27 Shape of Arches

8/9/2019 27 Shape of Arches

1/20

Curved beams

1. Introduction

Curved beams also called arches were invented about 2000 years ago. The purpose was to form

such a structure that would transfer loads, mainly the dead weight, to the ground by the elementsworking mostly or only in the state of compression. The reason for this was that the mainconstruction material in those times was the natural stone. It has relatively large compressivestrength and its tensile strength is about 10 times smaller. Hence, it was vital to avoid tension andbending involving tension.One may argue if the invention of arches was an engineering achievement or was it only a cleverimitation of nature, which uses arches and their 3D version concave shells successfully for amuch longer time. Just think of thin walled eggs, shells, turtles, etcHence, arches are perfect structures. Under a distributed loading they should exhibit mostly axialforces with very little bending and shear. The presence of concentrated forces disturbs this picture,what will be seen in the examples.

2. Statically determinate circular arches

Let us start with statically determinate circular arches. The internal forces: bending moments, axialforces and shear forces, can be determined using equilibrium equations.For the beginning we will consider a cantilever circular arch being one quarter of a circle with theradius R, loaded by the concentrated force Pat the tip. The arch is described in the Cartesian setof co-ordinates x,ywith the origin at the centre of the circle.

However, the calculations are easier, when carried out using the polar set of co-ordinates ,R. Therelations between the co-ordinates are:

sin

cos

Ry

Rx

=

=

To find functions of internal forces we introduce a cross-section at the point defined by the angle and write down the equilibrium equations for the fragment of arch to the left of the cross-section:

( ) 0:

0cos:

0sin:

=

=

=+

yRPMM

PTT

PNN

From these equations the functions of the internal forces can be found:

( )

sin1

cos

sin

=

=

=

PRM

PT

PN

P

R

x

yP

R

x

y

NT

M

y

x

NT

P

Ry


2/20

In the third equation the relation between yand was substituted. These functions can berepresented graphically. In the graphs values of the forces are presented for points spaced by

angles /12. Note, that = 0 corresponds to the support and = /2 to the cantilever tip.

As the next example let us consider a cantilever circular arch being one quarter of the circle withthe radius R, loaded by the uniformly distributed snow-type loading of the intensity q.

To find the function of internal forces we introduce a cross-section at the point defined by the angleand write down the equilibrium equations for the fragment of arch to the left of the cross-section.

In this case the part of loading acting on the considered fragment of the arch can be replaced by itsresultant, which is applied at the centre of length x. Thus, the equilibrium equations take the form:

02

:

0sin:

0cos:

=+

=+

=+

xqxMM

qxTT

qxNN

From these equations the functions of the internal forces can be found:

22

2

cos2

cossin

cos

qRM

qRT

qRN

=

=

=

In these equations the relation between xand was substituted. These functions can berepresented graphically. In the graphs values of the forces are presented for points spaced by

angles of /12.

0.0

T

q

R

x

y

R

x

y

NT

My

x

N

qx

qx x/2

0.258

0.0

0.500

0.7070.866

0.9661.0

N[P]

1.0

+

0.966

0.866

0.707

0.5000.258

T[P]0.741

0.500

0.2930.134

0.0340.0

M [PR]

1.0

0.933

1.0

0.750

0.500

0.2500.067

0.0

N[qR]

0.0

0.0

0.250

0.433

0.500

0.433

0.250

T[qR]0.467

0.5

0.375

0.2500.125

0.033

0.0

M[qR2]


3/20

Now let us consider a cantilever circular arch being one quarter of the circle with the radius R,loaded by the uniformly distributed self weight-type loading of the intensity q.

