Nonlinear analysis of RC members subjected to combined bending–shear–torsion stresses: a numerical multi-fiber displacement-based finite element model with warping

Abstract

An enhanced multi-fiber beam–column element suitable for the analysis of reinforced concrete members including the torsional effect is presented in this paper. The model is developed based on displacement formulation with the assumptions of small displacement. The section kinematics is based on the assumptions of a two-node Timoshenko beam and enhanced by introducing additional degrees of freedom at each section in order to take into account the warping phenomenon. For this, a system of fixed points is created and interpolated by Lagrange functions and polynomials. To take into account the effect of stirrups, a discretization of control sections into regions following its material response is applied. The basic assumption of the modified compression field theory with a secant-stiffness formulation is used to represent the constitutive material model for reinforced concrete. The model is validated by comparison with analytical solutions and several experimental tests. The simulations include a variety of monotonic load conditions under bending, shear and torsion for specimens with rectangular sections.

Appendices

Appendix A. Shape functions in two-node Timoshenko beam element

Shape functions for displacement vector:

$$\begin{aligned} N_1&= 1- \dfrac{x}{L}; \qquad N_2 = \dfrac{x}{L}, \\ N_{3y/z}&= \dfrac{1}{1 + \phi _{y/z}} \left[ 2 \left( \dfrac{x}{L} \right) ^3 - 3 \left( \dfrac{x}{L} \right) ^2 - \phi _{y/z} \dfrac{x}{L} + 1 + \phi _{y/z} \right] , \\ N_{4y/z}&= \dfrac{L}{1 + \phi _{y/z}} \left[ \left( \dfrac{x}{L} \right) ^3 - \left( 2 + \dfrac{\phi _{y/z}}{2} \right) \left( \dfrac{x}{L} \right) ^2 + \left( 1 + \dfrac{\phi _{y/z}}{2} \right) \dfrac{x}{L} \right] , \\ N_{5y/z}&= -\dfrac{1}{1 + \phi _{y/z}} \left[ 2 \left( \dfrac{x}{L} \right) ^3 - 3 \left( \dfrac{x}{L} \right) ^2 - \phi _{y/z} \dfrac{x}{L} \right] ,\\ N_{6y/z}&= \dfrac{L}{1 + \phi _{y/z}} \left[ \left( \dfrac{x}{L} \right) ^3 - \left( 1 - \dfrac{\phi _{y/z}}{2} \right) \left( \dfrac{x}{L} \right) ^2 - \dfrac{\phi _{y/z}}{2} \dfrac{x}{L} \right] ,\\ N_{7y/z}&= \dfrac{6}{\left( 1 + \phi _{y/z}\right) L} \left[ \left( \dfrac{x}{L} \right) ^2 - \dfrac{x}{L} \right] , \\ N_{8y/z}&= \dfrac{1}{1 + \phi _{y/z}} \left[ 3 \left( \dfrac{x}{L} \right) ^2 - \left( 4 + \phi _{y/z} \right) \dfrac{x}{L} + 1 + \phi _{y/z} \right] , \\ N_{9y/z}&= - \dfrac{6}{\left( 1 + \phi _{y/z}\right) L} \left[ \left( \dfrac{x}{L} \right) ^2 - \dfrac{x}{L} \right] , \\ N_{10y/z}&= \dfrac{1}{1 + \phi _{y/z}} \left[ 3 \left( \dfrac{x}{L} \right) ^2 - \left( 2 - \phi _{y/z} \right) \dfrac{x}{L} \right] \end{aligned}$$

with $\phi _{y/z}$ the ratio of rigidity between bending and shear with respect to y and z axis, respectively.

$$\begin{aligned} \phi _y = \dfrac{12}{L^2} \dfrac{ \displaystyle \int _S E y^2 \mathrm{d}A}{\kappa _y \displaystyle \int _S G \mathrm{d}S}; \qquad \phi _z = \dfrac{12}{L^2} \dfrac{ \displaystyle \int _S E z^2 \mathrm{d}A}{\kappa _z \displaystyle \int _S G \mathrm{d}S} \end{aligned}$$

$\kappa _y$ and $\kappa _z$ are the shear correction factors in y and z direction, respectively.

