41 lb)(3.5 ft + 5 ft) (49.15 lb)(yB) = 0

Solving gives

yB = 2.39 ft Ans (5)

+10

7

5 ft3.5 ft

3.5 ft


We still must find the tension in each cable segment.

A free-body consisting of support D will give the

tension in CD.

11

Summing x forces and then y forces gives

TCDx = 49.15 lb (6)

TCDy

= 34.41 lb (7)

The magnitude of the tension in CD is then

TCD = (TCDx)2 + (TCDy

)2

= (49.15 lb)2 + (34.41 lb)2

= 60.0 lb Ans.

D y 34.41 lb

D x 49.15 lb

T CDy

T CDx

T CD

12

D

Free-body diagram of D for determining tension TCD

80 lb

50 lbB

C

1.5 ftA

D

5 ft3.5 ft2 ft

3.5 ft

yB


50 lb

C

14

TCDx 49.15 lb (Eq. 7)

T CDy

34.41 lb (Eq. 6)

T BCx

T BCyT

BC

A free-body diagram of connection C will give

the tension in cable segment BC.13

Free-body diagram of C for determining tension TBC


TBCx = 49.15 lb (8)


Fy = 0: 34.41 lb TBCy

50 lb = 0

Solving gives

TBCy

= 15.59 lb (9)

+

80 lb

50 lbB

C

1.5 ftA

D

5 ft3.5 ft2 ft

3.5 ft

yB


15 Why did TBCy

turn out to be negative? Because

in drawing the tension

TBC on the free-body

diagram of C, we

assumed that B was

below C:

But this assumption was

wrong, as we should have

realized, since we have

already calculated yB and

found yB = 2.39 ft (Eq. 5).

The corrected free-body

diagram shows TBC pulling

up on C, which would lead

to a positive value of TBCy

16

Free-body diagram of C

Corrected free-body diagram of C

B lies

above C

80 lb50 lb

B

C

A

D

y B

3.5 ft

80 lb 50 lb

B

C

3.5 ft

A

D

2.39 ft

C

CD y

CD x

50 lb

T BC (Pulls down on C)

50 lb

C

CD y

CD x

T BC (Pulls up on C)


80 lb

B

T ABy

T BCx 49.15 lb

T ABx

T BCy

+15.59 lb

T AB

The magnitude of the tension in BC is

TBC = (TBCx)2 + (TBCy

)2

= (49.15 lb)2 + ( 15.59 lb)2

= 51.56 lb Ans.

17

The tension in AB can be found from a free-body

diagram of connector B.18

Free-body diagram of B for calculating TAB19

Since TBCy

points down, we are assuming correctly

that point C lies below B.

Summing x forces gives

TABx = 49.15 lb

Summing y forces gives

TABy

= 15.59 lb + 80 lb

= 95.59 lb

The magnitude of the tension in AB is then

TAB = (TABx)2 + (TABy

)2

= (49.15 lb)2 + (95.59 lb)2

= 107.5 lb Ans.

20


80 lb

B

A

A y

A x

TBC (unknown)

B (known)

The tension in the cable segments could have

been calculated by other approaches, for

example, by calculating the angle of inclination

of each segment and then summing moments

about one end of the segment.

Example:

21

MA = 0 gives an equation that will give us the

value of TBC. Proceeding to the next cable segment,

BC, and summing moments about end B of that

segment would give the value of the tension in that

segment, etc.

80 lb

50 lbB

C

1.5 ftA

D

5 ft3.5 ft2 ft

3.5 ft

yB


2 kip2 kip

2 kip

2 kip

A

F

18 ft 18 ft 18 ft 18 ft 18 ft

B

CD

E

15 ft

35 ft

7. The cable supports the four forces shown.

Determine the maximum tension in the cable.

Strategy: To find the maximum tension, Tmax in

the cable, we could find the tension in each of

the five segments and then pick the largest.

However, this is a tedious and time-consuming

approach, and we can find Tmax more easily if we

note two facts: 1) The horizontal component of

tension is the same in all cable segments, and 2)

the maximum tension occurs in the cable

segment with the maximum slope.

1


2 kip

B T BCxT

ABx

T BCy

T ABy

T BC

T AB

T horizontal

T

Demonstration that horizontal components

of tension are equal:2

Free-body diagram of B

Fx = 0: TABx + TBCx = 0

Thus

TABx = TBCx

That is, the horizontal components are

equal. A similar argument holds for points

C, D, and E.

+

Demonstration that Tmax occurs in the segment

with the greatest slope:3

For each cable segments,

T =

Since Thorizontal is the same for all segments, it

follow that T will be largest where cos is

smallest, that is, where is the largest the

steepest slope.

Thorizontal

cos


15 ft

2 kip2 kip2 kip

2 kip

A

B

EC

D

F

A x

A y

A

F y

F x

F

18 ft 18 ft 18 ft 18 ft 18 ft

35 ft

Thus the problem of finding Tmax has now

been reduced to determining whether the slope

is greater at A than at B.


4

5 To determine A, we can find Ax and Ay

first. To do this, sum moments about F for

a free-body consisting of the entire cable.

Equilibrium equation for cable

MF = 0: Ax(15 ft) y(5 18 ft)

+ (2 kip)(4 18 ft)

+ (2 kip)(3 18 ft)

+ (2 kip)(2 18 ft)

+ (2 kip)(1 18 ft) = 0 (1)

6

+


2 kip2 kip

2 kip

A

B

CD

A x

T DE

A y

18 ft 18 ft 18 ft

8

35 ft

Equilibrium equation for ABCD

MD = 0: Ax(35 ft) y(3 18 ft)

+ (2 kip) (2 18 ft)

+ (2 kip) (18 ft) = 0 (2)

Solving Eqs. 1 and 2 simultaneously gives

Ax = 4.154 kip (3)

y = 4.692 kip (4)

9

+

To obtain another equation for Ax and y, pass a

section through segment DE and consider a free

body ABCD.

7

Free-body diagram of ABCD


2 kip2 kip

2 kip

2 kip

A

FB

C

DE

F y

F x

A y 4.692 kip

A x 4.154 kip


10 To find the slope at F, consider a free body of the

entire cable.

11

12

+

+

Equilibrium equations for entire cable

Fx = 0: 4.154 kip + Fx = 0

Fy = 0: 4.692 kip + Fy (4 2 kip) = 0

Solving gives

Fx = 4.154 kip (5)

Fy = 3.308 kip (6)


A F

13

A y 4.692 kip

A x 4.154 kip

T AB

A

A

F y 3.308 kip

F x 4.154 kip

F

F

T FE

Compare slopes at A and F

Thus A is greater than F and it follows that Tmax occurs at A.

Tmax = magnitude of the reaction force at A

= (Ax)2 + (Ay)2

= (4.154 kip)2 + (4.692 kip)2

= 6.27 kip Ans.

14

A = tan ( ) = 48.5°1 4.692

4.154 4.1543.3081

F = tan ( ) = 38.5°


15 Of course, once we had calculated the components of the

reaction at F,

Fx = 4.154 kip (Eq. 5 repeated)

Fy = 3.308 kip (Eq. 6 repeated)

we could have computed the tension in EF:

TEF = (4.154 kip)2 + (3.308 kip)2

= 5.31 kip

and then we could have compared TEF with the tension in

TAB. That is, in this particular example, we didn't have to

compare slopes, but we did it to illustrate the principle

that Tmax occurs where the slope is a maximum.

6.5 Cables: Concentrated Loads - Civil Engineering · 6.5 Cables: Concentrated Loads Example 1,...

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