Chapter 5: Chapter 5:

Structural Analysis 7

Structural Analysis 7 ^{th th} Edition in SI Units Edition in SI Units

Chapter 5:

Chapter 5:

Cables and Arches

Cables and Arches

Cables Cables

• Assumptions when deriving the relations between Assumptions when deriving the relations between force in cable & its slope

force in cable & its slope

• Cable is perfectly flexible & inextensible Cable is perfectly flexible & inextensible

• Due to its flexibility, cable offers no resistance to Due to its flexibility, cable offers no resistance to shear or bending

shear or bending

• The force acting the cable is always tangent to the The force acting the cable is always tangent to the cable at points along its length

cable at points along its length

Cable subjected to concentrated loads Cable subjected to concentrated loads

• When a cable of negligible weight supports several When a cable of negligible weight supports several concentrated loads, the cable takes the form of

concentrated loads, the cable takes the form of several straight line segments

several straight line segments

• Each of the segment is subjected to a constant Each of the segment is subjected to a constant tensile force

tensile force

   specifies the angle specifies the angle of the chord AB

of the chord AB

• L = cable length L = cable length

Cable subjected to concentrated loads Cable subjected to concentrated loads

• If L If L _{1 1} , L , L _{2 2} & L & L _{3 3} and loads P and loads P _{1 1} & P & P _{2 2} are known, are known, determine the 9 unknowns consisting of the determine the 9 unknowns consisting of the

tension of in each of the 3 segments, the 4 tension of in each of the 3 segments, the 4

components of reactions at A & B and the sags y components of reactions at A & B and the sags y _{C C}

& y

& y _{D D}

• For solutions, we wrtite 2 eqns of equilibrium at For solutions, we wrtite 2 eqns of equilibrium at each of 4 points A, B, C & D

each of 4 points A, B, C & D

• Total 8 eqns Total 8 eqns

• The last eqn comes from the geometry of the The last eqn comes from the geometry of the cable

cable

Determine the tension in each segment of the cable. Also, what is the dimension h ?

Example 5.1

Example 5.1

By inspection, there are

 4 unknown external reactions (A _x , A _y , D _x and D _y )

 3 unknown cable tensions

These unknowns and sag, h can be determined from available equilibrium eqn applied to points A through D.

A more direct approach to the solution is to recognize that the slope of cable CD is specified.

Solution

Solution

Solution Solution

kN T

T kN

kN F

T kN

F

kN

T m T m kN m kN m

T M

o BC BC

BC BC

y

BC BC

x CD

CD CD

A

82 . 4

and

3 . 32

0 sin

8 ) 5 / 4 ( 79

. 6

0

0 cos

) 5 / 3 ( 79

. 6

0

5.2c) (Fig

C Point sequence.

in B and

C Now we can analyze the equilibriu m of points 79

.

6 5 )( 2 ) ( 4 / 5 )( 5 . 5 ) 3 ( 2 ) 8 ( 4 ) 0 /

3

( anti 0 - clockwise as ve With



    









 





     

 

 



Solution Solution

m m

h

kN T

kN kN

T

F

kN T

F

o o BA

BA

BA o BA

y

BA o BA

x

74 . 2 8

. 53 tan )

2 (

5.2(a) Fig

from Hence

90 . 6

and

8 . 53

0 3

3 . 32 sin 82

. 4 sin

0

0 3

. 32 cos 82

. 4 cos

0

5.2d) (Fig

B Point



































 



Cable subjected to a uniform distributed load

• The x,y axes have their origin located at the The x,y axes have their origin located at the

lowest point on the cable such that the slope is lowest point on the cable such that the slope is

zero at this point zero at this point

• Since the tensile force in the cable changes Since the tensile force in the cable changes

continuously in both magnitude & direction along continuously in both magnitude & direction along

the cable’s length, this load is denoted by

the cable’s length, this load is denoted by   T T

Cable subjected to a uniform distributed load

• The distributed load is represented by its resultant The distributed load is represented by its resultant force

force   o _o   x which acts at x which acts at   x/2 from point O x/2 from point O