Shape functions for strain vector:

$$\begin{aligned} B_1&= \dfrac{\partial N_1}{\partial x} = -\dfrac{1}{L}; \qquad B_2 = \dfrac{\partial N_2}{\partial x} = \dfrac{1}{L}, \\ B_{3y/z}&= \dfrac{\partial N_{3y/z}}{\partial x} - N_{7y/z} = - \dfrac{\phi _{y/z}}{L \left( 1 + \phi _{y/z} \right) }, \\ B_{4y/z}&= \dfrac{\partial N_{4y/z}}{\partial x} - N_{8y/z} = - \dfrac{\phi _{y/z}}{2 \left( 1 + \phi _{y/z} \right) }, \\ B_{5y/z}&= \dfrac{\partial N_{5y/z}}{\partial x} - N_{9y/z} = \dfrac{\phi _{y/z}}{L \left( 1 + \phi _{y/z} \right) }, \\ B_{6y/z}&= \dfrac{\partial N_{6y/z}}{\partial x} - N_{10y/z} = - \dfrac{\phi _{y/z}}{2 \left( 1 + \phi _{y/z} \right) }, \\ B_{7y/z}&= -\dfrac{\partial N_{7y/z}}{\partial x} = - \dfrac{6}{L \left( 1 + \phi _{y/z} \right) } \left( \dfrac{2x}{L^2} - \dfrac{1}{L}\right) , \\ B_{8y/z}&= -\dfrac{\partial N_{8y/z}}{\partial x} = \dfrac{1}{\left( 1 + \phi _{y/z} \right) } \left( \dfrac{6x}{L^2} - \dfrac{4 + \phi _{y/z}}{L}\right) , \\ B_{9y/z}&= -\dfrac{\partial N_{9y/z}}{\partial x} = - \dfrac{6}{L \left( 1 + \phi _{y/z} \right) } \left( \dfrac{2x}{L^2} - \dfrac{1}{L}\right) , \\ B_{10y/z}&= -\dfrac{\partial N_{10y/z}}{\partial x} =\dfrac{1}{\left( 1 + \phi _{y/z} \right) } \left( \dfrac{6x}{L^2} -\dfrac{2 - \phi _{y/z}}{L}\right) . \end{aligned}$$

Appendix B. Lagrange interpolation polynomial and enhanced compatibility matrix

Longitudinal interpolation matrix ${\mathbf {L}}(x)$ of ($3 \times 3.3.n_{\mathrm{w}}$):

$$\begin{aligned} {\mathbf {L}}(x) = \begin{bmatrix} {\mathbf {L}}_1(x)&\quad \dots&\quad {\mathbf {L}}_i(x)&\quad \dots {\mathbf {L}}_{nw}(x) \end{bmatrix}. \end{aligned}$$

(B.1)

${\mathbf {L}}_i(x)$ is a matrix of $3 \times 3.3$, containing the 1D Lagrange polynomial at section i:

$$\begin{aligned} {\mathbf {L}}_i(x) = \begin{bmatrix} L_i(x) &{}\quad 0 &{}\quad 0 &{}\quad L_i(x) &{}\quad 0 &{}\quad 0 &{}\quad L_i(x) &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad L_i(x) &{}\quad 0 &{}\quad 0 &{}\quad L_i(x) &{}\quad 0 &{}\quad 0 &{}\quad L_i(x) &{}\quad 0\\ 0 &{}\quad 0 &{}\quad L_i(x) &{}\quad 0 &{}\quad 0 &{}\quad L_i(x) &{}\quad 0 &{}\quad 0 &{}\quad L_i(x) \end{bmatrix}. \end{aligned}$$

(B.2)

Sectional interpolation matrices ${\mathbf {S}}_x(y,z)$ of ($3.3.n_{\mathrm{w}} \times 3.s_{\mathrm{w}}.n_{\mathrm{w}}$):

$$\begin{aligned} {\mathbf {S}}_x(y,z) = \begin{bmatrix} \hat{{\mathbf {S}}}_x(y,z) &{}\quad {\mathbf {0}}_{\mathrm{w}} &{}\quad {\mathbf {0}}_{\mathrm{w}} \\ {\mathbf {0}}_{\mathrm{w}} &{}\quad \hat{{\mathbf {S}}}_x(y,z) &{}\quad {\mathbf {0}}_{\mathrm{w}} \\ {\mathbf {0}}_{\mathrm{w}} &{}\quad {\mathbf {0}}_{\mathrm{w}} &{}\quad \hat{{\mathbf {S}}}_x(y,z) \end{bmatrix}; \quad \hat{{\mathbf {S}}}_x(y,z) = \begin{bmatrix} {{\mathbf {S}}_x^u}(y,z) &{}\quad {\mathbf {0}}_{3sw} &{}\quad {\mathbf {0}}_{3sw} \\ {\mathbf {0}}_{3sw} &{}\quad {{\mathbf {S}}_x^v}(y,z) &{}\quad {\mathbf {0}}_{3sw} \\ {\mathbf {0}}_{3sw} &{}\quad {\mathbf {0}}_{3sw} &{}\quad {{\mathbf {S}}_x^w}(y,z) \end{bmatrix}, \end{aligned}$$

