FRP Design Software - Flexural Strengthening Of RC Slabs - CFRP Wrap

Calculations of CFRP strengthened Slab-1 Section 1

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 moment : 25 kN∙m

Strengthened sec: Positive Moment

1.3.Basic Information

(1)Information of existing beam

Slab Thickness: 100mm

Concrete compressive strength: 45 MPa

Tensile reinforcement arrangement:

Effective depth of centroid of tensile reinforcement: 80 mm

Area of the longitudinal reinforcement in tension region: 785.4 mm²/m

Elastic modulus of tensile reinforcement: 200000 MPa

Yield strength of tensile reinforcement: 414 MPa

Compressive reinforcement arrangement:

Effective depth of centroid of compressive reinforcement: 20mm

Area of the longitudinal reinforcement in compressive region: 523.6mm²/m

Elastic modulus of compressive reinforcement: 200000MPa

Yield strength of compressive reinforcement: 414MPa 

(2)Loads

Existing load before strengthening: 15 kN∙m

Service Load: 15 kN∙m

(3)Strengthening System

Material: HM-30-I

Ultimate tensile strength of the FRP material as reported by the manufacturer: 4084 MPa

Ultimate rupture strain of FRP reinforcement: 0.0158

Thickness of FRP reinforcement: 0.167mm

Elastic modulus of CFRP: 259251 MPa

Environmental reduction factor: 0.95

Design ultimate tensile strength of FRP: f_fu=C_E f_fu^*=3879.8 MPa

Design rupture strain of FRP reinforcement: ε_fu=C_E ε_fu^*=0.015

Based on iteration calculation, propose following strengthening plan:

Width of FRP reinforcement: 100mm

Number of layers: 1 

Spacing of CFRP: 1660mm

Area of CFRP : 10.02 mm² 

Effective 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’=45MPa

β₁=0.72

Ultimate strain of concrete: ε_cu=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=414MPa

Force in tensile steel: T_s=A_sf_y=785.4×414= 325154.56N

Stress in compressive steel: f_s’=f_y’-0.85f_c’=414-0.85×45= 375.75MPa

Force in compressive steel: C_s=A_s’f_s’=523.6×375.75= 196742.07N

Compressive force in concrete: C_c=0.85f_c’bβ₁x

Equilibrium: T_s=C_c +C_s

325154.56= 0.85×45×1000×0.72x + 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-β₁x/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_ft_f))^0.5=0.41×(45/1 /259251/0.17)^0.5= 0.0132

ε_fd ≤ 0.9ε_fu, OK.

2.3. Check