Design of Circular Beams

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1 The Islamic University of Gaza Department of Civil Engineering Design of Special Reinforced Concrete Structures Design of Circular Beams ENGC 6353 Dr. Mohammed Arafa 1

2 Circular Beams They are most frequently used in circular reservoirs, spherical dome, curved balconies etc. Circular beams loaded and supported loads normal to their plans. The center of gravity of loads does not coincide with the centerline axis of the member. ENGC 6353 Dr. Mohammed Arafa

3 Circular Beams Circular beam are subjected to torsional moments in addition to shear and bending moment The torsional moment causes overturning of the beams unless the end supported of the beams are properly restrained. ENGC 6353 Dr. Mohammed Arafa 3

4 Semi-circular beam simply supported on three equally spaced supports ENGC 6353 Dr. Mohammed Arafa 4

5 Semi-circular beam supported on three supports Supports Reactions Let R A = R C = V 1 R B = V Taking the moment of the reaction about the line passing through B and parallel to AC M 0 V 1R wr R WR V 1 R V V F y 0 WR wr Rw ENGC 6353 Dr. Mohammed Arafa 5

6 S.F and B.M at Point P at angle from support C ENGC 6353 Dr. Mohammed Arafa 6

7 Semi-circular beam supported on three supports Taking the bending moment at point P at angle from the support say, C equal: R sin / M V 1R sin wr sin / / wr M R sin wr sin / M wr sin sin / ENGC 6353 Dr. Mohammed Arafa 7

8 Semi-circular beam supported on three supports Maximum negative moment occurs at point B M M ( at ) M max. ve B B 0.49wR Maximum Positive moment dm Max. M ve where ( 0) d max dm wr cos sin / cos / 0 d tan 9 43' M wR 8

9 Semi-circular beam supported on three supports Torsional Moment at point P R sin / T V 1 R R cos wr R cos / / T wr cos sin dt The section of maximum torsion 0 d dt 0 at 59 6' d T 0.104wR max. ENGC 6353 Dr. Mohammed Arafa 9

10 Circular beams loaded uniformly and supported on symmetrically placed columns A B ENGC 6353 Dr. Mohammed Arafa 10

11 Circular beams at symmetrically placed columns ENGC 6353 Dr. Mohammed Arafa 11

12 Circular beams at symmetrically placed columns Consider portion AB of beam between two consecutive columns A and B be with angle The load on portion AB of beam =wr The center of gravity of the load will lie at a distance R sin( /) ( /) from the center Let M 0 be the bending moment and V 0 be the shear at the supports ENGC 6353 Dr. Mohammed Arafa 1

13 Circular beams at symmetrically placed columns From geometry V 0 wr / The moment M 0 at the support can be resolved along the line AB and at right angle to it. The component along line AB is M 0 sin( /) and 0 at right angle of it. Taking the moment of all forces about line AB M cos( /) ENGC 6353 Dr. Mohammed Arafa 13

14 Circular beams at symmetrically placed columns Taking the moment of all forces about line AB R sin / M 0 sin / wr R cos / / M M 0 0 wr wr sin / cos / sin / / sin / 1 cot / ENGC 6353 Dr. Mohammed Arafa 14

15 S.F and B.M. at point P, at angle from the support Load between points A and P is wr Shearing force at point P wr V P wr wr ENGC 6353 Dr. Mohammed Arafa 15

16 Shear force and bending moment at point P The direction of vector representing bending moment at point P will act along line PO R sin / M V 0R sin M 0cos wr sin / / wr M wr wr M sin 1 cot / cos sin / wr sin cos cot / cos sin / since cos 1sin / M 1 sin cot / cos wr 16

17 Shear force and bending moment at point P The direction of vector representing torsion at point P will act at right angle to line PO wr R sin / T M sin R R cos wr R cos / 0 / since R R cos R 1 cos R sin / T wr 1 cot / sin wr sin / wr 1 sin / cos / 1 T wr 1 cot / sin wr sin / wr 1 sin T wr sin cot / sin sin / sin T wr cot / sin sin / since sin / 1cos T cos sin cot / sin 17 wr

18 Shear force and bending moment at point P To obtain maximum value of torsional moment wr R sin / T M 0 sin R R cos wr R cos / / dt wr 1 sin sin cot / cos 0 d M dt d 0 1 sin sin cot / cos 0 wr i.e. point of maximum torsion will be point of zero moment ENGC 6353 Dr. Mohammed Arafa 18

19 Circular beams loaded uniformly and supported on symmetrically n placed columns # of Supports R V K K` K`` Angle for Max Torsion 4 P/4 P/ P/5 P/ P/6 P/ P/7 P/ P/8 P/ P/9 P/ P/10 P/ P/1 P/ P M M T rw max support Midspan KwR K wr `` K wr ` T max M max(+ve) M max(-ve) V max T max ` / M max(-ve) V max O 19

20 Example 1 Design a semi-circular beam supported on three-equally spaced columns. The centers of the columns are on a circular curve of diameter 8m. The beam is support a uniformly distributed factored load of 5.14 t/m in addition to its own weight. f 350 kg / cm and f 400 kg / cm ' c y min L r 6.8m h L / / cm use Beam 40 70cm o. w. of the beam =0.4(0.7)(.5)(1.) = 0.84 t/m Total load= =5.98 t/m 0

21 Example 1 M max( ve ) 0.49wR 0.49(5.98)(4) t. m M max( ve ) wR (5.98)(4) t. m T max 0.104wR 0.104(5.98)(4) 9.97 t. m Reactions wr 5.98(4) R A 13.65ton R wr (5.98)(4) 47.84ton B wr Shear at point P: V wr 1

22 Example 1 Shear Force Diagram Bending Moment Diagram ENGC 6353

23 Example 1 Design for Reinforcement d= =63.95 A A ve s ( ve ) ve s ( ve ) cm cm min min ENGC 6353 Dr. Mohammed Arafa 3

24 Example 1 Design for Shear V 3.9 ton ( T 0.0at middle support) V max c (40)(63.95) 19.0ton V S 6.53 ton 0.75 AV (63.95) S AV cm / cm S ENGC 6353 Dr. Mohammed Arafa 4

25 Example 1 Design for Torsion if T u 0.65 p cp f A ' c cp cp cp 0h 0 0 h neglect torsion A bh, p b h, A x y, p x y and A 0.85x y The following equation has to be satisfied to check for ductlity V u Tu p h V c '.1 f c bw d 1.7Aoh bw d Reinforcement AT Tu S f A ys f A A p A p ' AT c cp AT l h and l,min h S f y S ENGC 6353 Dr. Mohammed Arafa 5

26 Example 1 Design for Torsion At section of maximum torsion is located at =59.43 T=9.97 t.m wr V u wr (4)( ) 11.6ton 360 M 0.0 V (40)(63.95) 19.0ton V No shear reinf.required c f ' ' 0.65 f c Acp c p cp 1.3 t. m T i.e torsion must be considered u u ENGC 6353 Dr. Mohammed Arafa 6

27 Example 1 Design for Torsion: Ductility Check The following equation has to be satisfied 0 0 V u Tu p h 0.63 f bw d 1.7Aoh x y p x y cm A h oh cm 3 5 V u Tu p h bw d 1.7Aoh ' c V c 19.0 bd w kg / cm '.1 f c kg / cm i.e section dimension are adequate for preventing brittle failure due to combined shear stresses. 7

28 Example 1 Design for Torsion 5 AT Tu S f A A S T min ys cm / 3.5b w f 400 ys cm / AV AT cm / cm S S A 1.13 Use (closed stirrups) with cm S 11.5 cm cm / cm 8

29 Example 1 Design for Torsion AT Al ph cm S l,min s, ve total s, ve total ' 1.3 f c Acp A T h f y S 400 A p cm Al 5.33cm 3 Use 0 top and 414skin reinforcement A A cm use 618mm cm use80mm 9

30 Example 1 Design for Torsion 30

31 Internal Forces in Circular beams Using polar coordinate dv w ds rd ds dv wr d dm ds dt dr dm ds dt V dr T r T V r as r= constant A B P dm T Vr d ENGC 6353 Dr. Mohammed Arafa 31

32 Internal Forces in Circular beams Using polar coordinate The component of the moment M along ds in the radial direction must be equal to the difference of the torsional moments along the same element dt Md dt M d Mds r This means that the torsional moment is maximum at point of zero bending moment ENGC 6353 Dr. Mohammed Arafa 3

33 Internal Forces in Circular beams Using polar coordinate Differentiating the equation dm T Vr d d M dt dv r d d d d d M M wr d Vr d r w The solution of this deferential equation gives M A sin B coswr ENGC 6353 Dr. Mohammed Arafa 33

34 Internal Forces in Circular beams Using polar coordinate M A sin B coswr The constants can be determined from conditions at supports The internal forces are in this manner statically indeterminate. In many cases, they can be determined from the conditions of equilibrium alone because the statically indeterminate values are either known or equal to zero. ENGC 6353 Dr. Mohammed Arafa 34

35 Circular beams loaded uniformly and supported on symmetrically n placed columns n rw R n 0 0 n rw V n The bending moment and shearing force V at any section at an angle from the centerline between two successive supports are given by: M r w n T r w n V cos 1 sin0 sin sin0 rw ENGC 6353 Dr. Mohammed Arafa 35

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