Plasticity-based criterion for confinement design of FRP jacketed concrete columns

Abstract

Fiber reinforced polymer confined concrete exhibits different stress–strain behavior under different confinement levels. Therefore, classification of confinement levels is often the first step of jacketing design. Currently, the classification criteria are empirical and strength-based. In this paper, a plasticity-based criterion for classification of confinement levels is developed in the framework of Drucker–Prager plasticity. The theoretically derived criterion is simple and can be conveniently used for engineering design. In addition, explicit relationships for calculating the strength and ductility ratios of FRP confined concrete columns are developed as functions of the confinement stiffness ratio and FRP rupture strain. These relationships can be conveniently used in the performance-based design of FRP jackets for concrete columns.

Appendices

Appendix A

Under the framework of DP plasticity, there are two fundamental components, that is, the yield function

$$F = \sqrt {3J_{2} } - \tan \varphi \frac{{I_{1} }}{3} - k = 0$$

(20)

and the plastic potential function

$$G = \sqrt {3J_{2} } - \tan \beta \frac{{I_{1} }}{3} + {\text{constant}}$$

(21)

where I ₁ is the 1st invariant of the stress tensor and J ₂ is the 2nd invariant of the deviatoric stress tensor; φ, k, and β are the material properties of the friction angle, cohesion and plastic dilation angle, respectively. Under the flow rule, one has

$${\text{d}}\varepsilon_{i}^{\text{p}} = {\text{d}}\lambda \frac{\partial G}{{\partial \sigma_{i} }} = {\text{d}}\lambda \left( {\frac{\sqrt 3 }{{6\sqrt {J_{2} } }}(2\sigma_{i} - (\sigma_{j} + \sigma_{k} )) - \frac{\tan \beta }{3}} \right)\,\,\,\,(i = 1,2,3,j = 2,3,1,k = 3,2,1)$$

(22)

where dλ is a non-negative scalar accounting for the magnitude of plastic shear strain increment.

Taking normalization, the plastic work increment can be written as follows

$${\text{d}}\bar{W}_{\text{p}} = \sum {\bar{\sigma }_{i} {\text{d}}\varepsilon_{i}^{\text{p}} = \bar{\sigma }_{1} {\text{d}}\varepsilon_{1}^{\text{p}} + \bar{\sigma }_{2} {\text{d}}\varepsilon_{2}^{\text{p}} + \bar{\sigma }_{3} {\text{d}}\varepsilon_{3}^{\text{p}} }$$

(23)

where the note “–” denotes the value normalized by f _c.

Substituting Eq. (22) into Eq. (23), one has

$$\begin{aligned} {\text{d}}\bar{W}_{\text{p}} = \sum\limits_{i = 1}^{3} {\bar{\sigma }_{i} {\text{d}}\lambda \frac{\partial G}{{\partial \sigma_{i} }} = {\text{d}}\lambda [\left( {\frac{{\sqrt 3 \bar{\sigma }_{1} }}{{6\sqrt {J_{2} } }}[2\bar{\sigma }_{1} - (\bar{\sigma }_{2} + \bar{\sigma }_{3} )] - \frac{\tan \beta }{3}\bar{\sigma }_{1} } \right)} \hfill \\ + \;\left( {\frac{{\sqrt 3 \bar{\sigma }_{2} }}{{6\sqrt {J_{2} } }}[2\bar{\sigma }_{2} - (\bar{\sigma }_{1} + \bar{\sigma }_{3} )] - \frac{\tan \beta }{3}\bar{\sigma }_{2} } \right) + \left( {\frac{{\sqrt 3 \bar{\sigma }_{3} }}{{6\sqrt {J_{2} } }}[2\bar{\sigma }_{3} - (\bar{\sigma }_{1} + \bar{\sigma }_{2} )] - \frac{\tan \beta }{3}\bar{\sigma }_{3} } \right)] \hfill \\ = {\text{d}}\lambda \frac{\sqrt 3 }{{\sqrt {J_{2} } }}(\frac{1}{6}((\bar{\sigma }_{1} - \bar{\sigma }_{2} )^{2} + (\bar{\sigma }_{2} - \bar{\sigma }_{3} )^{2} + (\bar{\sigma }_{3} - \bar{\sigma }_{1} )^{2} )) + {\text{d}}\lambda \tan \beta \sum\limits_{i = 1}^{3} {\frac{{\bar{\sigma }_{1} }}{3}} \hfill \\ = {\text{d}}\lambda \sqrt {3\bar{J}_{2} } - {\text{d}}\lambda \tan \beta \frac{{\bar{I}_{1} }}{3} \hfill \\ \end{aligned}$$

