PSI - Issue 2_A

Stefano Bennati et al. / Procedia Structural Integrity 2 (2016) 2682–2689 S. Bennati, D. Colonna and P.S. Valvo / Structural Integrity Procedia 00 (2016) 000–000

2688

7

By solving the differential problem defined by Eqs. (9) and (10) – as explained in detail in Bennati et al. (2016) – the following final expressions are obtained for the interfacial shear stress, ( ) ( ) exp( ), s q l s ql s τ ξ ξ λ ≅ − − − (11)

and internal forces in the beam,

l

1 2

  

 

  

  

[

]

, b Q ( ) N s qb s s l ξ = f

1 exp( ) , λ  s

− + − −

λ

, b Q V s q l s ( ) ( ), 1 = −

(12)

h

l

1 2

  

 

  

  

[

]

, b Q M s

( = + + − + q a s l a s )(2 )

ξ qb s s l

1 exp( ) . λ  s

− + − −

( )

b

f

λ

2

2

where

   

   

2

h

k h

1

1 and =

kb

=

+ + b

λ

ξ

.

b

(13)

f

2 λ

4 E A E I E A s b s b f

E I

2

f

s b

3.4. Stage 3 – Non-linear response and failure of the strengthened beam

Stage 3 corresponds to non-linear response due to plasticity of the steel beam or softening of the adhesive. This stage of behaviour cannot be described here because of length limits. See Bennati et al. (2016) for further details.

4. Illustrative example

As an illustration, we consider a S235 steel beam with IPE 600 cross section strengthened with a Sika ® Carbodur ® CFRP laminate. The material constants, factored according to the Eurocodes (EC 2005) and Italian regulations on FRP strengthening (CNR 2014), and geometric properties used in the example are the following: • steel: E s = 210 GPa, f yk = 235 MPa, γ s = 1.1, f yd = f yk / γ s = 213.64 MPa, A b = 15600 mm 2 , I b = 920800000 mm 4 ; • adhesive: G a = 4.923 GPa, τ k = 15 MPa, η a = 0.85, γ fd = 1.2, τ 0 = η a τ k / γ fd = 10.63 MPa, t a = 1 mm; • FRP laminate: E f = 165 GPa, f fk = 2800 MPa, γ f = 1.1, f fd = η a f fk / γ f = 2163.63 MPa, b f = 120 mm, t f = 2.6 mm. The span of the existing beam, 2 L , and permanent load due to non-structural elements, g 2 , are fixed imposing that

1 8



(

)

2 (2 ) 50% , yd b f W ≤

1 1 g

g L

γ

γ

+

   

G

G

2 2

(14)

g g L +

5

1

4

(2 ), L

≤

(2 )

1

2

E I

384

800

s b

where W b = 3069000 mm G 1 = 1.35 and γ G 2 = 1.50 (EC 2005). By assuming the equal sign in inequalities (14), the maximum theoretical values of 2 L and g 2 are first determined. Then, by rounding down such values to the nearest integer multiples of 500 mm and 0.50 kN/m, respectively, we obtain 2 L = 10500 mm and g 2 = 14.50 kN/m. The imposed load corresponding to the elastic limit state in the unstrengthened beam, q b 0 , is determined from: ( ) ( ) 2 0 1 1 2 2 1 2 , 8 γ γ γ + + = G G Q b yd b g g q L f W (15) 3 is the elastic section modulus and the partial factors are γ