Ultimate limit strength of rectangular RC section (2022.03.18)

A method of calculation of ultimate limit strength of rectangular RC member using rectangular stress block is described.
In this document, the positive sign of stress and strain for concretee and compressive reinforcement mean compression, and the positive sign of stress and asrein for tensile reinforcement means tension.
The calculation program is basically for symmetric rebar arrangement.

General formulas

A rectangular concrete cross section with double reinforcement is considered.

$\begin{align} &b & & \text{width of section} \\ &h & & \text{height of section} \\ \end{align}$

From Bernoulli-Euler hypothsis, the strains of tensile and compressive re-bars can be expressed as followings.

$\begin{align} & \epsilon_t=\cfrac{d-x}{x}\cdot \epsilon_{cu} & & +: \text{tensile}\\ & \epsilon_c=\cfrac{x-d'}{x}\cdot \epsilon_{cu} & & +: \text{compressive} \end{align}$

$\begin{align} &\epsilon_t & & \text{strain of tensile re-bar} \\ &\epsilon_c & & \text{strain of compressive re-bar} \\ &\epsilon_{cu} & & \text{ultimate strain of concrete} \\ &x & & \text{location of neutral axis from compressive edge}\\ &d & & \text{effective depth of tensile reinforcement} \\ &d' & & \text{cover of compressive reinforcement} \end{align}$

Using above, the stress of re-bar can be calculated as follows.

$\begin{align} &-f_y/\gamma_m \leq \sigma_s =E_s\cdot \epsilon_t \leq f_y/\gamma_m & & +:\text{tensille}\\ &-f_y/\gamma_m \leq \sigma_s'=E_s\cdot \epsilon_c \leq f_y/\gamma_m & & +:\text{compressive} \end{align}$

$\begin{align} &\sigma_s & & \text{sstress of tensile re-bar} \\ &\sigma_s' & & \text{stress of compressive re-bar} \\ &E_s & & \text{elastic modulus of reinforcement} \\ &f_y & & \text{yield strength of reinforcement} \\ &\gamma_m & & \text{partial safety factor for material} \end{align}$

The height of compressive stress block of cncrete of $a$ is defined as follow using the location of neutral axis of $x$ .

$\begin{equation} a=\beta_1\cdot x \leq h \end{equation}$

$\begin{align} &a & & \text{height of concrete stress block}\\ &\beta_1 & & \text{coefficient for height of concrete stress block} \end{align}$

The axsial force of $N$ and the bending moment of $M$ can be expressed as followings.

$\begin{equation} N=\beta_2\cdot f_{cu}\cdot b\cdot a - A_s\cdot \sigma_s + A_s'\cdot \sigma_s' \end{equation}$

$\begin{equation} M=N\cdot e=\beta_2\cdot f_{cu}\cdot b\cdot d^2 \left(\cfrac{a}{d}\right) \left\{1-\cfrac{1}{2}\left(\cfrac{a}{d}\right)\right\} + A_s'\cdot \sigma_s'\cdot (d-d') -N\cdot c \end{equation}$

$\begin{align} &N & & \text{axial force}\\ &M & & \text{bending moment}\\ &e & & \text{distance from centroid to loading point of $N$} \\ &c & & \text{distance from centroid to tensile re-bar}\\ &f_{cu} & & \text{concrete strength} \\ &\beta_2 & & \text{coefficient for width of concrete stress block} \\ &A_s & & \text{section area of tensile reinforcement} \\ &A_s' & & \text{section area of compressive reinforcement} \end{align}$

f:id:damyarou:20220318143524j:plain

Location of neutral axis

Balanced failure

In the case of balanced failure, following conditions are applied.

the concrete strain reaches the ultimate strain of $\epsilon_{cu}$ .
the same time, the tensile re-bar stress reaches the yield stress.

From above, the location of neutral axis of $x$ can be obtained as follow.

$\begin{gather} \epsilon_t=\cfrac{f_y}{\gamma_m\cdot E_s}=\cfrac{d-x}{x}\cdot \epsilon_{cu} \\ x=\cfrac{d}{1+\cfrac{f_y}{\gamma_m\cdot E_s\cdot \epsilon_{cu}}} \end{gather}$

When above $x$ is re-defined as $x_0$ , the axial force of $N_0$ for balanced failure can be calculated using $x_0$ .

Pure bending

In case of pure bending, following conditions are applied.

the axial force is zero ( $N=0$ ).
the concrete strain has reached the ultimate strain of $\epsilon_{cu}$ .
if $0 \geq N_0$ , the failure mode is tensile failure. This means the tensile re-bar stress has reached the yield stress.
if $N_0 \lt 0$ , the failure mode is compressive failure. This means the compressive re-bar stress has reached the yield stress.

From above, the location of neutral axis of $x$ can be obtained as a root of following quadratic eqiation which is derived from calculation formula of $N$ .

For tensile failure ( $N=0$ )

$\begin{equation} \beta_2\cdot f_{cu}\cdot b\cdot \beta_1\cdot x^2 +\left(A_s'\cdot E_s\cdot \epsilon_{cu}-A_s\cdot \cfrac{f_y}{\gamma_m}-N\right)\cdot x -A_s'\cdot E_s\cdot d'\cdot \epsilon_{cu}=0 \end{equation}$

$\begin{align} &c_1=\beta_2\cdot f_{cu}\cdot b\cdot \beta_1 \\ &c_2=A_s'\cdot E_s\cdot \epsilon_{cu}-A_s\cdot \cfrac{f_y}{\gamma_m}-N \\ &c_3=-A_s'\cdot E_s\cdot d'\cdot \epsilon_{cu} \end{align}$

For compressive failure ( $N=0$ )

$\begin{equation} \beta_2\cdot f_{cu}\cdot b\cdot \beta_1\cdot x^2 +\left(A_s\cdot E_s\cdot \epsilon_{cu}+A_s'\cdot \cfrac{f_y}{\gamma_m}-N\right)\cdot x -A_s\cdot E_s\cdot d\cdot \epsilon_{cu}=0 \end{equation}$

$\begin{align} &c_1=\beta_2\cdot f_{cu}\cdot b\cdot \beta_1 \\ &c_2=A_s\cdot E_s\cdot \epsilon_{cu}+A_s'\cdot \cfrac{f_y}{\gamma_m}-N \\ &c_3=-A_s\cdot E_s\cdot d\cdot \epsilon_{cu} \end{align}$

Root of quadratic equation

$\begin{equation} x=\cfrac{-c_2+\sqrt{c_2{}^2-4\cdot c_1\cdot c_3}}{2\cdot c_1} \end{equation}$

Calculation of ultimate strength

Constants based on BS 8110-1

$\begin{align} &\gamma_m=1.5 & & \text{for concrete} \\ &\gamma_m=1.05 & & \text{for reinforcement} \\ &\beta_1=0.9 & & \text{coefficient for height of concrete stress block} \\ &\beta_2=0.67 / \gamma_m & & =0.447 \quad (\gamma_m=1.5) \\ &E_s=200,000 & & \text{in MPa} \\ &\epsilon_{cu}=0.0035 & & \text{ultimate strain of concrete} \\ \end{align}$

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Calculation steps

Step	Description
1	Inpput parameters
3	Calculate neutral axis for balanced failure and pure bending
2	Specify neutral axis
3	Calculate axial force and moment
4	Draw M-N diagram

Program

import numpy as np
import matplotlib.pyplot as plt


def drawfig(a_ax,a_mo,tstr):
    fsz=10
    xmin=0 ; xmax=5; dx=1
    ymin=-10; ymax=30; dy=5
    iw,ih=5,5
    plt.figure(figsize=(iw,ih),facecolor='w')
    plt.rcParams['font.size']=fsz
    plt.rcParams['font.family']='sans-serif'
    plt.xlim([xmin,xmax])
    plt.ylim([ymin,ymax])
    plt.xlabel('Bending moment M/b/h**2 (MPa)')
    plt.ylabel('Axialforce N/b/h (MPa)')
    plt.xticks(np.arange(xmin,xmax+dx,dx))
    plt.yticks(np.arange(ymin,ymax+dy,dy))
    plt.grid(color='#999999',linestyle='solid')
    plt.title(tstr,loc='left',fontsize=fsz)
    plt.plot(a_mo,a_ax,'o-',ms=4,clip_on=False)
    plt.tight_layout()
    fnameF='fig_mn_rebar1.jpg'
    plt.savefig(fnameF, dpi=100, bbox_inches="tight", pad_inches=0.1)
    #plt.show()


