y R T However, the calculations are easier, when carried out using the polar set of co-ordinates ϕ,r. The relations between the co-ordinates are:

Transcription

1 Curved beams. Introduction Curved beams also called arches were invented about ears ago. he purpose was to form such a structure that would transfer loads, mainl the dead weight, to the ground b the elements working mostl or onl in the state of compression. he reason for this was that the main construction material in those times was the natural stone. It has relativel large compressive strength and its tensile strength is about times smaller. Hence, it was vital to avoid tension and bending involving tension. One ma argue if the invention of arches was an engineering achievement or was it onl a clever imitation of nature, which uses arches and their D version concave shells successfull for a much longer time. Just think of thin walled eggs, shells, turtles, etc Hence, arches are perfect structures. Under a distributed loading the should ehibit mostl aial forces with ver little bending and shear. he presence of concentrated forces disturbs this picture, what will be seen in the eamples.. Staticall determinate circular arches Let us start with staticall determinate circular arches. he internal forces: bending moments, aial 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, loaded b the concentrated force at the tip. he arch is described in the Cartesian set of co-ordinates, with the origin at the centre of the circle. However, the calculations are easier, when carried out using the polar set of co-ordinates,. he relations between the co-ordinates are: cos sin o find functions of internal forces we introduce a cross-section at the point defined b the angle and write down the equilibrium equations for the fragment of arch to the left of the cross-section: : + sin : cos : ( ) From these equations the functions of the internal forces can be found: sin cos sin)

2 In the third equation the relation between and was substituted. hese functions can be represented graphicall. In the graphs values of the forces are presented for points spaced b angles π/. ote, that corresponds to the support and π/ to the cantilever tip [ ] [ ] [ ] s the net eample let us consider a cantilever circular arch being one quarter of the circle with the radius, loaded b the uniforml distributed snow-tpe loading of the intensit q. q q / q o find the function of internal forces we introduce a cross-section at the point defined b the angle 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 b its resultant, which is applied at the centre of length. hus, the equilibrium equations take the form: : + q cos : + q sin : + q From these equations the functions of the internal forces can be found: q cos q q sin cos In these equations the relation between and was substituted. hese functions can be represented graphicall. In the graphs values of the forces are presented for points spaced b angles of π/. cos 67.5 [ q] [ q] [ q ]

3 ow let us consider a cantilever circular arch being one quarter of the circle with the radius, loaded b the uniforml distributed self weight-tpe loading of the intensit q. q a qds α a qds gain, to find the function of internal forces we introduce a cross-section at the point defined b the angle 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 onl b integration. he same concerns the bending moment, which can be obtained b the integration of moments due to elementar loads along the considered fragment of the arch. In order to carr out this integration an auiliar angular co-ordinate α is introduced. It will undergo integration in the limits from to π/. he elementar load resultant is equal to qds, where s is the curvilinear coordinate measured along the arch. It can be related to the angle via and the length a is given as ds a cosα With this in hand the equilibrium equations can be given as: : + : + π / π / : + qds cos qds sin π / ( a) qds Epressing all variables under integrals in terms of angles and noting, that the angle does not undergo integration, we epress the internal forces as: q cos q sin q π / [ ] q cos α [ ] q sin α ( cos cosα ) q [ α cos ] π / π / π / π / π q cos cos + sin π q cos π q sin hese functions can be represented graphicall. In the graphs values of the forces are presented for points spaced b angles of π/. π /

4 68.6 [ q] [ q] [ q ] he net tpe of loading to be considered is the uniforml distributed hdraulic pressure of the intensit q. First, let it be applied to a cantilever circular arch being one quarter of the circle with the radius. q b qds b a α a qds qdscosα qds qdscosα gain, to find the function of internal forces we introduce a cross-section at the point defined b the angle 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 onl b integration. he same concerns the bending moment, which can be obtained b the integration of the moments due to elementar loads along the considered fragment of the arch. In order to carr out this integration an auiliar angular co-ordinate α is introduced. his will undergo integration in the limits from to π/. he elementar load resultant is equal to qds, where s is the co-ordinate measured along the arch. his load now has a varing direction, so it should be projected on aes and. hese projections are: qdscosα and qds, respectivel. he value b is given b b With this in hand the equilibrium equations can be written down as: : + : + π / : + ( qds cos qds cosα sin) π / ( qds sin + qds cosα cos) π / [ qds ( a) + qds cosα( b )] Epressing all variables under integrals in terms of angles and noting, that the angle does not undergo integration, we epress the internal forces as: 4

