# Ch-23, Thermal properties & materials¶

## example-23.1 page no-681¶

In [5]:
from __future__ import division
#given
#length of glass rod and steel rod is equal at 273 K and differ by 1.2 mm at 373 K
T1=273  #K
T2=373  #K
#coefficients of linear expansion af glass and steel are
alphaG=8*10**-6  #per degree C
alphaS=12*10**-6   #per degrees C
#we know that
#lT2=lT1*(1+alpha(T2-T1))
#so for glass rod
#l100G=l0*(1+(alpha1)*(T2-T1))
#similarly for steel rod
#l100G=1.0008*l0   ----(1)
#l100S=lo*(1+(alpha2)*(T2-T1))
#l100S=1.0012*l0  ----------(2)
#we have given that
#l100S-l100G=1.2 mm  ---(3)
#from 1 and 2 put in 3, we get
#1.0012*l0-1.0008*l0=1.2
#so
l0=1.2*10**-3/(0.0012-.0008)  #m
print " the length of rod at 0 degrees celcius is %.f m"%(l0)
the length of rod at 0 degrees celcius is 3 m

## example-23.2 page no-684¶

In [7]:
#given
#coefficient of linear expansion of Cu and steel are
alphaCu=18*10**-6  #cm/cm/degree C
alphaSteel=14*10**-6  #cm/cm/degree C
#young's modulus of elasticity
ECu=106*10**9  #Pa
ESteel=200*10**9  #Pa
#part(a)
#since alphaCu>alphaSteel
#so steel will contract less.
print " Hence strip will bend in the direction of copper"
#part(b)
#annealing temp
T2=530  #degrees celcius
#room temp
T1=30  #degrees celcius
#difference in temp
T=T2-T1  #degrees celcius
#differnce in values of coefficient of linear expansion
alpha=alphaCu-alphaSteel  #cm/cm/degrees celcius
#differential contraction
contraction=T*alpha  #cm/cm
#since the two metals in the strip have equal dimension
#sigmaCu=sigmaSteel=sigma
#strain in copper
#eCu=sigma/Ecu    -----(1)
#strain in steel
#eSteel=sigma/ESteel   ---(2)
#sum of strains given by equation 1 and 2
sigma=contraction*ESteel*ECu/(ESteel+ECu)*10**-6  # MPa
print " the stresses in each metal is %0.3f MPa"%(sigma)
Hence strip will bend in the direction of copper
the stresses in each metal is 138.562 MPa