27 Shape of Arches

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 elements working mostly or only in the state of compression. The reason for this was that the main construction material in those times was the natural stone. It has relatively large compressive strength and its tensile strength is about 10 times smaller. Hence, it was vital to avoid tension and bending involving tension. One may argue if the invention of arches was an engineering achievement or was it only a clever imitation of nature, which uses arches and their 3D version – concave shells successfully for a much longer time. Just think of thin walled eggs, shells, turtles, etc… Hence, arches are perfect structures. Under a distributed loading they should exhibit mostly axial forces 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, axial forces 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 the radius R, loaded by the concentrated force P at the tip. The arch is described in the Cartesian set of co-ordinates x,y with the origin at the centre of the circle. y

y M

R–y

P

y

R

ϕ

ϕ

ϕ

P R T

P

N

ϕ

ϕ x

T

N

ϕ

x x

However, the calculations are easier, when carried out using the polar set of co-ordinates ϕ,R. The relations between the co-ordinates are: x = R cos ϕ y = R sin ϕ To find functions of internal forces we introduce a cross-section at the point defined by the angle ϕ and write down the equilibrium equations for the fragment of arch to the left of the cross-section:

∑ N : N + P sin ϕ = 0 ∑ T : T − P cos ϕ = 0 ∑ M ϕ : M − P (R − y ) = 0 From these equations the functions of the internal forces can be found:

N = −P sin ϕ T = P cos ϕ

M = PR (1 − sin ϕ )

1

In the third equation the relation between y and ϕ was substituted. These functions can be represented graphically. In the graphs values of the forces are presented for points spaced by angles π/12. Note, that ϕ = 0 corresponds to the support and ϕ = π/2 – to the cantilever tip. –1.0

0.0

–0.966 –0.866 – –0.707

0.0

–0.500 N [·P]

0.034 0.134 0.293

0.258 0.500 0.707

0.500

0.866

–0.258

M [·PR]

+

T [·P]

0.741

0.966 1.0

0.0

1.0

As the next example let us consider a cantilever circular arch being one quarter of the circle with the radius R, loaded by the uniformly distributed snow-type loading of the intensity q. y qx x/2

y q

M y

R

R T

N

ϕ

ϕ

ϕ

qx ϕ

ϕ

x

T

N

ϕ

x x

To find the function of internal forces we introduce a cross-section at the point defined by the angle ϕ and 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 its resultant, which is applied at the centre of length x. Thus, the equilibrium equations take the form:

∑ N : N + qx cos ϕ = 0 ∑ T : T + qx sin ϕ = 0 x

∑ Mϕ : M + qx 2

=0

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

N = −qR cos 2 ϕ T = −qR sin ϕ cos ϕ M=

− qR 2 cos 2 ϕ 2

In these equations the relation between x and ϕ was substituted. These functions can be represented graphically. In the graphs values of the forces are presented for points spaced by angles of π/12. 0.0

0.0

–0.067 –0.250

0.0

–0.250 –0.433

–0.500

–0.500 –

–0.750 N [·qR]

–

–0.933

T [·qR]

–0.433 –0.250 0.0

–1.0

2

0.033 0.125 0.250 0.375 M [·qR2] 0.467 0.5

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. y

x–a

qds y

q

α ϕ

N

T

ϕ

x

ϕ

M

dα

R

qds

ϕ

ϕ

T

N

ϕ

x x

a

Again, to find the function of internal forces we introduce a cross-section at the point defined by the angle ϕ and write down the equilibrium equations for the fragment of arch to the left of the crosssection. 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 integration of moments due to elementary loads along the considered fragment of the arch. In order to carry out this integration an auxiliary angular co-ordinate α is introduced. It will undergo integration in the limits from ϕ to π/2. The elementary load resultant is equal to qds, where s is the curvilinear coordinate measured along the arch. It can be related to the angle via ds = Rdα

and the length a is given as a = R cos α With this in hand the equilibrium equations can be given as: π /2

∑ N : N + ∫ qds cos ϕ = 0 ϕ

π /2

∑ T : T + ∫ qds sin ϕ = 0 ϕ

∑ Mϕ : M +

π /2

∫ (x − a )qds = 0

ϕ

Expressing all variables under integrals in terms of angles and noting, that the angle ϕ does not undergo integration, we express the internal forces as: N = −qR cos ϕ

