# Ch-14, Composite Material¶

## example-14.1 page no-429¶

In :
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)

flexure rigidity of sandwich beam is 733 times more


## example-14.2 page no-441¶

In :
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)

the volume ratio of aluminium and boron in aluminium boron composite 0.47


## example-14.3 page no-442¶

In :
#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)

the fraction of load carried by fibre in carbon-epoxy matrix composite containing 15 percent fibre by volume is 0.955
that by 65 percent fibre by volume is 0.996


## example-14.4 page no-442¶

In :
#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)

the longitudinal strength is 1.821 GPa
the logitudinal modulus is 85.725 GPa
the transvrse modulus is 9.524 GPa
the  poissons ratio is 0.347
shear modulus is 1.703 GPa


## example-14.5 page no-444¶

In :
#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)

longitudinal strength and longitudinal modulus for nu=1 is 0.936 GPa and 134.475 GPa
longitudinal strength and longitudinal modulus for nu=1/2 is 0.468 GPa and 67.237 GPa
longitudinal strength and longitudinal modulus for nu=1/4 is 0.234 GPa and 33.619 GPa
longitudinal strength and longitudinal modulus for nu=3/8 is 0.351 GPa and 50.428 GPa
longitudinal strength and longitudinal modulus for nu=1/6 is 0.156 GPa and 22.412 GPa