Influence of the prestress force on the secton forces - Aalto · Influence of the prestress force ... u⋅L2 is the moment due to uniformly ... (maximum distance of the tendon from

Rak 43.3110 Jännevoiman vaikutus voimasuureisiin 27.02.13 1

Influence of the prestress force on the secton forces

Simpy supported one-span beam with parabolic tendon profile

The eccentricity e of the prestress force has

−⋅=

2

2/L

x1f)x(y

f is the sag of the tendon

The bending moment diagram due to the prestress force )x(eP)x(Mp ⋅−= is also parabolic

At the midspan fPMp ⋅−=

Uniformly distributed load u causes also the parabolic bending moment diagram, which maximum

value is 8

LuM

2⋅

=

� The bending moment diagram due to prestress corresponds the negative uniformly

distributed load acting upward => load balancing force which reduce the bending moment

due to the imposed loads

8

LuMfPM

2

p

⋅==⋅−= =>

2L

fP8u

⋅⋅−=

At the support the derivative of the tendon profile is L

f4y ' ⋅

=

� The vertical component of the prestress force 2

Lu

L

f4PFF ba ⋅=

⋅⋅==

revese the shear force due to the load balancing force: 0=⋅++ LuFF ba

Equivalent forces which represents the influence of the prestress force


The tendon has been moved upward with the distance f so that the eccentricity at the midspan L/2

is e2=e(L/2)=0 and at the ends the eccentricities are e1 = e(0)=f ja e3 = e(L)=f.

The tendon profile is similar than before

Equation of the parabola

2

2/L

xf)x(y

⋅=

Second derivative of the parabola 2

''

L

f8y

⋅=

Load balancing force 2

8 P fu

L

⋅ ⋅= −

At the ends the eccentricities of the tendon are e1 and e2 => end moments f P e P M

f P e P M

2b

1a

⋅=⋅=

⋅=⋅=

Equivalent forces which represents the influence of the prestress force

Equilibrium: Fa+Fb+u L=0

Bending moment at the midspan 02

fPfP

L8

LfP8

2

MM

8

LuM

2

2

ba

2

=⋅+⋅

+⋅

⋅⋅⋅−=

++

⋅=

� eccentricity of the prestress e2=0


Rotation at the support for simply support beam:

uniformly distributed load 2L

f8u

⋅−=

EI3

LfP

EI24

Lu 3

01⋅

⋅⋅−=

⋅

⋅=α

end moments Ma = Mb = P f EI2

LfP

EI2

LMa02

⋅

⋅⋅=

⋅

⋅=α

sum of the rotations

EI3

L

2

fPfP

2

3fP

EI3

LM

2

3

8

Lu

EI3

LM

12

Lu

EI2

L

EI2

LM

EI24

Lua

2

a

2

a

3

02010⋅

⋅⋅

=

⋅⋅+⋅−⋅

⋅=

⋅+

⋅

⋅=

+

⋅

⋅=

⋅

⋅+

⋅

⋅=α+α=α


Beam with fixed ends at the both ends

Load balancing forces are the same as before:

Rotation at the support due to the load balancing forces for a simply supported beam

+

⋅

⋅=α a

2

0 M12

Lu

EI2

L

Fixed end => rotation at the support α=0

� moment ∆M, which kills the rotation α0:

∆M EI2

LM1

⋅

⋅∆=α

0EI2

LMM

12

Lu

EI2

La

2

10 =⋅

⋅∆+

+

⋅

⋅=α+α

3

fPfPfP

3

2fP

L12

LfP8M

12

LuM

2

2

a

2⋅−

=

⋅+⋅⋅−=

⋅+

⋅

⋅⋅⋅−−=

+

⋅−=∆

Total moment at the support L

EI2fP

3

2

12

LuMM 0

2

tbta

⋅⋅α−=⋅⋅=

⋅−==

The total moment is the same as the fixed end moment due uniformly distributed load u.

Moment due to the eccentricity of the prestress force fPMM ba ⋅==

Secondary moment

3

fPMMM

3

fPMMM

btbb

ataa

⋅−=−=∆

⋅−=−=∆


The secondary moment is induced because the rotation at the support due the eccentricity of the

prestress in the span is different than the rotation due to the eccentricity at the support.

If these rotations are the equal the secondary moment ∆M=0

(at the support e1=e3 = 2/3 f ; in the field e2 = 1/3 f ; the sag 231 e

2

eef +

+= )

The moment at the midspan

24

Lu

3

fP

2

3

fP

3

fP

2

fPfP

L8

LfP8

2

MM

2

MM

8

LuM

2

2

2

baba

2⋅

=⋅

=

÷−

⋅−

+⋅+⋅

+⋅

⋅⋅⋅−=

∆+∆+

++

⋅=

8

Lu 2⋅

is the moment due to the load balancing force for simply supported beam

24

Lu 2⋅

is the moment due to uniformly distributed load u for the beam with fixed end

Even if the eccentricity e2 = 0 at the midspan, the prestress force cause bending moment at this

point.

Influence of the fixed end moments to the field moment:

12

LufP

3

2

2

MM

2

MM 2

baba ⋅−=⋅⋅−=

∆+∆+

+


Beam with the foxed ends where the eccentricities of the prestress tendon are e1=e2 =0

Load balancing force 2L

fP8u

⋅⋅−=

The load balancing forces are the same as before:

The rotation at the support due to the load balancing forces for a simply supported beam

EI24

Lu 3

0⋅

⋅=α

When the rotation at the fixed support =0, so the rotation must nullify the moment at the supports

fP3

2

12

Lu

L

EI2M

2

0

at ⋅⋅=⋅

−=α⋅⋅−

=

Because the eccentricities at the supports are e1 =e3= 0, so the secondary moments are

=>

fP3

2M

fP3

2M

b

a

⋅⋅=∆

⋅⋅=∆

Moment at the midspan

24

Lu

3

fPfP

3

2fP

2

MM

8

LuM

2sek,bsek,a

2⋅

=⋅

−=−⋅⋅+⋅−=∆+∆

+⋅

=

corresponds the moment due to the uniformly distributed load u for the beam with the fixed ends

Beam with the foxed ends where the eccentricities of the prestress tendon are e1=e2 =0


The sag 2

eeef 21

2

++=

Load balancing force 2L

fP8u

⋅⋅−=

Load balancing forces:

The rotation at the support due to the load balancing forces for a simply supported beam

+⋅⋅+⋅⋅−

⋅=

+⋅+

⋅

⋅=α

3

)ee2(PfP

3

2

EI2

L

3

MM2

12

Lu

EI2

L 31ba

2

0

=>

2b

1a

ePfP3

2M

ePfP3

2M

⋅−⋅⋅=∆

⋅−⋅⋅=∆

Total moment at the support 12

LufP

3

2M

2

ta

⋅−=⋅⋅=

Secondary moment 1atasek,a ePfP3

2MMM ⋅−⋅⋅=−=∆

Moment at the midspan

24

Lu

3

fP

2

)ee(PfP

3

2

2

)ee(PfP

2

MM

2

MM

8

LuM

2

2121sek,bsek,atbta

2⋅

=⋅

−=+⋅

−⋅⋅++⋅

+⋅−=∆+∆

++

+⋅

=

corresponds the moment due to the uniformly distributed load u for the beam with the fixed ends