π /2

∫ dα = −qR cos ϕ [α ]ϕ

π /2

ϕ

T = −qR sin ϕ

π /2

∫ dα = −qR sin ϕ [α ]ϕ

π /2

ϕ

M = −qR 2

π  = −qR  − ϕ  cos ϕ 2  

π  = −qR  − ϕ  sin ϕ 2  

π /2

π /2 2 ∫ (cos ϕ − cos α )dα = −qR [α cos ϕ − sin α ]ϕ =

ϕ

π  = −qR 2  cos ϕ − 1 − ϕ cos ϕ + sin ϕ  2  These functions can be represented graphically. In the graphs values of the forces are presented for points spaced by angles of π/12.

3

0.0

0.0

–0.068 –0.262

0.0

–0.253

–0.555

–

0.128 0.262

–0.555 –

–0.907 N [·qR]

0.034

–0.453

T [·qR]

–1.264

0.407

–0.524 M [·qR2]

–0.339

0.523 0.571

0.0

–1.571

The next type of loading to be considered is the uniformly distributed hydraulic pressure of the intensity q. First, let it be applied to a cantilever circular arch being one quarter of the circle with the radius R. x–a qdssinα y b y dα

y q

b–y

R

ϕ

qds qdscosα M

ϕ

N

ϕ

αT

ϕ

x

a

qdssinα

T

ϕ

qdscosα N

ϕ

x x

Again, to find the function of internal forces we introduce a cross-section at the point defined by the angle ϕ and write down the equilibrium equations for the fragment of arch to the left of the crosssection. 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 integration of 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 α is introduced. This will undergo integration in the limits from ϕ to π/2. The elementary load resultant is equal to qds, where s is the co-ordinate measured along the arch. This load now has a varying direction, so it should be projected on axes x and y. These projections are: qdscosα and qdssinα, respectively. The value b is given by b = R sin α With this in hand the equilibrium equations can be written down as:

∑N : N +

π /2

∫ (qds sin α cos ϕ − qds cos α sin ϕ ) = 0

ϕ

∑T : T +

π /2

∫ (qds sin α sin ϕ + qds cos α cos ϕ ) = 0

ϕ

π /2

∑ Mϕ : M + ∫ [qds sin α (x − a ) + qds cos α (b − y )] = 0 ϕ

Expressing all variables under integrals in terms of angles and noting, that the angle ϕ does not undergo integration, we express the internal forces as:

4

N = −qR cos ϕ

π /2

π /2

ϕ

ϕ

∫ sin αdα ,+qR sin ϕ ∫ cos αdα = −qR cos ϕ [− cos α ]ϕ

(

)

π /2

+ qR sin ϕ [sin α ]ϕ

π /2

=

= qR − cos 2 ϕ + sin ϕ − sin 2 ϕ = qR (sin ϕ − 1) T = −qR sin ϕ

π /2

π /2

ϕ

ϕ

∫ sinαdα ,−qR cosϕ ∫ cos αdα = −qR sinϕ [− cosα ]ϕ

π /2

− qR cos ϕ [sinα ]ϕ

π /2

=

= qR (− sin ϕ cos ϕ − cos ϕ + sin ϕ cos ϕ ) = −qR cos ϕ M = −qR 2

π /2

∫ (sinα cos ϕ − sinα cos α + cos α sinα − cos α sinϕ )dα =

ϕ

= −qR 2 cos ϕ

(

π /2

∫ sinαdα + qR

ϕ

2

π /2

sin ϕ ∫ cos αdα = −qR 2 cos ϕ [− cos α ]ϕ

π /2

ϕ

)

+ qR 2 sin ϕ [sin α ]ϕ

π /2

=

= qR 2 − cos2 ϕ + sinϕ − sin2 ϕ = qR 2 (sin ϕ − 1) These functions can be represented graphically. In the graphs values of the forces are presented for points spaced by angles of π/12. 0.0

0.0

–0.034 –0.134

0.0

–0.258

–0.293

–

0.134 0.293

–0.707

–0.500 N [·qR]

0.034

–0.500

0.500

–0.866 T [·qR]

–0.741

–

M [·qR2]

–0.966

0.741

–1.0

–1.0

1.0

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

qdssinα qds y A qdscosα b

y q

A

dα

R

α

α x

x a

x

V

V

The first thing to calculate is the reaction in the roller support V. This can be done from the equilibrium of moments with respect to the hinge A for the right half of the arch. This equation reads: π /2