Again, to find the function of internal forces we introduce a cross-section at the point defined by the

angle and write down the equilibrium equations for the fragment of arch to the left of the cross-section. In this case the resultant of the fragment of loading under consideration can be found onlyby integration. The same concerns the bending moment, which can be obtained by the integration

of moments due to elementary loads along the considered fragment of the arch. In order to carryout this integration an auxiliary angular co-ordinate is introduced. It will undergo integration in the

limits from to /2. The elementary load resultant is equal to qds, where sis the curvilinear co-ordinate measured along the arch. It can be related to the angle via

Rdds=

and the length ais given as

cosRa=

With this in hand the equilibrium equations can be given as:

( ) 0:

0sin:

0cos:

2/

2/

2/

=+

=+

=+

qdsaxMM

qdsTT

qdsNN

Expressing all variables under integrals in terms of angles and noting, that the angle does notundergo integration, we express the internal forces as:

[ ]

[ ]

( ) [ ]

+=

===

===

===

sincos1cos2

sincoscoscos

sin2

sinsin

cos2

coscos

2

2/22/

2

2/2/

2/2/

qR

qRdqRM

qRqRdqRT

qRqRdqRN

These functions can be represented graphically. In the graphs values of the forces are presented

for points spaced by angles of /12.

NT

qds

q

R

x

y

x

y

NT

M

x

qds

a

xa

d


4/20

The next type of loading to be considered is the uniformly distributed hydraulic pressure of theintensity q. First, let it be applied to a cantilever circular arch being one quarter of the circle with theradius R.

Again, to find the function of internal forces we introduce a cross-section at the point defined by the

angle and write down the equilibrium equations for the fragment of arch to the left of the cross-section. In this case the resultant of the fragment of loading under consideration can be found only

by integration. The same concerns the bending moment, which can be obtained by the integrationof the moments due to elementary loads along the considered fragment of the arch. In order to

carry out this integration an auxiliary angular co-ordinate is introduced. This will undergo

integration in the limits from to /2. The elementary load resultant is equal to qds, where sis theco-ordinate measured along the arch. This load now has a varying direction, so it should be

projected on axes xand y. These projections are: qdscosand qdssin, respectively. The value bis given by

sinRb=

With this in hand the equilibrium equations can be written down as:

( )

( )

( ) ( )[ ] 0cossin:

0coscossinsin:

0sincoscossin:

2/

2/

2/

=++

=++

=+

ybqdsaxqdsMM

qdsqdsTT

qdsqdsNN

Expressing all variables under integrals in terms of angles and noting, that the angle does notundergo integration, we express the internal forces as:

qdscos

T

qdssin

bM

q

R

x

y

x

y

NT

x

N

qds

a

xa

dy

qdscos

qdssin

by

1.264

1.571

0.907

0.555

0.2620.068

0.0

N[qR]

0.0

0.0

0.339

0.524

0.555

0.453

0.253

T[qR] 0.523

0.571

0.407

0.2620.128

0.0340.0

M[qR2]


5/20

[ ] [ ]

( ) ( )1sinsinsincos

sinsincoscoscossin,sincos

22

2/2/2/2/

=+=

=+=+=

qRqR

qRqRdqRdqRN

[ ] [ ]

( )

( )

[ ] [ ]

( ) ( )1sinsinsincos

sinsincoscoscossinsincos

sincossincoscossincossin

coscossincoscossin

sincoscossincoscos,sinsin

2222

2/22/22/

22/

2

2/2

2/2/2/2/

=+=

=+=+=

=+=

=+=

===

qRqR

qRqRdqRdqR

dqRM

qRqR

qRqRdqRdqRT



Now let us solve an arch consisting of one quarter circle of radius Rwith a clamped support andanother quarter circle attached to the former by a hinge and supported by a roller. The arch isloaded by the uniformly distributed hydraulic pressure of the intensity q.

