PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
1 / 13
FRICTION LOSSES (REF a, b, c & f)
Conside...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
2 / 13
The above equation gives the frictional lo...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
3 / 13
In above equation m is approximately equal...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
4 / 13
SOLUTION
Given Data:
Wobble friction coe...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
5 / 13
0.75mm. If long shims are required to hold...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
6 / 13
x = Distance between two points
μ = Coeff...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
7 / 13
TWO-SEGMENT VARIATION (Ref. g)
For cases ...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
8 / 13
The figure above shows the losses that occ...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
9 / 13
Where, f1 = foe−kx for 0≤x≤L (Before ancho...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
10 / 13
Percentage loss of prestress due to ancho...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
11 / 13
α1 = 0 α2 = 8y22L2 = 8(0.150) 2(30) = 0.0...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
12 / 13
x f1 f2 Friction loss, fo – f1 Anchorage ...
PRESTRESSED CONCRETE
FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK
13 / 13
Percentage loss of prestress due to frict...
of 13

Prestress loss due to friction & anchorage take up

This document provides a detailed procedure for calculating prestress loss due to anchorage take-up. Prestress Loss due to friction is also discussed in detail.
Published on: Mar 4, 2016
Published in: Engineering      
Source: www.slideshare.net


Transcripts - Prestress loss due to friction & anchorage take up

  • 1. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 1 / 13 FRICTION LOSSES (REF a, b, c & f) Consider an infinitesimal length dx of a prestressing tendon whose centroid follows the arc of a circle of radius R, then the change in angle of the tendon as it goes around the length dx is, dα = dxR FRICTIONAL LOSS dF ALONG THE LENGTH dx DUE TO CURVATURE FRICTION (Ref. a) For this infinitesimal length dx, the stress in the tendon may be considered constant & equal to P; then the normal component of pressure produced by the stress P bending around an angle dα is, N = Pdα = PdxR Let, μ be the coefficient of curvature friction & K, the wobble friction coefficient. The amount of frictional loss dP around the length dx is given by, dP = −μN = − μPdxR = −μPdα dPP =−μdα Integrating on both sides with limits P1 & P2, ∫ 1PdPP2P1=−∫μdα ln|P2| − ln|P1|=− μα ln| P2P1|=− μα eln| P2P1 | = e−μα P2 = P1e− μα If L is the length of the curve with constant radius R, then α = L/ R, P2 = P1e− μLR dx P – dP P R dα dP dα P – dP P N = Pdα
  • 2. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 2 / 13 The above equation gives the frictional loss due to curvature. This equation can also be applied to compute the frictional loss due to wobble or length effect by substituting the loss KL for μα, P2 = P1e− KL PRESTRESS LOSS DUE TO WOBBLE FRICTION (Ref. b) To combine the wobble & curvature effect, we can simply write, P2 = P1e− μα−KL Or dividing by tendon area, the above equation can be written in terms of unit stresses, f2 = f1e− μα−KL The friction loss is obtained from above expression. Loss of steel stress is given as FR = f1 – f2, where f1 is the steel stress at the jacking end & L is length to the point under consideration. Thus, FR = f1−f2=f1−f1e− μα−KL=f1(1−e− μα−KL) For tendons with a succession of curves of varying radii, it is necessary to apply this formula from section to section. The reduced stress at the end of a segment can be used to compute the frictional loss for the next segment. Since practically for all prestressed-concrete members, the depth is small compared with the length, the curve is relatively flat. The angular change α is approximately given by the transverse deviation of the tendon divided by the projected length, both referred to the member axis. APPROXIMATE DETERMINATION OF CENTRAL ANGLE FOR A TENDON (Ref. a) From the figure above, we have, tanα2 = mx/2 = 2mx m y α α/2 x x/2 Actual profile due to wobbling Intended profile Tendon supports α (Intended angle change)