∑ M A : VR − ∫ [aqds sin α + (R − b )qds cos α ] = 0 0

5

After substitution for a and b and division by R the reaction can be calculated as: V = qR

π /2

π /2

0

0

∫ [sin α cos α + cos α (1 − sin α )]dα = qR ∫ cos αdα

y

T

ϕ qdssinα

M qdssinα qds

N y y–b

ϕ

ϕ

dα

α

π /2

= qR

ϕ T qdscosα

N

qdscosα

b

= qR [sin α ]0

qR

ϕ

x

x a–x a

V=qR

With the value of the reaction V the internal forces in the arch can be obtained from the equilibrium for the fragment of arch to the right of the cross-section ϕ. The adequate equations have the form: ϕ

∑ N : N + qR cos ϕ − ∫ (qds sin α cos ϕ − qds cos α sin ϕ ) = 0 0

ϕ

∑T : T + qR sin ϕ − ∫ (qds sin α sin ϕ + qds cos α cos ϕ ) = 0 0

ϕ

∑ Mϕ : M − qR (R − x ) + ∫ [qds sin α (a − x ) + qds cos α (y − b )] = 0 0

Note, that these equations are also valid for the angle ϕ greater than π/2, i.e. within the left-hand half of the arch. From these equations the functions of internal forces follow: ϕ

ϕ

N = −qR cos ϕ + qR cos ϕ ∫ sin αdα − qR sin ϕ ∫ cos αdα = 0

0

= −qR cos ϕ + qR cos ϕ [− cos α ]0 − qR sin ϕ [sin α ]0 = ϕ

ϕ

= −qR cos ϕ − qR cos 2 ϕ + qR cos ϕ − qR sin 2 ϕ = −qR ϕ

ϕ

T = −qR sin ϕ + qR sin ϕ ∫ sin αdα + qR cos ϕ ∫ cos αdα = 0

0

= −qR sin ϕ + qR sin ϕ [− cos α ]0 + qR cos ϕ [sin α ]0 = ϕ

ϕ

= −qR sin ϕ − qR sin ϕ cos ϕ + qR sin ϕ + qR cos ϕ sin ϕ = 0 ϕ

M = qR 2 (1 − cos ϕ ) − qR 2 ∫ [sin α cos α − sin α cos ϕ + cos α sin ϕ − cos α sin α ]dα = 0

{

}

= qR (1 − cos ϕ ) + qR cos ϕ [− cos α ]0 − sin ϕ [sin α ]0 = 2

2

(

ϕ

ϕ

)

= qR (1 − cos ϕ ) + qR − cos ϕ + cos ϕ − sin ϕ = 0 2

2

2

2

The most important thing to observe in these results are the vanishing values of the bending moment and the shear force. In this way we arrive at the notion of the optimal arch shape. It was mentioned in the introduction, that the arch was invented to carry mostly normal forces. It comes out, that for a given type of loading an arch shape can be found, which is characterized by the lack of 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, there must be one more condition fulfilled to get the optimal arch serving in the bending-free state.

6

Namely, its supports at the ends must provide reactions resulting in the axial force at the arch ends. The cantilever arch with the free end does not fulfil this condition. Hence, there was the bending moment present. It is also worth to note, that a circular arch with a hydraulic pressure in the bending-free state features the constant value of the normal force N equal to qR throughout 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 γ present only on its right-hand side. x–a qhdssinα qhds y qhdscosα b M y dα N αT

y b–y R

h

ϕ

ϕ

ϕ

ϕ

x

qhdssinα

ϕ

T

qhdscosα N

ϕ

x

a

x

This loading acts in the same direction, normal to the arch, as the previously considered hydraulic pressure. But the difference is in the fact, that its value is not constant but varies linearly with the increasing depth h measured from the liquid free surface. Thus, in the equations corresponding to the hydraulic pressure the load q must be replaced by q h = γh ⋅ 1m = γ (R − b ) = γR (1 − sin α ) Hence, the equilibrium equations for the cut-off segment of the arch read: π /2