Equivalent forces of th eprestress force at the folded points of a member

At the chamfer

Section forces due to prestress force

A simple model of the equavalent

forces

More refinement model

Compare fro instance the lifting force in a saddle beam


Balancing forces due to the curvature of a tendon

The balancing force is the 2.-derivative of the tendon curavture function

2

2

)(dx

ydPxu ⋅=

Circular arc with radius R: u(x) = P/R

Parabola arch: 2L

Pf8)x(u

⋅⋅−=

where f is the total sag (maximum distance of the tendon from the atraight line

between the inflection points

L is the distance of the inflection points of the tendon

or by derivating 2

2x

L

f4y ⋅

⋅−= => x

L

f8

dx

dy2

⋅⋅−

= => 22

2

L

f8

dx

yd ⋅−=

If the loss due to friction is taken into account P => P(x)

Tendon is defined by the levels of the tendon in different points and using the difference method

2

)1k(bvbvk)1k(bv

kx

zz2z)x(P)x(u

∆

+⋅−⋅=

+−

Geometry of the 2-degree parabola in a continuous beam


Separate coordinate systems at supports and the field:

at the support: x2; y2 at the field: x1; y1

Compatibility conditions at the inflection points: y1 + y2 = h and y1’ = y2’

Equation for a parabola y = k x2

=>

=⋅⋅−−⋅⋅

=⋅+−⋅

0ak2)ab(k2

hak)ab(k

21

2

2

2

1

=>

⋅=

−⋅=

ab

hk

)ab(b

hk

2

1

Parabola at the support region: 2

22 xab

hy ⋅

⋅= => the balancing force

ab

Ph2u2

⋅

⋅⋅=

Parabola in the field region: 2

11 x)ab(b

hy ⋅

−⋅= => the balancing force

)ab(b

Ph2u1

−⋅

⋅⋅=

If a length of the counter rounding is a = h and h =L/20 and the distance from the support to the

inflectionn point b=L/2 the equations of the parabolas and the balancing forces are:

at the support region L

x2y

2

22


L

P4u2

⋅

at the field region L9

x2y

2

11

⋅


L9

P4u1

⋅

⋅=


Equivalent forces of the prestress force in a continuous beam

General case

The equivalent forces of the prestress force:

)dc(d

)ee(P2u

)dc(c

)ee(P2u

)ba(b

)ee(P2u

)ba(a

)ee(P2u

32d

32c

21b

21a

+⋅

+⋅⋅=

+⋅

+⋅⋅−=

+⋅

+⋅⋅−=

+⋅

+⋅⋅=

Equilibrium:

0ducu

0buau

dc

ba

=⋅+⋅

=⋅+⋅

If the length of any part =0, the distributed load is substituted with point force F.

The edge span of a continuus beam

The equivalent forces of the prestress force:

)dc(d

)ee(P2u

)dc(c

)ee(P2u

b

eP2

)ba(b

)ee(P2u

b

eP2

)ba(

)ee(P2F

32d

32c

2

221b

221a

+⋅

+⋅⋅=

+⋅

+⋅⋅−=

⋅⋅−=

+⋅

+⋅⋅−=

⋅⋅=

+

⋅⋅= +

Equilibrium :

Fa + b⋅ub =0

c⋅uc + d⋅ud = 0


Load balancing forces due to the tendon consisting straight parts

c

eePF

c

ee

b

eePF

b

eePF

32c

3221b

21a

+⋅=

++

+⋅−=

+⋅−=


Example.

Let’ s see the beam which one end is simply supported and theother end is fixed (for instance the

edge span of the 2-span continuous beam)

Span of the beam L=20 m , beam cross-section: depth h=1,0 width b= 0.5 m

Assume that from the total uniformly distributed load of the beam is balanced a part which

corresponds the dead load g=25 kN/m .

Bending moment due to the imposed dead load g = 25 kN/m:

Max.field moment Mk = 703.1 kNm at the point xmax= 7,5 m from the edge support

Mm = 625 kNm at the point x=L/2=10 m

Support moment Mt = - 1250 kNm

Suppose the prestress force P=2000 kN

If the balancing force due to the prestress force is u=-25 kN/m, so the tendon geometry coincident

to the moment diagram due to the load g then the secondary moment = 0. Bending moment diagram

due to the prestress force is M=-P e

Suppose the prestress force P=2500 kN

Eccentricities at the simply support e1=0

at the point of the max. moment xmax = 7,5 m: e2=-703/-2500=0,2813 m

x=L/2= 10 m em=-625/-2500=0,25 m

at the middle support e3 = -1250/-2500=0.5 m

The lowest point of the tendon x=7,5 m => a=0 b=7,5 m c=20-7,5=12,5 m d=0

The sag f = em + (e1+e2)= 0.25+0.5/2=0.5 m = [Mm+(Mta +Mtb)/2]/P=(625+1250/2)/2500

The required value of the sag f depends on the prestress force. If P=2000 kN the sag f must be

f=0,625 m

The required minimum depth of the beam, when the prestress force P=2500 kN:

M1703.13

-1250.00


Suppose the centre distance of the tendon from the face c=0,1 m => h > 2*(0.5+0.1)m=1,2 m

For the prestress force P=2000 kN h > 1,45 m


2a

2b 2 2

2 3c 2 2

2 3d

2 P e 2 2500 0,28125F 187,5 kN

b 7,5

2 P e 2 2500 0,28125u 25 kN / m

b 7,5

2 P (e e ) 2 2500 (0.28125 0,5)u 25 kN / m

c 12,5

2 P (e e ) 2 2500 (0,28125 0,5)F 312,5 kN

c 12,5

⋅ ⋅ ⋅ ⋅= − = − =

⋅ ⋅ ⋅ ⋅= − = − = −

⋅ ⋅ + ⋅ ⋅ += − = − = −

⋅ ⋅ + ⋅ ⋅ += − = − =


The tendon is moved so that the straight line between the ends of the tendon is rotated at the simply

support downward an angle α= e3/L so that the eccentricity at the fixed end is e3= 0. The curve

formed by the tendon remains the same.

Eccentricity at the free support e1 = 0

at the point x= 7,5 m e2 =0,2813+0,5*7.5/20=0,4688 m

at the midspan x = 10 m em = 0.25+0.5/2=0.5 m

at the fixed end support e3 =0

Because the curvature of the tendon is the same, the load balancing force of the prestress force

u=-25 kN/m and the bending moments due to this load balancing force are the same as before.

The lowest point x=10 m => e2 = em => a=0 b=10 m c=10 m d=0

The sag f = 0.5 + 0/2 = 0.5 m is the same as before


2a

2b 2 2

2 3c 2 2

2 3d

2 P e 2 2500 0,5F 250 kN

b 10

2 P e 2 2500 0,5u 25 kN / m

b 10

2 P (e e ) 2 2500 (0.5 0)u 25 kN / m

c 10

2 P (e e ) 2 2500 (0,5 0)F 250 kN

c 10

⋅ ⋅ ⋅ ⋅= − = − =

⋅ ⋅ ⋅ ⋅= − = − = −

⋅ ⋅ + ⋅ ⋅ += − = − = −

⋅ ⋅ + ⋅ ⋅ += − = − =

Bending moment due to the prestress force is the same as before..

There exists the secondary moment because the tendon profile does not coincine to the bending

moment diagram due to the load balancing force..