The first thing to calculate is the reaction in the roller support V. This can be done from theequilibrium of moments with respect to the hinge A for the right half of the arch. This equationreads:

( )[ ] 0cossin:

2/

0

A =+

qdsbRaqdsVRM

V

qdscosbq

R

x

y

x

y

a

qds

x

d

qdssin

A

V

A

0.741

1.0

0.500

0.293

0.1340.034

0.0

N[qR]

1.0

0.0

0.966

0.866

0.707

0.500

0.258

T[qR] 0.741

1.0

0.500

0.293

0.1340.0340.0

M[qR2]


6/20

After substitution for aand band division by Rthe reaction can be calculated as:

( )[ ] [ ] qRqRdqRdqRV ===+= 2/

0

2/

0

2/

0

sincossin1coscossin

With the value of the reaction Vthe internal forces in the arch can be obtained from the equilibrium

for the fragment of arch to the right of the cross-section . The adequate equations have the form:

( )

( )

( ) ( ) ( )[ ] 0cossin:

0coscossinsinsin:

0sincoscossincos:

0

0

0

=++

=++

=+

byqdsxaqdsxRqRMM

qdsqdsqRTT

qdsqdsqRNN

Note, that these equations are also valid for the angle greater than /2, i.e. within the left-hand

half of the arch. From these equations the functions of internal forces follow:

[ ] [ ]

[ ] [ ]

( ) [ ]

( ) [ ] [ ]{ }( ) ( ) 0sincoscoscos1

sinsincoscoscos1

sincossincoscossincossincos1

0sincossincossinsin

sincoscossinsin

coscossinsinsin

sincoscoscos

sinsincoscoscos

cossinsincoscos

2222

0022

0

22

00

00

22

00

00

=++=

=+=

=+=

=++=

=++=

=++=

=+=

=+=

=+=

qRqR

qRqR

dqRqRM

qRqRqRqR

qRqRqR

dqRdqRqRT

qRqRqRqRqR

qRqRqR

dqRdqRqRN

The most important thing to observe in these results are the vanishing values of the bendingmoment and the shear force. In this way we arrive at the notion of the optimal arch shape. It wasmentioned in the introduction, that the arch was invented to carry mostly normal forces. It comesout, that for a given type of loading an arch shape can be found, which is characterized by the lackof bending. For the constant hydraulic pressure the corresponding shape is the circular arch,though the previous example with the cantilever arch might deny this statement. However, theremust be one more condition fulfilled to get the optimal arch serving in the bending-free state.

T

qdssinT

y

N

b

M

x

y

x

N

qds

aax

d

qdssin

qdscosqdscosyb

V=qR

qR


7/20

Namely, its supports at the ends must provide reactions resulting in the axial force at the archends. The cantilever arch with the free end does not fulfil this condition. Hence, there was thebending moment present.It is also worth to note, that a circular arch with a hydraulic pressure in the bending-free statefeatures the constant value of the normal force Nequal to qRthroughout the entire arch length.

As the last example in this group let us consider a cantilever arch being a quarter circle of radius R

The arch is loaded by the hydrostatic pressure due to the liquid of the specific weight presentonly on its right-hand side.

This loading acts in the same direction, normal to the arch, as the previously considered hydraulicpressure. But the difference is in the fact, that its value is not constant but varies linearly with theincreasing depth hmeasured from the liquid free surface. Thus, in the equations corresponding tothe hydraulic pressure the load qmust be replaced by

( ) ( ) sin1m1 === RbRhqh

Hence, the equilibrium equations for the cut-off segment of the arch read:

( )

( )

( ) ( )[ ] 0cossin:

0coscossinsin:

0sincoscossin:

2/

2/

2/

=++

=++

=+

ybdsqaxdsqMM

dsqdsqTT

dsqdsqNN

hh

hh

hh

Having expressed all the variables under integrals in terms of angles, introduced the notation for qh

and noted, that the angle does not undergo integration, the internal forces can be given as:

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

=

=

+=

2/3

2/3

2/

2

2/

2

2/2

2/2

coscossinsin1sinsincossin1

cossin1cossinsin1sin

cossin1sinsinsin1cos

dRdRM

dRdRT

dRdRN

These calculations involve the integrals of the functions sin2

and cossin, which can be easilyfound using the following trigonometric relations

qhdscos

T

qhdssin

bM

R

x

y

x

y

NT

x

N

qhds

a

xa

dy

qhdscos

qhdssin

by

h


8/20

2sin2

1sincos

2

12cos

2

1sin2

=

+=

Now the functions of the internal forces can be determined

( ) ( )