∑ N : N + ∫ (q hds sin α cos ϕ − q hds cos α sin ϕ ) = 0 ϕ

π /2

∑ T : T + ∫ (q hds sin α sin ϕ + qh ds cos α cos ϕ ) = 0 ϕ

∑ Mϕ : M +

π /2

∫ [q hds sin α (x − a ) + qh ds cos α (b − y )] = 0

ϕ

Having expressed all the variables under integrals in terms of angles, introduced the notation for qh and noted, that the angle ϕ does not undergo integration, the internal forces can be given as: N = −γR 2 cos ϕ

π /2

π /2

2 ∫ (1 − sin α ) sin αdα + γR sin ϕ ∫ (1 − sin α )cos αdα

ϕ

T = −γR 2 sin ϕ M = −γR

ϕ

π /2

∫ (1 − sin α ) sin αdα − γR ϕ

2

cos ϕ

π /2

∫ (1 − sin α ) cos αdα

ϕ

π /2 3

π /2

∫ (1 − sin α )cos α (sin α − sin ϕ )dα − γR ϕ∫ (1 − sin α ) sin α (cos ϕ − cos α )dα 3

ϕ

These calculations involve the integrals of the functions sin2α and cosαsinα, which can be easily found using the following trigonometric relations

7

1 1 sin 2 α = − cos 2α + 2 2 1 cos α sin α = sin 2α 2

Now the functions of the internal forces can be determined π /2

N = −γR cosϕ 2

π /2

∫ (sin α − sin α )dα + γR 2

2

∫ (cosα − cosα sin α )dα =

sin ϕ

ϕ

ϕ

π /2

1 1   = −γR cosϕ − cosα + sin 2α − α  4 2 ϕ  2

π /2

1   + γR sin ϕ sin α + cos 2α  4  ϕ

=

2

1 1 π ϕ  3  = −γR 2 cosϕ  cosϕ − sin 2ϕ − +  + γR 2 sin ϕ  − sin ϕ − cos 2ϕ  4 4 2 4 4     T = −γR sin ϕ 2

π /2

∫ (sin α − sin α )dα − γR 2

cos ϕ

2

π /2

ϕ

∫ (cos α − cos α sin α )dα =

ϕ π /2

1 1   = −γR 2 sin ϕ − cos α + sin 2α − α  4 2 ϕ 

π /2

1   − γR 2 cos ϕ sin α + cos 2α  4  ϕ

=

π ϕ 1 1  3  = −γR 2 sin ϕ  cos ϕ − sin 2ϕ − +  − γR 2 cos ϕ  − sin ϕ − cos 2ϕ  4 4 2 4 4     M = γR 3

π /2

π /2

3 2 ∫ (cos α − cos α sin α )(sin ϕ − sin α )dα + γR ∫ (sin α − sin α )(cos α − cos ϕ )dα =

ϕ

= γR 3 sin ϕ + γR

ϕ

π /2

π /2

∫ (cos α − cos α sin α )dα − γR ϕ∫ (cos α sin α − cos α sin α )dα ϕ 3

π /2 3

∫ (cos α sin α − sin ϕ

2

)

2

α cos α dα − γR cos ϕ π /2

1   = γR sin ϕ sin α + cos 2α  4  ϕ 3

3

π /2

∫ (sin α − sin α )dα = ϕ 2

π /2

1 1   − γR cos ϕ  − cos α + sin 2α − α  4 2 ϕ  3

=

1 1 π ϕ 3   = γR 3 sin ϕ  − sin ϕ − cos 2ϕ  − γR 3 cos ϕ  cos ϕ − sin 2ϕ − +  4 4 4 2 4   These functions can be represented graphically. In the graphs values of the forces are presented for points spaced by angles of π/12. 0.0

0.0

–0.0002 –0.0031

0.0

–0.003 –0.023

–0.0152

–0.1090

0.0466

–0.171 T [·γR ] 2

N [·γR ]

0.0152

–0.076

–0.0466 2

0.0002 0.0031

–

–

–0.314 –0.5

–0.2146

8

M [·γR ] 3

0.1090 0.2146

3. Statically indeterminate circular arches Statically indeterminate arches can be solved using the flexibility method. The shape of the arch determines 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 the principle of virtual work. The adequate well known formula reads: Mi Mk ds EI s

δ ik = ∫

The important issue to note is, that the integration must be carried out along the curve representing the arch. This opens a vast field of problems and different techniques invented to calculate the curvilinear integrals. In the case of circular arches this integration can be performed in a quite straightforward way introducing the polar set of co-ordinates and replacing the integral along the arch length with the one over the polar angle, as it was done in the examples of statically determinate arches to calculate 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 P at the tip. y

y

A

The modified system

A

P

P R

R

X1

B

ϕ

x

B

x

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

δ 11 X 1 + δ 1P = 0 For a circular arch the calculation of the flexibility coefficients in this equation can be carried out analytically after the change to the polar co-ordinates. The analytical functions of the bending moments in the basic states: X1 = 1 and P are necessary.

y

y

State X1 = 1

State P

P X1 = 1

R–y

y

R

ϕ

R

ϕ

x

x

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

9

M1 = x = R cos ϕ

MP = P (R − y ) = PR (1 − sin ϕ )

The flexibility coefficients can now be calculated from: M1M1 R3 ds = cos 2 ϕdϕ ∫ EI s s EI

δ11 = ∫

M1M P PR 3 ds = ∫ cos ϕ (1 − sin ϕ )dϕ EI EI s s

δ1P = ∫

The integral of the function cosαsinα was already discussed previously and the one for the function cos2ϕ can be found using the trigonometric relation

cos 2 ϕ =

1 1 cos 2ϕ + 2 2

With this in hand one gets:

δ11 =

R3 π / 2 R3 ϕ ϕ ( cos 2 + 1 ) d = ∫ 2EI 0 2EI

δ1P =

PR 3 2EI

π /2

π /2

 1  2 sin 2ϕ + ϕ   0 3

=

πR 3 4EI π /2

1   2 sin ϕ + cos 2ϕ  2EI  2 0

PR ∫ (2 cos ϕ − sin 2ϕ )dϕ =

0

=

PR 3 2EI

Then the redundant force can be obtained from the canonical equation as X1 = −

δ1P 2P =− δ11 π

The function of the bending moment in the statically indeterminate arch follows from the superposition rule: M (i ) = M1 X 1 + M P = −

2PR

π

2   cos ϕ + PR (1 − sin ϕ ) = PR  1 − sin ϕ − cos ϕ  π  

To check the correctness of the results a kinematic check should be performed. Let us check if the cross-section rotation at the clamped support B is zero. To this end the principle of virtual work and the reduction theorem are used: M (i )M (0 ) ds = 0 (? ) EI s

1ϕ 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 virtual loading in the form of the concentrated unit bending moment is applied there.

y A

R

M =1

VA B

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

10

∑ MB : VA R + 1 = 0

⇒ VA = −

1 R

and the function of the virtual bending moment is: M (0 ) = −

1 x = − 1 ⋅ cos ϕ R

The check of the cross-section rotation follows as: M (i )M (0 ) PR 2 ds = − EI EI s

1ϕB = ∫

PR 2 =− EI

π /2

∫ 0

2   cos ϕ  1 − sin ϕ − cos ϕ dϕ = π   π /2

1 1 1   sin ϕ + 4 cos 2ϕ − 2π sin 2ϕ − π ϕ   0

=0

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 arch central axis is expressed in the co-ordinates system x,y by a function

4f x (L − x ) L2

y= what for the data given in the figure yields

y =− y

1 2 4 x + x 9 3

6 kN/m 30 kN D E

C

f=4

A

x

B 3

3

3

3

[m]

L = 12

The arch consists of two segments connected by a hinge. Externally there are six reactions at the clamped supports A and B versus four equilibrium equations, the fourth is the equation of moments with respect to the hinge C. Hence, the system has two redundant forces. The modified system can be formed by the removal of the hinge C and then the horizontal and vertical internal forces therein are considered as the redundant forces X1 and X2. y 6 kN/m 30 kN C X1 A

D X2

X1

E

4

X2 3

B 3

3

11

3

x

The kinematic compatibility conditions require, that the relative displacements of arch fragments tips at the point C measured in the horizontal and vertical directions are zero. This leads to the canonical equations of the flexibility method in the form  δ 11 X 1 + δ 12 X 2 + δ 1P = 0  δ 21 X 1 + δ 22 X 2 + δ 2P = 0 and the flexibility coefficients are calculated from the curvilinear integrals Mi Mk ds EI s

δ ik = ∫

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

ds

ϕ

dx the following relation can be deduced ds =

dx cos ϕ

where ϕ is the slope of the tangent to the arch. Now the flexibility coefficients can be obtained from Mi Mk dx EI cos ϕ x

