Chapter9-Beams On Elastic Foundations

Ex1-pg273

In [3]:
import math
#calculate length and maxi force per unit of length between beam & foundation 
w=0.1 ##m
d=0.115 ##m
l=4. ##m
p=175. ##kN/m
k=14*10**6. ##Pa
E=200*10**9. ##Pa
I=(0.1*(0.15)**3.)

##deltav=(p/2*k)*derivative(x)*beta*exp**(betax)*(cos beta(x)+sin beta(x))
##vA=(p/2k)*(2-exp**(betaa)*cos betaa - exp**(betab)*cos betab)

beta=(k/(4.*E*I/12.))**(0.25)
print'%s %.2f %s'%("in meter inverse is= ",beta,"")

vmax=(p*(2-(-0.0345)-(0.0345)))/(2*14000.)
print'%s %.2f %s'%("in meter is= ",vmax,"")
z=k*vmax
print'%s %.2f %s'%("maxi force per unit of length between beam & foundation in kN/m is= ",z,"")

## Ans varies due to round of error
in meter inverse is=  0.89 
in meter is=  0.01 
maxi force per unit of length between beam & foundation in kN/m is=  175000.00 

Ex4-pg279

In [1]:
import math
#find foundation modulus of the equivalent continuous elastic support in Pa
a=1.5 ##m
E=206.8*10**9 ##Pa
K=10000. ##N/m
I=6*10**-6 ##m**4
P=6700 ##N
c=0.05

k=K/a
print'%s %.2f %s'%("foundation modulus of the equivalent continuous elastic support in Pa is=",k,"")

beta=(k/(4.*E*I))**(1/4.)
print(beta)

##sigmamax=(M*c/I)=(P*c/4*beta*I)
sigmamax=((P*c)/(4.*beta*I))
print'%s %.2f %s'%("in Pa is=",sigmamax,"")

vmax=(P*beta)/(2.*k)
print'%s %.2f %s'%("in meter is=",vmax,"")
foundation modulus of the equivalent continuous elastic support in Pa is= 6666.67 
0.191441787744
in Pa is= 72911632.81 
in meter is= 0.10