$T_f=k_G.u.{\tau }_{\mathit{max}}$

$l_{\mathit{ch}}=\frac{E_c.G_f}{f_{\mathit{ct}}^2}$

It has the following physical meaning: If a sample is under uniaxial tension and the material has a linear softening curve, the maximum length for getting stable fracture under displacement control over this length is equal to twice the characteristic length. This can be shown by comparing fracture work and elastic energy in a sample.

Torsion capacity model

Work by Gustaffson [2] and Hillerborg [3] indicates a dependency of shear capacity on the tensile strength of concrete ft , beam depth d, the reinforcement ratio ρ, span to depth ratio and the characteristic length lch . This also seems intuitively correct as shear cracks are the result of indirect tension stresses.

Torsion cracks are also the result of indirect tension stresses and the mechanism is arguably the same. Once micro-cracking commences stresses are aligned parallel and perpendicular to the cracks. The effect of fibres is to provide restraint in the perpendicular direction thus enhancing aggregate interlock that resists stresses parallel with the crack.

Steel fibres can therefore be said to enhance the toughness of concrete and it is possible to postulate a relationship between combined stresses, including indirect tension, and characteristic length. Hillerborg suggested a dependency of shear  capacity upon characteristic length as shown in the following equation.



With a bi-linear stress-displacement model (or softening curve) it is possible to calculate Gf and lch for steel fibre reinforced concrete. With these parameters a proportional function can be developed to calculate ultimate shear capacity.

Zink [4] developed a proportional function kG that correlated well with 233 test results on conventionally reinforced and prestressed concrete sections where:



In the absence of test results of SFRC members in torsion, this factor developed for shear capacity can be adopted for torsion. It would be a “first estimate” based on the similarity of mechanisms associated with shear and torsion, but would need to be verified by testing of a range of cross-sections, concrete strengths and fibre types.

Softening curve

A bi-linear softening curve (refer to figure 1) has been adopted to model concrete matrix behaviour when the concrete matrix passes through peak stress capacity. Three points on the bi-linear softening curve are defined based on test data. They will be derived in detail below, however referring to the chart below, in summary :

    • Point 1 represents the initial tensile stress limit of the concrete matrix.
     
    • Point 2 is reached at very small crack openings (micro-cracking) of 50μm. Until this point, only concrete is resisting tensile stresses. Steel fibres begin to be mobilised.
     
    • Point 3 is somewhat arbitrary, but is conservatively selected to be less than the opening at which fibres are effective. Based on direct tension testing at Victoria University this should, conservatively, be no greater than 5mm to 10mm to ensure a linear curve is valid.

Torsion Capacity of SFRC Based on Fracture Mechanics

Fracture energy and characteristic length

Fracture energy G_f
Characteristic length l_ch


           Bi-linear softening curve

Crack width range	Note
0 to 50μm ~ 200μm	Concrete only
200μm to wmax	Fibre only

Calculate point 1

Adopt

This is the average tensile stress and can be reduced to the 5% fractile (see notes below).


Calculate point 2

Fibre alone provides tensile capacity.



Where
        ◦ 0.3 is a measure of fibre effectiveness due to 3-D orientation.
        ◦ η_VOL is steel fibre proportion of matrix (volume).

        ◦ η is steel fibre dosage kg/m3
        ◦ ρ_s density of steel 7850 kg/m3
        ◦ σ_τ = fibre stress at bond failure (pullout)

Direct tension tests of steel fibre reinforced concrete indicate that moderate to high performance fibres will have tension stress exceeding 70% of their yield strength as they are pulled out post-crack. Therefore σ_τ would typically be around 800 to 1100 MPa. For very high performance steel fibres the stress would be higher.

Calculate point 3

For consistency, can adopt an approach that all steel fibres are pulled out of the matrix when :



however, in reality there will still be some fibres, less than half the dosage (say) that would be holding load. This will also depend on fibre type. Some fibres snap the end anchorage and easily slide out with very little residual capacity. Spade-end fibres have sufficient strength to work their way through the cement paste, deforming as they progress, and continue to hold considerable load. Hook-end fibres have sufficient strength to “un-kink” the hook-end during pull-out, but once the fibre has withdrawn past the last bend it has very little residual strength.

For these reasons, and because the final value of w_max beyond a certain value has little effect on the calculation of G_f, w_max can be a range of values. A conservative range for torsion would be :



For the purpose of calculating torsion capacity, one could adopt a crack width that reflects a best “guess” beyond which torsion is no longer viable in a semi-elastic manner. This is highly likely to depend on aggregate type and size and requires more research, but at this stage very small crack widths are recommended.

Fracture energy

Fracture energy G_f is the area under the softening curve. Characteristic length l_ch is calculated as follows :

Torsion capacity

Torsion capacity T_f can be calculated using an expression with area above the neutral axis, average tensile stress at the limit state and a multiplier effect based on the fracture zone where:

 $T_f=k_G.u.{\tau }_{\mathit{max}}$

$T_f={\left(\frac{5.l_{\mathit{ch}}}{d}\right)}^{\frac{1}{4}}.u.{\tau }_{\mathit{max}}$

${\tau }_{\mathit{max}}={\sigma }_{\mathit{ct}}=1.48.\ln .\left(1+\frac{f_c}{10}\right)$

$u=\alpha .x^{2}y$

$T_f=22.18x{10}^6.{\tau }_{\mathit{max}}$

${\tau }_{\mathit{max}}=1.48.\ln .\left(1+\frac{40}{10}\right)$

${\tau }_{\mathit{max}}=2.38\mathit{MPa}$

$T_f=52.83\mathit{kNm}$

$k_G=1.92$

$T_f=101.4\mathit{kNm}$

 

 

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