5 π / π / q cos, + q sin cosα q cos q ( cos + sin sin ) q( sin ) π / [ cosα ] + q sin[ ] π / q sin q q q q ( sin cos cos + sin cos) π / π / ( cos cosα + cosα cosα sin) cos, q cos π / + q sin cosα q sin π / q cos cosα q ( cos + sin sin ) q ( sin ) π / π / [ cosα ] q cos[ ] cos π / π / [ cosα ] + q sin[ ] hese functions can be represented graphicall. In the graphs values of the forces are presented for points spaced b angles of π/. π / [ q] [ q].74 [ q ] ow let us solve an arch consisting of one quarter circle of radius with a clamped support and another quarter circle attached to the former b a hinge and supported b a roller. he arch is loaded b the uniforml distributed hdraulic pressure of the intensit q. α q qds b a qds qdscosα α V V he first thing to calculate is the reaction in the roller support V. his can be done from the equilibrium of moments with respect to the hinge for the right half of the arch. his equation reads: π / [ aqds + ( b) qds cosα ] : V 5

6 fter substitution for a and b and division b the reaction can be calculated as: π / π / π / [ cosα + cosα )] q cosα q[ ] q V q b b qds qds qdscosα α a a Vq qds qdscosα q 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. he adequate equations have the form: : + q cos : + q sin : q ( qds cos qds cosα sin) ( qds sin + qds cosα cos) ( ) + [ qds ( a ) + qds cosα( b) ] ote, that these equations are also valid for the angle greater than π/, i.e. within the left-hand half of the arch. From these equations the functions of internal forces follow: q cos + q cos q sin cosα q cos + q cos q sin + q sin [ cosα] q sin[ ] q cos q cos + q cos q sin q [ cosα ] + q cos[ ] q sin q sin cos + q sin + q cos sin q q q 6 cos) q [ cosα cos + cosα sin cosα ] cos) + q { cos[ cosα ] sin[ ] } cos) + q ( cos + cos sin ) q sin + q sin + q cos cosα he most important thing to observe in these results are the vanishing values of the bending moment and the shear force. In this wa we arrive at the notion of the optimal arch shape. It was mentioned in the introduction, that the arch was invented to carr mostl normal forces. It comes out, that for a given tpe of loading an arch shape can be found, which is characterized b the lack of bending. For the constant hdraulic pressure the corresponding shape is the circular arch, though the previous eample with the cantilever arch might den this statement. However, there must be one more condition fulfilled to get the optimal arch serving in the bending-free state.

7 amel, its supports at the ends must provide reactions resulting in the aial force at the arch ends. he 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 hdraulic pressure in the bending-free state features the constant value of the normal force equal to q throughout the entire arch length. s the last eample in this group let us consider a cantilever arch being a quarter circle of radius he arch is loaded b the hdrostatic pressure due to the liquid of the specific weight γ present onl on its right-hand side. h b a q h ds q h ds q b h dscosα α a q h ds q h dscosα his loading acts in the same direction, normal to the arch, as the previousl considered hdraulic pressure. ut the difference is in the fact, that its value is not constant but varies linearl with the increasing depth h measured from the liquid free surface. hus, in the equations corresponding to the hdraulic pressure the load q must be replaced b q h ( b) γ ( ) γh m γ Hence, the equilibrium equations for the cut-off segment of the arch read: : + : + π / : + ( qhds cos qhds cosα sin) π / ( qhds sin + qhds cosα cos) π / [ qhds ( a) + qhds cosα( b )] Having epressed all the variables under integrals in terms of angles, introduced the notation for q h and noted, that the angle does not undergo integration, the internal forces can be given as: γ γ γ cos sin π / π / π / π / ) + γ sin ) ) γ cos ) ) cosα( sin) γ ) ( cos cosα ) π / π / cosα cosα hese calculations involve the integrals of the functions sin α and cosα, which can be easil found using the following trigonometric relations 7

8 sin α cos α + cosα ow the functions of the internal forces can be determined γ γ π / ( sin α ) + γ sin ( cosα cosα ) γ cos cos cosα + sin α α 4 π / + γ π cos cos sin + + γ 4 4 π / sin + cosα 4 π / sin sin cos 4 4 γ γ π / π / ( sin α ) γ cos ( cosα cosα ) γ sin sin cosα + sin α α 4 π / π sin cos sin + γ 4 4 γ cos + cos α 4 π / cos sin cos 4 4 γ γ + γ γ γ π π / sin ( cosα cosα )( sin ) + γ ( sin α )( cosα cos) / π / ( cosα cosα ) γ ( cosα cosα sin α ) / π / ( cosα sin α cosα ) γ cos ( sin α ) π sin + cos α 4 π / γ sin sin cos γ 4 4 cos cosα + sin α α 4 π / π / π cos cos sin hese functions can be represented graphicall. In the graphs values of the forces are presented for points spaced b angles of π/. [ γ ] [ γ ] [ γ.4 ]

9 . Staticall indeterminate circular arches Staticall indeterminate arches can be solved using the fleibilit method. he shape of the arch determines onl a technique to calculate coefficients in the canonical equations of the method. hese coefficients represent displacements in basic loading states and can be obtained from the principle of virtual work. he adequate well known formula reads: δ ik s i he important issue to note is, that the integration must be carried out along the curve representing the arch. his 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 wa 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 eamples of staticall determinate arches to calculate loading resultant and its moments. Let us consider a staticall indeterminate arch in the form of the propped cantilever being one quarter of the circle with the radius, loaded b the horizontal concentrated force at the tip. k ds X he modified sstem he modified sstem for the fleibilit method has a removed support at the point and the equivalent reaction at this support is considered as the redundant force X. he canonical equation of the fleibilit method resulting from the condition of zero vertical displacement of the point reads δ X + δ For a circular arch the calculation of the fleibilit coefficients in this equation can be carried out analticall after the change to the polar co-ordinates. he analtical functions of the bending moments in the basic states: X and are necessar. X State X State he corresponding functions of the bending moments, with a sign convention attributing positive values to the moments setting the internal side under tension (and the eternal one under compression), have the form: 9

10 cos ( ) sin) he fleibilit coefficients can now be calculated from: δ δ s s ds s ds cos d s cos sin) he integral of the function cosα was alread discussed previousl and the one for the function cos can be found using the trigonometric relation With this in hand one gets: δ δ π / ( cos + ) π / ( cos sin ) cos cos + d d sin + π / d π 4 sin + cos hen the redundant force can be obtained from the canonical equation as δ X δ π π / he function of the bending moment in the staticall indeterminate arch follows from the superposition rule: X + cos + sin sin cos π π ( i ) ( ) o check the correctness of the results a kinematic check should be performed. Let us check if the cross-section rotation at the clamped support is zero. o this end the principle of virtual work and the reduction theorem are used: ( i ) ( ) ds s he virtual bending moment comes from an appropriate modified sstem, different than the one used in the calculations above. Hence, the fied support is replaced with a hinge and the virtual loading in the form of the concentrated unit bending moment is applied there. (?) V he virtual reaction at the support results from the equilibrium of moments with respect to

11 : V + V and the function of the virtual bending moment is: ( ) cos he check of the cross-section rotation follows as: ( i ) ( ) π / ds s sin + cos sin 4 π π cos sin cos d π π / his proves the correctness of the calculations b the fleibilit method! s the net eample let us consider a staticall indeterminate arch of a parabolic shape. he arch central ais is epressed in the co-ordinates sstem, b a function what for the data given in the figure ields 4f L ( L ) k/m k C D E f 4 L [m] he arch consists of two segments connected b a hinge. Eternall there are si reactions at the clamped supports and versus four equilibrium equations, the fourth is the equation of moments with respect to the hinge C. Hence, the sstem has two redundant forces. he modified sstem can be formed b the removal of the hinge C and then the horizontal and vertical internal forces therein are considered as the redundant forces X and X. 6 k/m k X C X X D E 4 X

12 he kinematic compatibilit conditions require, that the relative displacements of arch fragments tips at the point C measured in the horizontal and vertical directions are zero. his leads to the canonical equations of the fleibilit method in the form δx δ X + δ X + δ X + δ + δ and the fleibilit coefficients are calculated from the curvilinear integrals δ ik s i In the case of the parabolic arch the transformation to the polar co-ordinates is not efficient. he integrals would be too comple. Instead, first the integrals are transformed to the straight-line ones. From the figure k ds ds d the following relation can be deduced ds d cos where is the slope of the tangent to the arch. ow the fleibilit coefficients can be obtained from δ ik ik d cos It must be noted, that the angle is variable along the arch, depending on the co-ordinate. It can be obtained from the first derivative of the arch centre line function. In our eample d tan d arctan Substitution of this relation to the formula for the coefficients δ ik leads to ver comple integrals, which cannot be solved analticall. he onl reasonable wa to solve them is the numerical integration. here are several methods available: rectangles, trapezium, Simpson, the famil of Gauss quadratures, etc. Some of these methods are based on a geometrical approach, where the entire area below the function graph is approimated b a finite number of basic areas. Here we will appl the trapezium method illustrated in the figure f() f n fn f f f a n b

13 he function domain a, b is divided into a finite number of n equal intervals and the determinate integral is represented as a sum of trapeziums areas b a ( ) d + + n f... + fter a substitution of the formulae for the trapezium area the integral can be finall epressed in terms of a finite number of function values at the ends of intervals b n a ( ) d ( f + f + f f f ) f + he method is approimate and its accurac depends on the densit of the function domain subdivision. he larger is the number of intervals, the more accurate results are obtained. In the considered eample the integration domain is, m and it will be subdivided into intervals of. m. o carr out the calculations the functions of bending moments in the basic states: X, X and are necessar. n X C X 4 C X 4 X 9 C 6 k/m D k E ( ) ( 4.5) ( 4.5) ( 9) on D on DE on E

