# Chapter 5:Bending Stresses in Beams¶

## Problem 5.1,Page no.121¶

In :
import math

#Initilization of variables

b=100 #mm #width of timber joist
d=200 #mm #depth of joist
L=3 #m #Length of beam
sigma=7 #KN/mm**2 #bending stress
w_1=5 #KN/mm**2 #unit weight of timber

#Calculations
w=0.1*0.2*1*5*100 #N/m #self weight of the joist
I_xx=1*12**-1*100*200**3 #mm**4 #M.I of section about N.A

#M=W*L+w*L**2*2**-1 #Max Bending moment
#Therefore,M=(3*W+450)

#using the relation M*I**-1=sigma*y**-1,we get

#Result
print"The Max value of Load applied is",round(W,2)

The Max value of Load applied is 1405.56


## Problem 5.2,Page no.122¶

In :
import math

#Initilization of variables

D=160 #mm #Overall Depth
B=150 #mm #Width of Flange
f_t=40 #mm #Flange thickness
W_t=50 #mm #Web thickness
sigma_t=20 #N/mm**2 #tension stress
sigma_c=75 #N/mm**2 #compression stress

#Calculations

#Rectangle-1
a_1=150*40 #mm**2 #Area of Rectangle-1
y_1=40*2**-1 #mm #C.G of Rectangle-1

#Rectangle-2
a_2=120*50 #mm**2 #Area of Rectangle-2
y_2=40+120*2**-1 #mm #C.G of Rectangle-2

Y_bar=(a_1*y_1+a_2*y_2)*(a_1+a_2)**-1 #mm #Distance of C.G from the bottom flange
I=1*12**-1*150*40**3+150*40*(60-40)**2+1*12**-1*50*120**3+50*120*(100-60)**2 #mm**4 #M.I of section about N.A
y_t=60 #mm #Permissible tensile stress at the bottom face of flange from N.A
y_c=100 #mm #Permissible tensile stress at the top face of flange from N.A

#M=W*L*4**-1 #Max bending mooment at the centre

#Using the relation M*I**-1=sigma_t*y_t**-1 we get

#Result
print"The Max Bending Moment at the centre is",round(W,2),"N"

The Max Bending Moment at the centre is 14492.16 N


## Problem 5.3,Page no.123¶

In :
import math

#Initilization of variables

b=10 #cm #width of beam
d=20 #cm depth of beam

#Calculations

#R_a and R_b are the reactions at A and B respectively.
#Moment of all forces about A

R_b=(4*4*4*2**-1-2*1.5)*(2)**-1 #KN
#R_a+R_b=18
R_a=18-R_b

#Consider a section at a distance x from A
#M_x=9.25*x-2(x-1.5)-4*x*x*2**-1=7.25*x+3-2*x**2

#Taking derivative of above equation to find max value of M_x we get
x=1.81 #m

M=7.25*x+3-2*x**2 #kN*m
I=b*d**3*12**-1 #cm**4 #M.I of the section
y=10
sigma=M*I**-1*y*10**8*(10**2)**-1 #Max bending stress

#Result
print"The Max Bending stress is",round(sigma,2),"KN/m**2"

The Max Bending stress is 14355.45 KN/m**2


## Problem 5.4,Page no.124¶

In :
import math

#Initilization of variables

sigma=8 #N/mm**2 #Max bending stress
A=450 #cm**2 #Area of joist
L=5 #m #span of floor

#Calculations
#Let x be the centre to centre spacingof the joists

#A_1=5*x**2 #m**2 #Area  of floor between any two joists
#W_1=5*x*W #N #total load supported by one interior joist
#M=W_1*L*8**-1 #Max bending moment
I=1*12**-1*b*(d*10**-2)**3*10**-2 #m**4 #M.I of joist
y=0.15 #cm #Distance of of farthest fibre
M=I*y**-1*sigma #N*m

#Now equating to max bending moment we get
x=(18000*8)*(25000*5)**-1

#Result
print"The Max Bending Moment is",round(x,2),"m"

The Max Bending Moment is 1.15 m


## Problem 5.7,Page no.126¶

In :
import math

#Initilization of variables

d=12 #m #depth of mast
D_1=20 #cm #diameter at the base
D_2=10 #cm #diameter at the top

#Calculations

#Consider section at a distance x cm below top of mast and y be the diameter at this section

#triangle OAB and ODC are similar,we get
#2*AB=x*120**-1
#EB=y=10+x*120**-1
#after simplifying we get, x=120*(y-10)

#Z=pi*64**-1*y**4)*(y*32**-1)**-1 #Section modulus
#After simplifying we get
#Z=(pi*y**3)*(32)**-1

#M=120*P(y-10) #bending moment at that section

#From flexural formula we get,
#sigma=M*Z**-1
#After substituting and simplifying above equation we get,
#sigma=3840*P*pi**-1*(1*y**2-1-10*y**3-1)

#To find max value of sigma taking derivative of above equation we get
y=15 #cm

#Now substituting value of y in all equations with variable y
x=120*(y-10)
#sigma=3840*P*(15-10)*(pi*15**3)**-1

