1 5.6 Stresses in the concrete and reinforcement during service

5.6.1 Loss of stress from relaxation, creep and contraction

cs cd caε =ε (t)+ε (t)

• εcs total deformation given by contraction (‰)

• εcd (t) total deformation due to drying contraction

• εca (t) total deformation due to endogene contraction

• t = 90 days concrete age when the checking is made

• Ac = A1+A2+A3 area of the concrete cross-section

• U perimeter of the concrete cross-section

• εcd,0 = 0,52 ‰ the nominal value of the drying contraction c0

2×Ah = (mm)

U

cd h cd,0ε (t)=k ×ε

0.5( 0.2 )

6

( ) ( ) ( )

( ) 1

( ) 2.5( 10) 10

ca as ca

t

as

ca ck

t t

t e

f

cs cd caε =ε (t)+ε (t)

2 5.6 Stresses in the concrete and reinforcement during service 5.6.1 Loss of stress due to relaxation, creep and contraction

• fck = 35 N/mm2 (C35/45) characteristic strength

• Ep = 180000 (N/mm2) elastic modulus of the prestressed reinforcement

• Δσpr loss of stress due to reinforcement relaxation

0.75(1 )

6.7 5 2

10005.39 10 ( / )1000

pr po

te N mm

• σpo stress in the reinforcement during transfer

• ρ1000 = 8% loss of stress at 1000 hours from the prestressing

• Ecm = 36000 (N/mm2) concrete elastic modulus

po

pk

σμ=

f

3 5.6.1 Loss of stress due to relaxation, creep and contraction

• φ(t,t0) creep coefficient

• φRH factor considering the humidity influence

• RH=50% humidity

• t0 = 28 days age of concrete during loading

• fcm = 43 N/mm2 (C35/45) average compressive strength

• Ac = A1+A2+A3 area of the concrete cross-section

• Ic = I1+I2+I3 inertia moment of the concrete cross-section

• zcp distance from the center of the reinforcement to the center of the cross-section 0.5

3

cm

35α =

f

0 0 c 0

0 RH cm 0

RH 1 23

0

φ(t,t )=φ ×β (t,t )

φ =φ ×β(f )×β(t )

1-RH/100φ = 1+ ×α ×α

0.1× h

0.7 0.2

1 2

cm cm

35 35α = α =

f f

4

• σc,QP stress in the concrete on the reinforcement direction in P+0.4Z combination

3 3 6

p0 p0 cp QP

c,QP cp cp

bi bi bi

2

T Z EQP

10 N 10 N ×z 10 Mσ = + z - z

A I I

(q +0.4×q )×LM = (kNm)

85.6.2 Stresses in the reinforcement during service

2

p po p,c+s+r

p p

p

σ =σ -Δσ (N/mm )

A ×σN = (kN)

1000

5.6.1 Loss of stress due to relaxation, creep and contraction

5 5.6.3 Stresses in the concrete during service

Span section

3 3 6p p cps s sext

b G G

bi bi bi

2

T Z Eext

10 N 10 N ×z 10 Mσ = - y + y

A I I

(1.35×q +1.5×q )×LM = (kNm)

8

3 3 6p p cpi i i 2ext

b G G

bi bi bi

10 N 10 N ×z 10 Mσ = + y - y (N/mm )

A I I

• if σbts≥fctk (tension) the concrete cracks

fctk = 4.2 N/mm2 (C35/45)

• if σbti≥ 0.6fck the concrete cross-section

or class must be increased

fck = 35 N/mm2

6 5.7 Strength checking 5.7.1 Strength checking during transfer

p0 c cd s ydN A ×f +A ×f

• Ac=A1+A2+A3 area of the concrete cross-section

• fcd =35/1.5=23.33 N/mm2 (C35/45) compressive design strength

• As mounting passive longitudinal reinforcement area (30Φ8 PC52)

5.7.2 Strength checking during service

Case 1 x ≤ hpga

22

s

π×8A =30 (mm )

4

pga cd p pd

p pd

pga cd

pd pk

b ×x×f =A ×f

A ×fx=

b ×f

f =0.9f /1.15

pga0pext cdM b ×x×f h -0.5×x (kNm)

7 5.7.2 Strength checking during service

Case 2 x > hpga

p pd

ga cd pga ga pga cd p pd

ga cd pga ga pga cd

A ×fb ×x×f +(b -b )h ×f =A ×f x=

b ×f +(b -b )h ×f

ga0p 0pext cd pga ga pga cd pgaM b ×x×f h -0.5×x +(b -b )h ×f h -0.5×h (kNm)

8 5.7.3 Shear force checking

2

4

bwSW r

dA n

max

, ( )SWRd s ywd Ed

AV z f ctg V kN

s

1 max

,max

(1.35 1.5 )( )

( ) 2

cw ga cd T Z ERd Ed

b z v f q q LV V kN ctg

ctg tg

σcp medium compressive strength in the concrete from prestressing

z = 0.9h0p

v1 = 0.6

max(ctgθ1; ctgθ2) ≥ 1 if the condition is not fulfilled the cross-section must be modified

ctgθ≤2.5 if it results a higher values in the computation will be considered ctgθ=2.5

(8,10)bwd mm

nr = 2

s = 100 mm

fywd = 0,8·fywk = 0,8·345 = 276 (N/mm2) PC52

1 0 0.25 ; 1.25 0.25 0.5

2.5 1 0.5 ;

cp

cw cp cd cw cd cp cd

cd

cp p

cw cd cp cd cp

cd bi

pentru f pentru f ff

Npentru f f

f A

for for

for