(24)

Adopting the yield function Eq. (20), Eq. (24) can be rewritten to

$${\text{d}}\bar{W}_{\text{p}} = {\text{d}}\lambda \left( {\frac{{\bar{I}_{1} }}{3}\tan \varphi + \bar{k} - \tan \beta \frac{{\bar{I}_{1} }}{3}} \right)$$

(25)

Let hydrostatic pressure p = I ₁/3 and axial plastic strain dɛ ^p₁  = αdλ, one has

$$\bar{W}_{\text{p}}^{\prime} = \frac{{{\text{d}}\bar{W}_{\text{p}} }}{\alpha \cdot {\text{d}}\lambda } = \frac{(\tan \varphi - \tan \beta )}{\alpha } \bar{p} + \frac{{\bar{k}}}{\alpha }$$

(26)

where parameter α can be determined in the case of uniaxial compression, when it has $\bar{\sigma }_{ 1} = \bar{\sigma }_{\text{c}}$, and $\bar{\sigma }_{ 2} = \bar{\sigma }_{ 3} = 0$. And the Drucker–Prager yield function can be expressed as follows

$$F = \bar{\sigma }_{c} - \tan \varphi \frac{{\bar{\sigma }_{c} }}{3} - \bar{k} = 0$$

(27)

It yields

$$\bar{k} = \bar{\sigma }_{c} - \tan \varphi \frac{{\bar{\sigma }_{c} }}{3}$$

(28)

Substituting Eq. (28) into Eq. (26), the plastic work modulus has the from for the case of uniaxial compression

$$\bar{W}_{\text{p}}^{\prime} = \frac{{(\tan \varphi - \tan \beta )\frac{{\bar{\sigma }_{c} }}{3} + \bar{\sigma }_{c} - \tan \varphi \frac{{\bar{\sigma }_{c} }}{3}}}{\alpha } = \frac{{1 - \frac{\tan \beta }{3}}}{\alpha }\bar{\sigma }_{c}$$

(29)

Since ${\text{d}}\bar{W}_{\text{p}} = \bar{\sigma }_{\text{c}} {\text{d}}\varepsilon_{\text{c}}^{\text{p}}$ in uniaxial case, $\bar{W}_{\text{p}}^{\prime}$ can also be expressed as

$$\bar{W}_{\text{p}}^{\prime} = \frac{{{\text{d}}\bar{W}_{\text{p}} }}{{{\text{d}}\varepsilon_{c}^{\text{p}} }} = \bar{\sigma }_{c}$$

(30)

the equality between Eq. (29) and Eq. (30) gives

$$\alpha = 1 - \frac{\tan \beta }{3}$$

(31)

Appendix B

See Fig. 9.

Fig. 9

Characteristic curves for β and φ

The parameter tanΦ is the function of the plastic dilation angle β and friction angle φ

$$\tan \Phi = \frac{\tan \varphi - \tan \beta }{{1 - \frac{\tan \beta }{3}}}$$

(32)

The characteristic curves of β and φ are illustrated in Fig. 9 by converting the equivalent plastic strain $\tilde{\varepsilon }^{\text{p}}$ into ɛ ^p₁ for φ. The conversion can be achieved through incremental computation at a fixed ρ (See Appendix C). As the range of β is from 61.32° to −41.30°, the denominator ranges from 0.39 to 1.29, remaining in positive. And the sign of tanΦ is governed by the relative value (tanφ–tanβ). Therefore the requirement tanΦ ≥ 0 can be guaranteed by the local extremum Φ_min ≥ 0, which is determined by the inequality as follows:

$$\left. {\beta_{\hbox{max} } \le \varphi } \right|_{{\varepsilon_{1}^{\text{p}} = \varepsilon_{\text{cr}}^{\text{p}} }}$$

(33)

$$\beta_{\hbox{max} } = \frac{ - 63.15\rho + 7268}{\rho + 113}.[37]$$

(33a)

$$\varepsilon_{\text{cr}}^{\text{p}} = \frac{0.00044\rho + 0.087}{\rho + 19.31}.[37]$$

(33b)

where ɛ ^p_cr , the axial plastic strain corresponding to β _max. It yields ρ ≥ 9.8 for tanΦ > 0. When ρ < 9.8, φ develops to be smaller than β after its peak β _max (Fig. 9). Since β and φ monotonically decrease to the asymptotic value, the negative sign of tanΦ afterwards can be ensured by the inequality as follows:

$$\beta_{\text{u}} (\rho ) = \frac{ - 41.30\rho + 1146}{\rho + 18.38} \le \varphi_{\text{r}} = 42.20$$

(34)

where β _μ and φ _r are the asymptotic values. It gives ρ ≤ 4.5 for tanΦ < 0 after its extremum. And for the ratio ρ in the range of 4.5 < ρ < 9.8, tanΦ will increase from negative to positive one.

