2013-08-21 1 Calculation Example Strengthening for flexure h w b w A s 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.
16
Embed
Design example flexure 2013-08-20.ppt [Kompatibilitetsläge]
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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
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: