1. CABLE LAYOUT CONTINUOUS BEAM LOAD BALANCING METHOD 1/24/20141

2. Group 5B Arafat Hossain Sadik Hasan Sadia Mahajabin Raihan Mannan Sadia Mannan Tanvir Alam Ikhtiar Khan Ifat Hasan Shawon Md. Asif Rahman Sarani Reza1/24/201410.01.03.096 10.01.03.097 10.01.03.098 10.01.03.101 10.01.03.102 10.01.03.104 10.01.03.106 10.01.03.107 10.01.03.108 10.01.03.110 2 3. 1/24/20143 4. What is Cable Layout Cable : A strong thick rope, usually twisted hemp or wire Layout : The arrangement or plan of something1/24/20144 5. Simple Beam Layout Simple Beam: a structural beam that rests on a supportat each end Simple Beam can be two types :1. Pretensioned 2. Posttensioned1/24/20145 6. Pre-tensioning : When the steel is tensioned before concrete placement, the process is called pre-tensioning. Post-tensioning : When the steel is tensioned after concrete placement, the process is called post-tensioning.1/24/20146 7. Layouts for pre-tensioned beam 1. Straight cables are preferred since they can be more easily tensioned between two abutments. 2. Such a section can not often economically designed because of conflicting requirements of the midspan and end section. 3. At the maximum moment section generally occuring at mid span, it is best to place the cable as near to the bottom as possible. 4. Since there is no external moment at the end it is best to arrange the tendons so that c.g.s will coincide with c.g.c1/24/20147 8. Bent SoffitCurved Soffit1. For both layouts c.g.s at mid span can be depressed as low as desired. 2. The end can be kept near c.g.c1/24/20148 9. Bent ExtradosCurved Extrados1. When it is possible to vary the extrados of concrete it can be advantageously used. 2. These will give a favorable height at mid span where it is most needed and yet yield a concentric or nearly concentric prestress at end section. 3. (d) is simpler in formwork than (e).1/24/20149 10. 1. Most pretensionning plans in USA have buried anchores along the stressing bed so that tendons for a pretensioned beam can be bent. 2. It may be economical to do so if the beam has to be of straight and uniform section.1/24/201410 11. Layouts For Post tensioned Beam1. Most of the layouts for pre tensioned beams can be used for post tensioned once as well. 2. For a beam of straight and uniform section the tendons are very often curved. 3. Curving the tendons will permit favorable position of c.g.s to be obtained at both the end and mid span sections and other points as well.1/24/201411 12. combination of curved tendons with curved soffits1. A combination of curved tendons with curved soffits is frequently used when straight soffits are not required. 2. This will permit a smaller curvature in tendons thus reducing the friction.1/24/201412 13. 1. Curved or bent cables are also combine with beams of variable depth.1. Combination of straight and curved tendons are sometimes found convenient.1/24/201413 14. 1. Some cables are bent upwards and anchored at the top flanges.1. Some cables are stopped part way in the bottom flange. 2. This arrangements will save some steel but may not be justified unless the saving is considerable as for very long span carrying heavy loads.1/24/201414 15. Cantilever Beam A cantilever is a beam anchored at only one end. The beam carries theload to the support where it is forced against by a moment and shear stress. Two general layouts are possible for cantilevers :1.Single Cantilevers 2. Double Cantilevers 1/24/201415 16. LAYOUTS FOR SINGLE CANTILEVER1. For a short span with a short cantilever a straight and uniform section may be the most economical. 2. It is only necessary to vary the c.g.s profile so that it will confirm with the requirements of the moment diagrams.1/24/201416 17. 1. When the cantilever span become longer it is advisable to taper the beam. 2. If the anchore span is short compared to the cantilever it may be entirely subjected to the negative moments and the c.g.s may have to be located above the c.g.c1/24/201417 18. 1. For longer span it may be desirable to haunched them. 2. The c.g.s profile can be properly curved or may remain practically straight. 1/24/201418 19. Layouts for double cantilever1. For short double cantilevers a straight and uniform section can be adopted.1. When the cantilevers are long they may be tapered. 1/24/201419 20. 1. If the anchor span is long it may be haunched.1. If the anchor span is short compared with the cantilevers the c.g.s line may lie near the top of the beam.1/24/201420 21. CONTINIOUS BEAM1/24/201421 22. CONTINUOUS BEAM A Continuous beam is one, which is supported on more than two supports. For usual loading on the beamhogging ( -ve ) moments causing convexity upwards at the supports and sagging ( +ve ) moments causing concavity upwards occur at mid span. 1/24/201422 23. ASSUMPTIONS OF CONTINUOUS BEAM 1. The ecentricities of the1/24/2014prestressing cable are small compared to length of the members. 2. Frictional loss of prestress is negligible. 3. The same tendons run through the entire length of the member.23 24. Advantages of Continuous Beam Over Simply Supported Beam The maximum bending moment in case ofcontinuous beam is much less than in case of simply supported beam of same span carrying same loads. In case of continuous beam, the average bending moment is lesser and hence lighter materials of construction can be used to resist the bending moment. 1/24/201424 25. COMPARING OF LOAD CARRYING CAPACITY BETWEEN SIMPLY SUPPORTED BEAM AND CONTINIUOUS BEAM1/24/201425 26. LOAD CARRYING CAPACITY OF SIMPLE BEAM1/24/201426 27. LOAD CARRYING CAPACITY OF SIMPLE BEAM1/24/201427 28. LOAD CARRYING CAPACITY OF CONTINUOUS BEAM1/24/201428 29. LOAD CARRYING CAPACITY OF CONTINUOUS BEAM1/24/201429 30. DISADVANTAGES OF CONTINUOUS BEAM1. Frictional loss is significant 2. Shortening of long continuous beam under prestress. 3. Concurrence of maximum moment and shear over support. 4. Difficulties in achiving continuity for precast elements. 1/24/201430 31. CABLE LAYOUT OF CONTINUOUS BEAM1/24/201431 32. Curbed Tendon In Straight BeamThis lay out is often used for slabs or short span beams The main objections here are the heavy frictional loss 1/24/201432 33. Straight Tendon In Curbed BeamUsed for longer span and heavier loads. Often difficult to get the optimum eccentricitiesalong the beam if the tendons are to remain entirely straight 1/24/201433 34. Curbed Tendon In Haunched Or Curbed BeamsThis would permit optimum depth of beam as well asideal position of steel at all points,aboidin excessive frictional loss. 1/24/201434 35. Overlapping TendonsOffer a possibility of varying prestressing force alongthe beam1/24/201435 36. Determination Of Resisting Moment Of Continuous Beam 1. Plot the primary moment diagram for the entirecontinuous beam as produced only by pre-stress eccentricity. As if there were no support to the beam. 2. Plot shear diagram. 3. Plot loading diagram. 4. Plot moment diagram corresponding to loading diagram considering all support.1/24/201436 37. 1/24/201437 38. 1/24/201438 39. Load Balancing Method Load in the concrete is balanced by stressing the steel.In the overall design of prestressed concrete structure, the effect of prestressing is viewed as the balancing of gravity load . This enables the transformation of a flexural member into a member under direct stress and thus greatly simplifies both the design and analysis of structure. The application of this method requires taking the concrete as a free body and replacing the tendons with forces acting on the concrete along the span.1/24/201439 40. Concept of Load Balancing Method There are three basic concepts in prestressed concretedesign 1. Stress concept : Treating prestressed concrete as an elastic material 2. Strength concept: Considering prestressed concrete as reinforced concrete dealing with ultimate strength. 3. Balanced load concept: Balancing a portion of the load on the structure. Load Balancing method follows the third one. 1/24/201440 41. Life History Of Prestressed Member Analysing the life history of the prestressed memberunder flexure leads to understanding the balanced load concept relative to other two concept. So lets find out the load deflection relationship of a member as a simple beam..1/24/201441 42. K1 = Factor ofsafety applied to working load to obtain minimum yield point. K2= Factor ofsafety applied to ultimate strength design to obtain minimum ultimate load. 1/24/201442 43. The load deflection relationship of the above figureleads to several critical points. Such as.. 1. Point of no deflection which indicates rectangular stress block. 2. Point of no tension which indicates triangular stressblock with zero stress at the bottom fiber. 1/24/201443 44. 