Secondary moment: at the point x=7,5 m ∆M= -703,1 + 2500*0,4688 = 468,8 kNm

at the point x= 10 m ∆M = -625 + 2500*0.5 = 625 kNm

at the fixed support ∆M = 1250 kNm

The required depth of the beam: h> 2⋅(0.5+0.1) = 1,2 m


The position of the tendon is moved so that the line between the ends of the tendon is rotated about

the anchor at the free support so that the max. eccentricities in the field and the fixed support are the

same so e2=e3=e. The shape and the curvature of the tendon are the same as before, so the balancing

force is also remains the same u=-25 kN/m.

m/kN25)xL(

)ee(P2u

m/kN25x

eP2u

2c

2b

−=−

+⋅⋅−=

−=⋅⋅

−=

=> 1x2

)xL(

2

2

=⋅

− => m2843,8L)12(x =⋅−=

m3431,025002

20)12(25

P2

L)12(ueee

2222

32 =⋅−

⋅−⋅−=

⋅−

⋅−⋅===

Eccentricities : at the free support e1 = 0

at the point x= 7,5 m e =0,4688 - 0,3431*7.5/20=0,340 m

at the point x=8.284 m e2 = 0.3431 m

at the point x = 10 m em = 0.5-0.3431/2=0.3285 m

at the fixed end e3 =0.3431 m

Because the curvature has not changed, the balancing force is also the same as before

u=-25 kN/m an dthe moments due to the balancing force are the same.

The lowest point x=10 m => e2 = em => a=0 b=10 m c=10 m d=0

The sag f = 0.3285+0.3431/2= 0.5 m is the same as before

Balancing forces:

2a

2b 2 2

2 3c 2 2

2 3d

2 P e 2 2500 0,3431F 207,1kN

b 8,2843

2 P e 2 2500 0,3431u 25 kN / m

b 8,2843

2 P (e e ) 2 2500 (0,3431 0,3431)u 25 kN / m

c (20 8,2843)

2 P (e e ) 2 2500 (0,3431 0,3431)F 292,85 k

c 20 8,2843

⋅ ⋅ ⋅ ⋅= − = − =

⋅ ⋅ ⋅ ⋅= − = − = −

⋅ ⋅ + ⋅ ⋅ += − = − = −

−

⋅ ⋅ + ⋅ ⋅ += − = − =

−N

The bending moment due to the prestress force is the same as before.

Secondary moment arises because the tendon profile does not coincides to the moment diagram due

to the balancing forces..


Secondary moment : at the point x=7,5 m ∆M= -703,1 + 2500⋅0,34 = 146,9 kNm

at the point x= 10 m ∆M = -625 + 2500⋅0.5 = 196,3 kNm

at the fixed support ∆M = 1250 - 2500⋅0,3431=392,3 kNm

Required depth of the beam: h > 2⋅(0,343+0,1)=0,886 m

If the prestress force is P=2000 kN the required depth of the beam should be h>1,110 m


The magnitude of the required balancing force

Starting from the requirements of the service limit state:

- the allowable tensile stress under the quasi-permanent and he frequent load combination

Stress in the tensile fibre: admct

balk

cW

MM

A

P,σσ ≤

++

−=

Mk is the bending moment under the quasi-permanent or the frequent load combination

Mbal is the moment of the balanced force due to the prestress force (opposite sign than the

bending moment due to the imposed loads )

σct,adm is the allowbale tensile stress of concrete under th quasi-permanent or the frequent load

combination;

for example: decompression limit state σct,adm = 0

cracking limit state σct,adm = fctm

If cracking is allowed, for a preliminary dimensioning when choosing the prestress force for the

allowable tensile stress can be used

σct,adm > fctm

Choose the balancing moment Mbal so that the above conditions fullfill

The balancing moment can be written as a function of the prestress force P and the eccentricity e..

The eccentricities of the tendon and the ratio of the eccentricities at different points of the beam can

be define from the moment diagram Mk due to the imposed loads.

Taking into account the depth of the beam and the required concrete covers for the tendons and the

ducts eccentricites ei can be define.

The minimum radius of the tendon is taken into account => rounding at the middle support

So the tendon profile has been defined.

The balancing forces u(P) are defined as function of prestress force P.

The moment diagram Mbal(P) due to the balancing force is calcuted as a function of P by loading

the structure with these balancing forces u(P) when P=1

Required prestress force P can be solved from the stress inequalities.

Then the final balancing forces and the balancing moment diagram are calculated.


Example: Load balancing by the prestress force

2-span continuous beam, span L

Bending moments due to uniformly distributed dead load in the 2-span continuous beam:

field: Mk = 0.07 q L2 at the point x= 0.375 L from the free support

middle support: Mt = -0.125 q L2

Choose that the eccentricities in the field and the middle supports are similar: e2=e3

Eccentricity at the free support e1 =0

The lowest point of the tendon is at the point of the max. field moment; eccenrticity e2

Eccentricity at the middle support e3

Near the middle support the inflection point of the tendon is about 1.2 m from the support; for

simplicity its is assumed that d~0

a=0 b = 0.375 L c=L-b –d~L-b = 0.625 L

Balancing forces:

2 2a

2 2 2b 2 2 2 2

2 3 2 3 2 3c 2 2 2

2 3 2 3 2 3d

2 P e 2 P e P eF 5,33

b 0.375 L L

2 P e 2 P e P eu 14,22

b 0,375 L L

2 P (e e ) 2 P (e e ) P (e e )u 5,12

c (c d) 0,625 L L

2 P (e e ) 2 P (e e ) P (e e )u 3,2 kN

c 0,625 L L

⋅ ⋅ ⋅ ⋅ ⋅= − = − = ⋅

⋅

⋅ ⋅ ⋅ ⋅ ⋅= − = − = − ⋅

⋅

⋅ ⋅ + ⋅ ⋅ + ⋅ += − = − = − ⋅

⋅ + ⋅

⋅ ⋅ + ⋅ ⋅ + ⋅ += − = − = ⋅

⋅

The balancing forces ub and uc are different. They are equal in the case where e2/e3 =Mk/Mt or

where the linear transformation from this state has been done to the tendon and the sag has

remainded unchanged as in above the examples.

M1 M2703.13-1250.00

703.12-1250.00


The balancing forces ub ja uc are different, but the enough exact result can be obtained by using as

an uniformly distributed load the average value um over the whole span.

( )322

322

2

cbm e56,2e67,9

L

P

2

e12,5e12,5e22,14

L

P

2

uuu ⋅+⋅−=

⋅+⋅+⋅⋅−=

+≈

The balancing moments due to the uniformly distributed load um:

Field: PeePeeLuM mbalk ⋅⋅+⋅−≈⋅⋅+⋅−=⋅⋅= )2,07,0()18,068,0(07,0 3232

2

,

Middle support: PeePeeLuM mbalt ⋅⋅+⋅≈⋅⋅+⋅=⋅⋅−= )3,02,1()32,021,1(125,0 3232

2

,

Stresses:

Field: admct

bbb

k

b

balk

cW

Lq

W

ee

AP

W

M

W

M

A

P,

2

32, 07,018,068,01σσ ≤

⋅⋅+

⋅+⋅+⋅−=++

−=

Support: admct

ttt

t

t

balt

cW

Lq

W

ee

AP

W

M

W

M

A

P,

2

32, 125,032,021,11σσ ≤

⋅⋅+

⋅+⋅+⋅−=−−

−=

For solving the prestress force P

( ) ( )

( ) ( )

⋅+⋅+

⋅−=

⋅+⋅+

⋅−⋅⋅

⋅+⋅+

⋅−=

⋅+⋅+

⋅−⋅⋅

≥

32

,

32

,

2

32

,

32

,

2

32,021,1

||

32,021,1

125,0

18,068,018,068,0

07,0

eeA

W

WM

eeA

W

WLq

eeA

W

WM

eeA

W

WLq

P

t

tadmctt

t

tadmct

b

badmctk

b

badmct

σσ

σσ


Example:

Beam h= 1000 mm b= 400 mm A=0,4 m2 Wb = Wt = 0,0667 m

3

Wb/A = Wt/A = h/6=167 mm

e = e1 = e2 = 450 mmm = 0,45 h

Decompression limit state σct,sall = 0

⋅⋅=

+⋅⋅+

⋅⋅

⋅⋅=

+⋅⋅+

⋅⋅

≥

h

Lq146,0

)32,021,1(h45.06

h

Lq125,0

h

Lq126,0

)18,068,0(h45.06

h

Lq07,0

P22

22

=> h

Lq146,0P

2⋅

⋅≥

The average balancing force (e = e1=e2)

( ) qh

e78,1q

h

e37,0

h

e41,1e56,2e67,9

L

Pu 32

322m ⋅⋅=⋅

⋅+⋅−=⋅+⋅−=

If e=0.45 h => um = 0,8 q

� The balancing force is 80 % of the imposed load

Balancing forces:

q67,0h

eeq75,0

L

)ee(P12,5u

q93,0h

eq08,2

L

eP22.14u

32

2

32c

2

2

2b

⋅=+

⋅⋅=+⋅

⋅=

⋅=⋅⋅=⋅

⋅=


Look at the beam with one fixed end presented before (or one span of the 2-span continuous beam)

h*b = 1,0*0.5 m span L=20 m

A=0,5 m2 Wb = Wt = 0,0833 m

3

Uniformly distribute imposed load q= 25 kN/m

Bending moment du eto the imposed load g = 25 kN/m:

Max. field moment Mk = 703.1 kNm at the point xmax= 7,5 m from the free support

Mm = 625 kNm at the midspan x=L/2=10 m

Fixed end moment Mt = - 1250 kNm

Choose te eccentricities

e3 = 0,45 h = 450 mm at fixed support

e2 = e3⋅Mk/|Mt| = 450⋅703,1/1250 = 253 mm at the point of the max. field moment

e1 = 0 at the free support

a=0 b= 7,5 m c=20-7,5=12,5 m d~0

The balancing forces can be calculated as function of prestress force P:

P009,05,12

)45,0253,0(P2

c

)ee(P2u

P009,05,7

253,0P2

b

eP2u

22

32c

22

2b

⋅−=+⋅⋅−

=+⋅⋅−

=

⋅−=⋅⋅−

=⋅⋅−

=

The average balancing force um = -0,009⋅P

Bending moments due to the balancing force um = -0,009⋅P :

field mPLuM mbalk ⋅⋅−=⋅⋅= 253,007.0 2

,

support mPLuM mbalt ⋅⋅=⋅⋅= 45,0125,0 2

,


Decompression limit state σct,adm = 0

Stresses:

Bottom fibre, field 00833,0

7031,0

0833,0

253,0

5,0 332

,=+

⋅−+

−=++

−=

m

MNm

m

mP

m

P

W

M

W

M

A

P

b

k

b

balk

cbotσ

� P=1676 kN

Top fibre, support 00833,0

250,1

0833,0

45,0

5,0 332

,=−

⋅−

−=−−

−=

m

MNm

m

mP

m

P

W

M

W

M

A

P

t

t

t

balt

opctσ

� P=2027 kN

The balancing force um = -0,009⋅P=-0,009⋅2027 = 18,2 kN/m => 73 % of the imposed load of the

beam

________________________________________________________________________

Let increase the eccentricity in field to the same magnitude as at the support so. e2 = 0,45 m = e3,

the lowest point of the tendon remains the same as before so b=7,5 m.

The balancing forces are calculated as function P:

P01152,05,12

)45,045,0(P2

c

)ee(P2u

P016,05,7

45,0P2

b

eP2u

22

32c

22

2b

⋅−=+⋅⋅−

=+⋅⋅−

=

⋅−=⋅⋅−

=⋅⋅−

=

The average balncing force P01376,02

P01152,0P016,0um ⋅−=

⋅−⋅−=

Bending moments due to the balancing force um = -0,1376⋅P :

field mPLuM mbalk ⋅⋅−=⋅⋅= 387,007.0 2

,

support mPLuM mbalt ⋅⋅=⋅⋅= 688,0125,0 2

,

Decompression limit state σct,adm = 0

Stresses:

Bottom fibre, field 00833,0

7031,0

0833,0

387,0

5,0 332

,=+

⋅−+

−=++

−=

m

MNm

m

mP

m

P

W

M

W

M

A

P

b

k

b

balk

cσ

� P=1270 kN

Top fibre, support 00833,0

250,1

0833,0

688,0

5,0 332

,=−

⋅−

−=−−

−=

m

MNm

m

mP

m

P

W

M

W

M

A

P

t

t

t

balt

ctopσ

� P=1462 kN


The balancingf force um = -0,01376⋅P=-0,01376⋅1462 = -20,1 kN/m => 80 % of the imposed load

of the beam

ub = -0,016⋅P = -0,016⋅1462 = - 23,39 kN/m 0...7.5 m

uc = -0,01152⋅P = -0,01152⋅1462 = -16,84 kN/m 7,5 m ....20 m

The balancing forces due to the average balancing force um:

field: kNm8,56520)1,20(07,0Lu07.0M 22

mbal,k −=⋅−⋅=⋅⋅=

support: kNm100520)1,20(125,0Lu125,0M 22

mbal,t =⋅−⋅−=⋅⋅−=

If the cracking limit state is allowed and for concrete is alloewd tensile stress σct,adm =fctm =3,5 MPa

(concrete grade C40/50):

Cracking moment Mr = fctm⋅W = 292 kNm

Prestress force P = 1120 kN is adequite. 1250

29212501462P

−⋅=


THE MAGNITUDE OF THE REQUIRED BALANCING MOMENT DUE TO THE

ECCENTRICITY OF THE PRESTRESS FORCE

Background:

The stress at the tension fibre fulfills the requirements given in the EC2 table 7.1N in order to restriction

of cracking

Concrete stress due to prestress force W

eP

A

Pc

⋅−−=σ

W is the section modulus about the tensile fibre

Allowable bending moment due to the imposed load

')1()()( rbaladmadmc MeA

WMW

W

eP

A

PWM ++

⋅=⋅+

⋅+=⋅+−= σσσ

where the balancing moment due to prestress force Mbal = P e

“Cracking moment” WM admr ⋅= σ'

=> The magnitude of the required balancing moment

ePk

keA

W

MMM p

p

rbal ⋅⋅=

⋅⋅+

−=

1

'

where kp = -1,0 for a simply supported beam

kp = -0,83 for the edge span of a continuous beam

kp = 1,48 for the middle support of a continuous beam

Exposure class Bending moment M ”Cracking moment ” Mr'

Requirement for cracking

X0, XC1

M=Mquasi-perm

Mr' = 0

Decompression limit state

M=Mquasi-perm

Mr' =0

Decompression

limit state

XC2

M=Mfrequent

Mr' =1,7k1fctmW

Crack width

wk ≤ 0,1 mm

M=Mquasi-perm

Mr' =2,1 k1fctm W

Crack width

wk ≤ 0,2 mm

XC3, XC4

M=Mfrequent

Mr' =2,4 k1fctm W

Crack width

wk ≤# 0,3 mm

1,1

7,0)m(h6,1k1

≤

≥−=

I

f in the tensile zone of the cross-section is used additional un-prestressed reinforcement, the allowable

tensile stress (1,7...2,4) k1 fctm can be increased with 0,3 MPa/As = 0,001 Ac. The upper limit of the

allowable tensile stress is 0,3 fck.