#After implifying above equation we get

P=(3500*pi*15**3)*(3840*5)**-1 #N #Magnitude of load causing failure

#Result

The Magnitude of Load is 1932.82 N


## Problem 5.8,Page no.127¶

In :
import math

#Initilization of variables

L=3 #m #span of beam
t=20 #mm #Thickness of steel
D=200 #mm #overall depth
B=140 #mm #overall width
b=100 #mm #width of inner rectangle
d=160 #mm #depth of inner rectangle
w=77 #KN/mm**2
sigma=100 #N/mm**2 #Bending stress
#Calculations
V=((D*10**-3*B*10**-3)-(d*10**-3*b*10**-3)) #m**3 #Volume of rectangular box
W=V*3*w #KN #Weight of Beam
I=(B*D**3-b*d**3)*12**-1 #mm**4 #M.I of beam section

#Now using the relation,M*I**-1=sigma*y**-1

y=200 #mm #distance from farthest fibre
M=I*sigma*2*y**-1 #N*mm

#M=3000*W+2772*3000*2**-1
#After sub values in above equation we get

W=((59.2*10**6-2772*3000*2**-1)*(3000)**-1)*10**-3 #KN #Max concentrated Load at free end

F=W+2.772*2**-1 #KN #shear force at half length

#Result
print"The shear force at half length is",round(F,2),"KN"

The shear force at half length is 19.73 KN


## Problem 5.9,Page no.128¶

In :
import math

#Initilization of variables

B=24 #mm #width of beam section
D=21.7 #mm #depth of beam section
E_1=11440 #MN/m**2 #Modulus of Elasticity parallel grain
E_2=2860 #MN/m**2 ##Modulus of Elasticity perpendicular grain
sigma_1=8.57 #MN/m**2
sigma_2=2.14 #MN/m**2
L=1.2 #m #span of beam

#Calculations

#Ratio of smaller modulus to larger modulus is E_2:E_1=1:4
#Dimension of transformed Beam section
b=18 #mm #width of Beam section
d=3.1 #mm #depth of beam section

I=(1*12**-1*B*10**-3*(D*10**-3)**3)-(3*(1*12**-1*b*10**-3*(d*10**-3)**3)) #m**4 #M.I of transformed section
y=21.7*10**-3*2**-1
M=I*sigma_1*10**6*y**-1 #N*m #Safe B.M
P=4*M*L**-1 #N

#Result

Safe value of Load is 53.45 N


## Problem 5.11,Page no.131¶

In :
import math

#Initilization of variables

D=4 #cm #Outside diameter
d=3 #cm #inside diamter
L=2 #m #span of beam

#Calculations

I=pi*64**-1*(D**4-d**4) #cm**4 #M.I
A=pi*4**-1*(D**2-d**2) #cm**2 #Area
y=2
Z=I*y**-1 #cm**3 #Section modulus

M=W*L*4**-1 #N*cm #Max bending moment

#From Flexural Formula
sigma=M*Z**-1 #N/cm**2

#For Tubes
I_1=4*(8.59+5.492*2**2) #cm**4

Z_1=122.32*4**-1 #cm**3

#M=W_1*200*4**-1 #N*cm
#After substituting values we get
#M=50*W_1 (equation 1)

#Again from Flexural Formula
M=sigma*Z_1

#substitute value of M in equation 1

W=11640*30.58*50**-1 #N

#Result

Max central load is 7119.02 N


## Problem 5.12,Page no.133¶

In :
import math

#Initilization of variables

b=200 #mm #width of beam
d=300 #mm #depth of beam
t=12 #mm #thickness of beam
E_s=220 #KN/m**2 #modulus of elasticity of steel
E_w=11 #KN/m**2 #modulus of elasticity of timber
sigma_s=115 #MN/m**2 #stress of steel
sigma_w=9.2 #MN/m**2 #stress of timber
L=2 #m #Span of beam

#Calculations

#E_w*E_s**-1=1*20**-1 #ratio of Modulus of elasticity of timber to steel

#(Part-1)
b_1=b*20**-1 #mm #web thickness of transformed section
stress=20*sigma_w #MN/m**2 #Allowable stress in web of equivalen beam
#But allowable stress in flanges is sigma_s is 115 KN/m**2 and therefore taken into consideration

d_1=324 #mm #depth of beam with thickness in consideration
I=1*12**-1*0.2*0.324**3-2*1*12**-1*0.095*0.3**3 #m**4 #M.I of transformed section

#Using Relation, M*I**-1=sigma*y**-1 we get

#Part-2
M_max=I*(324*10**-3*2**-1)**-1*sigma_s*10**6 #N*m #Max allowable Bending moment for steel section

#Part-3
#As beam is simply supported at the ends and the load is applied at the centre of beam
#M_max=W*L*4**-1 #Max Bending moment

#Result
print"Web thickness of Equivalent steel section is",round(b_1,2),"mm"
print"Max Allowable bending moment for section is",round(M_max,2),"N*m"


