2013-08-21

1

Calculation Example

Strengthening for flexure

hw

bw

As

1

1

Sektion 1-1 (Skala 3:1)

FRP

Last

L

beff

hf

The beam is a part of a slab in a parking garage and needs to be strengthened for additional load. Simply supported with L=8.0 m. Distributed load. Max moment due to service load 200 kNm and additional moment 430 kNm in ultimate limit state. The load during strengthening can be decreased to 170 kNm.

2013-08-21

2

Calculation Example

Geometrical PropertiesNotation

Value Unit Description

bf = beff= 2610 mm Effectiveflange(EC25.3.2.1)hf= 180 mm Heigthonflangehw= 520 mm Heigthonwebh= 700 mm Totalheigthc= 30 mm Concretecoverbw= 250 mm WidthwebAc= 599800 mm2 CrosssectionalareaconcreteAs= 1256.6 mm2 AreasteelreinfordementØt= 20 mm Diametersteelreinforcementd= 660 mm Level arm

L= 8000 mm Distance between supports

B = 5000 mm Distance between beams

Asw= 157.1 mm2 Area of stirrups

Øs= 10 mm Area shear reinforcement

s= 250 mm Internal distance shear 

reinforcement

Partical coeffecient factors

Concrete Steel FRPc =1.5 s =1.15 frp =1.2cc=0.85 ct=0.85φef=2.0cE =1.2

2013-08-21

3

Calculation ExampleStep 1: Investigate existing stage

Concrete

Characteristicvalues

Steel

Characteristicvaluesfck 40 MPa fyk 500 MPafctm 3.5 MPa Es 210 GPaEcm 35 GPa

Concrete Designvalues

Steel

Designvaluesfcd 22.6 MPa fyd 435 MPafctm 3.5 MPa Esd 183 GPa

Calculations in SLSProportional constants:

Calculation in (SLS):

• Investigate if the beam is cracked or not with consideration tothe original loading.

• Calculate the distance to the neutral axis.

12 2

1

180 5202610 180 250 520 180 18.0 1 1256.6 660

2 22610 180 250 520 18.0 1 1256.6

182.9

f weff f w w f s s

of f w w s s

h hb h b h h A d

yb h b h A

mm

,

1 210 1 218.0

35s efsd

sc eff cm

E φEα

E E

2013-08-21

4

Moment of inertia in stage I, I1, can then be calculated:

Calculation Example

2 23 3

1 0 0

2

0

23 3

22

10 4

112 2 12 2

1

2610 180 180 250 5202610 180 182.9

12 2 12

520250 520 182.9 180 18.0 1 1256.6 660 182.9

2

2.17 10

f f f w w wc s s f f w w f

s s

b h h b h hI I I b h y b h y h

A d y

mm

The maximum stress in the steel reinforcement and bottom fibresof the concrete beam can then be calculated:

0 00 0

1

6

10

1

200 10660 182.9 4.40

2.17 10

sc s s

M Md y d y

I I I

MPa

0 00 0

1

6

10

1

200 10700 182.9 4.76

2.17 10

cuc s s

M Mh y h y

I I I

MPa

The concrete can assumed being cracked when cu exceeds fctm =3.50 MPa. Since 4.76 > 3.50, the concrete is assumed crack andthe section is in stage II.

2013-08-21

5

Calculation Example

With the assumption that the neutral axis is in the flange the levelarm x can then be calculated:

22

2 2f f

s s s s s s

B CA

b x bA d x x A x A d

and x can then be calculated:

2 225581 25581 17061541

104.52 2 2 1305 2 1305 1305B B C

x mmA A A

The assumption that the neutral axis was in the flange was correct.We can then calculate the moment of inertia in stage II, I2:

232

2

23

2 10 4

112 2

2610 104.5 104.52610 104.5

12 2

20.6 1 1256.6 660 104.5 0.90 10

fc s s f s s

b x xI I I b x A d x

mm

2013-08-21

6

Calculation Example

Step 2: Calculate the initial strains and stresses

With M01=200 kNm (service load) the stresses in concreteand steel reinforcement can be calculated:

601

102

200 10104.5 2.33

0.90 10cö

Mx MPa

I

6

0110

2

200 1020.6 660 104.5 254.80

0.90 10s s

Md x MPa

I

With M02=170 kNm during strengthening, the followingstresses in concrete and steel are calculated:

602

102

170 10104.5 1.98

0.90 10cö

Mx MPa

I

6

0210

2

170 1020.6 660 104.5 216.58

0.90 10s s

Md x MPa

I

Corresponding strains can then also be calculated:

0 3

1.980.17

11.67 10cö

cö ceffE

0 3

216.581.18

183 10s

s ssdE

02,

1.08 700 104.51.16

660 104.5s

u s M

h x

d x

‰

‰

‰

2013-08-21

7

Calculation Example

Step 3: Calculate strengthening needsFRP

Characteristic values Design valuesfk 15 ‰ f 12.50 ‰Efk 160 GPa Ef 133.33 GPa

Check I.C debonding (often governing):

, 3

22.670.41 0.41 4.52

1 133.33 10 1.4cd

fd icfd f

fnE t

‰

Estimate the area of FRP

62

3 3

0.9 430 10 0.9 1256.6 434.78 660277

4.52 10 133.33 10 700d s yd

ff f

M A f dA mm

E h

This corresponds to two 100 x 1.4 mm2 CFRP laminates.Calculated the height of the compressive zone:

, 1256.6 434.78 4.52 133.33 280157.7

0.8 1.0 22.67 250s yd fd ic fd f

cd w

A f E Ax mm

f b

and then the moment capacity:

,2 2

0.8 0.81256.6 434.78 660 157.7 4.52 133.33 280 700 157.7

2 2

433.6

s yd fd ic fd fM A f d x E A h x

kNm

This exceeds the moment capacity asked for 430 kNm

2013-08-21

8

Calculation Example

Step 4: Check if the cross section isnormally reinforced (under reinforced)

The following should be fulfilled: bal

, 0

0.8 0.80.305

4.52 1.1611

3.5

balfd ic u

cu

, 1256.6 434.78 280 4.52 133.330.017

2610 700 22.67

s yd f fd ic fd

eff cd

A f A E

b hf

Maximum reinforced section:

and bal

Step 5: Calculate the anchor lengthCalculate the distance to the “last crack”, xcr, where the sectiontensile capacity corresponds to the cracking moment. On safeside the bending stiffness for the concrete only, neglect thesteel reinforcement, can be calculated:

0

2 2

180 5202610 180 250 520 180

2 2165.9

2610 180 250 520

f weff f w w f

eff f w w

h hb h b h h

yb h b h

mm

2013-08-21

9

Calculation Example

Step 5: Calculate the anchor length, cont.2 23 3

0 0

23 3

2

10 4

12 2 12 2

2610 180 180 250 5202610 180 165.9

12 2 12

520250 520 165.9 180

2

1.67 10

eff f f w w wc eff f w w f

b h h b h hI b h y b h y h

mm

Calculate the bending stiffness:

108 3

0

1.67 101.01 10

165.9c

c

IW mm

y

Calculate the cracking moment:

81.01 10 3.5 351.80crx c ctmM W f kNm

Calculate the distance to the “last crack”. The beam isplaced on free supports. We can then calculate:

2

( )2x A

xM x R x q ( )x AV x R qx

With

353.8 8 10215

2 2A

qLR kN

53.826.9 /

2 2q

kN m

and consequently x = 2294.2 mm

2013-08-21

10

Calculation Example

Calculate the displacement, al, and the bending moment Mxa insection xa:

0.45 0.45 660 297.0la d mm

Step 5: Calculate the anchor length, cont.

and:

376.66ax

M kNm

2013-08-21

11

Calculation Example

Step 5: Calculate the anchor length, cont.Calculate the tensile force in the FRP that together with thetensile reinforcement can carry the bending moment Mxa:

The force for yielding in the tensile reinforcement is calculated as:

1256.6 434.78 546.4s s ydF A f kN

2 23

3

376.660.9 0.9 700 85.60

200 10 1256.6 6601 1

133.33 10 280 700

axf

sd s

fd f

M hF kN

E A dE A h

Calculate the force in the FRP when the steel yields:

3376.66 660546.36 10 82.73

0.9 0.9 700 700ax

f s

M dF F kN

h h

Chose the largest load, i.e. Ff = 85,6 kN.