For the parameter $\bar{K}$

$$\bar{K} = \frac{{\bar{k}}}{{1 - \frac{\tan \beta }{3}}}$$

(35)

its post-peak behavior is monotonically developed. And hence the zero value of derivative of $\bar{K}$ at any ɛ ^p₁ larger than ɛ ^p_cr can confirm $\bar{K}^{\prime} = 0$. The boundary value to classify the “hardening” and “softening” in $\bar{K}$ can be obtained by setting

$$ \bar{K}^{\prime}\left|_{_{\varepsilon_{1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}}}}\right. \left. { = \frac{1}{\alpha}\frac{{{\text{d}}\bar{k}}}{{{\text{d}}\varepsilon_{1}^{\text{p}}}}}\right|_{{\varepsilon_{1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}}} }\left. { - \frac{{\bar{k}}}{{\alpha^{2} }}\frac{{{\text{d}}\alpha}}{{{\text{d}}\varepsilon_{1}^{\text{p}} }}} \right|_{{\varepsilon_{1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }} \left. { = \frac{1}{\alpha }\frac{{{\text{d}}\bar{k}}}{{{\text{d}}\varepsilon_{1}^{\text{p}} }}} \right|_{{\varepsilon_{1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }} \left. { + \frac{{\bar{k}}}{{\alpha^{2} }}\frac{{{\text{d}}\!\tan\beta }}{{3{\text{d}}\varepsilon_{ 1}^{\text{p}} }}} \right|_{{\varepsilon_{ 1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }} = 0 $$

(36)

where ɛ ^p_cu is plastic axial strain when β = β _μ. The plastic dilation angle can approach to constant value β _μ before failure if FRP has enough elongation capacity (Fig. 9), it has

$$\left. {\frac{{{\text{d}}\!\tan \,\beta }}{{{\text{d}}\varepsilon_{ 1}^{\text{p}} }}} \right|_{{\varepsilon_{ 1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }} = 0$$

(37)

And then the equilibrium Eq. (36) can be further expressed as

$$\bar{K}^{\prime}\Big|_{{\varepsilon_{1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }}=\frac{1}{\alpha }\frac{{{\text{d}}\bar{k}}}{{{\text{d}}\varepsilon_{1}^{\text{p}} }} \Big|_{{\varepsilon_{1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }} = 0$$

(38)

Determined by the normalized cohesion $\bar{k}$, the first order derivation with respect to ɛ ^p₁ goes to

$$\left. {\frac{{{\text{d}}\bar{k}}}{{{\text{d}}\varepsilon_{ 1}^{\text{p}} }}} \right|_{{\varepsilon_{ 1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }} \left. ={\left( {\frac{{E_{\text{p}} \tilde{\varepsilon }^{p} }}{{1 + \eta \tilde{\varepsilon }^{\text{p}} }}} \right)^{\prime } \frac{{{\text{d}}\tilde{\varepsilon }^{\text{p}} }}{{{\text{d}}\varepsilon_{1}^{\text{p}} }}} \right|_{{\varepsilon_{ 1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }} \left. { + \;2\tilde{\varepsilon }^{\text{p}} \zeta (\rho )\frac{{{\text{d}}\tilde{\varepsilon }^{\text{p}} }}{{{\text{d}}\varepsilon_{ 1}^{\text{p}} }}} \right|_{{\varepsilon_{ 1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }}$$

(39)

Under the current regulation of E _p and η, the rational function $\frac{{E_{\text{p}} \tilde{\varepsilon }^{\text{p}} }}{{1 + \eta \tilde{\varepsilon }^{\text{p}} }}$ monotonically increases towards its limit value, 0.41, when taking infinity of $\tilde{\varepsilon }^{\text{p}}$. When $\tilde{\varepsilon }^{\text{p}} \ge 0.01$, the difference to its limit already turns to be as minor as 1 %. As $\tilde{\varepsilon }_{1}^{\text{p}} < \tilde{\varepsilon }^{\text{p}} \le 0.0 1$, the corresponding value for $\tilde{\varepsilon }_{1}^{\text{p}}$ is far from reaching ɛ ^p_cu (Fig. 4a). So its first item in Eq. (39) can be neglected in the current analysis range, that is,

$$\bar{K}^{\prime}\left|_{_{\varepsilon_{1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}}}} {= \frac{{2\tilde{\varepsilon }^{\text{p}} }}{\alpha}\frac{{{\text{d}}\tilde{\varepsilon }^{\text{p}}}}{{{\text{d}}\varepsilon_{ 1}^{\text{p}} }}} \right|_{{\varepsilon_{ 1}^{\text{p}} \to \varepsilon_{\text{cu}}^{\text{p}} }} \zeta (\rho ) = 0$$

(40)

where

$$\zeta (\rho ) = 277 - 650.1\exp ( - 0.083\rho )$$

(41)

Therefore, zero value of $\bar{K}^{\prime}$ is determined by the function ζ. It gives the boundary value for ρ to be 10.3. When ρ > 10.3, $\bar{K}^{\prime} > 0$; otherwise, $\bar{K}^{\prime} < 0$ after reaching the extremum of $\bar{K}$.

Appendix C

The equivalent plastic strain $\tilde{\varepsilon }^{\text{p}}$ is defined as the norm of plastic strain tensor. Its increments has the following form in the case of concrete under uniform confinement

$${\text{d}}\tilde{\varepsilon }^{\text{p}} = \sqrt {{\text{d}}\varepsilon_{\text{1}}^{\text{p}} \cdot {\text{d}}\varepsilon_{\text{1}}^{\text{p}} + 2{\text{d}}\varepsilon_{ l}^{\text{p}} \cdot {\text{d}}\varepsilon_{ l}^{\text{p}} } = \sqrt {1 + 2\left( {\frac{{{\text{d}}\varepsilon_{ l}^{\text{p}} }}{{{\text{d}}\varepsilon_{\text{c}}^{\text{p}} }}} \right)^{2} } {\text{d}}\varepsilon_{\text{1}}^{\text{p}}$$

(42)

where the footnote “1” and “l” denote the axial direction and lateral direction.

Following the flow rule (Eq. 22), it has

$${\text{d}}\varepsilon_{\text{1}}^{\text{p}} = {\text{d}}\lambda \left( {\frac{\sqrt 3 }{{2\sqrt {\frac{{(\sigma_{\text{1}} - \sigma_{l} )^{2} }}{3}} }}2\frac{{(\sigma_{\text{1}} - \sigma_{l} )}}{3} + \frac{\tan \,\beta }{3}} \right) = {\text{d}}\lambda \left( { - 1 + \frac{\tan \,\beta }{3}} \right)$$

(43)

$${\text{d}}\varepsilon_{{1}}^{\text{p}} = {\text{d}}\lambda \left( {\frac{\sqrt 3 }{{2\sqrt {\frac{{(\sigma_{\text{1}} - \sigma_{l} )^{2} }}{3}} }}\frac{{(\sigma_{l} - \sigma_{1} )}}{3} + \frac{\tan \,\beta }{3}} \right) = {\text{d}}\lambda \left( {\frac{1}{2} + \frac{\tan \,\beta }{3}} \right)$$

(44)

It can resolve the ratio of lateral plastic strain-to-axial plastic strain

$$\frac{{{\text{d}}\varepsilon_{ l}^{\text{p}} }}{{{\text{d}}\varepsilon_{\text{1}}^{\text{p}} }} = \frac{2\,\tan \,\beta + 3}{2\,\tan \,\beta - 6}$$

(45)

Substituting Eq. (45) into Eq. (42), one has

$${\text{d}}\tilde{\varepsilon }^{\text{p}} = \sqrt {{\text{d}}\varepsilon_{\text{1}}^{\text{p}} \cdot {\text{d}}\varepsilon_{\text{1}}^{\text{p}} + 2{\text{d}}\varepsilon_{ l}^{\text{p}} \cdot {\text{d}}\varepsilon_{ l}^{\text{p}} } = \sqrt {1 + 2\left( {\frac{2\tan \beta + 3}{2\tan \beta - 6}} \right)^{2}} {\text{d}}\varepsilon_{\text{1}}^{\text{p}}$$

(46)

When the ratio ρ is fixed, the function of β (Table 3) can tell the value of tanβ at one fixed ɛ ^p_c . And hence $\tilde{\varepsilon }^{\text{p}}$ can be converted into ɛ ₁ ^p by incremental computation with the initial condition $\varepsilon_{\text{1,0}}^{\text{p}} = \tilde{\varepsilon }_{0}^{\text{p}} = 0$.