• Applying eqn of equilibrium yields: Applying eqn of equilibrium yields:

0 sin

cos )

2 / )(

( anti 0 - clockwise as ve With

0 )

sin(

) (

) (

sin

0

0 )

cos(

) (

cos 0













 

 

















   















  



x T

y T

x x

w M

T T

x w

T

F

T T

T F

o A

o y

x

















Cable subjected to a uniform distributed load

• Dividing each of these eqn by Dividing each of these eqn by   x and taking the x and taking the limit as

limit as   x x   0, hence, 0, hence,   y y   0 , 0 ,     0 and 0 and   T T   0 0 , we obtain:

, we obtain:

3 eqn

tan

2 eqn

) sin (

1 eqn

) 0 cos (













dx dy

dx w T d

dx T d

o

Cable subjected to a uniform distributed load

• Integrating Eqn 1 where T = F Integrating Eqn 1 where T = F _{H H} at x = 0, we have: at x = 0, we have:

• Which indicates the horizontal component of force Which indicates the horizontal component of force at any point along the cable remains constant

at any point along the cable remains constant

• Integrating Eqn 2 realizing that Tsin Integrating Eqn 2 realizing that Tsin   = 0 at x = 0, = 0 at x = 0, we have:

we have:

4 eqn

cos F _H T  

5 eqn

sin w x

T   _o

Cable subjected to a uniform distributed load

• Dividing Eqn 5 by Eqn 5.4 eliminates T Dividing Eqn 5 by Eqn 5.4 eliminates T

• Then using Eqn 3, we can obtain the slope at any Then using Eqn 3, we can obtain the slope at any point

point

• Performing a second integration with y = 0 at x = Performing a second integration with y = 0 at x = 0 yields

0 yields

7 eqn 2

x 2

F y w

H

 o

6 eqn

tan

H o

F x w dx dy 

 

Cable subjected to a uniform distributed load

• This is the eqn of parabola This is the eqn of parabola

• The constant F The constant F _{H H} may be obtained by using the may be obtained by using the boundary condition y = h at x = L

boundary condition y = h at x = L

• Thus Thus

• Substituting into Eqn 7 Substituting into Eqn 7

9 eqn

2

2

x L y  h

8 eqn

2

2

h

L

F _H  w ^o

Cable subjected to a uniform distributed load

• From Eqn 4, the max tension in the cable occurs From Eqn 4, the max tension in the cable occurs when

when   is max is max

• From Eqn 4 and 5 From Eqn 4 and 5

• Using Eqn 8 we can express T Using Eqn 8 we can express T _{max max} in terms of w in terms of w _{o o}

10 eqn

) (

2 ^{2 2}

max F w L

T  ^H  _o

11 eqn

) 2 / ( 1

2 ²

max w L L h

T  _o 

Cable subjected to a uniform distributed load

• We have neglect the weight of the cable which is We have neglect the weight of the cable which is uniform along the length

uniform along the length

• A cable subjected to its own weight will take the A cable subjected to its own weight will take the form of a catenary curve

form of a catenary curve

• If the sag-to-span ratio is small, this curve closely If the sag-to-span ratio is small, this curve closely approximates a parabolic shape

approximates a parabolic shape

The cable supports a girder which weighs 12kN/m. Determine the tension in the cable at points A, B & C.

Example 5.2

Example 5.2

The origin of the coordinate axes is established at point B, the lowest point on the cable where slope is zero,

Assuming point C is located x’ from B we have:

Solution Solution

(1) 6

2

12kN/m 2

2 2

2 x

x F x F

F y w

H H

H

o  



(2) ' 0 . 1 6 '

6 x ² F x ²

F _H  ^H 



For point A,

Thus from eqn 2 and 1, we have:

Solution Solution

m x

x x

x x F _H x

43 . 12 '

0 900

' 60 '

)]

' 30

( ' [

0 . 1 12 6

)]

' 30

( 6 [

12

2

2 2

2























(3) 7772 .

4 0 . 154

12

4 . 154 )

43 . 12 ( 0 .

1 ²

x dx x

dy

kN F _H









At point A,

We have,

Solution Solution

A o

x A dx

dy m

x

79 . 53

366 .

1 )

57 . 17 (

7772 .

0 tan

57 . 17 )

43 . 12 30

(

57 . 17

































F kN

T _o

A

A H 261 . 4

) 79 . 53 cos(

4 . 154

cos 

 

 

At point B, x = 0

At point C, x = 12.43m

Solution Solution

F kN T

dx dy

B o B H

B o x

B

4 . 0 154

cos 4 . 154 cos

0 0

tan

0























F kN T

dx dy

H C o

x C

6 . 4 214

. 154 0

. 44

9657 .

0 ) 43 . 12 ( 7772 .

0 tan

43 . 12





















Arches Arches

• An arch acts as inverted cable so it receives An arch acts as inverted cable so it receives loading in compression

loading in compression

• Because of its rigidity, it must also resist some Because of its rigidity, it must also resist some bending and shear depending upon how it is bending and shear depending upon how it is

loaded & shaped

loaded & shaped

Arches Arches

• Depending on its uses, several types of arches can Depending on its uses, several types of arches can be selected to support a loading

be selected to support a loading

Three-Hinged Arch Three-Hinged Arch

• The third hinge is located at the crown & the The third hinge is located at the crown & the supports are located at different elevations

supports are located at different elevations

• To determine the reactions at the supports, the To determine the reactions at the supports, the arch is disassembled

arch is disassembled

The three-hinged open-spandrel arch bridge has a parabolic shape and supports the uniform load. Show that the parabolic arch is subjected only to axial compression at an intermediate

point such as point D . Assume the load is uniformly transmitted to the arch ribs.

Example 5.4

Example 5.4

•Applying the eqn of equilibrium, we have:

Solution Solution

kN C

m kN

m C M

y y

A

160

0 )

20 ( 320

) 40

( 0

ve, as

moments direction

clockwise -

anti With Arch : Entire

   





Solution Solution

0

0 160

160

0

160 0

160 ( 10 ) 160 ( 20 ) ( 10 ) 0 160 0

ve, as

moments direction

clockwise -

anti

With segment BC : Arch

  

  















   

 





y y

y

x x

x

x B

B

kN kN

B

F

kN B

F

kN

C M kN m kN m C m

A section of the arch taken through point D

Solution Solution

5 . ) 0

20 ( tan 20

: is D at segment the

of slope The

5 . 2 )

20 /(

) 10 ( 10 10

10 2

2 2



 















 m x

dx x dy

m y

m x



Solution Solution

0 )

5 . 2 ( 160

) 5 (

80 0 - clockwise moments as ve : anti

With

0 6

. 26 cos 6

. 26 sin 80

0

0 6

. 26 sin 6

. 26 cos 160

0

: have we

5.10(d), Fig

, equlibrium of

eqn the

Applying





 

 





























m kN

m kN

M M

V N

kN F

V N

kN F

D D

D o D o

y

D o D o

x

0 0

9 . 178













D D

D

M V

kN

N

The three-hinged trussed arch supports the symmetric loading.

Determine the required height of the joints B and D , so that the arch takes a funicular shape. Member HG is intended to carry no force.

Example 5.6

Example 5.6

For a symmetric loading, the funicular shape for the arch must be parabolic as indicated by the dashed line. Here we must find the eqn which fits this shape.

With the x, y axes having an origin at C, the eqn is of the form of y = -cx ² . To obtain the constant c, we require:

Solution Solution

m m

m h

m m

m y

m c

m c m

D

375 . 3 125

. 1 5

. 4

5.12(a) Fig

From

125 . 1 )

3 )(

/ 125 . 0 ( Therefore,

/ 125 . 0

) 6 ( )

5 . 4 (

1

2 2





