def pure_bend(fcu,beta1,beta2,ecu,fy,rms,Es,ast,asc,dd,dc,bb,ax0):
    ax=0 # axial force
    if 0<ax0: # tensile failure
        c1=beta2*fcu*bb*beta1
        c2=asc*Es*ecu-ast*fy/rms-ax
        c3=-asc*Es*dc*ecu
    else: # compressive failure
        c1=beta2*fcu*bb*beta1
        c2=ast*Es*ecu+asc*fy/rms-ax
        c3=-ast*Es*dd*ecu
    x=(-c2+np.sqrt(c2**2-4*c1*c3))/2/c1 # neutral axis
    return x


def cal_mn(bb,hh,fcu,x,dd,dc,cc,beta1,beta2,ecu,Es,fy,rms,ast,asc):
    et=(dd-x)/x*ecu
    ec=(x-dc)/x*ecu
    sigt=Es*et
    sigc=Es*ec
    if fy/rms < sigt: sigt=fy/rms
    if sigt < -fy/rms: sigt=-fy/rms
    if fy/rms < sigc: sigc=fy/rms      
    if sigc < -fy/rms: sigc=-fy/rms
    aa=beta1*x
    if hh<aa: aa=hh
    ax=beta2*fcu*bb*aa-ast*sigt+asc*sigc
    mo=beta2*fcu*bb*dd**2*(aa/dd)*(1-1/2*(aa/dd))+asc*sigc*(dd-dc)-ax*cc
    return ax,mo


def calc(fcu,fy,bb,hh,ast,asc,dt,dc):
    # constants
    rmc=1.5 # partial safety factor for concrete
    rms=1.05 # partial safety factor for reinforcement
    beta1=0.9 # coefficient for height of concrete stress block
    beta2=0.67/rmc # coefficient for width of concrete stress block 
    Es=200000 # elastic modulus of reinforcement
    ecu=0.0035 # ultimate strain of concrete
    dd=hh-dt
    cc=hh/2-dt
    # Balanced failure
    x0=dd/(1+fy/rms/Es/ecu)
    ax0,mo0=cal_mn(bb,hh,fcu,x0,dd,dc,cc,beta1,beta2,ecu,Es,fy,rms,ast,asc)
    # Pure bending
    xb=pure_bend(fcu,beta1,beta2,ecu,fy,rms,Es,ast,asc,dd,dc,bb,ax0)
    # Max and Min axial force
    ax_max=beta2*fcu*bb*hh+ast*fy/rms+asc*fy/rms
    ax_min=-(ast*fy/rms+asc*fy/rms)
    # Definition of neutral axis for calculation
    xxx=np.linspace(1,1.2*hh,20)
    xxx=np.concatenate([xxx,[x0,xb]])
    xxx=np.sort(xxx)
    # calculation
    a_ax=np.zeros(len(xxx),dtype=np.float64)
    a_mo=np.zeros(len(xxx),dtype=np.float64)
    for i,x in enumerate(xxx):
        ax,mo=cal_mn(bb,hh,fcu,x,dd,dc,cc,beta1,beta2,ecu,Es,fy,rms,ast,asc)
        a_ax[i]=ax
        a_mo[i]=mo
    a_ax=a_ax/bb/hh
    a_mo=a_mo/bb/hh**2
    a_ax=np.concatenate([a_ax,[ax_max/bb/hh,ax_min/bb/hh]])
    a_mo=np.concatenate([a_mo,[0,0]])
    ii=np.argsort(a_ax)
    a_ax=a_ax[ii]
    a_mo=a_mo[ii]
    return a_ax,a_mo


def main():
    # parameteers
    fcu=30 # concrete strength
    fy=460 # yield strength of reinforcement
    bb=2000 # width of member
    hh=2000 # height of member
    ast=np.pi*32**2/4*20 # section area of tensile reinforcement
    asc=np.pi*32**2/4*20 # section area of compressive reinforcement
    dt=200 # cover of tensile reinforcement
    dc=200 # cover of compressive reinforcement
    a_ax,a_mo=calc(fcu,fy,bb,hh,ast,asc,dt,dc)    
    tstr='2T32ctc200x2layer,double,bxh=2000x2000'
    drawfig(a_ax,a_mo,tstr)
    
    
#---------------
# Execute
#---------------
if __name__ == '__main__': main()

Example of analysis result

f:id:damyarou:20220318143559j:plain

Thank you.