PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
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, is,
=
dα
dx R dx P P – dP
P dP
dα
P – dP N = Pdα
R
FRICTIONAL LOSS dF ALONG THE LENGTH dx DUE TO CURVATURE FRICTION (Ref. a) dα
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, is,
N
= = Pdα
Pdx R
Let, μ be the coefficient of curvature friction & K, K, the wobble friction coefficient. The amount of frictional loss dP around around the length dx is is given by,
= = = = ∫ = ∫ | | | |= = = − = − = − dP
μPdx
μN
dP
R
μPdα
μdα
P
Integrating on both sides with limits P 1 & P2, P2
P1
1
P
dP
μdα
ln P2 ln e
ln
P2
ln P1
P2
μα
P1 P2 P1
μα
e
P1 e
μα
μα
If L is the length of the curve with constant radius R, then α = L/ R, R,
P2
P1 e
1 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
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
=
P1 e
−
Actual profile due to
Tendon supports
Intended profile
α (Intended angle change)
PRESTRESS LOSS DUE TO WOBBLE FRICTION (Ref. b) To combine the wobble & curvature effect, we can simply write,
P = Pe− − f = f e− − FR = f f =f f e− − =f (1e− −)
Or dividing by tendon area, the above equation can be written in terms of unit stresses,
The friction loss is obtained from above expression. Loss of steel stress is given as FR = f 1 – f 2, where f 1 is the steel stress at the jacking end & L is length to the point under consideration. Thus,
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.
α/2
y
m
α x/2 x
APPROXIMATE DETERMINATION OF CENTRAL ANGLE FOR A TENDON (Ref. a) From the figure above, we have,
tan
α 2
= = m
2m
x/2
x
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PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
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,
= = 2x +Ry =R x42 +R2 +y2 2RyR =0 2y 2Ry+ x42 =0 α
2 2y
2
x
→
8y
α
x
(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,
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. Wobble Coefficient, K (per meter length x 10 -3)
Curvature Coefficient, μ
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
Type of Tendon & sheath
Tendons in flexible metal sheathing
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. R = 50m A
B
αDE
E C
D
αBC
400mm 150mm 800mm
F
150mm
R = 30m
IDEALIZED SECTION AT MID-SUPPORT 3.5m 5m
5m 3.5m 3m
20m
3 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
SOLUTION Given Data: Wobble friction coefficient, K
=
0.0018 m –1
Curvature friction coefficient, μ
=
0.2
Jacking force, F 1
=
1500 x 987.1
Geometric Properties:
= 1480650 N
Segment
L (m)
R (m)
AB
3.5
0
2y 2Ry+ L42 =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
-
=
α
8y L
(radians) 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
=
.000 9 . 36 × =
1480650
13449 1 8
1480650
100
9.162 %
ANCHORAGE T AKE-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
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PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
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 Δ a is ∆f s
=
∆a Es L
where, Δa
=
Amount of slip
L
=
Tendon length of the tendon
E s
=
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,
f x = f oe−μ α + K x =f oe−kx where,
α
=
Angle change
5 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS x
=
Distance between two points
μ
=
Coefficient of friction
K
=
Wobble coefficient
k
=
Friction index
BY : AYAZ M ALIK
From the figure above, the tendon stress before and after anchorage take-up is, respectively,
f 1 = f oe−kx− kb − x −kb kx f 2 = f be =f oe e b∫ f 1 f2 dx= 1 f o(1e−kb)2 k o Eskf o∆a =(1e−kb)2 b = 1k ln1√ Eskf o∆a
The area OAB is obtained by integration,
Therefore,
Solving the equation for b, we get
The following equation gives the anchorage loss at the end of the beam where jacking force is applied,
f o f a = f o(1e−2bk) f 1 f2 = f oe− k x(1e−2kb − x)
To find anchorage loss at any point from the end of the beam, following equation is used,
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 < l e, where l e 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.
6 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
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 k 1 and k 2 are involved, as shown in the figure. Tendon stress before anchoring is,
f x = f oe−k−k11 xb1 −kwhen 0 ≤ x ≤ b 1 f x = f oe e 2x − b1 when b1 ≤ x ≤ b1 +b f x = f oe−2− kk1 1b b1 1+ +22kk22bbekk 2 x2x−−bb11 when 0 ≤ x ≤ b1 f x = f oe e when b1 ≤ x ≤ b1 +b kk12 (1e−k1b1)(1e−k2 b) +2(1e−k1 b1)(1e−k2 b) e−k11 b1 Ekf o1∆ (1e−k1 b1)=0 After anchorage take-up losses, the tendon stress is,
Upon integration, the following equation is obtained,
This is a quadratic equation in terms of the unknown parameter (1 – e –k 2b).
