Chapter 4:Stresses in Beams¶
Example 4.4.1,Page no.130¶
In [1]:
import math
#Initilization of Variables
L=5000 #mm #Length of Beam
a=2000 #mm #Length of start of beam to Pt Load
b=3000 #mm #Length of Pt load to end of beam
A=150*250 #m**2 #Area of beam
b=150 #mm #Width of beam
d=250 #mm #Depth of beam
sigma=10#N/mm**2 #stress
l=2000 #m #Load applied from one end
#Calculations
#Moment of Inertia
I=1*12**-1*b*d**3 #m**4
#Distance from N.A to end
y_max=d*2**-1 #m
#Section Modulus
Z=1*6**-1*b*d**2 #mm**3
#Moment Carrying Capacity
M=sigma*Z #N-mm
#Let w be the Intensity of the Load in N/m,then Max moment
#M_max=w*L**2*8**-1 #N-mm
#After substituting values and further simplifying we get
#M_max=w*25*100*8**-1
#EQuating it to moment carrying capacity,we get max intensity load
w=M*(25*1000)**-1*8*10**-3
#Part-2
#Let P be the concentrated load,then max moment occurs under the load and its value
#M1=P*a*b*L**-1 #N-mm
#Equting it to moment carrying capacity we get
P=M*1200**-1*10**-3 #N
#Result
print"Max Intensity of u.d.l it can carry",round(w,3),"KN-m"
print"MAx concentrated Load P apllied at 2 m from one end is",round(P,3),"KN"
Example 4.4.2,Page no.131¶
In [1]:
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
D=70 #mm #External Diameter
t=8 #mm #Thickness of pipe
L=2500 #mm #span
sigma=150 #N/mm**2 #stress
#Calculations
#Internal Diameter
d=D-2*t #mm
#M.I Of Pipe
I=pi*64**-1*(D**4-d**4) #mm**4
y_max=D*2**-1 #mm
Z=I*(y_max)**-1 #mm**3
#Moment Carrying capacity
M=sigma*Z #N*mm
#Max moment int the beam occurs at the mid-span and is equal to
#m=P*L*4**-1
#Equating Max moment to moment carrying capacity we get,
#M=P*2.5*L*4**-1
#After substituting and simplifying we get
P=4*M*(L)**-1*10**-3 #N
#Result
print"Max concentrated load that can be applied at the centre of span is",round(P,3),"KN"
Example 4.4.3,Page no.132¶
In [32]:
import math
import numpy as np
#Initilization of Variables
#Plate dimensions
b1=240 #mm
d1=12 #mm
#Flange Dimensions
b2=180 #mm
d2=10 #mm
#web
b3=8 #mm
d3=480 #mm
D=500 #mm
sigma=150 #N/mm**2 #Stress
L=3000 #mm #span
#Calculations
#C.G of plate
y_bar1=(b1*d1*(d1*2**-1+D))*(b1*d1)**-1 #m
#C.G of top flange
y_bar2=(b2*d2*(D-d2*2**-1))*(b2*d2)**-1 #m
#C.G of web
y_bar3=(b3*d3*(d3*2**-1+d2))*(b3*d3)**-1 #m
#C.G of bottom flange
y_bar4=(b2*d2*(d2*2**-1))*(b2*d2)**-1 #m
#C.G of Body
Y=((b1*d1*(d1*2**-1+D))+(b2*d2*(D-d2*2**-1))+(b3*d3*(d3*2**-1+d2))+(b2*d2*(d2*2**-1)))*((b1*d1)+(b2*d2)+(b3*d3)+(b2*d2))**-1
#Moment of Inertia
I1=(1*12**-1*b1*d1**3+b1*d1*(d1*2**-1-round(Y,3)+D)**2) #mm**4
I2=(1*12**-1*b2*d2**3+b2*d2*(D-d2*2**-1-round(Y,3))**2) #mm**4
I3=(1*12**-1*b3*d3**3+b3*d3*(d3*2**-1-round(Y,3))**2) #mm**4
I4=(1*12**-1*b2*d2**3+b2*d2*(round(Y,3)-d2*2**-1)**2) #mm**4
I=(I1+I2+I3+I4)*10**-8 #mm*4
#Moment of resistance
MR=sigma*I*Y**-1
#MaX mOMENT PRODUCED after simplifying we get
#MM=4.5*w
#After equating Moment of resistance to max moment we get
w=198.769*4.5**-1 #KN-m
#Result
print"Moment of Resistance is",round(MR,2),"KN-mm"
print"Load the section can carry is",round(w,3),"KN/m"
Example 4.4.4,Page no.134¶
In [5]:
import math
#Initilization of Variables
#Flange (Top)
b1=80 #mm #Width
t1=40 #mm #Thickness
#Flange (Bottom)
b2=160 #mm #width
t2=40 #mm #Thickness
#web
d=120 #mm #Depth
t3=20 #mm #Thickness
D=200 #mm #Overall Depth
sigma1=30 #N/mm**2 #Tensile stress
sigma2=90 #N/mm**2 #Compressive stress
L=6000 #mm #Span
#Calculations
#Distance of centroid from bottom fibre
y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm
#Moment of Inertia
I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2
#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively
y_t=y_bar #mm
y_c=D-y_bar #mm
#Moment carrying capacity considering Tensile strength
M1=sigma1*I*y_t**-1*10**-6 #KN-m
#Moment carrying capacity considering compressive strength
M2=sigma2*I*y_c**-1*10**-6 #KN-m
#Max Bending moment in simply supported beam 6 m due to u.d.l
#M_max=w*L*10**-3*8**-1
#After simplifying further we get
#M_max=4.5*w
#Now Equating it to Moment carrying capacity, we get load carrying capacity
w=M1*4.5**-1 #KN/m
#Result
print"Max Uniformly Distributed Load is",round(w,3),"KN/m"
Example 4.4.5,Page no.136¶
In [26]:
import math
from scipy.integrate import *
#Initilization of Variables
#Flanges
b=200 #mm #Width
t=25 #mm #Thickness
D1=500 #mm #Overall Depth
t2=20 #mm #Thickness of web
d=450 #mm #Depth of web
#Calculations
#Consider,Element of Thickness "y" at Distance "dy" from N.A
#Let Bending stress "sigma_max"
#Stress on the element
#sigma=y*(D*2**-1)*sigma_max ..............(1)
#Area of Element
#A=b*dy .................................(2)
#Force on Element
#F=y*250**-1*sigma_max*b*dy
#Let M be the Moment of resistance
#M=y*250**-1*sigma_max*b*dy*y
#Moment of Resistance of top flange after simplification we gget
#M.R=2258333.3*f
#M.I of I section
I=1*12**-1*(b*D1**3-180*d**3)*10**-8
#Moment acting on section
#After simplifying we get
#M=2865833.3*f
#Percentage moment resistance
M1=2258333.3*2865833.3**-1*100
#Percentage moment resisted by web
M2=100-M1
#Result
print"Percentage Moment resisted by Flanges",round(M1,2),"%"
print"Percentage Moment resisted by web",round(M2,2),"%"
Example 4.4.6,Page no.137¶
In [2]:
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
#Flanges
b1=200 #mm #Width
t1=10 #mm #Thickness
#Web
d=380 #mm #Depth
t2=8 #mm #Thickness
D=400 #mm #Overall Depth
sigma=150 #N/mm**2
#Calculations
#Area
A=b1*t1+d*t2+b1*t1 #mm**2
#Moment of Inertia
I=1*12**-1*(b1*D**3-(b1-t2)*d**3)
#Bending Moment
M=sigma*I*(D*2**-1)**-1
#Square Section
#Let 'a' be the side
a=A**0.5
#Moment of Resistance of this section
M1=1*6**-1*a*a**2*sigma
X=M*M1**-1
#Rectangular section
#Let 'a' be the side and depth be 2*a
a=(A*2**-1)**0.5
#Moment of Rectangular secction
M2=1*6**-1*a*(2*a)**2*sigma
X2=M*M2**-1
#Circular section
#A=pi*d1**2*4**-1
d1=(A*4*pi**-1)**0.5
#Moment of circular section
M3=pi*32**-1*d1**3*sigma
X3=M*M3**-1
#Result
print"Moment of resistance of beam section",round(M,2),"mm"
print"Moment of resistance of square section",round(X,2),"mm"
print"Moment of resistance of rectangular section",round(X2,2),"mm"
print"Moment of resistance of circular section",round(X3,2),"mm"
Example 4.4.7,Page no.139¶
In [8]:
import math
#Initilization of Variables
F=12 #KN #Force at End of beam
L=2 #m #span
#Square section
b=d=200 #mm #Width and depth of beam
#Rectangular section
b1=150 #mm #Width
d1=300 #mm #Depth
#Calculations
#Max bending Moment
M=F*L*10**6 #N-mm
#M=sigma*b*d**2
sigma=M*6*(b*d**2)**-1 #N/mm**2
#Let W be the central concentrated Load in simply supported beam of span L1=3 m
#MAx Moment
#M1=W*L1*4**-1
#After Further simplifying we get
#M1=0.75*10**6 #N-mm
#The section has a moment of resistance
M1=sigma*1*6**-1*b1*d1**2
#Equating it to moment of resistance we get max load W
#0.75*10**6*W=M1
#After Further simplifying we get
W=M1*(0.75*10**6)**-1
#Result
print"Minimum Concentrated Load required to brek the beam",round(W,2),"KN"
Example 4.4.8,Page no.140¶
In [9]:
import math
#Initilization of Variables
L=3 #m #span
sigma_t=35 #N/mm**2 #Permissible stress in tension
sigma_c=90 #N/mm**2 #Permissible stress in compression
#Flanges
t=30 #mm #Thickness
d=250 #mm #Depth
#Web
t2=25 #mm #Thickness
b=600 #mm #Width
#Calculations
#Let y_bar be the Distance of N.A from Extreme Fibres
y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1
#Moment of Inertia
I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2
#Part-1
#If web is in Tension
y_t=y_bar #mm
y_c=d-y_bar #mm
#Moment carrying caryying capacity From consideration of tensile stress
M=sigma_t*I*(y_bar)**-1 #N-mm
#Moment carrying caryying capacity From consideration of compressive stress
M1=sigma_c*I*(y_c)**-1 #N-mm
#If w KN/m is u.d.l in beam,Max bending moment
#M=wl**2*8**-1
#After further simplifyng we get
#M=1.125*w*10**6 N-mm
w=M*(1.125*10**6)**-1 #KN
#Part-2
#If web is in compression
y_t2=178.299 #mm
y_c2=71.71 #mm
#Moment carrying caryying capacity From consideration of tensile stress
M2=sigma_t*I*(y_t2)**-1 #N-mm
#Moment carrying caryying capacity From consideration of compressive stress
M3=sigma_c*I*(y_c2)**-1 #N-mm
#Moment of resistance is M2
#Equating it to bending moment we get
#M2=1.125*10**6*w2
#After further simplifyng we get
w2=M2*(1.125*10**6)**-1
#Result
print"Uniformly Distributed Load carrying capacity if:web is in Tension",round(w,2),"KN"
print" :web is in compression",round(w2,3),"KN"
Example 4.4.9,Page no.141¶
In [3]:
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
b1=200 #mm #Width at base
b2=100 #mm #Width at top
L=8 #m Length
P=500 #N #Load
#Calculations
#Consider a section at y metres from top
#At this section diameter d is
#d=b2+y*L**-1*(b1-b2)
#After Further simplifying we get
#d=b2+12.5*y #mm
#Moment of Inertia
#I=pi*64**-1*d**4
#Section Modulus
#Z=pi*32**-1*(b1+12.5*y)**3
#Moment
#M=5*10**5*y #N-mm
#Let sigma be the fibre stress at this section then
#M=sigma*Z
#After sub values in above equation and further simplifying we get
#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1
#For sigma to be Max,d(sigma)*(dy)**-1=0
#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)
#After Further simplifying we get
#b2+12.5*y=37.5*y
#After Further simplifying we get
y=b2*25**-1 #m
#Stress at this section
sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1
#Result
print"Stress at Extreme Fibre is max",round(y,2),"m"
print"Max stress is",round(sigma,2),"N/mm**2"
Example 4.4.10,Page no.143¶
In [11]:
import math
#Initilization of Variables
H=10 #mm #Height
A1=160*160 #mm**2 #area of square section at bottom
L1=160 #mm #Length of square section at bottom
b1=160 #mm #width of square section at bottom
A2=80*80 #mm**2 #area of square section at top
L2=80 #mm #Length of square section at top
b2=80 #mm #Width of square section at top
P=100 #N #Pull
#Calculations
#Consider a section at distance y from top.
#Let the side of square bar be 'a'
#a=L2+y*(H)**-1*(b1-b2)
#After further simplifying we get
#a=L2+8*y
#Moment of Inertia
#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3
#After further simplifying we get
#I=a**4*12**-1
#Section Modulus
#Z=a**4*(12*a*(2)**0.5)**-1
#After further simplifying we get
#Z=2**0.5*a**3*(12)**-1 #mm**3
#Bending moment at this section=100*y N-mm
#M=100*10**3*y #N-mm
#But
#M=sigma*Z
#After sub values in above equation we get
#sigma=M*Z**-1
#After further simplifying we get
#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)
#For Max stress df*(dy)**-1=0
#After taking Derivative of above equation we get
#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)
#After further simplifying we get
y=80*16**-1 #m
#Max stress at this level is
sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1
#Result
print"Max Bending stress is Developed at",round(y,3),"m"
print"Value of Max Bending stress is",round(sigma,3),"N/mm**2"
Example 4.4.12,Page no.147¶
In [12]:
import math
#Initilization of Variables
b=200 #mm #Width of timber
d=400 #mm #Depth of timber
t=6 #mm #Thickness
b2=200 #mm #width of steel plate
t2=20 #mm #Thickness of steel plate
M=40*10**6 #KN-mm #Moment
#Let E_s*E_t**-1=X
X=20 #Ratio of Modulus of steel to timber
#Calculations
#let y_bar be the Distance of centroidfrom bottom most fibre
y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm
#Moment of Inertia
I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2
#distance of the top fibre from N-A
y_1=d+t-y_bar #mm
#Distance of the junction of timber and steel From N-A
y_2=y_bar-t #mm
#Stress in Timber at the top
Y=M*I**-1*y_1 #N/mm**2
#Stress in the Timber at the junction point
Z=M*I**-1*y_2
#Coressponding stress in steel at the junction point
Z2=X*Z #N/mm**2
#The stress in Extreme steel fibre
Z3=X*M*I**-1*y_bar
#Result
print"Stress in Extreme steel Fibre",round(Z3,2),"N/mm**2"
Example 4.4.13,Page no.149¶
In [13]:
import math
#Initilization of Variables
#Timber size
b=150 #mm #Width
d=300 #mm #Depth
t=6 #mm #Thickness of steel plate
l=6 #m #Span
#E_s*E_t**-1=20
#m=E_s*E_t**-1
m=20
sigma_timber=8 #N/mm**2 #Stress in timber
sigma_steel=150 #N/mm**2 #Stress in steel plate
#Let m*t=Y
Y=m*t #mm
L=(2*t+b)*m #mm #Width of flitched beam
#Calculations
#Due to synnetry cenroid,the neutral axis is half the depth
I=(1*12**-1*L*t**3+L*t*(b+t*2**-1)**2)*2+1*12**-1*(Y+b+Y)*d**3 #mm**4
y_max1=150 #mm #For timber
y_max2=156 #mm #For steel
#stress in steel
f_t1=1*m**-1*sigma_steel #N/mm**2
#Moment of resistance
M=f_t1*(I*y_max2**-1)
#load
w=8*M*(l**2)**-1*10**-6 #KN/m
#Result
print"Load beam can carry is",round(w,2),"KN/m"
Example 4.4.14,Page no.151¶
In [14]:
import math
#Initilization of Variables
L=6000 #mm #Span of beam
W=20*10**3 #N #Load
sigma=8 #N/mm**2 #Stress
b=200 #mm #Width of section
d=300 #mm #Depth of section
#Calculations
#let x be the distance from left side of beam
#Bending moment
#M=W*2**-1*x #Nmm .......(1)
#But M=sigma*Z ..........(2)
#Equating equation 1 and 2 we get
#W*2**-1*x=sigma*Z ............(3)
#Section Modulus
#Z=1*6*b*d**2 ...............(4)
#Equating equation 3 and 4 we get
#b*d**2=3*W*x*sigma**-1 .............(5)
#Beam of uniform strength of constant depth
#b=3*W*x*(sigma*d**2)
#When x=0
b=0
#When x=L*2**-1
b2=3*W*L*(2*sigma*d**2)**-1 #mm
#Beam with constant width of 200 mm
#We have
#d=(3*W*x*(sigma*d)**-1)**0.5
#thus depth varies as (x)**0.5
#when x=0
d1=0
#when x=L*2**-1
d2=(3*W*L*(2*sigma*200)**-1)**0.5 #mm
#Result
print"Cross section of rectangular beam is:",round(b2,2),"mm"
print" :",round(d2,2),"mm"
Example 4.4.15,Page no.154¶
In [15]:
import math
#Initilization of Variables
L=800 #mm #Span
n=5 #number of leaves
b=60 #mm #Width
t=10 #mm #thickness
sigma=250 #N/mm**2 #Stress
#Calculations
#section Modulus
