from __future__ import division
#given
#part(a)
#thickness of GRPS sheet
ts=3 #mm
#depth of this skin
tc=24 #mm
#breadth of skin
b=100 #mm
d=(ts+tc)/2 #mm
#moduli of polyster skin and foamare
Es=7000 #N/mm**2
Ec=20 #N/mm**2
#values of Is and Ic are
Is=2*(b*ts**3/12+b*ts*d**2) #mm**4
Ic=b*tc**3/12 #mmm**4
#as we know that flexure rigidity is D, flexure rigidity of skin is Ds and that of core is Dc
#D=Ds+Dc
D=2*Es*Is+Ec*Ic #mm**2 for two skins ---------------(1)
#part(b)
ts1=6 #mm
#Ds1=Es*Is1
Ds1=Es*b*ts1**2/12 #Nmm**2 --------------(2) (there is calculation mistake I myself have checked from calculator too)
#ratio of (1) and (2) is R (let)
R=D/Ds1
print "flexure rigidity of sandwich beam is %d times more"%(R)
from __future__ import division
#given
#young's modulus of aluminium, iron and boron are resp
#here we are representing a,b and c for aluminum, iron and boron resp.
Ea=71*10**9 #Pa
Eb=210*10**9 #Pa
Ec=440*10**9 #Pa
#as we know that Ec=Ef*Vf+Em*Vm where Ef and Em are the young's modulus of fibre and matrix resp.
#so we get
#210=71*Va+440*Vb --------------(1)
#assumin void volume is zero and we know that
#Va+Vb=1
#so Vb=1-Va ---------------(2)
#on solving 1 and 2 we get
Va=31.8
Vb=68.2
#ratio of Va and Vb
R=Va/Vb
print "the volume ratio of aluminium and boron in aluminium boron composite %0.2f"%(R)
#given
#elastic moduli of carbon fibre and epoxy resin
Ef=430 #GPa
E=3.6 #/GPa
#modular ratio
Pf=Ef/E
#part(a)
Vf1=0.15
Vm1=1-Vf1
R1=Vm1/Vf1
Pc1=Pf+R1
volume_fibre1=Pf/Pc1
#part(b)
Vf2=0.65
Vm2=1-Vf2
R2=Vm2/Vf2
Pc2=Pf+R2
volume_fibre2=Pf/Pc2
print """the fraction of load carried by fibre in carbon-epoxy matrix composite containing 15 percent fibre by volume is %0.3f
that by 65 percent fibre by volume is %0.3f"""%(volume_fibre1,volume_fibre2)
#given
Vf=.65
Vm=1-Vf
#part(a)
#longitudinal strength is determined by
#sigmac=sigmaf*Vf+sigmam*Vm
#here according to table given in question we have
sigmaf=2.8
sigmam=0.0025
sigmac=sigmaf*Vf+sigmam*Vm #GPa
#part(b)
#longitudinal modulus is given by
#Ec=Ef*Vf+Em*Vm
#here according to table
Ef=130
Em=3.5
Ec=(Ef*Vf+Em*Vm) #GPa
#part(c)
#transverse modulus is given by
#1/EC=Vf/Ef+Vm/Em
EC=(Vf/Ef+Vm/Em)**-1 #GPa
#part(d)
#poisson ratio
#nuLT=nuf*Vf+num*Vm
#here according to the table
nuf=0.34
num=0.36
nuLT=nuf*Vf+num*Vm
#part(e)
#shear modulus
#1/Glt=Vf/Gf+Vm/Gm
# here according to the table
Gf=2.2
Gm=1.2
GLT=(Vf/Gf+Vm/Gm)**-1 #GPa
print """the longitudinal strength is %0.3f GPa
the logitudinal modulus is %0.3f GPa
the transvrse modulus is %0.3f GPa
the poissons ratio is %0.3f
shear modulus is %0.3f GPa"""%(sigmac,Ec,EC,nuLT,GLT)
#given
#content of polyster
Vm=0.45
Vf=1-Vm
#longitudinal strength and modulus are calculated using following formulas
#sigmac=nu*sigmaf*Vf+singmam*Vm
#Ec=nu*Ef*Vf+Em*Vm
#according to the table given in the ques
Ef=240 #GPa
Em=4.5 #GPa
sigmaf=1.7 #GPa
sigmam=0.0029 #GPa
Gf=1.7
Gm=1.4
nuf=0.28
num=0.32
#values of nu
nu1=1
nu2=1/2
nu3=1/4
nu4=3/8
nu5=1/6
#now longitudinal strength for differnt nu
sigmac1=nu1*(sigmaf*Vf+sigmam*Vm) #GPa
sigmac2=nu2*(sigmaf*Vf+sigmam*Vm) #GPa
sigmac3=nu3*(sigmaf*Vf+sigmam*Vm) #GPa
sigmac4=nu4*(sigmaf*Vf+sigmam*Vm) #GPa
sigmac5=nu5*(sigmaf*Vf+sigmam*Vm) #GPa
#liongitudina modulus for differnt nu
EC1=nu1*(Ef*Vf+Em*Vf) #GPa
EC2=nu2*(Ef*Vf+Em*Vf) #GPa
EC3=nu3*(Ef*Vf+Em*Vf) #GPa
EC4=nu4*(Ef*Vf+Em*Vf) #GPa
EC5=nu5*(Ef*Vf+Em*Vf) #GPa
print """longitudinal strength and longitudinal modulus for nu=1 is %0.3f GPa and %0.3f GPa
longitudinal strength and longitudinal modulus for nu=1/2 is %0.3f GPa and %0.3f GPa
longitudinal strength and longitudinal modulus for nu=1/4 is %0.3f GPa and %0.3f GPa
longitudinal strength and longitudinal modulus for nu=3/8 is %0.3f GPa and %0.3f GPa
longitudinal strength and longitudinal modulus for nu=1/6 is %0.3f GPa and %0.3f GPa"""%(sigmac1,EC1,sigmac2,EC2,sigmac3,EC3,sigmac4,EC4,sigmac5,EC5)