STRESS VERSUS TIME IN THE STRANDS OF A PRETENSIONED CONCRETE GIRDER (Ref. l)
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PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
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 T AKE-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/mm 2.
150mm L = 12m
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 = μ αx + K = 0.3 80.1125⁄12 + 0.004 = 0.0065 per m Back-slip penetration Length:
The length of back sliding is calculated using the following equation;
00650.0015=6.618 m b = 1k ln1√ Eskf o∆a=0.01065ln1 √ 2000000.1100 Friction & Anchorage Take-Up Loss: The following equation gives the anchorage loss at the end of the beam where jacking force is applied,
f o f a = f o(1e−2bk)=1100(1e−..)=90.678 MPa f 1 f2 = f oe− k x(1e−2kb − x)
To find anchorage loss at any point from the end of the beam, following equation is used,
8 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
Where,
f 1 = f oe−kx− kb − xfor 0≤x≤L Bef o re anchorage l o ss f 2 = f be =f oe−kb ekx for 0≤x≤b After anchorage loss
These equations are used to generate a table showing loss of prestress due to friction (f 1) and due to anchorage (f 2). The results are plotted on the graph. x
f 1
f 2
Friction loss, f o – f 1
Anchorage loss, f 1 – f 2
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
-
PRESTRESS LOSS DUE TO FRICTION & ANCHORAGE 1120 f1 1100 ) a P 1080 M ( s s 1060 e r t s 1040 e r P 1020
f2
f o – f 1 f o – f a
le
f 1 – f 2
1000 0
2
4
6
8
10
12
14
Distance 'x' from the jacking end of the beam (m)
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
= 1100.82.503900 ×100 = 7.504 %
9 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
Percentage loss of prestress due to anchorage
BY : AYAZ M ALIK
= 1100.90.607800 ×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 T AKE-UP LOSS – V ARIABLE 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/mm 2.
R = 750m
L2
L1
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
=
0m
R2
=
750.00 m
Geometric Properties: Depression of the middle portion of the tendon can be calculated using the following equation
2y2 2R2y2 + L422 =0 2y2 2750y2 + 304 =0
Solving the above quadratic equation, we get,
y2 =0.150 m Therefore,
10 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
α1 = 08y 80.150 α2 = 2L2 = 230 = 0.02 radians k1 = μ αL11 + K = 0.3 03 + 0.002 = 0.0020 per m k2 = μ Lα⁄22 + K = 0.3 0.15020 + 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,
Eskf o1∆1 =(1e−k1b1)2 1200 −. 1 e [ ] ∆= 2000000.002 =0.00011 m Solving the above equation we get,
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;
kk12 (1e−k1b1)(1e−k2 b) +2(1e−k1 b1)(1e−k2 b) e−k11 b1 Ekfo 1∆ (1e−k1 b1)=0 −k2 b) (1e 1e−k2 b =0.0176 b=7.400 m This equation is quadratic in terms of
. Solving the above equation yields,
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,
f 1 = f oe−k−k11 xb1 −kwhen 0 ≤ x ≤ b 1 f 1 = f oe e 2x − b1 when b1 ≤ x ≤ b1 +b f 2 = f oe−2− kk1 1b 1b 1+ +22kk22bbekk 2 x2x−−b1b1 when 0 ≤ x ≤ b1 f 2 = f oe e when b1 ≤ x ≤ b1 +b
After anchorage take-up losses, the tendon stress is given by the follo wing equations,
These equations are used to generate a table showing loss of prestress due to friction (f 1) and due to anchorage (f 2). The results are plotted on the graph.
11 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
BY : AYAZ M ALIK
x
f 1
f 2
Friction loss, f o – f 1
Anchorage loss, f 1 – f 2
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
-
PRESTRESS LOSS DUE TO FRICTION & ANCHORAGE 1220
) 1200 a P M1180 ( s s e r t 1160 s e r P 1140
f1
b1 + b
b1
f2
f o – f 1 f 1 – f 2
1120 0
2
4
6
8
10
12
14
16
Distance 'x' from the Jacking End of the Beam (m)
The total losses of prestress due to friction and anchorage are,
18
12 / 13
PRESTRESSED CONCRETE FRICTION & ANCHORAGE T AKE-UP LOSS
Percentage loss of prestress due to friction
Percentage loss of prestress due to anchorage
BY : AYAZ M ALIK
= 1200.49.305700 ×100 = 4.113 % = 1200.63.900200 ×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. b. c. d. e.
T. Y. Lin, Ned H. Burns, “Design of Prestressed Concrete Structures”, 3 rd Edition, 1981 Arthur H. Nilson, “Design of Prestressed Concrete”, 2 nd Edition, 1987 Cement Association of Canada, “Concrete Design Handbook”, 3 rd Edition, 2012 Canadian Standards Association, “CAN/CSA-A23.3-04–Design of Concrete Structures”, 2007 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
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