Z=n*6**-1*b*t**2 #mm**3
#from the relation
#sigma*Z=M ...................(1)
#M=P*L*4**-1
#sub values of M in equation 1 we get
P=sigma*Z*4*L**-1*10**-3 #KN #Load
#Length of Leaves
L1=0.2*L #mm
L2=0.4*L #mm
L3=0.6*L #mm
L4=0.8*L #mm
L5=L #mm
#Result
print"Max Load it can take is",round(P,2),"KN"
print"Length of leaves:L1",round(L1,2),"mm"
print" :L2",round(L2,2),"mm"
Example 4.4.16,Page no.161¶
In [16]:
import math
import matplotlib.pyplot as plt
#Initilization of Variables
F=20*10**3 #N #Shear Force
#Tee section
#Flange
b=100 #mm #Width
t=12 #mm #Thickness
#Web
d=88 #mm #Depth
t2=12 #mm #Thicknes
D=100 #mm #Overall Depth
#Calculations
#Distance of C.G from Top Fibre
y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm
#Moment Of Inertia
I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4
#shear stress at bottom Flange
#Area above this level
A=b*t #mm**2
#C.G of this area from N-A
y2=y-t*2**-1
#Stress at bottom of flange
sigma=F*A*y2*(b*I)**-1 #N/mm**2
#sigma2 at same level but in web where width is 12 mm
sigma2=F*A*y2*(t2*I)**-1 #N/mm**2
#To find shear stress at N-A
X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3
sigma3=F*X*(t2*I)**-1 #N/mm**2
#Shear stress at top and bottom fibre is zero
#sigma4 and sigma5 are top and bottom fibre shear stress
sigma4=sigma5=0
#Result
print "The Shear Force and Bending Moment Diagrams are the results"
#Plotting the Shear Force Diagram
X1=[0,t,t,y,D]
Y1=[sigma4,sigma,sigma2,sigma3,sigma5]
Z1=[0,0,0,0,0]
plt.plot(X1,Y1,X1,Z1)
plt.xlabel("Length x in m")
plt.ylabel("Shear Force in kN")
plt.show()
Example 4.4.17,Page no.163¶
In [17]:
import math
import matplotlib.pyplot as plt
#Initilization of Variables
F=40*10**3 #N #shear Force
#I-section
#Flanges
b=80 #mm #Width of flange
t=20 #mm #Thickness
#Web
d=200 #mm #Depth
t2=20 #mm #Thickness
#Flange-2
b2=160 #mm #Width
t3=20 #mm #Thickness
D=240 #mm #Overall Depth
#Calculations
#Distance of N-A from Top Fibre
y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm
#Moment of Inertia
I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4
#Shear stress bottom of flange
sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2
#At same Level but in web
sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2
#for shear stress at N.A
X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3
sigma3=F*X*(t2*I)**-1 #N/mm**2
#Shear stress at bottom of web
X=b2*t3*((D-y)-t3*2**-1) #mm**3
#Stress at bottom of web
sigma4=F*X*(t2*I)**-1 #N/mm**2
#Stress at Lower flange
sigma5=F*X*(b2*I)**-1 #N/mm**2
#Result
print "The Shear Force Diagram is the result"
#Plotting the Shear Force Diagram
X1=[0,20,20,140,220,220,240]
Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]
Z1=[0,0,0,0,0,0,0]
plt.plot(X1,Y1,X1,Z1)
plt.xlabel("Length in mm")
plt.ylabel("Shear Force in N")
plt.show()
Example 4.4.18,Page no.164¶
In [18]:
import math
#Initilization of Variables
F=30*10**3 #N #Shear Force
#Channel Section
d=400 #mm #Depth of web
t=10 #mm #THickness of web
t2=15 #mm #Thickness of flange
b=100 #mm #Width of flange
#Rectangular Welded section
b2=80 #mm #Width
d2=60 #mm #Depth
#Calculations
#Distance of Centroid From Top Fibre
y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm
#Moment Of Inertia of the section about N-A
I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2
#Shear stress at level of weld
sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2
#Max Shear Stress occurs at Neutral Axis
X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1
sigma_max=F*X*((b+t)*I)**-1
#Result
print"Shear stress in the weld is",round(sigma,2),"N/mm**2"
print"Max shear stress is",round(sigma_max,2),"N/mm**2"
Example 4.4.19,Page no.165¶
In [4]:
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
#Wooden Section
b=300 #mm #Width
d=300 #mm #Depth
D=100 #mm #Diameter of Bore
