Unit C8: Prestressed Concrete

VOL-C8 Prestressed Concrete GENERAL PRINCIPLE Wires are stretched & the anchored against a concrete member Applies a net compressive stress to the member. used to counteract tensile bending stresses ➤ Concrete is much stronger in compression than tension PRE-TENSIONING Used mainly for precast members manufactured in factory lines of moulds with prestressing cables passed Cables are Stressed & anchored to blocks at either end Concrete is poured & allowed to cure Cables are released & cut at the edge of each member Anchor block Debonding sleeve mould pre-stressing cable Common per-tensioned members Precast bridge beams (Y&T beams) Precast voided slab units for building structures Precast Tee (or double Tee) beams for carparks & other building structures POST-TENSIONING Most common for members cast on site cast with dusts. once the concrete has gained sufficient strength, the cables are threaded through the ducts Cable stressed & anchored against the concrete. Common applications post-tensioned bridges, including glued segmental post-tensioned flat slabs PRESTRESSING DESIGN PRINCIPLES Designed firstly to satisfy Serviceability stress limits Both min & max moment cases need to be checked. Transfer case when the dead load is mim. the concrete strength is below its final strength Concrete 75 assumed linear elastic for the serviceability stress checks STRESS, COMBINATIONS Consider a rectangular beam with a straight prestressing cable at the centroid, which is anchored at the ends & applies prestress force P to the concrete P Comp. Comp. comp. NA. tension tension tension Bending stress Prestress Combined P Eccentric pre-stress, force Consider the case where a member has min. & max. loading the pre-stressing cable is not at the centroid but rather at some eccentricity of the member e from the N.A P ↑e (tue up) e (-ve up) M mm M max Bending moment envelope P Comp Comp. Tension Comp. Ν.Α. Tension Comp. B_{b} = (My)/I = M Z t,b D PA = P A B rs = (P e )y I = P e Z tb 3=3 0 +3 rA + 3rs Prestress Bending Prestress Axial Bending Combined The Stress on the top & bottom faces of the beam for an top+ Zt + M Ze applied moment of M Where M = applied service ability moment (sagging tue ) Prestress force (the for applied comp.) 12. always tue P = A = Cross-sectional area of member Section modulus at bottom face take as a tue no. (yb length from N.A. to the bottom edge) Z_{b} = = I / y_{6} Zt = section modulus at top face, taken as a tue no. (Y+= length from N.A. to the top edge) = I / y_{t} Cable eccentricity (tue upward) e = Stress in member (axial stress), compressive tue EXAMPLE bottom A lom simply supported beam of section b = 200 , h=600 has an axial prestress of loookN applied at an eccentricity of 100 mm below the N.A. level. Compute the stresses in the top & fibres of the beam at the mid-span location due to this prestress & dead weight only. Assume a weight density of 25KN/m³ for the concrete W = (0.2 * 0.6) * 25 = 3kN / m A = 200 * 0.6 = 1.2 * 10 ^ 5 * m * m ^ 2 Z_{t} = Z_{b} = (b * h ^ a)/b = (200bo * s ^ 2)/b = 1.2 * 10 ^ 5 = (1000 * 10 ^ 3)/(1.2 * 10 ^ 5) + ((1000 * 10 ^ 3)(- 100))/(1.2 * 10 ^ 5) +* 37.5*10^ 8 1*2*10^ 5 M= omega L^ 2 8 = 3*10^ 2 8 *= 37.5kNm = 37.5 * 10 ^ 6 * Nmm partial tp = p A + p.e Zt + M/(Zt) = 3.125N / m * m ^ 2 13.5N / m * m ^ 2 = 1000*10^ 5 1.2*10^ 5 - ((1000 * 10 ^ 3)(- 100))/(1.2 * 10 ^ 5) *-* 37.5*10^ 6 1.2*10^ 5 B boffon = p A - p.e Z_{b} - M/Z_{b} REAL PRE STRESSED STRUCTURES Normal to vary the level of prestress along a member (by debonding cables) Normal to vary the eccentricity by having non- straight cables DEBONDING sleeves are put over cables & only become effective at the end of the sleeve . All transfer is bond Method in bridge beams CURVED CABLES lateral