1 / 11

e3 0.3 m⋅:=Right support

e2 0.3 m⋅:=At the point of the max. field moment

e1 0.3 m⋅:=Left support A

Eccentricities of the tendons:

P 2500 kN⋅:=Prestress force

α1 L⋅ 8 m=

α1 0.5:=The maximum point of the field moment

R 8.0 m⋅:=Curvature radius at the supports

L 16.0 m⋅:=Span

Symmterical beam fixed at the both ends the curvature and the eccentricity of the tendon are similar,tendons are horizontal at the supports A and B.

Example about the equivalent (balancing) forces due to prestress force

2 / 11

d 1.2 m=d α3 L⋅:=Region E...B

c 6.8 m=c α1 L⋅ α3 L⋅−:=Region D...E:

α3 0.075=α32 8.0⋅ m⋅ 0.3 m⋅ 0.3 m⋅+( )⋅[ ]

0.5 16 m⋅( )2

⋅

:=

α3

2 R⋅ e2 e3+( )⋅

α1 L2

⋅

:=Inflection point; distance from the right support

b 6.8 m=b 1 α1−( ) L⋅ a−:=region C...D

Region A...Ca 1.2 m=a α2 L⋅:=

α2 0.075=α22 8.0⋅ m⋅ 0.3 m⋅ 0.3 m⋅+( )⋅[ ]

1 0.5−( ) 16 m⋅( )2

⋅

:=

α2

2 R⋅ e1 e2+( )⋅

1 α1−( ) L2

⋅

:=Inflection point; distance from the left support

3 / 11

Checking the equilibrium: ua a⋅ 375 kN= uc c⋅ 375− kN=

ub b⋅ 375− kN= ud d⋅ 375 kN=

ua a⋅ ub b⋅+ 0 kN= uc c⋅ ud d⋅+ 0 kN=

The equivalent loads due to prestress are presented in the picture (b):

Centric cpmpressive normal force N P−:= N 2.5− 103

× kN=

Moments at the ends MPa P e1⋅:= MPa 750 kNm=

MPb P e3⋅:= MPb 750 kNm=

Load balancing forces

Region A...C ua

2 P⋅ e1 e2+( )⋅ a a b+( )⋅

:=

ua2 2500⋅ kN⋅ 0.3 m⋅ 0.3 m⋅+( )⋅[ ]

1.2 m⋅ 1.2 m⋅ 6.8 m⋅+( )⋅:= ua 312.5

kN

m=

Region C...D

ub

2 P⋅ e1 e2+( )⋅ −

b a b+( )⋅:=

ub2 2500⋅ kN⋅ 0.3 m⋅ 0.3 m⋅+( )⋅[ ]−

6.8 m⋅ 1.2 m⋅ 6.8 m⋅+( )⋅:= ub 55.15−

kN

m=

Region D...E

uc

2 P⋅ e2 e3+( )⋅ −

c c d+( )⋅:=

uc2 2500⋅ kN⋅ 0.3 m⋅ 0.3 m⋅+( )⋅[ ]−

6.8 m⋅ 1.2 m⋅ 6.8 m⋅+( )⋅:= uc 55.15−

kN

m=

Region E....B

ud

2 P⋅ e2 e3+( )⋅ d c d+( )⋅

:=

ud2 2500⋅ kN⋅ 0.3 m⋅ 0.3 m⋅+( )⋅[ ]

1.2 m⋅ 1.2 m⋅ 6.8 m⋅+( )⋅:= ud 312.5

kN

m=

4 / 11

ub 55.147−kN

m= at the distance b 6.8 m= and

at the distance c 6.8 m=uc 55.147−

kN

m=

Support moments:

Ma2 ub− L⋅b c+( )

24⋅ 3

b c+( )

L

2

−

⋅:=

Ma2 55.15−kN

m⋅

− 16⋅ m⋅6.8 m⋅ 6.8 m⋅+( )

24⋅ 3

6.8 m⋅ 6.8 m⋅+( )

16 m⋅

2

−

⋅:= Ma2 1.139 103

× m2 kN

m=

Mb2 Ma2:= Mb2 1.139 103

× kNm=

Field moment

Mk2 ub L⋅b c+( )

8⋅ 2

b c+( )

L

−

⋅Ma2 Mb2+( )

2+:=

Mk2 55.15−kN

m⋅

16⋅ m⋅6.8 m⋅ 6.8 m⋅+( )

8⋅ 2

6.8 m⋅ 6.8 m⋅+( )

16 m⋅

−

⋅1139 kNm⋅ 1139 kNm⋅+( )

2+:=

Mk2 586.092− kNm=

Moments due to the balancing forces together (case 1+ case 2):

Support moment

MA Ma1 Ma2+:= MA 925.061 kNm=

Moment diagram due to the prestressing force

The moment diagram (fig. (c)) due to the prestressing force can be obtained by calculating the bending moments due to the balancing loads ua, ub,uc ja ud for the beam with the fixed supoorts at the both ends.

- 1. Balancing forces ua 312.5kN

m= at the distance a 1.2 m= and

ud 312.5kN

m= at the distance d 1.2 m=

Support moments

Ma1 ua−a2

3⋅ 1.5

a

L−

⋅:= Ma1 312.5−kN

m⋅

1.2 m⋅( )2

3⋅ 1.5

1.2 m⋅

16 m⋅−

⋅:=

Ma1 213.75− kNm=

Mb1 Ma1:= Mb1 213.75− kNm=

Field moment

Mk1 uaa2

2⋅

Ma1 Mb1+( )2

+:= Mk1 312.5kN

m⋅

1.2 m⋅( )2

2⋅

213.75− kNm⋅ 213.75 kNm⋅−( )

2+:=

Mk1 11.25 kNm=

- 2. Balancing forces

5 / 11

Msec_b 175 kNm=Msec_b 925 kNm⋅ 750 kNm⋅−:=Msec_b MB MPb−:=

Msec_a 175 kNm=Msec_a 925 kNm⋅ 750 kNm⋅−:=Msec_a MA MPa−:=

MPb 750 kNm=ja MPa 750 kNm=End moments

Secondary moments at the support is obtained by subtracting the end moments P*e from the support moments due the balancing forces.

Mk 574.842− kNm=Mk Mk1 Mk2+:=

Field moment

MB 925.061 kNm=MB Mb1 Mb2+:=

MA 925.061 kNm=MA Ma1 Ma2+:=

6 / 11

Region E...B d α3 L⋅:= d 0.9 m=

Balancing forcesua

2 P⋅ e1 e2+( )⋅ a a b+( )⋅

:= ua 312.5kN

m=

Region A...C

Region C...D

ub

2 P⋅ e1 e2+( )⋅ −

b a b+( )⋅:= ub 39.61−

kN

m=

Region D...E uc

2 P⋅ e2 e3+( )⋅ −

c c d+( )⋅:= uc 39.613−

kN

m=

Region E....B ud

2 P⋅ e2 e3+( )⋅ d c d+( )⋅

:= ud 312.5kN

m=

Checking the equilibrium: ua a⋅ 281.25 kN= uc c⋅ 281.25− kN=

ub b⋅ 281.25− kN= ud d⋅ 281.25 kN=

ua a⋅ ub b⋅+ 0 kN= uc c⋅ ud d⋅+ 0 kN=

The eccentricities are changed so that the eccentricity at the point of the max. field moment is only e2=0.15 m

Eccentricities of the tendon:

Left supporti A e1 0.3 m⋅:=

At the point of the max. field moment e2 0.15 m⋅:=

Right support e3 0.3 m⋅:=

Inflection point from the left support α2

2 R⋅ e1 e2+( )⋅

1 α1−( ) L2

⋅

:=α2 0.056=

a α2 L⋅:= a 0.9 m=Region A...C

Region C...D b 1 α1−( ) L⋅ a−:= b 7.1 m=

Inflection point from th eright supportα3

2 R⋅ e2 e3+( )⋅

α1 L2

⋅

:= α3 0.056=

Region D...E: c α1 L⋅ α3 L⋅−:= c 7.1 m=

7 / 11

Support moments:

Ma2 ub− L⋅b c+( )

24⋅ 3

b c+( )

L

2

−

⋅:= Ma2 829.6 kNm=

Mb2 Ma2:= Mb2 829.6 kNm=

Field moment

Mk2 ub L⋅b c+( )

8⋅ 2

b c+( )

L

−

⋅Ma2 Mb2+( )

2+:= Mk2 421.934− kNm=

Moments due to the balancing forces together (case 1+ case 2):

Support moment

MA Ma1 Ma2+:= MA 707.813 kNm=

MB Mb1 Mb2+:= MB 707.813 kNm=

Field moment

Mk Mk1 Mk2+:= Mk 417.188− kNm=

Secondary moments at the support is obtained by subtracting the end moments P*e from the support moments due the balancing forces.

End moments MPa 750 kNm= ja MPb 750 kNm=

Msec_a MA MPa−:= Msec_a 42.188− kNm=

Msec_b MB MPb−:= Msec_b 42.188− kNm=

The smaller eccentricity in the field leads also the smaller support moments due to prestress force and so the smaller load balancing effect.

Moment diagram due to prestress force

The moment diagram (fig. (c)) due to the prestressing force can be obtained by calculating the bending moments due to the balancing loads ua, ub,uc ja ud for the beam with the fixed supoorts at the both ends.

- 1. Balancing forces ua 312.5kN

m= at the distance a 0.9 m= and

ud 312.5kN

m= at the distance d 0.9 m=

Support moments

Ma1 ua−a2

3⋅ 1.5

a

L−

⋅:= Ma1 121.816− kNm=

Mb1 Ma1:= Mb1 121.816− kNm=

Field moment

Mk1 uaa2

2⋅

Ma1 Mb1+( )2

+:= Mk1 4.746 kNm=

- 2. Balancing forces ub 39.613−kN

m= at the distance b 7.1 m= and

at the distance c 7.1 m=uc 39.613−

kN

m=

Rak. 43-3111 Immediate losses of post-tension members 1

Prestress loss due to friction and anchorage slip

Prestress loss due to friction between the tendon and the duct

θ⋅µ−

µ ⋅=θ eP)(P o

whrere

µ is the friction coefficient

θ is the change of the angle between the point Po and P(θ)

Stress σpx in the tendon at the distance x from the active (jacking)

anchor

)x(

popx e ⋅β+αΣ⋅µ−⋅σ=σ

where

σpx is the stress at the distance x from the active anchor or

the point where the stress is σpo

σpo is the stress at the active anchor

µ is the specific friction coefficient between the tendon and the duct for the specific post-

tension system; typically µ = 0,15...0.3

β is the specific wobble coefficient for the spesific post-tension system; typically

β=0,005... 0,02 rad/m

Σα is the sum of the angular changes measured from the active anchor (or the point of σpo)

at the distance x; so Σα is the total change of the angle at the distance x

Σα is putted in radians and x in metres to the equation.

Friction is caused by the friction between the tendon and the duct. Friction is induced at the points

where due to the change of the direction of the tendon (like at curved parts of the tendon profile) the

tendon is pressed against the duct.

Friction is induced also at the straight part of the tendon profile when the tendon touch the wall of the

duct because of the duct is not precisely straight but is a little wavy due to small changes of the

angles due to deflection of the duct between the support points or inaccurancies of the support levels.

This is taken into account with the wobble coefficient β which gives straight the changes of angle in

radians at length unit m. The total change of angle due to these waves at the distance x is βx.

The sum of the angle changes can be defined from the tendon geometry:


- tendon profile y=f(x)

- inclination of the tendon dy/dx

- the change of the inclination at the length unit d2y/dx

2

- sum of the angular changes at the distance x from the active anchor ∫ −==Σx

x yxydxdx

yd

0

2

2

)0(')('α

- if the tendon profile is composed with different curved parts , the interpolation is done for pieces at

the distance 0…x.

- sum of the angular changes can be also obtained from the balancing forces P

dxxu

x

x

∫=Σ

0

)(

α

The straight tendon: Σα = 0

The parabola tendon: Σαx=2 a/b2 x

If the angle changes are small (α1 < 0,36 rad ~20o ) the sum of the angle changes Σα can be calculated

according the notes in the figure from the equation

∑ ∑⋅

=αi i i

ii

a8

l i=1,2,….

If the exponent 3,0)x( ≤⋅β+αΣ⋅µ the stress at the distance x from the active anchor can be

calculated for the equation

( )[ ]x1popx ⋅β+αΣ⋅µ−⋅σ=σ

The tendon profile in the point of view the friction losses can be judge with the integral

0.1...9.0LP

dxP

0

L

0

x

=⋅

⋅∫


Loss due to the anchorage slip δδδδ

When the tendon is fixed with wedges to the anchor there is small amount (few mm) of split δ at the

anchorages and the tendon will shorten and so the force in the tendon decrease => prestress loss.

Friction between the tendon and the duct decrease gradually the loss due to anchorage slip when

going forward from the jacking anchor. The loss due to anchorage slippage has it’s the greatest value

at the jacking anchor. If there is no friction (like pre-tensioned tendons) the tendon shortens at the

same amount at the whole distance between the anchors and the loss due to anchorage slippage is the

same at the whole distance between the anchors. Due to friction the loss due to the anchorage slippage

extends only a certain distance from the jacking anchor.

The stress in the tendon decrease due to friction when going forward from the jacking anchor and the

loss of the anchorage slippage decrease and so the stress increase when going from the jacking

anchor. The change of the loss due to anchorage slippage at the unit length is equal to the change of

the stress due to friction at the unit length. So the tangent of the curve of the loss due to anchorage

slippage (the inclination of the line) is equal to the tangent of the curve of the loss due friction but

with the opposite sign.

The aim is to search the point x1, where the loss due to anchorage slip is decreased as much as the

loss due to friction; in other words to search the point where the curve of the loss due to anchorage

slip intersects the curve of he loss due to friction.

The anchorage slip δ can obtained by integrating the change of the strain ∆ε due to the slip at the

distance x1 :

∫ ∫∫ ⋅⋅

∆

∆=⋅

⋅⋅

−⋅≈

⋅

−=

⋅

∆=ε∆=δ

1x

0

1x

0 pp

2

11

pp

1x0

pp

0xx

1x

0 pp AE

x

x

Px

AE2

)PP(2dx

AE

PPdx

AE

P dx


1

1x0

x

PP

x

P −=

∆

∆

By solving x1 β⋅µ⋅σ

⋅δ=

∆

∆

⋅⋅δ=

p

ppp

1

E

x

P

AEx

Loss due to anchorage slip )(2 1, xxxanc PPP −⋅=∆

The force in the tendon at the point x the loss due to the anchorage slip taken into account is

xxxxxxancxx PPPPPPPP −⋅=−⋅−=∆−= 11,0 2)(2

Straight tendon: µβ = 0

Parabola tendon: µβ −> 2µ a/b2

+ µβ

Circular tendon: µβ −> µ/R + µβ

Procedure for defining the loss due to anchorage slip is following:

- Suppose the length of the influence of the loss due to anchorage slip the distance x1 from he

jacking anchor

- Calculate the prestress force Px1 at the distance x1 from the jacking anchor taking into account

the loss due to friction

- Calculate the average change of the prestress force per the unit length at the distance x1

∆P/∆x=[P0-Px1]/x1

- Calculate the influence length of the loss duet to anchorage slip the new value

x

P

AEx

pp

new

∆

∆

⋅=

δ,1

- If the new value differs from he assuming value, the prestress force Px1,new and the change

∆P/∆x and then the new x1,new

- Assume the change of the prestress force due to friction and correspondingly the change of

loss due to anchorage slip as linear, so we get the triangular drawn in the figure.

- The loss due to the anchorage slip is ∆Panc,0=2 [P0 – Px1]

Prestress force at the point 0 ≤ x ≤ x1 is Px0=Px - 2⋅(Px – Px1)

- Anchorage slip is x

P

∆

∆=δ

The anchorage slip is typically 1...12 mm depending on the post-tensioning system. The influence

length of the loss due to the anchorage slip is about 10...20 m.


Compensation of the loss due to the anchorage

1. The tendons are overstressed the value σpo,max and the tendons are locked

2. The tendons are retensioned again to the value σpo

3. Disks are installed to the anchor so that no anchorage slippage does not happen



Elastic shortening in post-tensioned members

The immediate change of the force in the tendon due to the elastic shortening of concrete

cmi

p

pcpeE

EAP ⋅⋅σ=∆

where

σcp is the concrete compressive stress at the centroid of the tendon ; σcp is calculated by

using the net cross-section opf concrete where the afread of the ducts are subtracted

because there are holes at the location of the ducts (the ducts are not yeat injected

Ap is the area of the tendon

Ep is the elastic modulus of the tendon

Ecmi is the average elastic modulus of concrete at the time of tensioning

When several tendons are tensioned separately at turn without retensioning afterwards, the immediate

change of the force in each tendon is calculated from the equation

cmi

p

pcpeiE

E

n2

1nAP ⋅

⋅

−⋅⋅σ=∆

where

σcp is concrete compressive stress at the centroid of the tendons represents the force of

Ap is the area of one tendon

n is the number of the tendons or tensioning stages if each tendon is tensioned at several

stages

The equation is got by the following:

Suppose there are 3 tendons in a member:

- The fist tendon is tensioned to the value P1 = σpoAp = P/3: total prestress force is then P/3 and the

concrete compressive stress is σcp/3 and the elastic shorteninmg of concrete is εcp=1/3 σcp/Ecmi. When

the first tendon is in tensioning so the elastic shortening during the tensioning is taken into account the

strain corresponding the force P1 ;so the force P1 is left after tensioning the first tendon.

- The second tendon is tensioned to the value P2 = σpoAp = P/3, which is left in the tendon after

tensioning this tendon. The force of the second tendon increase the concrete compressive stress by

amount σcp/3. Concrete is after the tensioning the first tendon shortened by amount ∆εcp =

=1/3 σcp/Ecmi , which reduce the foprce in the forst tendon by amount ∆Pe12 = 1/3 σcp Ap Ep/Ecmi

(because the distance between the anchors of the fiorst tendon is shortened). The total prestress force

is now P1 - ∆Pe1 + P2.

- The third starnd is tensioned alos to the value P3 = σpoAp = P/3, which is left in the tendon after

tensioning this tendon. The force of the third tendon increase the concrete compressive stress by

amount σcp/3 causing to the shortening ∆εcp = =1/3 σcp/Ecmi. of concrete. The distance between the

anchors of the first tendon is further shortened and the force in the first tendon is furher reduced by

amount ∆Pe13 = 1/3 σcp Ap Ep/Ecmi. The force in the first tendon is now P1 - ∆Pe12-∆Pe13.


The distance between the anchors of the second tendon is also shortened during tensiong the third

tendon and so the force in the second tendon is correspondingly decreased by anount

∆Pe23 = 1/3 σcp Ap Ep/Ecmi. and the force in the second tendon is now P2 - ∆Pe23.

The total prestress is now

P0 = P1 - ∆Pe12-∆Pe13+ P2 - ∆Pe23+P3 =

= P/3-(1/3 σcp Ap Ep/Ecmi +1/3 σcp Ap Ep/Ecmi)+P/3- 1/3 σcp Ap Ep/Ecmi + P/3

=P-3⋅1/3 σcp Ap Ep/Ecmi=

= P – (n-1)⋅ 1/n σcp Ap Ep/Ecmi - (n-2)⋅ 1/n σcp Ap Ep/Ecmi- …-1/n σcp Ap Ep/Ecmi

= P-n/2⋅(n-1)⋅ 1/n σcp Ap Ep/Ecmi = P-(n-1)/2⋅σcp Ap Ep/Ecmi

The average prestress force is Pmi =P0/n = P/n-(n-1)/2n⋅σcp Ap Ep/Ecmi

The effect of the elastic shortening can be reduced by retensioning (tightening ) the loosed tendons.

The first tendon is retensioned with the force 2/3 σcp Ap Ep/Ecmi and the second tendon is retensioned

with the force 1/3 σcp Ap Ep/Ecmi and the possible further reduction due to pretensioning of the first

tendon. The third tendon is also retensioned wuth the force which corresponds the loss due to

retensiong of the forst and the second tendons. Then the final loss due to elastic shortening is only the

loss in the first tendon due to retensioning of the second and the third tendon and the loss in the

second tendon due to retensioning the third tendon which are quite small.

Rak.43-3110 Ankkurijännepalkin kitka- ja lukitushäviöt

1

Curved length which can lengthen is assumed ~50 m

L 50 m=L 2 25⋅ m⋅:=Length of the beam

Jacking is dine only at one end

Po 1383.8 kN=Po σpo Ap⋅:=Prestress force

σpo 1240 MPa⋅:=Initial prestress at the jacking anchor

β 0.005rad

m⋅:=wobble coefficientµ 0.2:=Friction coefficient

Ep 195000MPa:=Elastic modulus

St 1600/1800Prestyressing steel grade

Ap 1116 mm2

=Ap 12 Ap1⋅:=Ap1 93 mm2

⋅:=12 φ12,5Strands

VSL post-tension system

Input data:

EXCAMPLE ABOUT THE PRETRESS LOSSES DUE FRICTION AND

ANCHORAGE SLIPPAGE IN A POST-TENSIONED BEAM


2

=> P10left 1370.1 kN=

Right side of the inflection point after the change of the angle x 10 m= right

angular change α15

180π⋅ rad⋅:= α 0.262 rad=

µ α β x⋅+( )⋅ 0.0624=

P10right Po eµ− α β x⋅+( )⋅

⋅:=P10right 1300.2 kN= (loss 5,6 %)

The force does not change suddenly but the curvature of the tendon equals the force.At the inflection point the prestress force should be about the average value of thevalues at the leftside and the right side of the inflection point; so

P10

P10left P10right+

2:= P10 1335.1 kN= (loss 3,2 %)

Prestress loss due to friction

x 0.4L

2⋅:= x 10 m= (x should be measure along the curved tendon, but the straight length

is uswed because the error is small) ___________________

The angular change at the distance 0 ... x=10 m is concentrated at the inflection point

Left side of the inflection point before the change of the angle

Sum of the angular changes:

Σα 0:= µ Σα β x⋅+( )⋅ 0.01=

P10left Po eµ− Σα β x⋅+( )⋅

⋅:=P10left 1370.1 kN= (loss 0,7 %)

x ... 10 m prestress force changes Po 1383.8 kN=


3

µ Σα β x⋅+( )⋅ 0.188=

P40 Po eµ− Σα β x⋅+( )⋅

⋅:= P40 1146.3 kN= (loss 15,7 %)

At the passive anchor x 50 m⋅:=

Sum of the anmgularchanges

Σα 15 20+ 15+( ) deg⋅:= Σα 50 deg= Σα 0.873 rad=

µ Σα β x⋅+( )⋅ 0.1950

180π⋅ rad⋅ 0.004

rad

m⋅ 40⋅ m⋅+

⋅:= µ Σα β x⋅+( )⋅ 0.225=

P50 Po eµ− Σα β x⋅+( )⋅

⋅:= P50 1105.5 kN= (häviöt 18,4 %)

If the jacking is done from the both ends, the left and the right span are similar.