  • 3. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 3 / 13 In above equation m is approximately equal to twice the sag y. Also, for small angles, the tangent of an angle is nearly equal to the angle itself, measured in radians. Therefore, α2 = 2(2y) x → α = 8yx (radians The value of y can be obtained from the arc geometry. Let R be the known radius of the arc, then using the Pythagoras’ theorem, we get, ( x2) 2+(R−y)2=R2 x24+R2+y2−2Ry−R2=0 y2−2Ry+ x24=0 Solving the above quadratic equation, y can be obtained, which can be used to find the value of α. Typical values of wobble friction coefficient & curvature friction coefficient are listed below. These values are taken from Ref. c. Type of Tendon & sheath Wobble Coefficient, K (per meter length x 10-3) Curvature Coefficient, μ Tendons in flexible metal sheathing Wire tendons 3.3 – 5.0 0.15 – 0.25 7 wire strands 1.6 – 6.5 0.15 – 0.25 High strength bars 0.3 – 2.0 0.08 – 0.30 Tendons in rigid metal sheath 7 wire strand 0.70 0.15 – 0.25 Pre-greased tendons Wire tendons & 7 wire strand 1.0 – 6.5 0.05 – 0.15 Mastic coated tendons Wire tendons & 7 wire strand 3.3 – 6.6 0.05 – 0.15 EXAMPLE 1 – FRICTIONAL LOSS A concrete beam, continuous over two spans is post-tensioned at both ends on a flat base. The prestress applied is 1500MPa. The modulus of elasticity of steel & concrete are 200000MPa & 33100 MPa respectively. The beam is prestressed using 7-wire strand. The idealized sections are shown below. Compute the percentage loss of prestress due to friction at middle support. IDEALIZED SECTION AT MID-SUPPORT αDE 3.5m 20m A B C D E 800mm 150mm 150mm αBC 5m 5m 3.5m 3m 400mm R = 50m R = 30m F
  • 4. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 4 / 13 SOLUTION Given Data: Wobble friction coefficient, K = 0.0018 m –1 Curvature friction coefficient, μ = 0.2 Jacking force, F1 = 1500 x 987.1 = 1480650 N Geometric Properties: Segment L (m) R (m) y2−2Ry+ L24=0 α = 8yL (radians) AB 3.5 0 - 0 BC 10.0 50 0.2506 0.2005 CD 3.5 0 - 0 DF 6.0 30 0.1504 0.200 DE 3.0 0.100 (αDF/2) Frictional Loss: To take into account the gradual reduction of stress from A towards E, the tendon is divided into 4 portions from A to E. The reduced prestress force obtained at the end of each segment is used as the starting stress for the next segment. Results are shown below in tabulated form. Segment L (m) KL μα e – KL – μα Reduced Force F1e – KL – μα (N) Remarks AB 3.5 0.0063 0.0000 0.9937 1471321.905 F1 = 1480650.000 N BC 10.0 0.0180 0.0401 0.9436 1388367.018 F1 = 1471351.227 N CD 3.5 0.0063 0.0000 0.9937 1379620.306 F1 = 1388367.018 N DE 3.0 0.0054 0.0200 0.9749 1344991.836 F1 = 1379620.306 N Total frictional loss from A to E = 1480650.000−1344991.8361480650 × 100 = 9.162 % ANCHORAGE TAKE-UP LOSS (REF. a, b, e & g) For most systems of posttensioning, when a tendon is tensioned to its full value, the jack is released & the prestress is transferred to the anchorage. The anchorage fixtures that are subject to stresses at this transfer will tend to deform, thus allowing the tendon to slacken slightly. Friction wedges employed to hold the wires will slip a little distance before the wires can be firmly gripped. The amount of slippage depends on the type of wedge & the stress in the wires, but it is typically between 3mm to 9mm. For direct bearing anchorages, the heads & nuts are subject to a slight deformation at the release of the jack. An average value for such deformations may be only about
