design example flexure 2013-08-20.ppt …€¦ · proportional constants: ... ax l mm fe e, 1980.3...
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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
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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
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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.
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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
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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
‰
‰
‰
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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
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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
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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
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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
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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
‰
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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
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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
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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
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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
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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
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