δ ik = ∫

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

tan ϕ =

dy dx

In our example 4  2 x+  3  9

ϕ = arctan −

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

f(x)

fn–1

fn

f2 f0

f1 A1

a

∆x

A2

∆x

An

∆x 12

b

x

The function domain x ∈ a, b is divided into a finite number of n equal intervals ∆x and the determinate integral is represented as a sum of trapeziums areas b

∫ f (x )dx = A1 + A2 + ... + An

a

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 b

∫ f (x )dx = a

∆x (f0 + 2f1 + 2f2 + ... + 2fn −1 + fn ) 2

The method is approximate and its accuracy depends on the density of the function domain subdivision. The larger is the number of intervals, the more accurate results are obtained. In the considered example the integration domain is x ∈ 0,12 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 = 1 and P are necessary.

y

M1 = 3 − y

X1=1 C X1=1 3 A

B 3

3

3

–1

4 x

3

3

3

y

M2 = x − 3

C X2=1 A

4

X2=1 3

B 3

3

x

6 kN/m 30 kN D E

4

A

B 3

3

3

9

3

y

C

–3

 6 2 on AD − 2 (x − 3 )  MP = − 6 ⋅ 3(x − 4.5 ) on DE  − 6 ⋅ 3(x − 4.5 ) − 30(x − 9 ) on EB 

x

–225 –27

3

13

–27

–81

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

y

tanϕ

cosϕ

M1

M2

MP

M 1M 1 cos ϕ

M1M 2 cos ϕ

M 2M 2 cos ϕ

M1M P cos ϕ

m 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0

m 0.0 1.2222 2.2222 3.0 3.5556 3.8889 4.0 3.8889 3.5556 3.0 2.2222 1.2222 0.0

– 1.3333 1.1111 0.8889 0.6667 0.4444 0.2222 0.0 –0.2222 –0.4444 –0.6667 –0.8889 –1.1111 –1.3333

– 0.6 0.6690 0.7474 0.8320 0.9138 0.9762 1.0 0.9762 0.9138 0.8320 0.7474 0.6690 0.6

m 3.0 1.7778 0.7778 0.0 –0.5556 –0.8889 –1.0 –0.8889 –0.5556 0.0 0.7778 1.7778 3.0

m –3.0 –2.0 –1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0

kNm –27.0 –12.0 –3.0 0.0 –3.0 –12.0 –27.0 –45.0 –63.0 –81.0 –129.0 –177.0 –225.0

m2 15.0 4.7243 0.8094 0.0 0.3378 0.8094 1.0 0.8094 0.3378 0.0 0.8094 4.7243 15.0

m2 –15.0 –5.3148 –1.0407 0.0 –0.6080 –1.8211 –3.0 –3.6423 –3.0401 0.0 7.2847 21.259 45.0

m2 15.0 5.9791 1.3380 0.0 1.0943 4.0975 9.0 16.390 27.358 43.269 65.561 95.665 135.0

kNm2 –135.0 –31.892 –3.1220 0.0 1.8240 10.927 27.0 40.976 38.305 0.0 –134.25 –470.36 –1125.0

M 2MP cos ϕ

M(i)

kNm2 kNm 135.0 2.391 35.874 1.894 4.0139 1.097 0.0 0.0 –3.2830 –1.397 –24.585 –3.091 –81.0 –5.085 –184.39 –4.379 –344.71 2.027 –584.13 14.136 –1208.2 1.945 –2116.7 –-4.546 –3375.0 –5.337

Applying the trapezium formula for the coefficients yields the following results

δ11 =

29.362 25.077 344.75 − 1150.6 − 6127.1 , δ12 = , δ 22 = , δ1P = , δ1P = EI EI EI EI EI

With these coefficients found the canonical equations can be solved to get the values of the redundant forces X 1 = 25.598 kN ,

X 2 = 15.856 kN ,

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

M (i ) = M1 X 1 + M 2 X 2 + M P 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 crosssection rotation at the clamped support A will be calculated from the principle of virtual work and the reduction theorem as

M (i )M (0 ) M (i )M (0 ) ds = ∫ dx = 0 (? ) EI s x EI cos ϕ

1ϕ A = ∫

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

y D E

C

1

f=4 B

A 1 4

3 1 12

3

3

L = 12

3

[m] 1 12

x 1 4

Having calculated the reactions at the supports in the usual way the function of the virtual bending moment in the modified system can be expressed as M (0 ) = −

1 1 x− y+1 12 4

14

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

y

M(i)

M (0 )

cosϕ

m 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0

m 0.0 1.2222 2.2222 3.0 3.5556 3.8889 4.0 3.8889 3.5556 3.0 2.2222 1.2222 0.0

kNm 2.391 1.894 1.097 0.0 –1.397 –3.091 –5.085 –4.379 2.027 14.136 1.945 –-4.546 –5.337

– 1.0 0.6111 0.2778 0.0 –0.2222 –0.3889 –0.50 –0.5556 –0.5556 –0.50 –0.3889 –0.2222 0.0

– 0.6 0.6690 0.7474 0.8320 0.9138 0.9762 1.0 0.9762 0.9138 0.8320 0.7474 0.6690 0.6

M ( i ) M (0 ) cos ϕ kNm 3.985 1.730 0.408 0.0 0.340 1.231 2.543 2.492 –1.232 –8.495 –1.012 1.510 0.0

− 0.016 EI This result related to the maximum absolute value component in the integral calculation, i.e. 8.495/EI is only 0.19%, what allows to conclude, that the angle ϕA is indeed zero and that the calculated 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 final result of the kinematic check is ϕ A =

y 6 kN/m 30 kN C 25.65 15.86

25.65 A

3

D 25.65 15.86

E

4 B

3

3

25.65

3

2.4 32.14

33.86

x

5.3

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

y 6 kN/m

N

α

M T

25.65 A 2.4

x

33.86

the equilibrium equations of forces in the direction N and T yield

15

N = −25.65 cos α − (33.86 − 6 x ) sin α T = −25.65 sin α + (33.86 − 6 x ) cos α

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

y

T

α

30 kN

N E

B

x 25.65

3 5.3

32.14

y

α

M N

T

B

x 25.65

3 32.14

5.3

The equilibrium equations yield: for ED section: N = (30 − 32.14 ) sin α − 25.65 cos α T = (30 − 32.14 ) cos α + 25.65 sin α

for EB section: N = −32.14 sin α − 25.65 cos α T = −32.14 cos α + 25.65 sin α It must be noted, that in all the analyzed cases the angle α is an angle from the first quarter and is always positive. So it is related to the oriented angle ϕ as α = ϕ The values of shear and normal forces are given in the following table x m 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0

y m 0.0 1.2222 2.2222 3.0 3.5556 3.8889 4.0 3.8889 3.5556 3.0 2.2222 1.2222 0.0

cosα – 0.6 0.6690 0.7474 0.8320 0.9138 0.9762 1.0 0.9762 0.9138 0.8320 0.7474 0.6690 0.6

sinα – 0.8 0.743 0.664 0.556 0.406 0.217 0.0 0.217 0.406 0.556 0.664 0.743 0.8

N kN –42.5 –37.9 –33.7 –30.1 –27.4 –25.9 –25.7 –25.5 –24.3 –22.5 / –39.2 –40.5 –41.0 –41.1

T kN –0.20 –0.42 –0.69 –1.04 –1.40 –-1.79 –2.14 3.47 8.46 12.50 / –12.5 –6.99 –2.44 1.24

16

Note, that for x = 9.0 m, i.e. at the point E, where the point load is applied, the shear and axial forces are discontinuous. The respective “jumps” in the functions are equal to the respective components of the concentrated force along tangent and normal to the tangent. The values of the internal forces can now be presented graphically. 5.1 + – M [kNm]

T [kN]

5.3

–12.5

–0.2

14.1

2.4

–2.14

–25.7

N [kN]

12.5

– 1.24

–22.5 –39.2

– –41.1

–42.5

It is worth to observe, that the differential relation between the bending moment and the shear force is also valid for curved beams. So, the zero value of the shear force corresponds to the local extreme value of the bending moment. It is also interesting to observe the disturbing influence of the point load. The left half of the arch loaded by the distributed force is characterized by a very small bending (and shear), while the right half 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. y

6 kN/m

C

f=4

A

B 9

3

x [m]

L = 12

For the convenience of the calculations the same modified system is adopted y 6 kN/m C X2 X1