14 he results of calculations for the fleibilit coefficients are presented in the table. tan cos cos cos cos cos cos m m m m km m m m km km km ppling the trapezium formula for the coefficients ields the following results δ 9.6, δ 5.77, δ 44.75, 5.6 δ, δ 67. With these coefficients found the canonical equations can be solved to get the values of the redundant forces X 5.598k, X 5.856k, he final values of the bending moments in the arch are obtained from the superposition rule ( i ) X + X + and their values are given in the last column in the table above. he correctness of these calculations should be verified b the kinematic check. Here the crosssection rotation at the clamped support will be calculated from the principle of virtual work and the reduction theorem as ( i ) ( ) ( i ) ( ) ds d cos s o this end another modified sstem is introduced with the appropriate virtual loading in the form of the concentrated unit moment at the point. (?) (i) C D E f 4 4 [m] L 4 Having calculated the reactions at the supports in the usual wa the function of the virtual bending moment in the modified sstem can be epressed as ( ) + 4 4

15 he 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 (i) ( ) cos ( ) i () cos m m km km he final result of the kinematic check is his result related to the maimum absolute value component in the integral calculation, i.e / is onl.9%, what allows to conclude, that the angle is indeed zero and that the calculated bending moments in the staticall indeterminate arch are correct. o complete the calculations let us also find the functions of shear and aial forces in the arch C k/m D k E he values of reactions in supports and can be obtained from the equilibrium equations for the parts C and C 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 D 6 k/m α the equilibrium equations of forces in the direction and ield 5

16 5.65 cosα (.86 6) (.86 6)cosα For the right half of the arch D two cross-sections must be considered with the separation point E. α k E α he equilibrium equations ield: for ED section: for E section: (.4) 5.65 cosα (.4) cosα cosα.4 cosα It must be noted, that in all the analzed cases the angle α is an angle from the first quarter and is alwas positive. So it is related to the oriented angle asα he values of shear and normal forces are given in the following table cosα m m k k / 9..5 /

17 ote, that for 9. m, i.e. at the point E, where the point load is applied, the shear and aial forces are discontinuous. he respective jumps in the functions are equal to the respective components of the concentrated force along tangent and normal to the tangent. he values of the internal forces can now be presented graphicall [km] [k] [k] 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 etreme value of the bending moment. It is also interesting to observe the disturbing influence of the point load. he left half of the arch loaded b the distributed force is characterized b a ver small bending (and shear), while the right half with the point load is subjected to a relativel larger bending (and shear). ow let us consider the same arch subjected to the uniforml distributed load of the snow-tpe along the entire length. 6 k/m C f 4 9 L [m] For the convenience of the calculations the same modified sstem is adopted 6 k/m C X X X 4 X 9 7

18 with the sstem of canonical equations of the similar form as previousl: δx δ X + δ X + δ X + δ + δ he states X, X are identical and the fleibilit coefficients are δ 9.6, δ 5.77, δ he difference is onl in the state C 6 k/m 4 6 ( ) Calculation of the coefficients related to this state is given in the table cos cos 8 cos m m m m km km km km Using the trapezium formula the remaining fleibilit coefficients are found: 44 δ, δ 688 ow the canonical equations can be solved to get the values of the redundant forces (i) X 6.99k, X 8.k, he final values of the bending moments in the arch are obtained from the superposition rule ( i ) X + X + and their values are given in the last column in the table above. hese values are ver close to zero. Indeed, the should be eactl zero, because the parabolic arch is the optimal one for the snow-tpe loading and it is in the bending-free state provided the supports allow for the reactions transmitting the aial force. his conditions are fulfilled in the considered eample. Hence, the

19 small non-zero values of the final bending moments are due to the round-up errors and have to be neglected. In this situation ever kinematic check is fulfilled. Finall let us find the functions of the shear forces (which should be zero, too) and the non-zero aial forces. Having found the reactions 6 k/m C 8. D k/m α α 6 k/m the following equilibrium equations can be written for the left (from to D) 6.99 cosα and the right half (from to D) of the arch ( 6. 6) (6. 6)cosα 6.99 cosα 6.99 [ 6. 6( ) ] sin [ 6. 6( ) ] cosα It must be noted again, that in these cases the angle α is an angle from the first quarter and is alwas positive. So it is related to the oriented angle asα α 9

20 he values of shear and normal forces are given in the following table cosα m m k k he shear force values are indeed zero, the results in the last column are onl due to round-up errors. he graphical representation of the aial forces concludes the eample [k] 45.