F=10*10**3 #N #Shear Force
#Calculations
#Moment Of Inertia Of Section
I=1*12**-1*b*d**3-pi*64**-1*D**4
#Shear stress at crown of circle
sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1
#Let a*y_bar=X
X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3
#Shear Stress at Neutral Axis
sigma2=F*X*((b-D)*I)**-1 #N/mm**2
#Result
print"Shearing Stress at Crown of Bore",round(sigma,3),"N/mm**2"
print"Shear Stress at Neutral Axis",round(sigma2,3),"N/mm**2"
Example 4.4.20,Page no.166¶
In [20]:
import math
#Initilization of Variables
#flanges
b=200 #mm #width
t1=25 #mm #Thickness
#web
d=450 #mm #Depth
t2=20 #mm #thickness
D=500 #mm #Total Depth of section
#Calculations
#Moment Of Inertia of the section about N-A
I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4
#Consider an element in the web at distance y from y from N-A
#Depth of web section=225-y
#C.G From N-A
#y2=y+(((D*2**-1-t)-y)*2**-1)
#ay_bar for section at y
#Let ay_bar be X
#X=X1 be of Flange + X2 be of web above y
#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1
#After Sub values and Further simplifying we get
#X=1187500+10*(225**2-y**2)
#Shear stress at y
#sigma_y=F*(X)*(t2*I)**-1
#Shear Force resisted by the Element
#F1=F*X*t2*dy*(t2*I)**-1
#Shear stress resisted by web
#sigma=2*F*I**-1*(X)*dy
#After Integrating above equation and further simplifying we get
#sigma=0.9578*F
sigma=0.9578*100
#Result
print"Shear Resisted by web",round(sigma,2),"%"
Example 4.4.21,Page no.167¶
In [21]:
import math
#Initilization of Variables
#Wooden Beam
b=150 #mm #width
d=250 #mm #Depth
L=5000 #mm #span
m=11.2 #N/mm**2 #Max Bending stress
sigma=0.7 #N/mm**2 #Max shear stress
#Calculations
#Let 'a' be the distance from left support
#Max shear force
#F=R_A=W*(L-a)*L**-1
#Max Moment
#M=W*(L-a)*a*L**-1
#But M=sigma*Z
#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2 .....................(1)
#In Rectangular Section MAx stress is 1.5 times Avg shear stress
F=sigma*b*d*1.5**-1
#W*(L-a)*L**-1=F .....................(2)
#Dividing Equation 1 nad 2 we get
a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1
#Sub above value in equation 2 we get
W=(L-a)**-1*L*F*10**-3 #KN
#Result
print"Load is",round(W,2),"KN"
print"Distance from Left support is",round(a,2),"mm"
Example 4.4.22,Page no.168¶
In [22]:
import math
#Initilization of Variables
L=1000 #mm #span
#Rectangular Section
b=200 #mm #width
d=400 #mm #depth
sigma=1.5 #N/mm**2 #Shear stress
#Calculations
#Let AB be the cantilever beam subjected to load W KN at free end
#MAx shear Force
#F=W*10**3 #KN
#Since Max shear stress in Rectangular section
#sigma_max=1.5*F*A**-1
#After sub values and further simplifyng we get
W=1.5*b*d*(1.5*1000)**-1 #KN
#Moment at fixwed end
M=W*1 #KN-m
y_max=d*2**-1 #mm
#M.I
I=1*12**-1*b*d**3 #mm**3
#MAx Stress
sigma_max=M*10**6*I**-1*y_max
#Result
print"Concentrated Load is",round(sigma_max,2),"N/mm**2"
Example 4.4.24,Page no.170¶
In [23]:
import math
#Initilization of Variables
L=4000 #mm #span
#Rectangular Cross-section
b=100 #mm #Width
d=200 #mm #Thickness
F_per=10 #N/mm**2 #Max Bending stress
q_max=0.6 #N/mm**2 #Shear stress
#Calculations
#If the Load W is in KN/m
#Max shear Force
#F=w*l*2**-1 #KN
#After substituting values and further simplifying we get
#M=2*w #KN-m
#Max Load from Consideration of moment
#M=1*6**-1*b*d**2*F_per
#After substituting values and further simplifying we get
w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m
#Max Load from Consideration of shear stress
#q_max=1.5*F*(b*d)**-1 #N
#After substituting values and further simplifying we get
F=q_max*(1.5)*b*d #N
#If w is Max Load in KN/m,then
#2*w*1000=8000
#After Rearranging and Further simplifying we get
w2=8000*(2*1000)**-1 #KN/m
#Result
print"Uniformly Distributed Load Beam can carry is",round(w,2),"KN/m"