X1 A

4

X2

B

3

9

17

x

with the system of canonical equations of the similar form as previously:  δ 11 X 1 + δ 12 X 2 + δ 1P = 0  δ 21 X 1 + δ 22 X 2 + δ 2P = 0 The states X1 = 1, X2 = 1 are identical and the flexibility coefficients are

δ11 =

29.362 25.077 344.75 , δ12 = , δ 22 = EI EI EI

The difference is only in the state P y 6 kN/m

MP = − C

6 (x − 3 )2 2

4 A

B 3

x

3

–243 –27

–27

–108

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

y

cosϕ

M1

M2

MP

M1M P cos ϕ

M 2MP cos ϕ

M(i)

m 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0

m 0.0 1.2222 2.2222 3.0 3.5556 3.8889 4.0 3.8889 3.5556 3.0 2.2222 1.2222 0.0

– 0.6 0.6690 0.7474 0.8320 0.9138 0.9762 1.0 0.9762 0.9138 0.8320 0.7474 0.6690 0.6

m 3.0 1.7778 0.7778 0.0 –0.5556 –0.8889 –1.0 –0.8889 –0.5556 0.0 0.7778 1.7778 3.0

m –3.0 –2.0 –1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0

kNm –27.0 –12.0 –3.0 0.0 –3.0 –12.0 –27.0 –48.0 –75.0 –108.0 –147.0 –192.0 –243.0

kNm2 –135.0 –31.89 –3.122 0.0 1.824 10.93 27.0 43.71 45.60 0.0 –153.0 –510.2 –1215

kNm2 135.0 35.87 4.014 0.0 –3.283 –24.59 -81.0 –196.7 –410.4 –778.8 –1377 –2296 –3645

kNm –0.030 –0.017 –0.007 0.0 0.004 0.009 0.010 0.009 0.004 0.000 –0.007 –0.017 –0.030

Using the trapezium formula the remaining flexibility coefficients are found:

δ 1P =

− 1244 − 6883 , δ 2P = EI EI

Now the canonical equations can be solved to get the values of the redundant forces X 1 = 26.99kN ,

X 2 = 18.00kN ,

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

M (i ) = M1 X 1 + M 2 X 2 + M P and their values are given in the last column in the table above. These values are very close to zero. Indeed, they should be exactly zero, because the parabolic arch is the optimal one for the snow-type loading and it is in the bending-free state provided the supports allow for the reactions transmitting the axial force. This conditions are fulfilled in the considered example. Hence, the

18

small non-zero values of the final bending moments are due to the round-up errors and have to be neglected. 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-zero axial forces. Having found the reactions y 6 kN/m C

D 26.99 18.00

26.99 26.99 A

18.00

4

x

B

3

3

3

26.99

3

0.0

0.0

36.00

36.00

y 6 kN/m

N

α

M T

26.99 A

x

36.00

0.0

6 kN/m

y

α

M N

T

B

x 26.99

12 – x 0.0 36.00

the following equilibrium equations can be written for the left (from A to D) N = −26.99 cos α − (36.00 − 6 x ) sin α T = −26.99 sin α + (36.00 − 6 x ) cos α and the right half (from B to D) of the arch N = −26.99 cos α − [36.00 − 6(12 − x )]sin α T = 26.99 sin α − [36.00 − 6(12 − x )]cos α It must be noted again, that in these cases the angle α is an angle from the first quarter and is always positive. So it is related to the oriented angle ϕ as α = ϕ

19

The values of shear and normal forces are given in the following table x m 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0

y m 0.0 1.2222 2.2222 3.0 3.5556 3.8889 4.0 3.8889 3.5556 3.0 2.2222 1.2222 0.0

cosα – 0.6 0.6690 0.7474 0.8320 0.9138 0.9762 1.0 0.9762 0.9138 0.8320 0.7474 0.6690 0.6

sinα – 0.8 0.743 0.664 0.556 0.406 0.217 0.0 0.217 0.406 0.556 0.664 0.743 0.8

N kN –45.0 –40.4 –36.1 –32.4 –29.5 –27.6 –27.0 –27.6 –29.5 –32.4 –36.1 –40.4 –45.0

T kN 0.008 0.007 0.007 0.006 0.004 0.002 0.000 0.002 0.004 0.006 0.007 0.007 0.008

The shear force values are indeed zero, the results in the last column are only due to round-up errors. The graphical representation of the axial forces concludes the example –27.0

N [kN]

–

–45.0

–45.0

20