  • 5. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 5 / 13 0.75mm. If long shims are required to hold the elongated wires in place, there will be a deformation in the shims at transfer of prestress. As an example, a shim 0.3m long may deform 0.25mm. Wide variation can occur & large anchorage “set” or “take-up” is possible due to the fact that the hard, smooth wires may not immediately grip the steel before it has slipped through. A general formula for computing the loss of prestress due to anchorage deformation Δa is Δfs = ΔaEsL where, Δa = Amount of slip L = Tendon length of the tendon Es = Elastic modulus of the prestressing steel The above equation is based on the assumption that the slip is uniformly distributed over the length of the tendon. This is approximately so for pretension, & may apply for posttensioning, if the tendon is well greased or encased in low-friction plastic sheathing, & if wobble & curvature are small. For many post-tensioned beams, however, the anchorage slip loss is mostly confined to a region close to the jacking anchorage. Distribution along the tendon is prevented by reverse friction as the tendon slips inward, & the steel stress throughout much of the tendon length may be unaffected by anchorage slip. PRESTRESS VARIATION BEFORE & AFTER ANCHORAGE (Ref. e & g) Referring to the figure above, curves OB and AB are both characterized by the frictional parameters of the prestressing system. Once these parameters are known together with the anchorage take-up distance a , the length b of the back sliding segment and the stress loss at any location can be calculated. The basic relationship for loss of prestress due to friction is, fx = foe−(μ α + K x)=foe−kx where, α = Angle change
  • 6. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 6 / 13 x = Distance between two points μ = Coefficient of friction K = Wobble coefficient k = Friction index From the figure above, the tendon stress before and after anchorage take-up is, respectively, f1 = foe−kx f2 = fbe− k(b − x)=foe−2kb ekx The area OAB is obtained by integration, ∫(f1−f2) bodx= 1kfo(1−e−kb) 2 Therefore, Esk Δafo=(1−e−kb) 2 Solving the equation for b, we get b = −( 1k) ln(1−√ Esk Δafo) The following equation gives the anchorage loss at the end of the beam where jacking force is applied, fo−fa = fo(1−e−2bk) To find anchorage loss at any point from the end of the beam, following equation is used, f1−f2 = foe− k x(1−e−2k(b − x)) where, x is the distance from jacking end to the point under consideration. For a location outside the anchorage length (x > b), the steel stress is not affected by the anchorage losses. For a location inside the anchorage length (x < b), loss due to both, friction & anchorage seating occurs & is given by the above equation. It should be remember that the above equation is valid only if b < le, where le is effective beam length or the maximum length available for distribution of anchorage seating losses; one half of the member length if tensioning is done from both ends simultaneously; length of the member if post-tensioning is done from one end only. The importance of anchorage slip also depends on the length of the member or casting bed. For very short tendons, anchorage set will produce high slip losses. For long tendons or casting beds, slip becomes insignificant. The above equations deal with tendon profiles in a single uniform curvature only.
  • 7. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 7 / 13 TWO-SEGMENT VARIATION (Ref. g) For cases where the back sliding takes place over several segments of different curvatures, the problem becomes a little more complicated. Consider the case where two segments with friction indices k1 and k2 are involved, as shown in the figure. Tendon stress before anchoring is, fx = foe−k1 x when 0 ≤ x ≤ b1 fx = foe−k1 b1 e−k2(x − b1) when b1 ≤ x ≤ b1+b After anchorage take-up losses, the tendon stress is, fx = foe−(2 k1 b1 + 2 k2 b)ek 2(x − b1) when 0 ≤ x ≤ b1 fx = foe−( k1 b1 + 2 k2 b)ek 2(x − b1) when b1 ≤ x ≤ b1+b Upon integration, the following equation is obtained, [ k1k2−(1−e−k1b1)](1−e−k2 b) 2+2(1−e−k1 b1)(1−e−k2 b)− 1e−k1 b1[ Ek1Δfo−(1−e−k1 b1) 2]=0 This is a quadratic equation in terms of the unknown parameter (1 – e –k2b). STRESS VERSUS TIME IN THE STRANDS OF A PRETENSIONED CONCRETE GIRDER (Ref. l)
