cable and arches 4

Module 5

Cables and Arches Version 2 CE IIT, Kharagpur

Lesson 34

Symmetrical Hingeless Arch

Version 2 CE IIT, Kharagpur

Instructional Objectives: After reading this chapter the student will be able to 1. Analyse hingeless arch by the method of least work. 2. Analyse the fixed-fixed arch by the elastic-centre method. 3. Compute reactions and stresses in hingeless arch due to temperature

change. 34.1 Introduction As stated in the previous lesson, two-hinged and three-hinged arches are commonly used in practice. The deflection and the moment at the center of the hingeless arch are somewhat smaller than that of the two-hinged arch. However, the hingeless arch has to be designed for support moment. A hingeless arch (fixed–fixed arch) is a statically redundant structure having three redundant reactions. In the case of fixed–fixed arch there are six reaction components; three at each fixed end. Apart from three equilibrium equations three more equations are required to calculate bending moment, shear force and horizontal thrust at any cross section of the arch. These three extra equations may be set up from the geometry deformation of the arch. 34.2 Analysis of Symmetrical Hingeless Arch

Consider a symmetrical arch of span L and central rise of Let the loading on the arch is also symmetrical as shown in Fig 34.1. Consider reaction components

ch


at the left support A i.e., bending moment , vertical reaction and horizontal thrust as redundants.

aM ayR

aH Considering only the strain energy due to axial compression and bending, the strain energy U of the arch may be written as

∫∫ +=ss

EAdsN

EIdsMU

0

2

0

2

22 (34.1)

where M and are respectively the bending moment and axial force of the arch rib. Since the support

NA is fixed, one could write following three equations at

that point.

0=∂∂

aMU (34.2a)

0=∂∂

aHU (34.2b)

0=∂∂

ayRU (34.2c)

Knowing dimensions of the arch and loading, using the above three equations, the unknown redundant reactions and may be evaluated. aa HM , ayRSince the arch and the loading are symmetrical, the shear force at the crown is zero. Hence, at the crown we have only two unknowns. Hence, if we take the internal forces at the crown as the redundant, the problem gets simplified.


Hence, consider bending moment and the axial force at the crown as the redundant. Since the arch and the loading is symmetrical, we can write from the principle of least work

cM cN

0=∂∂

cMU (34.3a)

0=∂∂

cNU (34.3b)

000

=∂∂

+∂∂

=∂∂

∫∫s

c

s

cc

dsMN

EANds

MM

EIM

MU (34.4a)

000

=∂∂

+∂∂

=∂∂

∫∫s

c

s

cc

dsNN

EANds

NM

EIM

NU (34.4b)

Where, is the length of centerline of the arch,s I is the moment of inertia of the cross section and A is the area of the cross section of the arch. Let and be the bending moment and the axial force at any cross section due to external loading. Now the bending moment and the axial force at any section is given by

0M 0N


0MyNMM cc ++= (34.5a)

0cos NNN c += θ (34.5b)

1=∂∂

cMM ; y

NM

c

=∂∂ ; θcos=

∂∂

cNN ; 0=

∂∂

cMN . (34.6)

Equation (34.4a) and (34.4b) may be simplified as,

0)0()1(00

=+ ∫∫ss

dsEANds

EIM

0

0 0 0

s s s

c cMds y dsM N ds

EI EI EI+ = −∫ ∫ ∫ (34.7a)

0cos00

=+ ∫∫ss

dsEANyds

EIM θ

2

2 0 0

0 0 0 0 0

cos coss s s s s

c c cM y N y N M y Nds ds ds ds dsEI EI EA EI EA

θ θ+ + = − −∫ ∫ ∫ ∫ ∫ (34.7b)

From equations 34.7a and 34.7b, the redundant and may be calculated provided arch geometry and loading are defined. If the loading is unsymmetrical or the arch is unsymmetrical, then the problem becomes more complex. For such problems either column analogy or elastic center method must be adopted. However, one could still get the answer from the method of least work with little more effort.

