Chap 6 (1)

7/23/2019 Chap 6 (1)

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CHAPTER 6

THREE PINNED ARCH

EN MOHAMMAD AMIRULKHAIRIBIN ZUBIR

BAA2113THEORY OF STRUCTURE

SESSION 2013/2014 SEMESTER I

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Learning outcomes.

At the e! "# the $e%%"& %t'!et%h"'$! (e )($e t"*

E+,$)- the #'.t-" "# ).h

De%.-(e the #'.t-" "# ).h Dete-e the e).t-" )t %',,"t

#" thee ).h %t'.t'e

Dete-e the -te)$ #".e% )t ),"-t )t ).h %t'.t'e

D) %he) #".e& )+-)$ #".e )!(e!- "et !-))

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What is the arch ?

•Sh),e•F'.t-"•T,e%

A.he% .) (e ,"$")$& .-.'$)& ,))("$-.&e$-,t-.)$ )! ) "the .'e! %h),e

A.he% )e %h),e! t" t)5e $")! )("e the)! !ee$",e! "$ .",e%%-"

A 3 h-e ).h '%')$ h)% h-e% )t 2 %,--)! )t the ." -t -% %t)t-.)$$ !ete-)te

2 h-e! ).he% )e "e ,).t-.)$ ('t -t -%

%t)t-.)$$ -!ete-)te

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hinge

Springingline

crown

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FUNCTIONS OF ARC Arches ha!e "een use# $or a !er% long

time to span large #istance i.e

"ri#ges&"uil#ing to carr% trans!erseloa#ing e'cientl%.

Arch carries most o$ the loa# a(iall% with"en#ing moment greatl% re#uce# #ue tothe cur!ature o$ the arch

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T)*+S OF ARC

(a) Fixed arch

(a) Two-hinge arch

(a) Three-hinge arch

(a) Tied arch

,ue to supports%stem

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T)*+S OF ARC

(a) parabolic arch (a) Triangular arch

,ue to itsshape

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+-uation o$ para"olic arch

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y h

x

L

y c=h

L/2 L/2

% /( 0 L1 (2When ( L&3 4 % %c su" intoen-.%c/ 0L&32 0 L1 L&3 2 /L3&5

There$ore / 5%c&L3 5h&L3then su" into e-n.

% 05h&L3 2( 0 L1 ( 2

% 05%c&L3 2 0 L(1 (3 2

% 05h&L3 20 L(1 (32

simpli$%

6666*ara"olic e-uation

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slopetan Where,

)2(4

84

)4(24Slope

44

44

)(4 but

Slope

1-

2

2

2

2

2

2

2

2

2

=

−=

−=

−==

−=

−=

−=

=

θ

x L L

h

L

hx

L

h

L

xh

L

h

dx

dy

Lhx

Lhx

L

hx

L

hxL

x L Lhx y

dx

dy

At any point of the arch (parabolic)

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+(ample 7

Calculate the reaction at support Aan# 8 as shown in 9gure "elow.

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Solution6

,raw F8,

kN V

V

F

kN V

V

M

A

A

y

B

B

A

5.312

5.1!151413

5.1!

)4()15(15)1(14)5(13

structure"holehe#onsi$er t

=

=+−−−

=

=

=−++

=

∑

∑

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kN H Hence

kN H H

M

A

B

B

C

!5.2%8,

!5.2%8)8()2(5.1!

=

==+

=∑

Now4 consi#er segment

C8

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The a(ial $orce4 N shear $orce4 : an#"en#ing moment4 ; in the arch ri"

Shear $orce must "e parallel to the crosssection sur$ace4 whilst the a(ial $orce must"e perpen#icular to the shear $orce. Thepositi!e were shown in 9gure "elow.

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+(ample 3

,etermine the internal $orces at thepoints , an# + in the three hingepara"olic arch as shown in 9gure"elow.

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Solution6

Appl% e-uation o$ e-uili"rium4 consi#erthe whole structure then ta/ingmoment a"out A an# 8.

Consi#er RS or LS4 ta/e moment a"outC is e-ual to =ero

kN V kN V B A 2,% ==

kN H 4=

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Consi#er segment C,

5.

))5.2(21(1

)5.2(4

)2(4

8!5.1

)5.21(1

)5.2)(5.2(4

)(4

2

2

2

2

=

−=

−=

=

−=

−=

x L L

hSlope

m

x L

L

hx y D

kN Q

Q

o

3.

)5!.2%cos(4)5!.2%cos(%)5!.2%sin(4

$irection&in'esolin5!.2%

)5.(tan 1

−=

=+−+

=

= −

θ

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Consi#er segment C,

5.

))5.2(21(1

)5.2(4

)2(4

8!5.1

)5.21(1

)5.2)(5.2(4

)(4

2

2

2

2

=

−=

−=

=

−=

−=

x L L

hSlope

m

x L

L

hx y D

kN N

N

o

!2.44

)5!.2%sin(4)5!.2%sin(%)5!.2%cos(4

$irection *in'esolin

5!.2%

)5.(tan 1

=

=+−−

=

= −

θ

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Consi#er segment C,

5.

))5.2(21(1

)5.2(4

)2(4

8!5.1

)5.21(1

)5.2)(5.2(4

)(4

2

2

2

2

=

−=

−=

=

−=

−=

x L L

hSlope

m

x L

L

hx y D

kNm M

M

D

D

o

25

)8!5.1(4)5.2(%)2

5.2(4

+

,o,ent-en$in

5!.2%

)5.(tan

1

=

=−+−−

=

=

=

∑

−θ

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+(ample >

,etermine the "en#ing moment at 3m $rom the right han# support 8 an#a(ial an# shear $orce at point , an# +.

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Solution6

Calculate )+

m

y E

33.8

)1%(%

)1)(15(42

=

−=

kN V

V

F

kN V

V

M

A

A

y

B

B

A

3%.1!

)2(18%4.1

%4.1

)33.8(5)21)(2(1)1(8)%(

=

=−−+

=

=

=−+++−

=

∑

∑

• Consi#er whole structure

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Consi#er RS o$ the arch4 ta/ing moment at Cis e-ual to =ero.

kN H

H

F

kN H

H

M

A

A

x

B

B

C

!2.1!

5!2.1%5

!2.1%5

)15()33.815(5)5)(1(1)3(%4.1

=

=−−

=

=

=+−++−

=

∑

∑

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Calculate the "en#ing moment at 3 m $romsupport 4 sa% point F

m

y F

58.14

)25%(%

)25)(15(42

=

−=

kNm M

M

M

F

F

F

45.5%

)5.2)(5(1)33.858.14(5

)58.14(!2.1%5)25(%4.1

−=

=+−+

+−

=∑

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At point , 0with point loa#2

kN N

N

kN Q

Q

1%.1/!

)%!.33sin(8

)%!.33sin(3%.1!)%!.33cos(!2.1!


3.12

)%!.33cos(8

)%!.33cos(3%.1!)%!.33sin(!2.1!

$irection&in'esolin

=

=+

−−

−=

=+

−+

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At point F 0with point loa#2

kN N

N

kN Q

Q

8!.1/!)%!.33cos(5

)%!.33cos(!2.1%5)%!.33sin(%4.1


/%.1

)%!.33sin(5

)%!.33sin(!2.1%5)%!.33cos(%4.1

$irection&in'esolin

==−

−−

=

=−

−+

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+(ample 5

,etermine the reactions at supports an#"en#ing moment un#er the loa#

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Solution6

m ymr

r

r

r

l

l

A 1241%4

1%4

2

1

1

2

1

2

1

=−==∴

=

=

m y

y

x L

L

hx y

A

A

A

12

)14(4

)1)(1%(4

)(4

2

2

=

−=

−=

O7

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Consi#er the whole structure. Ta/ing momentat A is e-ual =ero

)(.........................24123

)12()3()2(1)5(8

i H V

H V M

B B

B B

A

=−

=+−+=∑

• Consi#er RS4 ta/ing moment at C is e-ual to=ero

)...(....................11%2

)1%()2()1(1

ii H V

H V M

B B

B B

C

=−

=+−=∑

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Resol!ing "% using calculator

kN H

kN V

B

B

!5

11

=

=

kN H H

F

kN V

V

F

B A

x

A

A

y

!5

!

1811

==

=

=

=−−+

=

∑

∑

• Appl% static e-uation

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Consi#er LS4 ta/ing moment un#er loa#@/N

m y

m

x L L

hx y

31215

15

)154(4

)15)(1%(4

)(4

0

8

2

28

=−=

=

−=

−=

kNm M

M M

125

)3(!5)5(!

8

8

8

=

=−+−

=

∑

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Consi#er RS4 ta/ing moment un#er loa#7/N

m

x L L

hx y

12

)14(4

)1)(1%(4

)(4

2

21

=

−=

−=

kNm M

M

M

2

)1(11)12(!5

1

1

1

=

=−+

=∑

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+(ample

,raw shear $orce4 "en#ing moment an# a(ial$orce #iagram $or the para"olic arch shown in9gure "elow.

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Solution6

P"-t

A+-)$ F".e&

N

85N9

She)

F".e& :

85N9

Be!-

M"et

85N9A 7B.>3 37.7

,

D.3B3

0without

loa#2

5D.B>

0without

loa#2 >D.

BB.D

0With loa#2

15D.B

0With loa#2C [email protected] 13

+ @3.3D 1.B 1>73.

8 D.3D@ 37.B

Summar%

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AF,

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SF,

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8;,

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Chap 6 (1)