# Chapter 25: Laminated Composite Structures¶

## Example 25.1 Pg.No.653¶

In [10]:
from __future__ import division
import math

#variable declaration
Ee=5000               #youngs modulus of epoxy (N/mm^2)
Ec=200000             #youngs modulus of carbon (N/mm^2)

# E1=Ef*Af/A+Em+Am/A        equation 25.4
A=50*80               #total area (mm^2)
Ae=40*80              #area of epoxy (mm^2)
Ac=10*80              #area of carbon(mm^2)
L=500                  #length of bar(mm)
vc=0.3                 #poisson ratio
ve=0.2

#effective youngs modulus
E1=(Ee*Ae+Ec*Ac)/A

epsilon1=sigma1/E1

delta1=epsilon1*L
v1t=(ve*Ae+vc*Ac)/A
epsilon_t=-v1t*epsilon1

thickness=L
delta_t=-epsilon_t*(thickness)

sigma_m=Ee*epsilon1
sigma_f=Ec*epsilon1
print "stresses in epoxy = %2.2f N/mm^2\n"%(sigma_m)
print "stresses in carbon = %2.2f N/mm^2\n"%(sigma_f)

stresses in epoxy = 2.84 N/mm^2

stresses in carbon = 113.64 N/mm^2



## Example 25.2 Pg.No.657¶

In [12]:
from __future__ import division
import math

tau=40            #shear stress(N/mm^2)
G=5000            #shear modulus(N/mm^2)
v1t=0.3            #poisson ratio
Et=80000            #youngs modulus in transverse direction(N/mm^2)
E1=120000           #youngs modulus in longitudinal direction(N/mm^2)
sigma_x=50          #direction stress in x direciton
sigma_y=25                           #in y direction

vt1=v1t*Et/E1 #minor poisson ratio

epsilon1=sigma_x/E1-vt1*sigma_y/Et
epsilont=sigma_y/Et-v1t*sigma_x/E1
print "direct strain in x direction = %2.2e \n"%(epsilon1)
print "direct strain in y direction = %2.2e \n"%(epsilont)

gama1t=tau/G
print "shear strain in the ply = %2.2e \n"%(gama1t)

direct strain in x direction = 3.54e-04

direct strain in y direction = 1.88e-04

shear strain in the ply = 8.00e-03



## Example 25.3 Pg.No.661¶

In [23]:
from __future__ import division
import math
import numpy as np

El=150000          #youngs modulus(N/mm^2)
Et=90000          #youngs modulus(N/mm^2)
Glt=5000          #shear modulus(N/mm^2)
vlt=0.3            #poisson ratio
theta=45             #longitudinal plane inclination

s11=1/El
s22=1/Et
s12=-vlt/El
s33=1/Glt

a=np.array([[53.45,-46.55,1],[-46.55,53.45,0],[-2.2,-2.2,0]])
b=np.array([60,40,0])
x=np.dot(a,b)
print "strain in the x direction = %4.4e\n"%(x[0]*10**-6)
print "strain in the y direction = %4.4e\n"%(x[1]*10**-6)
print "shear strain in the xy plane = %4.4e\n"%(x[2]*10**-6)

strain in the x direction = 1.3450e-03

strain in the y direction = -6.5500e-04

shear strain in the xy plane = -2.2000e-04



## Example 25.4 Pg.No.664¶

In [28]:
from __future__ import division
import math
import numpy as np

l1=150
t1=1
l2=100            #dimensions shown in Fig 25.10 (mm)
t2=2
Ef=60000            #youngs modulus of flanges(N/mm^2)
Ew=20000            #youngs modulus of web(N/mm^2)

#sum of b*t*E
sum_btE=2*l2*t2*Ef+l1*t1*Ew

epsilon_z=P/sum_btE           #equ 25.37
P_flange=epsilon_z*l2*t2*Ef
P_web=epsilon_z*l1*t1*Ew
print "axial load in flange = %2.2f kN\n"%(P_flange/1000)
print "axial load in web = %2.2f kN\n"%(P_web/1000)

axial load in flange = 17.78 kN

axial load in web = 4.44 kN



## Example 25.5 Pg.No.666¶

In [39]:
from __future__ import division
import math
import numpy as np

l1=100
l2=50
t1=1         #dimension shown in Fig 25.11 (mm)
t2=2
Ef=50000         #youngs modulus of flange(N/mm^2)
Ew=15000          #youngs modulus of web(N/mm^2)
Mx=10**6           #bending moment (N.mm)
My=0

Ixx=2*Ef*l2*t2*l2**2+Ew*t1*l1**3/12
Iyy=Ef*t2*l1**3/12
Ixy=Ef*l2*t2*(50)*(50)+Ef*l2*t2*(-50)*(-50)

x=50
y=50      #point 1
Ez=50000

sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)
print "direct stress at point 1 = %3.1f N/mm^2\n"%(sigma_z)
x=0
y=50
sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)
print "direct stress at point 2 = %3.1f N/mm^2\n"%(sigma_z)

x=0
y=50
Ez=15000
sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)
print "direct stress at point 2 in the web = %3.1f N/mm^2\n"%(sigma_z)

direct stress at point 1 = 102.6 N/mm^2

direct stress at point 2 = -51.3 N/mm^2

direct stress at point 2 in the web = -15.4 N/mm^2



## Example 25.6 Pg.No.668¶

In [57]:
from __future__ import division
import math
import numpy as np
from sympy import symbols, integrate

s=symbols('s')

l12=250
l23=300
t31=t12=2                 #dimensions shown in Fig 25.12 (mm)
t23=1.5

E12=E31=45000          #youngs modulus of sides given in name(N/mm^2)
E23=20000

Ixx=2*E12*t12*l12**3*(l23/2/l12)**2/12+E23*t23*l23**3/12
alpha=math.asin(l23/2/l12)

Sx=0
Sy=2*10**3
q12=-E12*Sy/Ixx*integrate(-2*s,(s,0,250))*math.sin(alpha)
print "shear flow at point 2 in the flange 12 = %2.2f N/mm\n"%(q12)

q23=-E23*Sy/Ixx*integrate(-225+1.5*s,(s,0,300))+22.2
print "shear flow at point 3 in the flange 23 = %2.2f N/mm\n"%(q23)
q0=14.2

q12=-E12*Sy/Ixx*integrate(-2*s,s)*math.sin(alpha)-q0
print "shear flow in the flange 12 and 31 "
print q12

q23=-E23*Sy/Ixx*integrate(-225+1.5*s,s,)+22.2-q0
print "\nshear flow in the web 23"
print q23

shear flow at point 2 in the flange 12 = 22.22 N/mm

shear flow at point 3 in the flange 23 = 22.20 N/mm

shear flow in the flange 12 and 31
0.000355555555555556*s**2 - 14.2

shear flow in the web 23
-0.000197530864197531*s**2 + 0.0592592592592593*s + 8.0


## Example 25.7 Pg.No.671¶

In [66]:
from __future__ import division
import math
import numpy as np
from sympy import symbols, integrate

l1=200
t1=2
l2=100         #dimensions shown in Fig 25.13 (mm)
t2=1
T=10*10**6       #torque applied (N.mm)
Gl=20000          #laminate shear modulus(N/mm^2)
Gw=35000            #web shear modulus(N/mm^2)
A=l1*l2

q=T/2/A

#from eqn 25.47
int_ds_by_Gt=2*l1/Gl/t1+2*l2/Gw/t2

#Let's say we want to calculate warping at point 1
#warping at mid of web is zero (W0=0) integrate eqn 25.47 from
#mid of web to point 1
W0=0
A0s=50*100
W1=W0+q*(l2/2/Gw/t2-int_ds_by_Gt/A*A0s)
print "warping at point 1 is = %2.2f mm\n"%(W1)

warping at point 1 is = -0.63 mm



## Example 25.8 Pg.No.673¶

In [74]:
from __future__ import division
import math
import numpy as np
from sympy import symbols, integrate

l1=50
t1=2.5
l2=25             #dimension shown in Fig 18.12(mm)
t2=1.5

Gl=20000
Gw=15000           #shear modulus (N/mm^2)
T=10*10**3         #torque applied (N.mm)
AR=25*25/2
GJ=2*Gl*l2*t2**3/3+Gw*l1*t1**3/3
#eqn 25.49
dO_dz=T/GJ

t_max12=2*Gl*(t2/2)*dO_dz
t_max23=2*Gw*(t1/2)*dO_dz
print "maximum shear stress in the web = %2.2f N/mm^2\n"%(t_max12)
print "maximum shear stress in the laminate = %2.2f N/mm^2\n"%(t_max23)

W1=-2*AR*dO_dz
print "warping at point 1 = %2.2f mm\n"%(W1)

maximum shear stress in the web = 59.63 N/mm^2

maximum shear stress in the laminate = 74.53 N/mm^2

warping at point 1 = -1.24 mm


In [ ]: