Relationship between Shear Strength and Uniaxial Strength

The definitions of symbols are shown below in this discussion considering compression members such as concrete or rock material.

$c$	cohesion of material
$\phi$	internal friction angle of material
$\sigma_c$	uniaxial compressive strength of material (negative value)
$\sigma_t$	uniaxial tensile strength of material (positive value)

f:id:damyarou:20190522083439j:plain

Considering a parameter $a$ , a cohesion $c$ can be expressed as follow.

$\begin{equation} c=a\cdot \tan\phi \end{equation}$

A radius of Mohr circle for a compressive strength $\sigma_c$ can be expressed as follow.

$\begin{equation} \cfrac{|\sigma_c|}{2}=\left(a+\cfrac{|\sigma_c|}{2}\right)\cdot \sin\phi \end{equation}$

Using same manner, a radius of Mohr circle for a tensile strength $\sigma_t$ can be expressed as follow.

$\begin{equation} \cfrac{\sigma_t}{2}=\left(a-\cfrac{\sigma_t}{2}\right)\cdot \sin\phi \end{equation}$

From above relationships, a compressive strength $\sigma_c$ and a tensile strength $\sigma_t$ can be obtained as functions of $c$ and $\phi$ by erasing parameter $a$ .

$\begin{equation} |\sigma_c|=\cfrac{2\cdot c\cdot \cos\phi}{1-\sin\phi} \qquad \sigma_t=\cfrac{2\cdot c\cdot \cos\phi}{1+\sin\phi} \end{equation}$

Next, using above relations, following equation can be obtained.

$\begin{gather} 2\cdot c\cdot \cos\phi=|\sigma_c|\cdot (1-\sin\phi)=\sigma_t\cdot (1+\sin\phi) \\ |\sigma_c|-\sigma_t=\sin\phi\cdot (|\sigma_c|+\sigma_t) \\ \sin\phi=\cfrac{|\sigma_c|-\sigma_t}{|\sigma_c|+\sigma_t} \end{gather}$

Using above, a cohesion $c$ can be obtained as follow.

$\begin{equation} c=\cfrac{|\sigma_c|\cdot \sigma_t}{\cos\phi\cdot (|\sigma_c|+\sigma_t)} \end{equation}$

作図プログラム

import numpy as np
import matplotlib.pyplot as plt


# global variables
sigt=3
sigc=-8
phi=np.arcsin((np.abs(sigc)-sigt)/(np.abs(sigc)+sigt))
cc=np.abs(sigc)*sigt/(np.abs(sigc)+sigt)/np.cos(phi)
aa=cc/np.tan(phi)


def barrow(x1,y1,x2,y2,ss,dd,ii):
    # (x1,y1),(x2,y2)：両矢印始点と終点
    # 両矢印始点と終点座標は文字列の傾きを考慮して決定する
    # ss：描画文字列
    # dd：文字列中央の寸法線からの距離
    # ii：１なら寸法線の上側，２なら寸法線の下側にテキストを描画
    col='#333333'
    arst='<->,head_width=3,head_length=10'
    aprop=dict(arrowstyle=arst,connectionstyle='arc3',facecolor=col,edgecolor=col,shrinkA=0,shrinkB=0,lw=1)
    plt.annotate('',
        xy=(x1,y1), xycoords='data',
        xytext=(x2,y2), textcoords='data', fontsize=0, arrowprops=aprop)
    al=np.sqrt((x2-x1)**2+(y2-y1)**2)
    cs=(x2-x1)/al
    sn=(y2-y1)/al
    rt=np.degrees(np.arccos(cs))
    if y2-y1<0: rt=-rt
    if ii==1:
        xs=0.5*(x1+x2)-dd*sn
        ys=0.5*(y1+y2)+dd*cs
        plt.text(xs,ys,ss,ha='center',va='center',rotation=rt)
    else:
        xs=0.5*(x1+x2)+dd*sn
        ys=0.5*(y1+y2)-dd*cs
        plt.text(xs,ys,ss,ha='center',va='center',rotation=rt)
        
    
def draw_axes(xmin,xmax,ymin,ymax):
    # 座標軸描画
    col='#000000'
    arst='<|-,head_width=0.3,head_length=0.5'
    aprop=dict(arrowstyle=arst,connectionstyle='arc3',facecolor=col,edgecolor=col,lw=1)
    plt.annotate(r'$\sigma$',
        xy=(xmin,0), xycoords='data',
        xytext=(xmax,0), textcoords='data', fontsize=20, ha='left',va='center', arrowprops=aprop)
    plt.annotate(r'$\tau$',
        xy=(0,0), xycoords='data',
        xytext=(0,ymax), textcoords='data', fontsize=20, ha='center',va='bottom', arrowprops=aprop)
    

def drawfig(xmin,xmax,ymin,ymax,fsz):
    # x-y axses
    draw_axes(xmin,xmax,ymin,ymax)
    # failure criterion
    x1=aa   ; y1=0
    x2=sigc ; y2=cc-sigc*np.tan(phi)
    plt.plot([x1,x2],[y1,y2],'-',color='#ff0000',lw=3)
    xs=0.5*sigc; ys=cc-xs*np.tan(phi)
    strs="Mohr-Coulomb's Failure Criteria\n$\\tau=c-\sigma\cdot\\tan{\phi}$"
    plt.text(xs,ys+0.08*(ymax-ymin),strs,ha='center',va='center',rotation=-np.degrees(phi),fontsize=fsz-4)    
    theta=np.linspace(np.pi-phi,np.pi,20,endpoint=True)
    r1=0.07*(xmax-xmin)
    x=r1*np.cos(theta)+aa
    y=r1*np.sin(theta)
    plt.plot(x,y,'-',color='#000000',lw=1)
    r2=r1*0.9
    x=r2*np.cos(theta)+aa
    y=r2*np.sin(theta)
    plt.plot(x,y,'-',color='#000000',lw=1)
    plt.text(aa-r1,0,r'$\phi$',ha='right',va='bottom',color='#000000')
    # Mohr circle
    theta=np.linspace(0,np.pi,180,endpoint=True)
    rr=sigt/2
    x=rr*np.cos(theta)+rr
    y=rr*np.sin(theta)
    plt.plot(x,y,'-',color='#0000ff',lw=2)
    rr=-sigc/2
    x=rr*np.cos(theta)-rr
    y=rr*np.sin(theta)
    plt.plot(x,y,'-',color='#0000ff',lw=2)
    # points
    x=np.array([sigc,sigc/2,0,sigt/2,sigt,aa])
    y=np.array([0,0,0,0,0,0])
    plt.plot(x,y,'o',color='#000000',clip_on=False)

    ds=0.02*(ymax-ymin)
    ss=[r'$\sigma_c$',r'$\frac{\sigma_c}{2}$',r'$0$',r'$\frac{\sigma_t}{2}$',r'$\sigma_t$',r'$a$']
    for i in range(len(ss)):
        plt.text(x[i],-ds,ss[i],va='top',ha='center',fontsize=fsz)
    plt.plot([0],[cc],'o',color='#000000')
    plt.text(0.1,cc,r'$c$',va='bottom',ha='left',fontsize=fsz)

    ds=0.08*(ymax-ymin)
    x2=-cc*np.cos(phi); y2=cc+cc*np.sin(phi); x1=sigc/2; y1=0; barrow(x1,y1,x2,y2,r'$\frac{|\sigma_c|}{2}$',ds,1)
    x2=cc*np.cos(phi); y2=cc-cc*np.sin(phi); x1=sigt/2; y1=0; barrow(x1,y1,x2,y2,r'$\frac{\sigma_t}{2}$',ds,1)
    
    
    
def main():
    xmin=sigc-1; xmax=aa+1
    ymin=0     ; ymax=cc-xmin*np.tan(phi)
    fsz=20   # fontsize
    iw=10     # image width
    ih=iw*(ymax-ymin)/(xmax-xmin)
    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.axis('off')
    plt.gca().set_aspect('equal',adjustable='box')

    drawfig(xmin,xmax,ymin,ymax,fsz)

    plt.tight_layout()
    fnameF='fig_mohr1.jpg'
    plt.savefig(fnameF, dpi=200, bbox_inches="tight", pad_inches=0.1)
    plt.show()


#==============
# Execution
#==============
if __name__ == '__main__': main()