load transferred prestress force x Change in angle of the cable over the section results in a ull for a parabolic cable moment at any point due to the stresses = Pe STRESS LIMITS & DESIGN CRITERIA To obtain an economical level of phe-stress while ensuring that all the stress criteria are satisfled Jacking force (P) the reference prestressed force the max. ever applied . As soon as the jacking process. TS finished the force in the prestress = represents the transfer case alpha_{t} = 0.85 to 0.95 Due to Creep & shrinkage of the concrete & the relaxation of the steel tendons (cables), the force reduces to dup alpha_{w} = 0.654 * 0.15 working (long-tenm) case Tendons are normally made up of 7-seven wire strands Revelant criteria (alpha_{t}*p)/A + (alpha_{t}*pe)/Z_{t} + M tr Z t >=-f tt (1) Top transfer gamma t = p/A + (pe)/z_{t} +* M 2 epsilon (alpha_{t}*p)/A - (alpha_{t}*pe)/Z_{b} - M_{t}/Z_{b} <= f ct (2) Bottom transfer delta b = p A - p.e Z_{b} - M/Z_{b} (alpha_{omega}*p)/A + alpha omega p.e Z t +* M max Z t <= f i omega (3) Top compression working (alpha_{w}*p)/A - (alpha_{w}*pe)/Z_{b} - M wax Z b >=-f tw (4) Bottom tension working (alpha_{w}*p)/A - (alpha_{w}*pe)/Z_{b} - M mn Z b *<=* f cw (6) Bottom Compression working where f tt = tensile limit at transfer Compressive limit at transfer f ct = tensile limit at working f tw = f cw = Compressive limit at working Equations (1) to (6) can be used to Compute min moduli (3)-(5), Z t >= M omega*omega*x -M omega n f omega omega +f t omega *-Q f epsilon*omega +f t omega i >= M max -M max Z epsilon f c*omega *+f t omega >=* M max -M mn Z b Z b *>= M max -M omega n f omega +f i omega =0 (6) - (4) H is advisable to increase the mộn 2 values by 30%, to ensure sensible prestress levels Only the working level Case will be considered [ equation (3) t_{0}(6) ] Zt fow () A (): > e (9) Top compression Zb fow ()+-Mmak (UP) ze (10) Bottom compression -Z t * f c*omega *( 1 alpha omega p )- Z 4 A -M max *( 1 alpha omega p )<= epsilon (11) Top tension - Z_{b} * f_{omega}(i/(alpha_{omega}*p)) + Z_{b}/A -M max *( i alpha omega p )<= e (12) Bottom compression Procedure 1) Compute section load prop. & Serviceability moments (partial factor of I for both permanent & Variable 2) Choose an eccentricity for the tendon In the lower flange 4) Check all working criteria ie. (3) to (6) EXAMPLE For the beam shown below, select an appropriate level of prestress by eccentricity & check whether all working stress criteria are satisfied. The transfer condition is neglected here... Assume f cc =40 N/mm^ 2 f cw =24 N/mm^ 2 , few 3.5N/mm² alpha_{w} = 0.15 min. working load = lokN/m max (mcludes self weight) = 20kN / m 14m N.A. 400 150 150 400 e 150 A= 400×150× 3 = 180000 mm² e = - (200 + 75) = - 275mm = - 0.275m I 400x700 12 3 125x400 12 3 x 2 = 1.01 x 100 mm 4 = 0.0101 m4 2 = 0.18 m² In the middle of bottom flange l.e. e = - 0.275m Select e min. w man =10 kN/m M min := W mm L^ 2 8 = 10* i4^ 2 8 =245 kNm max. = 20 kN/m M max :=: W max L^ 2 8 *= 20* i varphi^ 2 8 =490KNm Rearrange equation (4) (4) bottom tension working (alpha_{w}*p)/A *-* alpha w p e Z b - M max Z b >=-f t omega Z b = I y b = 0.010| 30 * 10 ^ - 3 = 0.02 * 0.886 p= -F sut + M sun 2 b alpha w ( 1 A - e Z w ) = -3500+ 4 pi10 0.0276 0.15( 1 0.0.0215 - -0.215 0.02176 ) = 13479 12.82 =io51kd Say P = 1100KN Check all working Stresses f f cw =24 N/cn^ 2 , f tw =3.5 N/mm^ 2 ) STRESS LIMITS eq * 3 = 13264kNl * m ^ 2 <= f iw (alpha_{omega}*p)/A + alpha omega p.l Z t +* M max Z t <= f epsilon omega :-(z) (alpha_{w}*p)/A - (alpha_{w}*pe)/Z_{b} - M max Z b >=-f t :-(4) e 9 4= - 287bkv / (m ^ 2) ...