+

+=

=

++

+=

=+=

2cos4

1sin

4

3sin

242sin

4

1coscos

2cos4

1sinsin

2

12sin

4

1coscos

sincoscossinsinsincos

22

2/

2

2/

2

2/

2

2/

22

RR

RR

dRdRN

( ) ( )

+=

=

+

+=

==

2cos4

1sin

4

3cos

242sin

4

1cossin

2cos4

1sincos

2

12sin

4

1cossin

sincoscoscossinsinsin

22

2/

2

2/

2

2/

2

2/

22

RR

RR

dRdRT

( )( ) ( )( )

( ) ( )

( ) ( )

+

=

=

+

+=

=+

=

=+=

242sin

4

1coscos2cos

4

1sin

4

3sin

2

12sin

4

1coscos2cos

4

1sinsin

sinsincoscossinsincos

sincossincossincoscossin

coscossinsinsinsinsincoscos

33

2/

3

2/

3

2/

23

2/

23

2/

23

2/

3

2/

23

2/

3

RR

RR

dRdR

dRdR

dRdRM



0.1090

0.2146

0.0466

0.0152

0.00310.0002

0.0

N[R2]

0.5

0.0

0.314

0.171

0.076

0.023

0.003

T[R2] 0.1090

0.2146

0.0466

0.0152

0.00310.00020.0

M[R3]


9/20

3. Statically indeterminate circular arches

Statically indeterminate arches can be solved using the flexibility method. The shape of the archdetermines only a technique to calculate coefficients in the canonical equations of the method.These coefficients represent displacements in basic loading states and can be obtained from theprinciple of virtual work. The adequate well known formula reads:

dsEI

MM

s

kiik =

The important issue to note is, that the integration must be carried out along the curve representingthe arch. This opens a vast field of problems and different techniques invented to calculate thecurvilinear integrals.In the case of circular arches this integration can be performed in a quite straightforward wayintroducing the polar set of co-ordinates and replacing the integral along the arch length with theone over the polar angle, as it was done in the examples of statically determinate arches tocalculate loading resultant and its moments.Let us consider a statically indeterminate arch in the form of the propped cantilever being one

quarter of the circle with the radius R, loaded by the horizontal concentrated force Pat the tip.

The modified system for the flexibility method has a removed support at the point A and theequivalent reaction at this support is considered as the redundant force X1.The canonical equation of the flexibility method resulting from the condition of zero verticaldisplacement of the point A reads

01111 =+ PX

For a circular arch the calculation of the flexibility coefficients in this equation can be carried outanalytically after the change to the polar co-ordinates. The analytical functions of the bendingmoments in the basic states: X1= 1 and Pare necessary.

The corresponding functions of the bending moments, with a sign convention attributing positivevalues to the moments setting the internal side under tension (and the external one undercompression), have the form:

P

R

x

y

PR

x

y

X1

The modified system

R

x

y

X1= 1

State X1= 1

x

P

R

x

y State P

yR y

A

B

A

B


10/20

( ) ( )

sin1

cos1

==

==

PRyRPM

RxM

P

The flexibility coefficients can now be calculated from:

( )

==

==

ss

PP

ss

dEI

PRds

EI

MM

dEI

R

dsEI

MM

sin1cos

cos

31

1

23

11

11

The integral of the function cossinwas already discussed previously and the one for the function

cos2 can be found using the trigonometric relation

2

12cos

2

1cos2 +=

With this in hand one gets:

( )

( )EI

PR

EI

PRd

EI

PR

EI

R

EI

Rd

EI

R

P2

2cos2

1sin2

22sincos2

2

42sin

2

1

212cos

2

32/

0

32/

0

3

1

32/

0

32/

0

3

11

=

+==

=

+=+=

Then the redundant force can be obtained from the canonical equation as

PX P

2

11

11 ==

The function of the bending moment in the statically indeterminate arch follows from the

superposition rule:

( ) ( )

=+=+=

cos2

sin1sin1cos2

11 PRPRPR

MXMM Pi

To check the correctness of the results a kinematic check should be performed. Let us check if thecross-section rotation at the clamped support B is zero. To this end the principle of virtual work andthe reduction theorem are used:

( ) ( )

( )?010

== dsEI

MM

s

i

B

The virtual bending moment comes from an appropriate modified system, different than the one

used in the calculations above. Hence, the fixed support B is replaced with a hinge and the virtualloading in the form of the concentrated unit bending moment is applied there.

The virtual reaction at the support A results from the equilibrium of moments with respect to B

R

x

yA

B

1=M

AV


11/20

RVRVMB

101: AA ==+

and the function of the virtual bending moment is:

( ) cos110

== xR

M

The check of the cross-section rotation follows as:

( ) ( )

01

2sin2

12cos

4

1sin

cos2

sin1cos1

2/

0

2

2/

0

20

=

+=

=

==

EI

PR

dEI

PRds

EI

MM

s

i

B

This proves the correctness of the calculations by the flexibility method!

As the next example let us consider a statically indeterminate arch of a parabolic shape. The archcentral axis is expressed in the co-ordinates system x,yby a function

( )xLxL

fy =

2

4

what for the data given in the figure yields

xxy3

4

9

1 2+=

The arch consists of two segments connected by a hinge. Externally there are six reactions at theclamped supports A and B versus four equilibrium equations, the fourth is the equation of momentswith respect to the hinge C. Hence, the system has two redundant forces. The modified systemcan be formed by the removal of the hinge C and then the horizontal and vertical internal forcestherein are considered as the redundant forces X1and X2.

y

x

30 kN

6 kN/m

A B

C

DE f= 4

L= 12

3 3 3 3[m]

y

x

30 kN

6 kN/m

A B

DE 4

3 3 3 3

CX1X1

X2

X2


12/20

The kinematic compatibility conditions require, that the relative displacements of arch fragmentstips at the point C measured in the horizontal and vertical directions are zero. This leads to thecanonical equations of the flexibility method in the form

=++

=++

0

0

2222121

1212111

P

P

XX

XX

and the flexibility coefficients are calculated from the curvilinear integrals

dsEI

MM

s

kiik =

In the case of the parabolic arch the transformation to the polar co-ordinates is not efficient. Theintegrals would be too complex.Instead, first the integrals are transformed to the straight-line ones. From the figure

the following relation can be deduced

cos

dxds=

where is the slope of the tangent to the arch. Now the flexibility coefficients can be obtained from

dx

EI

MM

x

kiik =

cos

It must be noted, that the angle is variable along the arch, depending on the co-ordinate x. It canbe obtained from the first derivative of the arch centre line function.

dx

dy=tan

In our example

+=

3

4

9

2arctan x

Substitution of this relation to the formula for the coefficients ikleads to very complex integrals,which cannot be solved analytically. The only reasonable way to solve them is the numericalintegration. There are several methods available: rectangles, trapezium, Simpson, the family ofGauss quadratures, etc. Some of these methods are based on a geometrical approach, where theentire area below the function graph is approximated by a finite number of basic areas. Here wewill apply the trapezium method illustrated in the figure

ds

dx

x

y

a b

f(x)

A1 A2 An

f1

f2f0

fn1 fn

x x x


13/20

The function domain

bax ,

is divided into a finite number of nequal intervals xand the determinate integral is represented asa sum of trapeziums areas

( ) nb

a

AAAdxxf +++= ...21

After a substitution of the formulae for the trapezium area the integral can be finally expressed in

terms of a finite number of function values at the ends of intervals x

( ) ( )nnb

a

fffffx

dxxf +++++

= 1210 2...222

The method is approximate and its accuracy depends on the density of the function domainsubdivision. The larger is the number of intervals, the more accurate results are obtained.In the considered example the integration domain is