P40 P10:= ja P50 Po:=

x 25 m⋅:= at the middle support

Take at the bendig point the half of the change of the angle α2 10 deg⋅:=

Sum of tge angularchanges

Σα 15 10+( ) deg⋅:= Σα 25 deg= Σα 0.436 rad=

µ Σα β x⋅+( )⋅ 0.1925

180π⋅ rad⋅ 0.004

rad

m⋅ 25⋅ m⋅+

⋅:= µ Σα β x⋅+( )⋅ 0.112=

P25 Po eµ− Σα β x⋅+( )⋅

⋅:= P25 1236.9 kN= (loss 9,7 %)

x 40 m⋅:= at the max. field moment in the right span

Sum of the angularchanges

Σα 15 20+15

2+

deg⋅:= Σα 42.5 deg= Σα 0.742 rad=

µ Σα β x⋅+( )⋅ 0.1942.5

180π⋅ rad⋅ 0.004

rad

m⋅ 40⋅ m⋅+

⋅:=


4

P15 1293.7 kN=P15 Po eµ− Σα15 β x1⋅+( )⋅

⋅:=

µ Σα15 β x1⋅+( )⋅ 0.0674=

Σα15 0.262 rad=Σα15 15 deg⋅:=is the same as at the point x=10 m rightside :.

x1 15 m=The angular changes are concentrated at the inflection point

x1 15 m=Prestress for at the point

x1 15 m⋅:=take a new

x1 10 m=>x1new 16.372 m=x1new

δ Ep⋅ Ap⋅

∆p0...10

:=

(N/mm) ∆p0...10 4.871kN

m=∆p0...10

Po P10−( )x1

:=

Loss due to friction per unit length at the distance 10 m

x1 10 m⋅:=

Suppose, that the loss due to the anchorage slip influence upto 10 m from the edge support

δ 6 mm⋅:=Anchorage slip

Prestress loss due to the anchorage slip


5

Because the anchorage slip influences still at the point of the max. field moment it is useful to eliminate at least partly.

∆Pluk 180.29 kN=∆Pluk 2 ∆p0...15⋅ x1⋅:=Loss due to the anchorage slipat the jacking anchor

x11 14.574 m⋅:=By continuing the iteration the inluence length of the loss due to anchorage slip is

The result is enough close to the suppoesed value

x1 15 m=~x1new 14.74 m=x1new

δ Ep⋅ Ap⋅

∆p0...15

:=

(N/mm) ∆p0...15 6.01kN

m=∆p0...15

Po P15−( )x1

:=

Loss due to friction per unit length at the distance 0...15 m


6

jacking anchor passive anchor

Average prestress force in one tendon at the whole length of the beam

Pm0 m⋅

L

xP1 x( )⌠⌡

d

L:=

P1

Approximately

Likimäärin ΣPL1204 1294+( )

214.74⋅

1294 1236+( )

225 14.74−( )⋅+

1236 1146+( )

240 25−( )⋅+:=

ΣPL ΣPL1146 1106+( )

250 40−( )⋅+:=

PmΣPL kN⋅ m⋅

L:= Pm 1210.3 kN=


7

Elastic modulus of concrete at the timeof tensioning

Ecmi

fcmi

fcm

0.3

22000⋅ MPa⋅fcm

10 MPa⋅

0.3

⋅:=

Ecmi 32837 MPa=

Initial presrtess force in one tendon Pm 1210.3 kN= the averagfe force at the whole length

Compression stress in concrete due toone tendon

σcp

Pm−

Aco

:= σcp 2.051− MPa=

Elastic strain of concrete εcp

σcp

Ecmi

:= εcp 0.062− %o=

Loss due to the elastic shortening in the first strand due to tensioning the second and third strand

∆Pe1 2 εcp⋅ Ep⋅ Ap⋅:= ∆Pe1 27.188− kN=

Force in the first strand P1 Pm ∆Pe1+:= P1 1183.1 kN=

Elastic shortening in a post-tension member

Suppose that there are 3 ducts (tendon) with 12 strands in each duct. All straands in one duct is tensioned simultaneously..

Diameter of the duct φputki 65 mm⋅:=

Number of the tensionings n 3:=

Height of the beam h 1.2 m⋅:= width b 0.5 m⋅:=

Concrete area at the tensioning stage when the ducts are not injected; there are empty holes at the location of the ducts.

Before the injection of the ducts there is no bond between the strandsand the concrete. In this stage the strands are not included to the effective cross-section. The cross-section consist only the net are of the concrete.

After the ducts have been injected the the injection concrete and the starnds are included to the effective transformed cross-section in the same way than for the pre-tensioned members.

Aco b h⋅ nπ φputki

2⋅

4⋅−:= Aco 590045 mm

2=

Concrete strength at the timeof the tensioning C20/37 fcmi 38 MPa⋅:=

Final concrete grade C40/45 fcm 48 MPa⋅:=


8

∆Pem 13.594− kN=∆Pem

σcptot

Ecmi

Ep⋅ Ap⋅n 1−( )

2 n⋅⋅:=

Average loss due elastic shortening per one tendon

σcptot 6.154− MPa=σcptot

n Pm−( )⋅

Aco

:=

Concrete compression stress due to 3 tendon:

Remark! In EC2 equation ∆σc means the concrete compression stress due to one tendon

∆Pem 13.594− kN=∆Pem Ap Ep⋅n 1−( )

2 n⋅

σcp

Ecmi

⋅n 1−( )

2 n⋅

σcp

Ecmi

⋅+n 1−( )

2n

σcp

Ecmi

⋅+

⋅:=

Average loss due to elastic shortening (EC2 equation (5.44))

Pom 1196.7 kN=PomP

n:=Average prestress force in one tendon

∆Pem 13.594− kN=∆Pem

∆Pe

n:=Average loss in one tendon:

∆Pe 40.782− kN=∆Pe ∆Pe1 ∆Pe2+:=

Total loss due to the elastic shortening

P 3590.1 kN=P P1 P2+ P3+:=Total prestress forcejust after the tensioning

P3 1210.3 kN=P3 Pm:=

The third strand does not get the loss due to the elastic shortening because the shortening of the 3. tendon is eleminated by controlling the jacking force; so the force in the thrid strand is

P2 1196.7 kN=P2 Pm ∆Pe2+:=Force in the second strand

∆Pe2 13.594− kN=∆Pe2 εcp Ep⋅ Ap⋅:=

Loss due to the elastic shortening in the second strand due to tensioning the third strand

Documents

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