theory of structures chapter 5 : three pin...

by Saffuan Wan Ahmad

THEORY OF STRUCTURESCHAPTER 5 : THREE PIN

ARCH

bySaffuan Wan Ahmad

Faculty of Civil Engineering & Earth [email protected]

For updated version, please click on http://ocw.ump.edu.my


Chapter 5 – Three Pin Arch

• Aims– Determine internal forces, shear forces and bending moments in arch member

• Expected Outcomes :– Able to explain the function of arch– Able to describe the function of arch– Able to determine the reaction at support for three arch structure – Able to determine the internal forces at any point at arch structure – Able to draw shear force, axial force and bending moment diagram

• References– Mechanics of Materials, R.C. Hibbeler, 7th Edition, Prentice Hall – Structural Analysis, Hibbeler, 7th Edition, Prentice Hall– Structural Analysis, SI Edition by Aslam Kassimali,Cengage Learning– Structural Analysis, Coates, Coatie and Kong – Structural Analysis - A Classical and Matrix Approach, Jack C. McCormac and James K.

Nelson, Jr., 4th Edition, John Wiley


hinge

Springingline

crown

5.1 WHAT IS ARCH??


5.1 WHAT IS ARCH??


• Arches have been used for a very long time to spanlarge distance i.e bridges/building to carrytransverse loading efficiently.

• Arch carries most of the load axially with bendingmoment greatly reduced due to the curvature ofthe arch

5.2 FUNCTIONS OF ARCH


(a) Fixed arch(a) Two-hinge arch

(a) Three-hinge arch(a) Tied arch

#Due to support system

5.3 TYPES OF ARCH


h= height of the arch

B (L,0)A (0,0)

P (x,y)C (L/2,h)

x

y

L

5.4 EQUATION OF PARABOLIC ARCH


• Shear force must be parallel to the cross section surface, whilst the axial force must be perpendicular to the shear force. The positive were shown in figure below.

The axial force, N shear force, Q and bending moment, M in the arch rib

5.5 FORCES IN ARCH


yh

x

L

yc=h

L/2 L/2

y= kx ( L- x )

When x = L/2 , y = yc sub into enq.

yc=k (L/2) ( L- L/2 ) = kL2/4

Therefore k= 4yc/L2 = 4h/L2 then sub into eqn.

y= (4h/L2 )x ( L- x )

y= (4yc/L2 ) ( Lx- x2 )

y= (4h/L2 )( Lx- x2) simplify

…………Parabolic equation


slopetan Where,

)2(4

84

)4(24Slope

44

44

)(4but

Slope

1-

2

2

2

2

2

2

2

2

2

xLL

hLhx

Lh

Lxh

Lh

dxdy

Lhx

Lhx

Lhx

LhxL

xLLhxy

dxdy

At any point of the arch (parabolic)


EXAMPLE 1

• Calculate the reaction at support A and B as shown in figure below.


Solution Example 1…

Draw FBD

kNVV

F

kNVV

M

A

A

y

B

B

A

5.31205.107150140130

0

5.1070)40()15(150)10(140)5(130

0structure wholeheConsider t

kNV

kNV

A

B

5.312

5.107


kNHHence

kNHH

M

A

B

B

C

75.268,

75.2680)8()20(5.107

0

Now, consider segment CB



EXAMPLE 2

• Determine the internal forces at the points D and E in the three hinge parabolic arch as shown in figure below.

A B

CD E

2.5 m 2.5 m 3 m 2 m

2.5 m

16 kN/m


Apply equation of equilibrium, consider the whole structure then taking moment about A and B.