Web thickness of Equivalent steel section is 10.0 mm
Max Allowable bending moment for section is 98935.78 N*m
Allowable safe Load is 197871.56 N


## Problem 5.13,Page no.135¶

In :
import math

#Initilization of variables

d=10 #cm #distance between joists
t=2 #cm #thickness of steel plate
d_2=20 #cm #depth of beam
sigma_t=8.5 #N/mm**2 #stress in timber
E_s=2*10**5 #N/mm**2 #Modulus of elasticity of steel
E_t=10**4 #N/mm**2 ##Modulus of elasticity of timber
L=5 #cm #span of beam

#calculation
sigma=10*15**-1*sigma_t #stress in timber at distance of 10 cm from XX

dell=sigma*E_t**-1 #strain in timber at 10 cm from XX

sigma_s=dell*E_s #N/mm**2 #Max stress

#For Timber
Z_w=1*6**-1*10*30**2*2 #cm**3 #section modulus of timber
M_w=sigma_t*100*Z_w #moment of resistance of timber

#For steel
Z_s=1*6**-1*2*20**2 #cm**3 #section modulus of steel
M_s=sigma_s*Z_s*100  #moment of resistance of steel

M=(M_w+M_s)*10**-5 #total moment of resistance

#M=w*L**2*8**-1 #N*m #Max bending moment
w=8*M*(L**2)**-1 #N/m #Max uniform distributed Load

#Result
print"Moment of resistance is",round(M,3),"N*cm"

Moment of resistance is 40.611 N*cm
Max uniform distributed Load 12.996 N/m


## Problem 5.14,Page no.136¶

In :
import math

#Initilization of variables

B=10 #cm #width of timber section
D=15 #cm #depth of timber section
b=10 #cm #width of steel plate
t=12 #mm #thickness
L=4 #m #Span of beam
m=20 #Ratio of modulus of elasticity of steel to timber

#Calculations

y_1=15*2**-1 #C.G of timber
y_2=1.2*2**-1 #C.G of steel plate
b_s=10*m**-1 #cm #Equivalent width of steel
Y_bar=(10*1.2*0.6+15*0.5*8.7)*(10*1.2+15*0.5)**-1 #cm #distance of C.G from bottom edge

I=1*12**-1*10*(1.2)**3+10*1.2*(3.72-0.6)**2+1*12**-1*0.5*(15)**3+0.5*15*(7.5-3.72)**2
M=W*10**5*L**2*8**-1 #N*m

Y_bar_1=3.72 #cm #C.G from bottom edge
Y_bar_2=16.2-Y_bar #cm #C.G from top edge

sigma_1=(M*I**-1*Y_bar_1)*10**-2 #N/mm**2 #stress at bottom

sigma_2=(M*I**-1*Y_bar_2)*10**-2 #N/mm**2 #stress at top

sigma_max=sigma_2*m**-1

#The Answers in book for Moment of Inertia about x-x axis onwards are incorrect

#Result
print M
print"The Max Stress in steel is",round(sigma_1,2),"N/mm**2"
print"The Max Stress in timber is",round(sigma_max,2),"N/mm**2"

600000.0
The Max Stress in steel is 60.98 N/mm**2
The Max Stress in timber is 10.23 N/mm**2


## Problem 5.15,Page no.137¶

In :
import math

#Initilization of variables

B=20 #cm #width of timber
D=30 #cm #depth of timber
d=25 #cm #depth of steel plate
b=1.2 #cm #width of steel plate
sigma_s=90 #N/mm**2 #Bending stress in steel
sigma_t=6 #N/mm**2 #Bending stress in timber
m=20 #Ratio of modulus of elasticity of of steel to timber

#Calculation

#Equivalent width of wood section,when 1.2 cm wide steel plate is replaced by steel plate is
b_1=1.2*20 #cm
d_1=25 #cm #depth of wood section
y_1=d*2**-1 #cm #C.G of timber section
y_2=D*2**-1 #cm #C.G of steel section

Y_bar=(2*d*b_1*y_1+D*B*y_2)*(2*d*b_1+D*B)**-1 #cm #Distance of C.G from Bottom edge
I=B*D**3*12**-1+B*D*(y_2-Y_bar)**2+2*(b_1*d_1**3*12**-1+b_1*d_1*(Y_bar-y_1)**2) #M.I of equivalent timber section about N.A
Y=30-Y_bar #distance of C.G from top of equivalent wood section

#Thus max stress will occur at top and that in steel will occur at bottom
#sigma_s=m*Y_bar*Y**-1*sigma_t

#After simplifying we get
#sigma_s=15.99*sigma_t

sigma_t=sigma_s*15.99**-1 #N/mm**2 #Max stress in Equivalent timber section

Z_t=I*Y**-1 #Section modulus of equivalent section
M=sigma_t*Z_t*10**-5*100 #Moment of resistance of beam

#Result
print"Position of N.A is",round(Y_bar,2),"cm"
print"Moment of Resistance of beam is",round(M,2),"KN*m"

Position of N.A is 13.33 cm
Moment of Resistance of beam is 37.15 KN*m