  • 8. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 8 / 13 The figure above shows the losses that occur with time in the strands in a pretensioned concrete girder. By the end of this lecture only friction and anchorage seating loss, & elastic shortening have been discussed. Other losses, which are time-dependent losses such as creep, shrinkage and relaxation of steel, will be discussed in next lecture. EXAMPLE 2 – FRICTION & ANCHORAGE TAKE-UP LOSS A 12m long beam is posttensioned from one end. The tendon has a parabolic profile as shown in the figure with a constant curvature. Compute the percentage loss of prestress due to friction and anchorage take-up if the jacking stress is 1100 N/mm2. IDEALIZED SECTION AT MID-SECTION SOLUTION Given Data: Wobble friction coefficient, K = 0.004 m –1 Curvature friction coefficient, μ = 0.3 Amount of slip, Δa = 1.5 mm Tendon depression, y = 150 mm Therefore, k = μαx + K = 0.38(0.15)12⁄ 12 + 0.004 = 0.0065 per m Back-slip penetration Length: The length of back sliding is calculated using the following equation; b = −( 1k) ln(1−√ Esk Δafo)=−( 10.0065)ln(1−√ 200000(0.0065)(0.0015) 1100)=6.618 m Friction & Anchorage Take-Up Loss: The following equation gives the anchorage loss at the end of the beam where jacking force is applied, fo−fa = fo(1−e−2bk)=1100(1−e−2(6.618)(0.0065))=90.678 MPa To find anchorage loss at any point from the end of the beam, following equation is used, f1−f2 = foe− k x(1−e−2k(b − x)) 150mm L = 12m
  • 9. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 9 / 13 Where, f1 = foe−kx for 0≤x≤L (Before anchorage loss) f2 = fbe− k(b − x)=foe−2kb ekx for 0≤x≤b (After anchorage loss) These equations are used to generate a table showing loss of prestress due to friction (f1) and due to anchorage (f2). The results are plotted on the graph. x f1 f2 Friction loss, fo – f1 Anchorage loss, f1 – f2 m MPa MPa MPa MPa 0 1100.000 1009.322 0 90.678 2 1085.793 1022.529 7.127 63.264 4 1071.769 1035.908 28.231 35.860 6 1057.926 1049.463 42.074 8.463 b = 6.618 1053.686 1053.686 46.314 0.000 8 1044.262 - 55.738 - 10 1030.774 - 69.226 - 12 1017.461 - 82.539 - The above results show that the anchorage loss is significant near the jacking end of the beam and reduces to zero at distance b from the jacking end. Also, back slip takes place over more than half of the tendon. The total losses of prestress due to friction and anchorage are, Percentage loss of prestress due to friction = 82.5391100.000×100 = 7.504 % 1000 1020 1040 1060 1080 1100 1120 0 2 4 6 8 10 12 14 Prestress (MPa) Distance 'x' from the jacking end of the beam (m) PRESTRESS LOSS DUE TO FRICTION & ANCHORAGE f1 f2 fo – fa f1 – f2 le fo – f1
  • 10. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 10 / 13 Percentage loss of prestress due to anchorage = 90.6781100.000×100 = 8.243 % Note that the anchorage loss is calculated at the jacking end since it has maximum value at this location, while friction loss is calculated for the whole length of the tendon. EXAMPLE 3 – FRICTION & ANCHORAGE TAKE-UP LOSS – VARIABLE CURVATURE A 36m long beam is posttensioned from both ends. The tendon has a parabolic profile in the middle 30m, with a radius of 750m as shown in the figure. Tendon is straight in the 3m regions near each end. Compute the percentage loss of prestress due to friction and anchorage take-up if the jacking stress is 1200 N/mm2. SOLUTION Given Data: Wobble friction coefficient, K = 0.002 m –1 Curvature friction coefficient, μ = 0.30 Amount of slip, Δa = 1.50 mm b1 or L1 = 3.00 m L2 = 30.00 m R1 = 0 m R2 = 750.00 m Geometric Properties: Depression of the middle portion of the tendon can be calculated using the following equation y22−2R2y2+ L224=0 y22−2(750)y2+ (30)24=0 Solving the above quadratic equation, we get, y2=0.150 m Therefore, L1 R = 750m L2
  • 11. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 11 / 13 α1 = 0 α2 = 8y22L2 = 8(0.150) 2(30) = 0.02 radians k1 = μα1L1 + K = 0.303 + 0.002 = 0.0020 per m k2 = μα2L22⁄ + K = 0.30.02015 + 0.002 = 0.0024 per m Back-slip penetration Length: Assuming that the back-slip is restricted to the first segment, the maximum anchorage take-up is calculated using the following equation, Esk 1Δ1fo=(1−e−k1b1) 2 Solving the above equation we get, Δ1= [1−e−(0.002)(3)] 2(1200) 200000(0.002) =0.00011 m Since 1 is less than the specified anchorage take-up, back-slip penetrates beyond the first segment and equation for two segments has to be used to calculate the back-slip penetration length. The length of back sliding is calculated using the following equation; [ k1k2−(1−e−k1b1)](1−e−k2 b) 2+2(1−e−k1 b1)(1−e−k2 b)− 1e−k1 b1[ Ek1Δfo−(1−e−k1 b1) 2]=0 This equation is quadratic in terms of (1−e−k2 b). Solving the above equation yields, 1−e−k2 b=0.0176 b=7.400 m Friction & Anchorage Take-Up Loss: The following equations give the prestress loss before the anchorage loss occurs i.e., the loss is only due to friction, f1 = foe−k1 x when 0 ≤ x ≤ b1 f1 = foe−k1 b1 e−k2(x − b1) when b1 ≤ x ≤ b1+b After anchorage take-up losses, the tendon stress is given by the following equations, f2 = foe−(2 k1 b1 + 2 k2 b)ek 2(x − b1) when 0 ≤ x ≤ b1 f2 = foe−( k1 b1 + 2 k2 b)ek 2(x − b1) when b1 ≤ x ≤ b1+b These equations are used to generate a table showing loss of prestress due to friction (f1) and due to anchorage (f2). The results are plotted on the graph.
  • 12. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 12 / 13 x f1 f2 Friction loss, fo – f1 Anchorage loss, f1 – f2 m MPa MPa MPa MPa 0 1200.000 1136.098 0.000 63.902 1 1197.602 1138.828 2.398 58.774 2 1195.210 1141.565 4.790 53.645 b1 = 3 1192.822 1144.308 7.178 48.514 4 1189.962 1153.961 10.038 36.002 5 1187.110 1156.733 12.890 30.376 6 1184.264 1159.513 15.736 24.751 7 1181.425 1162.299 18.575 19.126 8 1178.593 1165.092 21.407 13.501 b + b1 = 10.4 1171.823 1171.823 28.177 0.000 12 1167.332 - 32.668 - 14 1161.743 - 38.257 - 16 1156.180 - 43.820 - 18 1150.643 - 49.357 - The total losses of prestress due to friction and anchorage are, 1120 1140 1160 1180 1200 1220 0 2 4 6 8 10 12 14 16 18 Prestress (MPa) Distance 'x' from the Jacking End of the Beam (m) PRESTRESS LOSS DUE TO FRICTION & ANCHORAGE f1 f2 f1 – f2 b1 b1 + b fo – f1
  • 13. PRESTRESSED CONCRETE FRICTION & ANCHORAGE TAKE-UP LOSS BY: AYAZ MALIK 13 / 13 Percentage loss of prestress due to friction = 49.3571200.000×100 = 4.113 % Percentage loss of prestress due to anchorage = 63.9021200.000×100 = 5.325 % Note that the anchorage loss is calculated at the jacking end since it has maximum value at this location, while friction loss is calculated for the half length of the tendon (Since jacking force is applied at both ends). REFERENCES a. T. Y. Lin, Ned H. Burns, “Design of Prestressed Concrete Structures”, 3rd Edition, 1981 b. Arthur H. Nilson, “Design of Prestressed Concrete”, 2nd Edition, 1987 c. Cement Association of Canada, “Concrete Design Handbook”, 3rd Edition, 2012 d. Canadian Standards Association, “CAN/CSA-A23.3-04–Design of Concrete Structures”, 2007 e. Ti Huang, Burt Hoffman, “Prediction of Prestress Losses in Posttensioned Members”, Department of Transportation, Commonwealth of Pennsylvania, 1978 f. Gail S. Kelly, “Prestress Losses in Posttensioned Structures”, PTI Technical Notes, 2000 g. Ti Huang, “Anchorage take-up loss in Posttensioned Members”, 1969 h. PCI, “Post-Tensioning Manual”, 1972 i. Maher K. Tadros, Nabil Al-Omaishi, Stephen J. Seguirant, James G. Gallt, “Prestress Losses in Pretensioned High-strength Concrete Bridge Girders”, NCHRP Report 496, 2003

Related Documents