(B.3)

where ${\mathbf {0}}_{\mathrm{w}}$ is a zero matrix of ($9 \times 3.s_{\mathrm{w}}$) columns; ${\mathbf {0}}_{3sw}$ is a zero matrix of ($3 \times s_{\mathrm{w}}$); ${\mathbf {S}}_x^u(y,z)$, ${\mathbf {S}}_x^v(y,z)$ and ${\mathbf {S}}_x^w(y,z)$ are three matrices of ($3 \times s_{\mathrm{w}}$) containing the row vector $\bar{{\mathbf {S}}}(y,z)$ in Eq. (8) and the zero row vector of $s_{\mathrm{w}}$ columns:

$$\begin{aligned} {{\mathbf {S}}_x^u}(y,z) = \begin{bmatrix} \bar{{\mathbf {S}}}(y,z) \\ {\mathbf {0}}_{sw} \\ {\mathbf {0}}_{sw} \end{bmatrix}; \quad {{\mathbf {S}}_x^v}(y,z) = \begin{bmatrix} {\mathbf {0}}_{sw} \\ \bar{{\mathbf {S}}}(y,z) \\ {\mathbf {0}}_{sw} \end{bmatrix}; \quad {{\mathbf {S}}_x^w}(y,z) = \begin{bmatrix} {\mathbf {0}}_{sw} \\ {\mathbf {0}}_{sw} \\ \bar{{\mathbf {S}}}(y,z) \end{bmatrix}. \qquad \end{aligned}$$

(B.4)

Sectional interpolation matrices ${\mathbf {S}}_{yz}(y,z)$:

$$\begin{aligned} {\mathbf {S}}_{yz}(y,z) = \begin{bmatrix} \hat{{\mathbf {S}}}_{yz}(y,z) &{}\quad {\mathbf {0}}_{\mathrm{w}} &{}\quad {\mathbf {0}}_{\mathrm{w}} \\ {\mathbf {0}}_{\mathrm{w}} &{}\quad \hat{{\mathbf {S}}}_{yz}(y,z) &{}\quad {\mathbf {0}}_{\mathrm{w}} \\ {\mathbf {0}}_{\mathrm{w}} &{}\quad {\mathbf {0}}_{\mathrm{w}} &{}\quad \hat{{\mathbf {S}}}_{yz}(y,z) \end{bmatrix}; \quad \hat{{\mathbf {S}}}_{yz}(y,z) = \begin{bmatrix} {\mathbf {S}}_{yz}^u(y,z) &{}\quad {\mathbf {0}}_{3sw} &{}\quad {\mathbf {0}}_{3sw} \\ {\mathbf {0}}_{3sw} &{}\quad {\mathbf {S}}_{yz}^v(y,z) &{}\quad {\mathbf {0}}_{3sw} \\ {\mathbf {0}}_{3sw} &{}\quad {\mathbf {0}}_{3sw} &{}\quad {\mathbf {S}}_{yz}^w(y,z) \end{bmatrix}.\nonumber \\ \end{aligned}$$

(B.5)

${\mathbf {S}}_{yz}^u(y,z)$, ${\mathbf {S}}_{yz}^v(y,z)$ and ${\mathbf {S}}_{yz}^w(y,z)$ are three matrices of ($3 \times s_{\mathrm{w}}$) containing the derivation with respect to y and z of the row vector $\bar{{\mathbf {S}}}(y,z)$ in Eq. (8) and the zero row vector of $s_{\mathrm{w}}$ columns:

$$\begin{aligned} {\mathbf {S}}_{yz}^u(y,z) = \begin{bmatrix} {\mathbf {0}}_{sw} \\ \dfrac{\partial \bar{{\mathbf {S}}}(y,z)}{\partial y} \\ \dfrac{\partial \bar{{\mathbf {S}}}(y,z)}{\partial z} \\ \end{bmatrix}; \quad {\mathbf {S}}_{yz}^v(y,z) = \begin{bmatrix} {\mathbf {0}}_{sw} \\ {\mathbf {0}}_{sw} \\ {\mathbf {0}}_{sw} \end{bmatrix}; \quad {\mathbf {S}}_{yz}^w(y,z) = \begin{bmatrix} {\mathbf {0}}_{sw} \\ {\mathbf {0}}_{sw} \\ {\mathbf {0}}_{sw} \end{bmatrix} \end{aligned}$$

(B.6)