3. Point of cracking which occurs when the extreme fiber is stressed to the modulus of rupture. 4. Point of yielding at which steel is stressed beyond its yield point so that complete recovery is not possible. 5. Point of ultimate load which represents the maximum load carried by the member at failure.1/24/201444 45. Mechanism & Explanation Of Balanced load Concept According to figure there are three stages of beam Behavior : Applied LoadingsStages of beam behavior DL+k3LL No deflection DL+LL No tension K2(DL+LL) Ultimate1/24/201445 46. Where, DL+LL is the stress concept with some allowabletension on beam or no tension. K2(DL+LL) is the strength concept consists with the ultimate strength of the beam. DL+K3LL is the balanced load concept with the point of no deflection where k3 is zero or some value much less than one.1/24/201446 47. In balanced load concept design is done by point of nodeflection. So prestressing is done in such a way so that effective prestress balances the sustained loading & beam remain perfectly level without deflecting.1/24/201447 48. ADVANTAGES OF LOAD BALANCING METHOD 1.simplest approach to prestressed design and analysis for statically indeterminate structures. 2.It has advantages both in calculating and in visualizing. 3.Convenience in the computation of deflections.1/24/201448 49. SIMPLE & CANTILEVER BEAM WITH LOAD BALANCING METHODFigure illustrates how to balance a concentrated load by sharply bending the c.g.s. at midspan , thus creating an upward component V=2Fsin1/24/201449 50. If V exactly balances a concentrated load PApplied at midspan the beam is not subjected to any transverse load. The stresses in the beam at any section are simply given byAny loading addition to P will cause bending and additional stresses computed by1/24/201450 51. Figure illustrates the balancing of a uniformlydistributed load by means of a parabolic cable whose upward component is given by1/24/201451 52. If the externally applied load w is exactly balanced bythe component Wb there is no bending in the beam. The beam is again under a uniform compression with stressExternal load produced moment M and corresponding stresses1/24/201452 53. Figure represents a cantilever beam. Any verticalcomponent at the cantilever end C will upset the balance, unless there is an externally applied load at that tip. 1/24/201453 54. To balance a uniform load w, the tangent to the c.g.s. atC will have to be horizontal. Then the parabola for the cantilever portion can best be located by computingAnd the parabola for the anchor arm by1/24/201454 55. EXAMPLE PROBLEM OF CANTILEVER BEAM A double cantilever beam is to be designed so that itsprestress will exactly balance the total uniform load of 23.3 KN/m normally carried on the beam. Design the beam using the least amount of prestress, assuming that the c.g.s must have a concrete protection of at least 76.2mm. If a concentrated load of 62KN is added at the mid span compute the maximum top and bottom fiber stresses1/24/201455 56. SOLUTION: To use the least amount of prestress, the ecentricity over the support should be a maximum that is , h=300mm or o.3m. The prestress required is F = wL2 / 2h =(23.3x62)/(2x0.300)=1395KN The sag for the parabola must be h1 = w(L1)2/8F =(23.3x14.82)/(8*1395) =0.46m1/24/201456 57. Uniform compressive stress f= F/Ac = 1395/(2.28x105) = 6.12 Mpa Moment M at the mid span due to P=62KN M=PL/4=(62x14.8)/4=229KN-m Extreme fiber stresses are f=Mc/I=6M/bd2 = (6x229x106)/(300x7602) = 7.93MPa Stress at mid span are ftop = -6.12-7.93=-14.05MPa compression f bottom = -6.12+7.93 =+1.81 MPa tension 1/24/201457 58. EXAMPLE PROBLEM OF CONTINUOUS BEAMS: For the symmetrical continuous beam prestressed with F= 1420KN along a parabolic cable as shown, compute the extreme fiber stresses over thecenter support DL+LL=23.0KN/m1/24/201458 59. SOLUTION: The upward transverse component of prestress is Wb= 8Fh/L2 = (8x1420x0.3)/(15x15)=15.1KN/m The beam is balanced under uniform stress f= (1420x103)/(300x760)= -6.2MPa For applied load w=23.0 KN/m, the unbalanced downward load = (23.0-15.1) = 7.9 KN/m This load produces a negative moment over the center support, M= wl2/8 =(7.9x152)/8 = 222KN-m And fiber stresses, f = Mc/I = (6x222x106)/(300x7602) = 7.68 Mpa ftop = -6.2+7.68=+1.48MPa tension f bottom = -6.2-7.68 =+13.88 MPa compression 1/24/201459 60. 1/24/201460