Structure Analysis IStructure Analysis IChapter 5Chapter 5

C bl A hCables & Arches

CablesCables

• Cables are often used in engineering structure to• Cables are often used in engineering structure to support and or transmit loads from one member to

thanother

C bl bj d dCables subjected to concentrated loads

Example 1Example 1Determine the tension in cables and what is the dimension h ?

( ) ( )M A 0

43

=∑( ) ( )

kNTTT

CD

CDCD

79.60)4(8)2(35.52 5

453

=

=−−+

0)(960 3∑ θ

BCBCy

BCBCx

T

TF

TF

824

0sin8)(79.60

0cos)(79.60

54

53

=+−→=

=−→=

∑∑

θ

θ

oBC

BCT

3.32

82.4

=

=

θ

BABAo

y

BABAo

x

TF

TF

0sin3)3.32(sin82.40

0cos)3.32(cos82.40

=+−−→=

=−→=

∑∑

θ

θ

oBA

BAT

8.53

90.6

=

=

θ

( )( ) mh 74.28.53tan2 ==

Cables subjected to a Uniform Distributed Load

Analysis ProcedureAnalysis Procedure

Analysis ProcedureAnalysis Procedure

( ) ( )( ) ( )

0 cos cos 0

0 i i 0xF T T T

F T T T

θ θ θ

θ θ θ

= − + + ∆ + ∆ =

∆ ∆ ∆∑∑ ( ) ( )0

0

0 sin sin 0

0 cos sin 0

y

y

F T w x T T

xF w x T y T x

θ θ θ

θ θ

= − − ∆ + + ∆ + ∆ =

∆= ∆ ⋅ − ⋅∆ + ⋅∆ =

∑∑ 00 cos s 0

2Dividing each eq. by∆x and taking the limit as ∆x 0

d h 0 0 d T 0

y w x y xθ θ

θ→

∆ ∆ ∆

∑

and hence y 0, 0 and T 0θ∆ → ∆ → ∆ →

Analysis Procedure

( )d T θ

Analysis Procedure

( )cos0 (1)

d Tdx

θ=

( )0

sin(2)

d Tw

θ= 0 ( )

t (3)

dxdy θtan (3)ydx

θ=

Analysis Procedure

Hat x=0 T=F⇒

Analysis Procedure

( ) ( )Hat x 0 T F

Integrate eqs. 1 and 2⇒

cos (4)sin (5)

HT FT w x

θθ== 0sin (5)

dividing 4 by 5T w xθ =

0tan = w xdydx F

θ =Hdx F

Analysis Procedure2

0 0

2w x w xdy y

d F F= ⇒ =

Analysis Procedure

20

2

@ ,2

H H

H

dx F F

w Lx L y h Fh

= = =

22

2hhy xL

⇒ =

max max is at where is maximumcos

H

LFT Tθθ

=

( )22 2 2max 0

2

H HT F V F w L

L

= + = +

⎛ ⎞2

0 12Lw Lh

⎛ ⎞= + ⎜ ⎟⎝ ⎠

Example 2Example 2Determine the tension of the cable at points A, B, C Assume the girder weight is 850 lb/ft

Example 3Example 3Determine the tension of the cable at points A, B

2

HF

AV V

HF

2

20

2Hw LF

h=

⋅

AVBV

( )2500 157031.25

2 8HF = =⋅

500 30 7500 Ib=7.5k2 2A B

WLV V ×= = = =

2 2 2 2max 7.031 7.5 10.280A BT T T H V= = = + = + =

Example 4

ArchesArches

Arches TypesArches Types

Example of Fixed ArchExample of Fixed Arch

Three Hinge Arches

Example 5Determine the internal forces at Section D

∑

BM

kNB

BM

y

yA

0)20(67)8(60)6(0

67

0)28(60)10(100)40( 0

=+−−⇒=

=

=−−⇒=

∑

∑

( )

FkNB

BM

x

x

xRightC

086

0)20(67)8(60)6(0

=

=

=+⇒=

∑

∑

kNA

kNAx

93

0F86

y =

=

∑kNAy 93=

86 20242

86

86 cos20

93 sin20

86 i 2093

86 sin20

93 cos20 242

112.6

58

Example 6

Determine the internal forces at Section D

0 25A yM C kips= ⇒ =∑yF 0

25

0 25 (50) (25) 25 (25) 0yA kips

M C

=

=∑

∑ ( ) 0 25 (50) (25) 25 (25) 0

250

B xRight

x

x

M C

C kipsF

= ⇒ − − =

=

=

∑

∑25

x

xA kips=∑

yF 0

0B

=∑0

0 25

y

x

x

B

FB kips

=

=

=∑

x p

26 6oθ = 26.6θ

22

2550

y x−=

25 12 5 i 27 95N θ θ+( )

( )

2

1

25 250

25tan 2 25 0.5

dy xdx

dy−

−=

−= = × = −

25cos 12.5sin 27.9525 12.5cos 0.0

D

D

NV sin

θ θθ θ

= + == − =

( )225

tan 2 25 0.550

26.6x

o

dx

θ=

×

=25(6.25) 12.5(12.5) 0DM = − + =

Example 7

Problem 1Problem 1Determine the internal forces at B

Problem 2Determine the Tension at A & B