Check that the load in the FRP in the studied section does notexceed:

, ,f e f x f fdF A E

2 10022 250 0.82

2 1001 1250

f cb

f c

b bk

b b

20.03 0.03 1.0 40 3.5 0.35 /f b ck ctmG k f f Nmm mm

, 3

2 2 0.351.95

133.33 10 1.4f

f xfd f

G

E t

‰

2013-08-21

12

Calculation Example

Step 5: Calculate of the anchor length, cont.

, , 1.95 280 133.33 72.81f e f x f fdF A E kN

Which is less then Ff. Calculate a new bending moment, Mf,e:2

, ,

233

3

0.9 1

200 10 1256.6 6600.9 700 72.81 10 1 320.37

133.33 10 280 700

sd sf e f e

fd f

E A dM hF

E A h

kNm

and:

, ,

3 3

0.9

6600.9 700 72.81 10 546.36 10 370.41

700

f e f e s

dM h F F

h

kNm

Choose Mf,e=320.37 kNm, and:

,

2

, , 1980.32f ex A f e f e

qxM R x x mm

3133.33 10 1.4163.3

2 2 3.5

fd fe

ctm

E tl mm

f

Anchor length:

Choose 250 mm.

a la

xf,e

FRP

Ra

, 1980.3 250 1730.3f e ea x l mm

2013-08-21

13

Calculation Example

Choose a=100 mm (anchor the laminate as close as possibleto the support). The maximum shear stress for a simplysupported beam for a distributed load can then be calculated:

2

max 2

23

3 8 2

2

2

100 2 100 4000 0.109 40004.7 1026.9 0.16

2 29.17 10 1.00 10 0.109

a

cd c

a al lGqsE W

MPa

Where l = L/2 and z0 = h - x and

0

3 3 3

3 8

1 1

1 14.7 10 2 100 133.33 10 280 29.17 10 599800 0.109

2 595.5229.17 10 1.00 10

a f

fd f cd c cd c

G b zs E A E A E W

Step 6: Calculate shear- and peeling stresses

2013-08-21

14

Calculation Example

Step 6: Calculate shear- and peeling stresses

Shear and normal stresses is only affected by additionalloads, i.e. the loads that are added after strengthening.

Calculate the normal stress.

53.75 21.25 32.5 /after beforeq q q kN m

Support force:32.5 8000

1302 2A

qLR kN

Bending moment at a=100 mm

2 23 3 100( ) 130 10 100 32.5 10 12.7

2 2x A

aM x R a q kNm

Normal stress calculation

6

0 101

12.7 10700 182.9 0.30

2.17 10x

x

Mh y MPa

I

Failure criteria: 1 ctmf

2

21

22

2 2

0.30 0.16 0.30 0.160.16 0.40

2 2

x y x yxy

MPa

Which is less than fctm = 3.5 MPa

2013-08-21

15

Calculation Example

Design for strengthening, calculation steps

Step 1: Investigate existing stage

Step 2: Calculate the initial strains and stresses

Step 3: Calculate strengthening needs

Step 4: Check if the cross section isnormally reinforced (under reinforced)

Step 5: Calculate the anchor length

Step 6: Calculate shear and peeling stresses

2013-08-21

16

Calculation steps

• If the section is cracked• Stresses and strains• Existing strain fields

Calculation in (SLS)

Design for strengthening (ULS)

• Estimate the area of the FRP• Calculated the moment capacity (iteration)

Calculate the anchorage length

Calculate the peeling stresses

Check failure criteria

Calculation Example