cM cN


Consider an unloaded fixed-fixed arch of span L . The rise in temperature, would introduce a horizontal thrust and a moment at the supports. Now due to rise in temperature, the moment at any cross-section of the arch

tH tM

tHMM tt −= (34.8)

Now strain energy stored in the arch

∫=s

EIdsMU

0

2

2

Now applying the Castigliano’s first theorem,

0

s

t t

U ML T dsH EI

α∂= =

∂ ∂∫MH∂

2

0 0

s st

tM y yLT ds H dsEI E

α == −∫ ∫ I (34.9)

Also,

dsMM

EIM

MU s

tt∫ ∂

∂==

∂∂

0

0

0)(

0

=−

∫s

tt dsEI

yHM

∫ ∫ =−s s

tt EIydsH

EIdsM

0 0

0 (34.10)


34.3 Temperature stresses

Solving equations 34.9 and 34.10, and may be calculated. tM tH Example 34.1 A semicircular fixed-fixed arch of constant cross section is subjected to symmetrical concentrated load as shown in Fig 34.4. Determine the reactions of the arch.


Solution: Since, the arch is symmetrical and the loading is also symmetrical,

kN40== byay RR (1) Now the strain energy of the arch is given by,


∫∫ +=ss

EAdsN

EIdsMU

0

2

0

2

22 (2)

Let us choose and as redundants. Then we have, aH aM

0=∂∂

aMU and 0=

∂∂

aHU (3)

The bending moment at any cross section is given by,

Daaay yHMxRM θθ ≤≤−−= 0 (4)

2/)10(40 πθθ ≤≤−−−−= Daaay xyHMxRM

θθ cos)90cos( aa RHN +−=

Daa RHN θθθθ ≤≤+= 0cossin (5)

sin ( 40)cosa aN H Rθ θ= + − / 2θ θ π≤ ≤ (6)

θsinRy =

)cos1( θ−= Rx

And θRdds = See Fig 34.5.

0)0()1(00

=+−=∂∂

∫∫ss

a

dsEANds

EIM

MU

00

=∫s

dsEIM Since the arch is symmetrical, integration need to be carried out

between limits 2/0 πto and the result is multiplied by two.

022/

0

=∫π

dsEIM

0]10)cos1([40sin)cos1(402/

552.2/

2/

0

2/

0

2/

0

=−−−−−− ∫∫∫∫ θθθθθθθπ

π

πππ

RdRRdRHRdMRdR aa


092.135304.41571.18310.22 222 =+−−− RRRHRMR aa

092.13556.16915571.1477.342 =+−−− aa HM

0837.30815571.1 =−+ aa HM

0)(sin)(00

=+−=∂∂

∫∫ss

a

dsEANdsy

EIM

HU θ

0cos40)(sin1)(sin)cossin(

]]}10)cos1([40){[sin(1)}sin()]cos1(40){[sin(1

2/

552.2/

2/

0

2/

552.2/

2/

0

=−+

+−−−−−−−−

∫∫

∫∫π

π

π

π

π

π

θθθθθθθ

θθθθθθθ

RdEA

RdEA

RH

RdRREI

RdRHMRREI

aa

aa

∫

∫

=−−−

+−++++−

2/

552.2/

233

22/

0

23233

0}cossin40sin400cossin40sin40{

}cossin)(

sinsinsincossin40sin40{

π

π

π

θθθθθθθ

θθθ

θθθθθθ

dEA

REI

REIR

EIR

dEA

RREA

RHEI

RHEI

RMEIR

EIR ayaaa

0)0555.0(40)333.0(400)0554.0(40)333.0(40

)21(40)785.0()785.0()1()

21(40)1(40

2

22

=−−−

+−++++−

ARRIII

ARARH

IH

IRM

IIaaa

0258.23266 =++− aa MH (7)

Solving equations (6) and (7), and are evaluated. Thus, aH aM

kN 28.28=aH kN 42.466−=aM (8)


Equations (34.7a) and (34.7b) are quite difficult to solve. However, they can be further simplified if the origin of co-ordinates is moved from to in Fig. 34.3. The distance d is chosen such that

C O1 ( )y y d= − satisfies the following condition.

