The Hashemite University Department of Civil Engineering. Dr. Hazim Dwairi 1

Transcription

1 Department of Civil Engineering Lecture 8 Deflection and Camber Introduction Prestressed concrete beams are more slender than R.C. beams, high span/depth ratios; thus, more deflection. Camber may be important. Camber may increase, with concrete creep and with time. Bridge camber may cause pavement to be uneven, even dangerous. Excessive roof camber may create drainage problems. Excessive floor camber partition cracking and other non-structural cracking. 1

2 Introduction The total deflection is a resultant of the upward deflection due to prestressing force and downward deflection due to the gravity loads. Only the flexural deformation is considered and any shear deformation is neglected in the calculation of deflection. The deflection of a member is calculated at least for two cases: Short term deflection at transfer Long term at service loading Introduction The short term deflection at transfer is due to the initial prestressing force and self-weight without the effect of creep and shrinkage of concrete. The long term deflection under service loads is due to the effective prestressing force and the total gravity loads. The deflection of a flexural member is calculated to satisfy a limit state of serviceability. 2

3 Deflection due to Gravity Loads The methods of calculation of deflection are taught in structural analysis-i course. Such methods used are: Double integration method Moment-area method Conjugate beam method Principle of virtual work Students are expected to review at least one of the above mentioned methods. Deflection due to Gravity Loads 3

4 Deflection due to Gravity Loads Deflection due to Prestressing Force The prestressing force causes a deflection only if the CGS is eccentric to the CGC. Deflection due to prestressing force is calculated by the load-balancing method. 4

5 e Parabola 2 5 PeL = 48 EI L e 1 PeL = 12 EI 2 L/2 L/2 e 2 2 (3 4 a ) PeL = 24 EI al (1-2a)L al e 2 1 PeL = 8 EI L 5

6 e 1 Parabola e 2 e+e e 1 2 L Pe2L 1 Pe L 5 = + 8 EI 48 EI Moment of Inertia ' Class U: f t 0.62 f c Use gross section moment of inertia, I g ' ' Class T: 0.62 f c ft fc Use effective moment of inertia, I e ' Class C: f t > f c Use effective moment of inertia, I e 6

7 Effective Moment of Inertia M I = I + ( I I ) I 3 cr e cr g cr g M a M cr ftl f r = 1 M a f L M Max. service unfactored live load moment f f f tl r L a total service load concrete stress modulus of rupture service live load concrete stress Cracked Moment of Inertia The PCI Approach: 2 2 cr = ( p ps p + s s )( pρ p + sρs ) I n A d n A d n n n n p s E = E E = E s c ps c 7

8 Long-term Deflection Approximate Method: Due to prestress: pi + pe Final = pe Cu 2 Pe pe = i Pi Due to prestress & Self weight: pi + pe Final = pe Cu + (1 + Cu ) 2 To account for the effect of creep on self weight D Long-term Deflection Due to prestress, Self weight, sustained dead load & live load pi + pe Final = pe Cu + (1 + Cu )( 2 + )+ Alternatively, use long-term multipliers from PCI (Table 4.8.2) Deflection limits in ACI (Table 9.5-b) PCI design aids and for typical elastic deflections D SD L 8

9 Example The simply supported I-beam shown in cross- section and elevation is to carry a uniform service live load totaling 8kN/m over 12m span, in addition to its own weight. The beam will be pretensioned using multiple seven-wire strands, eccentricity is 130mm and constant. The P/S force immediately after transfer is 750kN, reducing to 530kN effective. The 28 day compressive strength of concrete is 40 MPa. Calculate deflections and check with allowable values. 8 kn/m 130mm 300mm 125mm 350mm 100mm 130mm A c = 110,000 mm 2 I c = x 10 9 mm 4 S = x 107 mm 4 r 2 = 42,595 mm 2 125mm 9

10 Compute Stresses at Transfer and Service -3.75MPa MPa P i + M D +3.16MPa -9.89MPa 2 P e + M D + M L Approximste Method: f = MPa < f = = 3.92 MPa t Class U: use I E c g = = 29,725 MPa r PeL i P i = = 8 EI 8 29, = 12.6 mm P i 530 P e = 12.6 = 8.9 mm

11 Long-term deflection at 360 days Creep Coefficient at 360 days: C C t t t 360 = C (2.35) 0.6 u = 0.6 t = = 1.82 P + i Pe 360 = P C e t = 8.9 ( 1.82 ) = 28.5 mm 2 Long-term deflection at full service load: P + i Pe = C + ( + )(1 + C ) + e 2 Net P t D SD t L Instantaneuos deflection due to selfweight 4 4 5w D L D = = 384 EI , = mm D Instantaneuos deflection due to Live load 4 4 5w LL D = = 384 EI , = mm L There is no superimposed dead load, = 0 = ( ) Net = mm Net 9 9 SD 11

12 PCI multipliers for long-term deflection and camber At Erection Deflection (downward) component apply to the elastic deflection due to the member weight at release of prestress Camber (upward) component apply to the elastic camber due to prestress at the time of release of prestress Without composite topping With composite deflection Final Without With Deflection (downward) component apply to the elastic deflection due to the member weight at release of prestress Camber (upward) component apply to the elastic camber due to prestress at the time of release of prestress Deflection (downward) apply to the elastic deflection due to the superimposed dead load only Deflection (downward) apply to the elastic deflection caused by the composite topping

13 Use PCI multipliers in previous example Selfweight multiplier = 2.7 Camber due to P multiplier = 2.45 i = ( 2.45 )( 12.6 ) + ( 2.7 )( 5.3 ) + ( 15.5 ) Net = 106mm 1.06 Net ACI maximum permissible deflections 13

14 AASHTO maximum permissible deflections 14