FRP Design Software - Flexural Strengthening Of RC Slabs With CFRP Plate
1. Basic Information
1 .1 . Design Criteria
ACI 318M-14 Building Code Requirements for Structural Concrete
ACI-440-2R-17 Guide for the Design and Construction of Externally Bonded FRP Systems for Strengthening Concrete Structures
1.2. Design Requirement
Required load mom ent : 25 k N ∙ m
Strengthened sec : Positive Moment
1. 3. Basic Information
(1) Information of existing beam
S lab Thickness: 100mm
Concrete compressive strength : 45 MPa
Tensile reinforcement arrangement :
E ffective depth of centroid of tensi le reinforcement : 80 mm
A rea of the longitudinal reinforcement in tension region : 785.4 mm 2 /m
Elastic modulus of tensi le reinforcement : 200000 MPa
Yield strength of tensi le reinforcement : 414 MPa
Compressive reinforcement arrangement:
E ffective depth of centroid of compressive reinforcement : 20mm
A rea of the longitudinal reinforcement in compressive region : 523.6mm 2 /m
Elastic modulus of compressive reinforcement : 200000MPa
Yield strength of compressive reinforcement : 414 MPa
(2) Loads
Existing load before strengthening : 15 kN ∙ m
Service Load : 15 kN ∙ m
(3) Strengthening System
Material : HM-2.0T-I
U ltimate tensile strength of the FRP material as reported by the manufacturer: 2499 MPa
U ltimate rupture strain of FRP reinforcement : 0.014
T hickness of FRP reinforcement: 2mm
Elastic modulus of CFRP : 178477 MPa
Environmental reduction factor : 0.95
D esign ultimate tensile strength of FRP : f fu = C E f fu * = 2374.05 MPa
D esign rupture strain of FRP reinforcement : ε fu = C E ε fu * =0.0133
Based on iteration calculation, propose following strengthening plan:
W idth of FRP reinforcement: 50 mm
Number of layers : 1
Spacing of CFRP: 1660mm
Area of CFRP : 60 mm 2
E ffective depth of centroid of CFRP: d f = 100 mm
2. Design Calculations
2.1 Capacity of existing beam
2. 1.1 Original section properties
a) Concrete :
Compressive strength of concrete : f c ’ = 45 MPa
β 1 = 0.72
Ultimate strain of concrete : ε c u =0.003
Elastic modulus of concrete : E c = 4700 f c ’ 0.5 =31528.56MPa
b ) Steel reinforcement :
Yield strain of tensile steel : ε y = f y / E s =0.0021
Yield strain of compressive steel : ε y ’ = f y ’ / E s ’ =0.0021
2. 1.2 Capacity of existing beam
Assume compressive steel yield
Stress in tensile steel : f s= f y =414 MPa
Force in tensile steel : T s =A s f y = 785.4 × 414= 325154.56N
Stress in compressive steel : f s ’ = f y ’ -0.85 f c ’ = 414-0.85 × 45= 375.75 MPa
Force in compressive steel : C s = A s ’ f s ’ = 523.6×375.75= 196742.07 N
Compressive force in concrete : C c = 0.85 f c ’bβ 1 x
Equilibrium : T s = C c + C s
325154.56= 0.85 × 45 × 1000 × 0.72 x + 196742.07
Depth of compressive zone :
x= 4.64mm
Strain in compressive steel : ε s ’ = ( x - d’ ) ε cu /x =(4.64-20) ×0.003/ 4.64= 0.0099
ε s ’ ≥ ε y ’ , good.
M n ,existing =C c ( d-β 1 x/ 2) + C s ( d-d’ )
= 0.85 × 45 × 1000 × 0.72 × 4.64 × (80-0.72 × 4.64 /2 )+ 216769.71 ×( 80-20 )= 21.86 kN ∙ m
Strain in tensile steel : ε t = ( d -x ) ε cu /x =(80-4.64) ×0.003/ 4.64= 0.0487
Reduction factor: ϕ = 0.9
Ultimate capacity of existing beam : M u ,existing = ϕ M n ,existing = 0.9 × 21.86 = 19.68kN ∙ m
2.2 . Check design strain of CFRP
Determine design strain of CFRP
ε fd= 0.41( f c ’ /( nE f t f )) 0.5 =0.41× ( 45/1 /178477/2) 0.5 = 0.0046
ε fd ≤ 0.9 ε f u , OK .
Language: English
