Chapter 16: Bending of open and closed thin-walled beams

Example 16.1 Pg. No.456

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# Fig 16.6 Dimensions
from __future__ import division
import math

#variable declaration
l=300            #length of I section (mm)
b=200            #width of I section beam (mm)
w1=25            #width of center section of I beam (mm)
w2=20            #width of upper and lower section of I beam (mm)
M=100*10**6      #moment applied in vertical plane (N*mm)

#second moment of area Ixx
# Ixx=b*l^3/12
Ixx=b*l**3/12-(b-w1)*(l-w2-w2)**3/12
print "\nSecond moment of area of I section beam = %5.3e mm^4"%(Ixx)

#sigma_z=My/I  reference 16.9
# @ 
y=150
sigma_z=M*y/Ixx
print "\ndirect stress at the top of the I section (y=150) = %3.2f N/mm^2 (compression)"%(sigma_z)

# @
y=-150
sigma_z=M*y/Ixx
print "\ndirect stress at the bottom of the I section (y=-150) = %3.2f N/mm^2 (tension)"%(sigma_z)

Second moment of area of I section beam = 1.937e+08 mm^4

direct stress at the top of the I section (y=150) = 77.45 N/mm^2 (compression)

direct stress at the bottom of the I section (y=-150) = -77.45 N/mm^2 (tension)

Example 16.2 Pg. No.457

In [17]:
# Fig 16.6 reference
from __future__ import division
import math

#variable declaration
l=300            #length of I section (mm)
b=200            #width of I section beam (mm)
w1=25            #width of center section of I beam (mm)
w2=20            #width of upper and lower section of I beam (mm)
M=100*10**6      #moment applied in vertical plane (N*mm)

# second moment of area
# Iyy=wb^3/12
Iyy=2*20*200**3/12+260*25**3/12
print "\nSecond moment of area of I section beam = %5.3e mm^4"%(Iyy)

#sigma_z=Mx/I  
# @ 
x=100
sigma_z=M*x/Iyy
print "\ndirect stress at the top of the I section (y=150) = %3.2f N/mm^2 (compression)"%(sigma_z)

# @
x=-100
sigma_z=M*x/Iyy
print "\ndirect stress at the bottom of the I section (y=-150) = %3.2f N/mm^2 (tension)"%(sigma_z)

Second moment of area of I section beam = 2.701e+07 mm^4

direct stress at the top of the I section (y=150) = 370.30 N/mm^2 (compression)

direct stress at the bottom of the I section (y=-150) = -370.30 N/mm^2 (tension)

Example 16.3 Pg.No.458

In [38]:
from __future__ import division
import math

#variable declaration
l=300            #length of I section (mm)
b=200            #width of I section beam (mm)
w1=25            #width of center section of I beam (mm)
w2=20            #width of upper and lower section of I beam (mm)
M=100*10**6      #moment applied in vertical plane (N*mm)
theta=30         #angle at which bending moment is applied(degree)

Mx=M*math.cos(math.radians(30))
My=M*math.sin(math.radians(30))

# sigma_z=Mx/Ixx*y+My/Iyy*x
# @top left hand corner
y=150
x=-100
sigma_z=Mx/Ixx*y-My/Iyy*x  
print "\ndirect stress at the top left hand corner = %3.1f N/mm^2 (tension)"%(sigma_z)

# @ top right hand corner
x=100
y=150
sigma_z=Mx/Ixx*y-My/Iyy*x  
print "\ndirect stress at the top right hand corner = %3.1f N/mm^2 (compression)"%(sigma_z)

alpha=math.atan(My*Ixx/Mx/Iyy)
print "\ninclination = %3.1f degree\n"%(alpha*180/math.pi)

direct stress at the top left hand corner = 252.2 N/mm^2 (tension)

direct stress at the top right hand corner = -118.1 N/mm^2 (compression)

inclination = 76.4 degree

Example 16.4 Pg.No.466

In [62]:
# reference Fig 16.13
import math
from __future__ import division
#variable declaration
l=80           #length of one section (mm)
w=8            #thickness of each section (mm)
b1=40          #width of one section (mm)
b2=80          #width of other section (mm)
Mx=1500*10*3   #bending moment (N.mm)
My=0

#centroid
# in this example C is taken at top surface x axis aligned to AB and y axis is 
# aligned to E surface 
y_bar=((b1+b2)*w*w/2+l*w*(w+l/2))/(l*w+(b1+b2)*w)
print "\ny coordinate of centroid = %3.1f mm"%(y_bar)

x_bar=((b1+b2)*w*((b1+b2)/2-b1+w/2)+l*w*(w/2))/(l*w+(b1+b2)*w)
print "\nx coordinate of centroid = %3.1f mm"%(x_bar)

#second area of moment
#IB=IC+Ab^2              IB=moment about any point,IC=moment about centroid, b=distace between both points
Ixx=(b1+b2)*w**3/12+(b1+b2)*w*(y_bar-w/2)**2+w*l**3/12+l*w*(w+l/2-y_bar)**2
print "\nsecond moment of area about x axis = %3.2e mm^4"%(Ixx)

Iyy=w*(b1+b2)**3/12+(b1+b2)*w*((b1+b2)/2-b1+w/2-x_bar)**2+l*w**3/12+l*w*(x_bar-w/2)**2
print "\nsecond moment of area about y axis = %3.2e mm^4"%(Iyy)

Ixy=(b1+b2)*w*(y_bar-w/2)*((b1+b2)/2-b1+w/2-x_bar)+l*w*(w+l/2-y_bar)*(x_bar-w/2)
print "\nsecond moment of area about x and y axis = %3.2e mm^4"%(Ixy)

Mx=15*10**5
def f(x,y):
        return Mx*(Iyy*y-Ixy*x)/(Ixx*Iyy-Ixy**2)
sigma_z_max= f(-8,-66.4)
print "\nmaximum direct shear stress = %3.0f N/mm^2 (compressive)\n"%(sigma_z_max)

y coordinate of centroid = 21.6 mm

x coordinate of centroid = 16.0 mm

second moment of area about x axis = 1.09e+06 mm^4

second moment of area about y axis = 1.31e+06 mm^4

second moment of area about x and y axis = 3.38e+05 mm^4

maximum direct shear stress = -96 N/mm^2 (compressive)
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