>=-f tw e*q_{5} = 4774kN / (w ^ 2) >=-f tw (alpha_{u}*P)/A + alpha u P.e Z t + M un Z t >=-f tu -*(5) e qb =5615 kv/ kappa^ 2 <= f cw . (alpha_{w}*p)/A - (alpha_{w}*pe)/z_{b} - M wn z b <= f cw -(6) Comp. strength Imits 0.6 fak under characteristic loads & prestress (unfactored) 0.45 fck under quasi-permanent loads k prestress Tensile strength Imits feem may be taken as 0.3 f cx ^ 2/ s Other Checks not considered ULS Flexure. ULS Shear ULS Shear flow Deflection Stresses at anchorages in post-tension members LOSS OF PRESTRESS 1) Anchorage slip (post-tensioned only) . when a prestress jack releases a tendon after stressing the cable draws-in' a little before the wedges lock the tendon off More important in Short cables Draw-in is around 5mm 2) Elastic shortening Pre-tensioned concrete An immediate loss of pre-stress when cables are released from the anchorage blocks due to elastic shortening of the concrete Post tensioned members Cables are Jacked against the concrete ➤ If only one cable is present 1 the elastic shortening loss is zero ➤ If more than one In each cable according to the Cable, there will be different losses 3) Friction losses occur due to friction between ducts (or deviators) & tendons 4) Creep Creep strain (Ecrp) depends on the ultimate creep factor ($) Short term elastic modulus of the concrete (Ec) & the stress level in the concrete at the position being considered Oc Φ Ec Eup=E 00 Φa varies with concrete mix & enviromental conditions vary between 2 44 5) Shrinkage The earlier the prestressing is applied the greater the F loss due to shrinkage Shrinkage strain (Eus) = autogenous component (Ecd) + drying component. (Eco). Use the long term value (after 30 years) Ecd varies the type of concrete (10%-20% of the total shrinkage) Eca varies with type of concrete & environmental conditions Depends on the notional size (2Ac/u) [u = exposed perimeter] Typizal values Ecn = 280 × 10-6 to 330x106 for outdoor exposure. Em 300x106 to 600 x 106 for indoor exposure 6) Relaxation of steel within tendons accounts for 1% to 5% loss of pre-stress Depends on the Stress level & stress history in the tendon EXAMPLE A lom simply supported beam of section b=200, h=600 has an at an eccentricity of 100 mm below the N.A. level. axial prestress of loookN applied The beam has long term loading (g.p.) of 6kN/m (this includes its self-weight). Ecs = 300x106, Assume fpk 1890 N/mm², Ec = 25000 N/mm², Es 200000 N/mm², Φα=3 tendon area = Ap=556mm² (four 7-wire strands), & the stress at jacking is 70% of the characteristic Steel Strength (ie. 0.7 x 1860 N/mm²) Calculate the total losses due to the following (i) anchorage slip of 5 mem one end 5 == L = 5×10-4 10000 AB dram in = Edrawin Es = (5×10-4) (200000) 100 N/mm² (ii) Shrinkage (iii) Creep strain at the level of the prestressing tendon Section I_{c} = (b * h ^ 3)/12 = (200 * 500)/12 ^ 3 = 2.08 * 10 ^ 9 * m * m ^ 4 A_{c} = bh = 200 * 500 =1*10^ 5 mm^ * Applied M pp = w q rho L^ 2 8 = 6*10^ 2 8 =.751Nm Tendon Jacking force (P) matrix p&=&0.7 f pkl &A p. &=&0.7*&i860*556&=&44 4* io^ 3 &N&=&4i4 kN matrix Stress at Jacking Compute the stress in the concrete at the level of the tendon (allowing for the draw-in loss) P feadar. = 12 * 2 * 556 = 668 * 10 ^ 3 * N = 669kN = Ptenden Ac + = 5.76N / m * m ^ 2 Y tendon Prendem ecc Ic Z + Map Y tenden Ic Z = (669 * 10 ^ 3)/(10 ^ 5) + ((669 * 10 ^ 3)(- 40)(- 40)(- 40))/(2delta * 10 ^ 4) + (75*10^ 4 )(-40) 2.0 delta*10^ 9 . sigma_{b} = p/A - p_{e}/z_{b} - M/z_{b} O t = p A +. p_{e}/Z_{t} + M/Z_{t} vec S toalm = 0.7 * 186o - 100 = 1202N / m * m ^ 2 compute the Creep strain in the concrete at the level for the tendon Sigma op = delta c E e phi= 5.7 delta 250 00 *3=6.9|* is^ -4 Compute the Charge In Stress in the tendon due to the creep Strain Delta delta op = Sigma op E s. =6.91*10^ -4 *200000.=138N/m^ * Total loss Delta F tot =i Delta0+b phi+1.38.=298 N/mm^ 2 298 % loss = 22.9% 0.7×1860×100%