1

Prestress Losses

Dr Antonis Michael

Department of Civil Engineering

Frederick University

Losses Due to Elastic Deformation of

Pre-Tensioned Concrete

Consider a pre-tensioned beam with a tendon at

the c.g.c:

L

ESES

∆=ε

Pi Pi

L

∆ES

2

For the pre-tensioned tendon:

)(

)(

tEA

P

E

t

cmc

i

c

cES

⋅==

σε

( ) c

i

cm

p

pelA

P⋅⋅=∆

tE

EAP

( ))(

tE

EAP tc

cm

p

pel σ∆⋅⋅=∆

For an Eccentric Tendon and Accounting

Self-Weight

From stress superposition the

stress at the c.g.s is:( )

c

cssw

c

csPi

c

ic

I

yM

I

yyP

A

P ⋅+

⋅⋅−−=∆ tσ

Pi has a lower value after transfer of prestress

compared to the initial jacking force Pj (due to elastic shortening)

Pi is difficult to accurately determine and the reduction

is only a small percentage of Pi

Therefore assume Pi = Pj

3

Example

L = 19.52m

102 mm

h

b

Given:

fck = 41.4 MPa, fci = 31 MPa

fpk = 1862 MPa, fp0,1k = 1551.7 MPa, Ep = 186.2 GPa

b = 381 mm, h = 762 mm

Ten 12.7 mm tendons are used to prestress the beam (Atendon = 98.7 mm2)

Find:

(a) Prestress loss due to elastic shortening

(b) Check stresses in concrete

kNMPamAP ppi 13785.1396000987.02

max, =⋅=⋅= σ

222000987.09877.9810 mmmmmAp ==⋅=

{ } { } MPaff kppkp 5.13967.15519.0;18628.0min9.0;8.0min 1,0max, =⋅⋅=⋅⋅=σ

2222903.0381.0762.0 mmmAc =⋅=

4

3

01405.012

)762.0(381.0m

mmIc =

⋅=

mmm

y p 279.0102.02

762.0=−=

4

Solution

( )kNm

LwM SW

SW 2.1978

2.1583.6

8

22

=⋅

=⋅

=

m

kN

m

kNmmwSW 83.6)5.23(381.0762.0

3=⋅⋅=

( ) ( )GPa

fE ci

ci 89.3010

3122

1022

3.03.0

=

=

=

( ) MPakPac 47.8846501405.0

279.02.197

01405.0

279.0279.01378

2903.0

1378t −=−=

⋅+

⋅⋅−−=∆σ

csp yy =

Stress at centroid of tendon at mid-span:

( )kNtc

cm

p

pel 4.50847089.30

2.186000987.0)(

tE

EAP =⋅⋅=∆⋅⋅=∆ σ

Force in tendon after elastic shortening of concrete:

kNPPP elii 6.13274.5013780,

=−=∆−=

Stresses at top and bottom immediately after prestress:

c

bsw

c

bPi

c

i

botI

yM

I

yyP

A

P ⋅+

⋅⋅−−=

0,0,σ

MPakPabot 27.9927001405.0

381.02.197

01405.0

381.0279.06.1327

2903.0

6.1327−=−=

⋅+

⋅⋅−−=σ

5

c

tsw

c

tPi

c

i

topI

yM

I

yyP

A

P ⋅−

⋅⋅+−=

0,0,σ

MPakPatop 124.05.12301405.0

381.02.197

01405.0

381.0279.06.1327

2903.0

6.1327==

⋅−

⋅⋅+−=σ

Allowable stresses as per EC2:

cic f⋅≤ 6.0σ

MPac

c

6.18

316.0

≤

⋅≤

σ

σ

OKMPa

OKMPa

top

bot

⇒=

⇒−=

124.0

27.9

σ

σ

What if the Beam was Post-Tensioned?

Same data as before but for post-tensioned beams EC2:

( )( )∑

⋅⋅⋅=

tE

tjEAP

cm

cppel

σ∆∆

kNel 2.2589.30

47.85.0186200000987.0P =

⋅⋅⋅=∆

45.0102

110

2

1=

⋅

−=

−=

n

nj

kNel 7.2289.30

47.845.0186200000987.0P =

⋅⋅⋅=∆

The of 0.5 instead of the actual value for j is conservative.

6

Loss Due to Anchorage Slip

Data on anchorage slip must be supplied by the manufacturer. For wedge type strand anchors the slip is 3 to 10 mm.

L

Appsl

∆⋅⋅=∆ EAP

∆A is the anchor slip

L is the length of the tendon

Losses due to Friction

In the absence of data in a European Technical Approval, values for

unintended regular displacements for internal tendons will generally be in the

range 0,005 < k < 0,01 per metre.

)1(P)(P)k(

maxµ

xex +−−=∆ θµ

Table 5.1: Coefficients of friction µµµµ of post-tensioned internal tendons and external unbonded tendons

External unbonded tendons Internal tendons

1) Steel duct/ non

lubricated HDPE duct/ non

lubricated Steel duct/ lubricated

HDPE duct/ lubricated

Cold drawn wire 0,17 0,25 0,14 0,18 0,12 Strand 0,19 0,24 0,12 0,16 0,10 Deformed bar 0,65 - - - -

Smooth round bar 0,33 - - - - 1)

for tendons which fill about half of the duct

Note: HPDE - High density polyethylene

7

Example for Friction Losses

Post-tensioned concrete beam with the

tendon profile shown

L = 21.41m L = 21.41m

0.167

0.100

A

B C

D

E

Assume:

Pmax = 1400 MPa, k = 0.005

Internal tendons (7-wire strands)

Find:

Prestress losses due to friction.

)1(P)(P)k(

maxµ

xex +−−=∆ θµ

kNex 04.96)1(1400)(P))41.21(0.005267.0(19.0

µ=−=∆ +−

The friction loss from A to E is approximately -7%