# Chapter14-Stress Concentrations¶

## Ex1-pg521¶

In [2]:
import math
## initialization of variables
#calculate the  orientain of the hole and value of this tensile stress
## part (a)
ab_r=100.
Sigma_1=-20. ##MPa
Sigma_2=-75. ##MPa
##calculations
A=(Sigma_1+Sigma_2)/(Sigma_1-Sigma_2)
th=1/2.*math.acos((A*math.sinh(2*alphao)-1/2.*(math.sinh(2.*alphao)+math.cosh(2.*alphao)))/A)
print('pat (a)')
print'%s %.2f %s'%('\n theta = ',th,' rad')
##part (b)
S_bb=-(Sigma_1-Sigma_2)**2./(2.*(Sigma_1+Sigma_2))*(1.+math.cosh(2.*alphao)/math.sinh(2.*alphao))
print('\n part (b)')
print'%s %.2f %s'%('\n Maximum tensile stress =',S_bb,' MPa')
##part (c)
Beta=math.exp(2*alphao)*math.cosh(2*alphao)-2*A**2*(math.sinh(2*alphao))**2
Beta=1/2.*math.acos(Beta/(math.exp(2*alphao)))
print('\n part (c)')
print'%s %.2f %s'%('\n Beta = ',Beta,' rad')

pat (a)

part (b)

Maximum tensile stress = 812.08  MPa

part (c)



## Ex2-pg529¶

In [3]:
#calculate the  magnitude of the tension working load
## initialization of variables
import math
S_u=420. ##MPa
SF=4.00
D=110. ##mm
d=50.0 ##mm
w=20. ##mm
rho=10.0 ##mm
SF=4.0
##calculations
t=(D-d)/2.
tr=t/rho
rd=rho/d
S_cs=1+2.*math.sqrt(tr)
A=w*d
Pf=S_u*A/1.83
P=Pf/SF
print'%s %.2f %s'%('P = ',P/10**3,' kN')

P =  57.38  kN


## Ex3-pg529¶

In [4]:
#calculate the safety factor and maximum stress in the beam at the fillet
## initialization of variables
import math
##part(a)
H=200. ##mm
h=100. ##mm
rho=10. ##mm
Sigma_u=250. ##MPa
P=1.5 ##kN
L=1.4 ##m
b=40 ##mm
P=P*10**3
L=L*10**3
Hr=H/h
rh=rho/h
S_cc=1.77
c=h/2.
I=b*h**3./12.
S_max=S_cc*P*L*c/I
print('part (a)')
print'%s %.2f %s'%('\n Flexural design stress = ',S_max,' MPa')
##part (b)
SF=Sigma_u*I/(S_cc*P*L*c)
print('\n part (b)')
print'%s %.2f %s'%('\n SF = ',SF,'')

part (a)

Flexural design stress =  55.75  MPa

part (b)

SF =  4.48