Consider RHS or LHS, take moment about C is equal to zero

kNVkNV BA 20,60

kNH 40



• Consider segment CD

5.0

))5.2(210(10

)5.2(4

)2(4875.1

)5.210(10

)5.2)(5.2(4

)(4

2

2

2

2

xLLhSlope

m

xLLhxyD

kNQQ

o

003.00)57.26cos(40)57.26cos(60)57.26sin(40

direction Qin Resolving57.26

)5.0(tan 1



5.0

))5.2(210(10

)5.2(4

)2(4875.1

)5.210(10

)5.2)(5.2(4

)(4

2

2

2

2

xLLhSlope

m

xLLhxyD

kNNN

o

72.440)57.26sin(40)57.26sin(60)57.26cos(40

direction Nin Resolving57.26

)5.0(tan 1



5.0

))5.2(210(10

)5.2(4

)2(4875.1

)5.210(10

)5.2)(5.2(4

)(4

2

2

2

2

xLLhSlope

m

xLLhxyD

kNmM

M

D

D

o

25

0)875.1(40)5.2(60)25.2(40

0Mmoment Bending

57.26)5.0(tan

D

1


EXAMPLE 3

• Determine the bending moment at 25 m fromthe right hand support B and axial and shearforce at point D and E.

30 m

15 m

A B

C

10 kN/m

30 m

8 kN

5 kN

10 m 10 m

D E

VB

HB

VA

HA

20 m


• Calculate YE

m

yE

33.8

)1060(60

)10)(15(42

kNVV

F

kNVV

M

A

A

y

B

B

A

36.1070)20(10864.100

0

64.1000)33.8(5)2010)(20(10)10(8)60(

0

• Consider whole structure



• Consider RHS of the arch, taking moment at C is equal to zero.

10 m 10 m 10 m

15 m8.33 m

10 kN/m

5 kN

100.64 kN

HB

kNHH

F

kNHH

M

A

A

x

B

B

C

72.1700572.165

0

72.1650)15()33.815(5)5)(10(10)30(64.100

0


• Calculate the bending moment at 25 m from support , say point F

m

yF

58.14

)2560(60

)25)(15(42

kNmM

MM

F

F

F

45.560)5.2)(5(10)33.858.14(5

)58.14(72.165)25(64.1000


• At point D (with point load)

kNN

N

kNQ

Q

16.1970)67.33sin(8

)67.33sin(36.107)67.33cos(72.170direction Nin Resolving

03.120)67.33cos(8

)67.33cos(36.107)67.33sin(72.170direction Qin Resolving


• At point F (with point load)

kNN

N

kNQ

Q

87.1970)67.33cos(5

)67.33cos(72.165)67.33sin(64.100direction Nin Resolving

96.100)67.33sin(5

)67.33sin(72.165)67.33cos(64.100direction Qin Resolving


EXAMPLE 4

Determine the reactions at supports and bending momentunder the load :


mymr

r

rr

ll

A 1241641640

20

1

1

2

1

2

1

my

y

xLLhxy

A

A

A

12

)1040(40

)10)(16(4

)(4

2

2

Or…



• Consider the whole structure. Taking moment at A is equal zero

)(.........................240012300)12()30()20(100)5(80

0

iHVHV

M

BB

BB

A

• Consider RHS, taking moment at C is equal to zero

)...(....................100016200)16()20()10(100

0

iiHVHV

M

BB

BB

C


• Resolving by using calculator

kNHkNV

B

B

75110

kNHHF

kNVV

F

BA

x

A

A

y

750

70010080110

0

• Apply static equation


• Consider LHS, taking moment under load 80kN

mym

xLLhxy

3121515

)1540(40

)15)(16(4

)(4

'80

2

280

kNmMMM

1250)3(75)5(70

0

80

80

80


• Consider RHS, taking moment under load 100kN

m

xLLhxy

12

)1040(40

)10)(16(4

)(4

2

2100

kNmMM

M

2000)10(110)12(75

0

100

100

100


THANKS


Author Information

Mohd Arif Bin SulaimanMohd Faizal Bin Md. Jaafar

Mohammad Amirulkhairi Bin ZubirRokiah Binti Othman

Norhaiza Binti GhazaliShariza Binti Mat Aris

theory of structures chapter 5 : three pin...

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