12,0x m

and it will be subdivided into 12 intervals of x= 1.0 m.To carry out the calculations the functions of bending moments in the basic states: X1= 1, X2= 1and Pare necessary.

yM = 31

32 = xM

( )

( )

( ) ( )

=

EBon

DEon

ADon

9305.436

5.436

32

6 2

xx

x

x

MP

y

x

30 kN

6 kN/m

A B

DE 4

3 3 3 3

C

y

xA B

4

3 3 3 3

CX1=1

X1=1

33 3

1

3

9xA B

4

3 3 3 3

C

X2=1

X2=1

y

27 27 81

225


14/20

The results of calculations for the flexibility coefficients are presented in the table.

x y tan cos M1 M2 MP

cos11MM

cos

21MM cos

22MM cos

1 PMM cos

2 PMM M

(i)

m m m m kNm m2

m2 m

2 kNm

2 kNm

2 kNm

0.0 0.0 1.3333 0.6 3.0 3.0 27.0 15.0 15.0 15.0 135.0 135.0 2.391

1.0 1.2222 1.1111 0.6690 1.7778 2.0 12.0 4.7243 5.3148 5.9791 31.892 35.874 1.8942.0 2.2222 0.8889 0.7474 0.7778 1.0 3.0 0.8094 1.0407 1.3380 3.1220 4.0139 1.097

3.0 3.0 0.6667 0.8320 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.04.0 3.5556 0.4444 0.9138 0.5556 1.0 3.0 0.3378 0.6080 1.0943 1.8240 3.2830 1.397

5.0 3.8889 0.2222 0.9762 0.8889 2.0 12.0 0.8094 1.8211 4.0975 10.927 24.585 3.091

6.0 4.0 0.0 1.0 1.0 3.0 27.0 1.0 3.0 9.0 27.0 81.0 5.0857.0 3.8889 0.2222 0.9762 0.8889 4.0 45.0 0.8094 3.6423 16.390 40.976 184.39 4.3798.0 3.5556 0.4444 0.9138 0.5556 5.0 63.0 0.3378 3.0401 27.358 38.305 344.71 2.027

9.0 3.0 0.6667 0.8320 0.0 6.0 81.0 0.0 0.0 43.269 0.0 584.13 14.136

10.0 2.2222 0.8889 0.7474 0.7778 7.0 129.0 0.8094 7.2847 65.561 134.25 1208.2 1.94511.0 1.2222 1.1111 0.6690 1.7778 8.0 177.0 4.7243 21.259 95.665 470.36 2116.7 -4.546

12.0 0.0 1.3333 0.6 3.0 9.0 225.0 15.0 45.0 135.0 1125.0 3375.0 5.337

Applying the trapezium formula for the coefficients yields the following results

EI362.29

11 = ,EI077.25

12 = ,EI

75.34422 = ,

EIP

6.11501

= ,EI

P1.6127

1=

With these coefficients found the canonical equations can be solved to get the values of theredundant forces

kN598.251 =X , kN856.152 =X ,

The final values of the bending moments in the arch are obtained from the superposition rule

( )P

i MXMXMM ++= 2211

and their values are given in the last column in the table above.The correctness of these calculations should be verified by the kinematic check. Here the cross-section rotation at the clamped support A will be calculated from the principle of virtual work andthe reduction theorem as

( ) ( ) ( ) ( )

( )?0cos

100

=== dxEIMM

dsEI

MM

x

i

s

i

A

To this end another modified system is introduced with the appropriate virtual loading in the form ofthe concentrated unit moment at the point A.