( ) 0

00

1 =−

= ∫∫ss

dsEI

dydsEIy

(34.10a)

Solving which, the distance d may be computed as

∫

∫= s

s

EIds

dsEIy

d

0

0 34.10b)

The point is known as the elastic centre of the arch. Now equation (34.7a) can be written with respect to new origin . Towards this, substitute

OO dyy += 1 in

equation (34.7a).


34.4 Elastic centre method

∫∫∫ −=+

+ss

c

s

c dsEIM

dsEI

dyN

EIdsM

0

0

0

1

0

)( (34.11)

In the above equation, ∫s

dsEIy

0

1 is zero. Hence the above equation is rewritten as

∫

∫−=+ s

s

cc

EIds

dsEIM

dNM

0

0

0

(34.12)

Now, is the moment )( dNM cc + 0

~M at (see Fig. 34.3). Similarly the equation (34.7b) is also simplified. Thus we obtain,

O

∫

∫−=+= s

s

cc

EIds

dsEIM

dNMM

0

0

0

0~

(34.13)

and,

∫ ∫

∫ ∫

+

+−==

s s

s s

c

dsEA

dsEIy

dsEA

Nds

EIyM

NH

0 0

21

20 0

010

0cos

cos~

θ

θ

(34.14)

34.4.1Temperature stresses Consider a symmetrical hinge less arch of span L , subjected to a temperature rise of . Let elastic centre O be the origin of co-ordinates and be the redundants. The magnitude of horizontal force be such as to counteract the

increase in the span

CT ° 00~,~ MH

0~H

2LTα

due to rise in temperature Also from Symmetry,

there must not be any rotation at the crown. Hence,

.T

∫ =∂∂

==∂∂ s

OOds

MM

EIM

MU

00~0 (34.15)

2~~~00

TLds

HN

EANds

HM

EIM

HU s

O

s

OO

α∫∫ =

∂∂

+∂∂

=∂∂ (34.16)


Moment at any section is calculated by,

yHMM OO~~ +=

θcos~

OHN =

0~

0

=∫s

O dsEIM

0~=OM (34.17)

and

2cos

cos~~

0 01

1 LTdsEA

Hdsy

EIyHs s

OO αθθ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛∫ ∫

Simplifying the above equation,

∫ ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛=

s sO

dsEA

dsEIy

LT

H

0 0

21

2 cos2~

θ

α

(34.18)

Using equation (34.18), the horizontal thrust due to uniform temperature rise in the arch can be easily calculated provided the dimensions of the arch are known. Usually the area of the cross section and moment of inertia of the arch vary along the arch axis.

OH~

Example 30.2 A symmetrical hinge less circular arch of constant cross section is subjected to a uniformly distributed load of . The arch dimensions are shown in Fig. 34.7a. Calculate the horizontal thrust and moment at

kN/m10A .


The distance of the elastic centre from the crown is calculated by equation, d C


∫

∫= s

s

EIds

dsEIy

d

0

0 (1)

From Fig.34.7b, the ordinate at , is given by d

( )θcos150 −=y

( )

∫

∫−

= 6/

0

6/

0

50

50cos150

π

π

θ

θθ

EId

dEI

d

150

6 2 2.2535 m.

6

d

π

π

⎛ ⎞−⎜ ⎟⎝ ⎠= = (2)

The elastic centre O lies at a distance of from the crown. The moment at the elastic centre may be calculated by equation (34.12). Now the bending moment at any section of the arch due to applied loading at a distance

m2535.2

x from elastic centre is

∫

∫−= s

s

O

EIds

dsEIx

M

0

0

25

~ (3)

In the present case, θsin50=x and θdds 50= , constant=EI

/ 63 2

0/ 6

0

5 50 sin

50O

dM

d

π

π

θ θ

θ

− ×=

∫

∫%

2

1 sin6 2 35 50 1081.29 kN.m

26

C CM N d

π π

π

⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟× ⎝ ⎠⎝ ⎠+ = − = − (4)


10 cos2OLN x θ⎛ ⎞= −⎜ ⎟

⎝ ⎠

And.

1y y d= −


( )1 50 1 cos 2.25y θ= − − 1 47.75 50cosy θ= −

Now OH~ is given by equation (34.14). Thus

∫ ∫

∫ ∫

+

+

−==s s

s s

c

dsEA

dsEIy

dsEA

Nds

EIyM

NH

0 0

21

20 0

010

0cos

cos~

θ

θ

(5)

( )/ 6

20 1

0 0

1 5 47.75 50cos 50s M y ds x d

EI EI

π

θ θ= −∫ ∫

( ) ( )/ 6

2

0

250 50sin 47.75 50cos dEI

π

θ θ θ= −∫

( )( )/ 6

2

0

625000 23.875 1 cos 2 50cos sin dEI

π

θ θ θ= − −∫ θ

( ) ( )/ 6

0

625000 123.875 1 cos 2 25 cos cos3 cos2

dEI

π

θ θ θ θ⎛ ⎞⎛ ⎞= − − − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫ θ

49630.735EI

= (6)

( )/ 6

20

0 0

cos 1 10 25 coss N ds x d

EA EA

πθ θ θ= −∫ ∫

/ 6

2

0

10 1 cos 225 50sin cos2

dEA

π θ θ θ θ⎛ + ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫

( ) ( )( )/ 6

0

10 12.5 1 cos 2 25 sin sin cos 2 dEA

π

θ θ θ θ= + − +∫ θ

( )( )/ 6

/ 6 / 600

0

10 1 112.5 sin 2 25 (cos ) cos3 cos2 3EA

ππ πθ θ θ θ θ

⎛ ⎞⎛ ⎞= + − − + − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠


81.795EA

= (7)

( )/ 62

21

0 0

1 47.75 50cos 50s y ds d

EI EI

π

θ θ= −∫ ∫

( )/ 6

2

0

50 2280.06 2500cos 4775cos dEI

π

θ θ θ= + −∫

50 12280.06 1250 sin 4775sin6 6 2 3EI 6π π π⎛ ⎞⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

π

105.046EI

= (8)

/ 62

0 0

cos 50 (1 cos 2 )2

s

ds dEA EA

πθ θ θ= +∫ ∫

25 1 sin 23.9156 2 3EAπ π⎛ ⎞= + =⎜ ⎟⎝ ⎠

(9)

⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ +−

−=

EAEI

EAEIH915.23046.105

795.81735.49630~

0 (10)

Consider an arch cross section of 300 500 mm× ; and 3 43.125 10 mI −= ×

20.15 mA = . Then,

( )( )0

15881835.2 545.3470.25 kN

33614.72 159.43H

− += − = −

+% (11)

In equation (5), if the second term in the numerator and the second term in the denominator were neglected then, we get,


0

49630.735

472.67 kN105.046

EIH

EI

⎛ ⎞−⎜ ⎟⎝ ⎠= − = −⎛ ⎞⎜ ⎟⎝ ⎠

% (12)

Thus calculating OH~ by neglecting second term in the numerator and denominator induces an error which is less than 0.5%. Hence for all practical purposes one could simplify the expression for OH~ as,

∫

∫−=

s

s

dsEIy

dsEI

yM

H

0

12

0

10

0~

(13)

Now we have,

1081.29C CM N d+ = −

470.25CN = −

23.22 kN.mCM = − (14)

Moment at , B2252510 ××+= CB MM

225251022.23 ××+−=

(15) 3101.78 kN.m=Also . CB NH =

Since the arch and the loading are symmetrical, BA MM = and . BA HH = Summary In this lesson, hingeless arches are considered. The analysis of hingeless arch by the method of least work is given in the beginning. This is followed by the analysis of hingeless arch by the elastic centre method. The procedure to compute stresses developed in the hingeless arch due to temperature change is discussed. A few problems are solved illustrate the various issues involved in the analysis of hingeless arches.


cable and arches 4

Documents