Having calculated the reactions at the supports in the usual way the function of the virtual bendingmoment in the modified system can be expressed as

( )

14

1

12

10+= yxM

y

xA B

C

D

E f= 4

L= 12

3 3 3 3 [m]

1

4

1

4

1

12

1

12

1


15/20

The calculation of the integral in the kinematic check is also carried out using the trapeziummethod and the results are given in the following table

x y M(i) ( )0M cos

cos

)0()( MM i

m m kNm kNm

0.0 0.0 2.391 1.0 0.6 3.9851.0 1.2222 1.894 0.6111 0.6690 1.7302.0 2.2222 1.097 0.2778 0.7474 0.408

3.0 3.0 0.0 0.0 0.8320 0.04.0 3.5556 1.397 0.2222 0.9138 0.340

5.0 3.8889 3.091 0.3889 0.9762 1.2316.0 4.0 5.085 0.50 1.0 2.543

7.0 3.8889 4.379 0.5556 0.9762 2.492

8.0 3.5556 2.027 0.5556 0.9138 1.2329.0 3.0 14.136 0.50 0.8320 8.495

10.0 2.2222 1.945 0.3889 0.7474 1.012

11.0 1.2222 -4.546 0.2222 0.6690 1.51012.0 0.0 5.337 0.0 0.6 0.0

The final result of the kinematic check is EIA

016.0=

This result related to the maximum absolute value component in the integral calculation, i.e.8.495/EIis only 0.19%, what allows to conclude, that the angle Ais indeed zero and that thecalculated bending moments in the statically indeterminate arch are correct.To complete the calculations let us also find the functions of shear and axial forces in the arch.

The values of reactions in supports A and B can be obtained from the equilibrium equations for theparts AC and CB of the arch. With these values in hand one can write down the appropriateequations of equilibrium to get the functions of internal forces. For the left half of the arch AD

the equilibrium equations of forces in the direction Nand Tyield

y

x

30 kN

6 kN/m

A B

DE 4

3 3 3 3

C25.65

15.8615.86

25.65

33.86 32.14

25.65

25.65

2.4 5.3

y

xA

33.86

25.65

2.4

N

T

M

6 kN/m


16/20

( )

cos)686.33(sin65.25

sin686.33cos65.25

xT

xN

+=

=

For the right half of the arch DB two cross-sections must be considered with the separation point E.

The equilibrium equations yield:

for ED section:

( )

( )

sin65.25cos14.3230

cos65.25sin14.3230

+=

=

T

N

for EB section:

sin65.25cos14.32

cos65.25sin14.32

+=

=

T

N

It must be noted, that in all the analyzed cases the angle is an angle from the first quarter and is

always positive. So it is related to the oriented angle as =

The values of shear and normal forces are given in the following table

x y cos sin N T

m m kN kN

0.0 0.0 0.6 0.8 42.5 0.201.0 1.2222 0.6690 0.743 37.9 0.422.0 2.2222 0.7474 0.664 33.7 0.69

3.0 3.0 0.8320 0.556 30.1 1.04

4.0 3.5556 0.9138 0.406 27.4 1.405.0 3.8889 0.9762 0.217 25.9 -1.79

6.0 4.0 1.0 0.0 25.7 2.147.0 3.8889 0.9762 0.217 25.5 3.478.0 3.5556 0.9138 0.406 24.3 8.46

9.0 3.0 0.8320 0.556 22.5 / 39.2 12.50 / 12.5

10.0 2.2222 0.7474 0.664 40.5 6.9911.0 1.2222 0.6690 0.743 41.0 2.44

12.0 0.0 0.6 0.8 41.1 1.24

xB

3

32.14

25.65

5.3

y N T

M

x

30 kN

B

E

3

32.14

25.65

5.3

yN

T

M


17/20

Note, that for x = 9.0 m, i.e. at the point E, where the point load is applied, the shear and axialforces are discontinuous. The respective jumps in the functions are equal to the respectivecomponents of the concentrated force along tangent and normal to the tangent.The values of the internal forces can now be presented graphically.

It is worth to observe, that the differential relation between the bending moment and the shearforce is also valid for curved beams. So, the zero value of the shear force corresponds to the localextreme value of the bending moment.It is also interesting to observe the disturbing influence of the point load. The left half of the archloaded by the distributed force is characterized by a very small bending (and shear), while the righthalf with the point load is subjected to a relatively larger bending (and shear).

Now let us consider the same arch subjected to the uniformly distributed load of the snow-type

along the entire length.

For the convenience of the calculations the same modified system is adopted

2.4

5.1

14.15.3

M[kNm]

0.2

2.14

12.5 1.24

T[kN]

12.5

+

42.5

25.7

39.2

41.1

N

[kN]

22.5

y

x

6 kN/m

A B

C f= 4

L= 12

3 9[m]

y

x

6 kN/m

A B

4

3 9

CX1X1

X2

X2


18/20

with the system of canonical equations of the similar form as previously:

=++

=++

0

0

2222121

1212111

P

P

XX

XX

The states X1= 1, X2= 1 are identical and the flexibility coefficients are

EI

362.2911 = ,

EI

077.2512 = ,

EI

75.34422 =

The difference is only in the state P

( )232

6= xMP

Calculation of the coefficients related to this state is given in the table

x y cos M1 M2 MP

cos

1 PMM cos

2 PMM M

(i)

m m m m kNm kNm2 kNm

2 kNm

0.0 0.0 0.6 3.0 3.0 27.0 135.0 135.0 0.0301.0 1.2222 0.6690 1.7778 2.0 12.0 31.89 35.87 0.0172.0 2.2222 0.7474 0.7778 1.0 3.0 3.122 4.014 0.0073.0 3.0 0.8320 0.0 0.0 0.0 0.0 0.0 0.0

4.0 3.5556 0.9138 0.5556 1.0 3.0 1.824 3.283 0.004

5.0 3.8889 0.9762 0.8889 2.0 12.0 10.93 24.59 0.0096.0 4.0 1.0 1.0 3.0 27.0 27.0 -81.0 0.010

7.0 3.8889 0.9762 0.8889 4.0 48.0 43.71 196.7 0.0098.0 3.5556 0.9138 0.5556 5.0 75.0 45.60 410.4 0.004

9.0 3.0 0.8320 0.0 6.0 108.0 0.0 778.8 0.00010.0 2.2222 0.7474 0.7778 7.0 147.0 153.0 1377 0.007

11.0 1.2222 0.6690 1.7778 8.0 192.0 510.2 2296 0.017

12.0 0.0 0.6 3.0 9.0 243.0 1215 3645 0.030

Using the trapezium formula the remaining flexibility coefficients are found:

EIP

12441

= ,

EIP

68832

=

Now the canonical equations can be solved to get the values of the redundant forces

kN99.261 =X , kN00.182 =X ,

The final values of the bending moments in the arch are obtained from the superposition rule

( )P

i MXMXMM ++= 2211

and their values are given in the last column in the table above. These values are very close tozero. Indeed, they should be exactly zero, because the parabolic arch is the optimal one for thesnow-type loading and it is in the bending-free state provided the supports allow for the reactionstransmitting the axial force. This conditions are fulfilled in the considered example. Hence, the

y

x

6 kN/m

A B

4

3 3

C

27 27 108

243


19/20

small non-zero values of the final bending moments are due to the round-up errors and have to beneglected.In this situation every kinematic check is fulfilled.Finally let us find the functions of the shear forces (which should be zero, too) and the non-zeroaxial forces.Having found the reactions

the following equilibrium equations can be written for the left (from A to D)

( )

cos)600.36(sin99.26

sin600.36cos99.26

xT

xN

+=

=

and the right half (from B to D) of the arch

( )[ ]

( )[ ]

cos12600.36sin99.26

sin12600.36cos99.26

xT

xN

=

=

It must be noted again, that in these cases the angle is an angle from the first quarter and is

always positive. So it is related to the oriented angle as =

y

xA

36.00

26.99

0.0

N

T

M

6 kN/m

xB

36.00

26.99

0.0

y

N T

M

6 kN/m

12 x

y

x

6 kN/m

A B

D

4

3 3 3 3

C26.99

18.0018.00

26.99

36.00 36.00

26.99

26.99

0.0 0.0


20/20

Documents

27 Shape of Arches