# Chapter no 11:Combined Direct And Bending Stresses¶

## Problem 11.1,Page no.273¶

In [8]:
import math

#Initilization of variables

e=0.06 #m #eccentricity
b=0.240 #m #width of column
d=0.150 #m #depth of column

#Calculations

sigma_d=P*(b*d)**-1 #KN/m**2
M=P*e #KN*m #Moment due to eccentricity
Z=(d*(b)**2)*6**-1 #mm**3

sigma_b=M*Z**-1 #KN/m**2

sigma_CD=sigma_d+sigma_b
sigma_AB=sigma_d-sigma_b

#Result
print"Stress at face CD is",round(sigma_CD,2),"KN/m**2"
print"Stress at face AB is",round(sigma_AB,2),"KN/m**2"

Stress at face CD is 694.44 KN/m**2
Stress at face AB is -138.89 KN/m**2


## Problem 11.2,Page no.274¶

In [1]:
import math

#Initilization of variables

d=2 #cm #Diameter of specimen

#Calculations

#Let P be the Load on the section

A=pi*4**-1*d**2 #cm**2 #Area of section
I=pi*64**-1*d**4 #cm**4 #M.I of the section
y=d*2**-1 #cm
Z=I*y**-1 #cm**3 #Section modulus
#M=P.e #Moment

#sigma_d=(4*P)*(pi*d**2)**-1 #N/cm**2

#stress due to moment
#sigma_b=(32*P*e)*(pi*d**3)**-1 N/cm**2

#Maximum stress
#sigma_r_max=(((4*P)*(pi*d**2)**-1)+((32*P*e)*(pi*d**3)**-1))

#Mean stress
#sigma_r_mean=((4*P)*(pi*d**2)**-1)

#Since the maximum stress is 20% greater than the mean stress
#(((4*P)*(pi*d**2))+((32*P*e)*(pi*d**3)))=1.2*4*P*(pi*d**2)**-1

#After substituing values and simplifyinf we get

e=0.2*d*8**-1 #cm #distance of line of thrust from the axis

#Result
print"The distance of line of thrust from the axis is",round(e,2),"cm"

The distance of line of thrust from the axis is 0.05 cm


## Problem 11.3,Page no.274¶

In [3]:
import math

#Initilization of variables

A=300 #cm**2 #Area of column
e=5 #cm #eccentricity

#Calculations

#sigma_d=P*A**-1  #Direct compressive stress
#M=P*e #Bending Moment
Z=((20**4-10**4)*(6*20)**-1) #cm**3 #Section modulus

#sigma_b=M*Z**-1=P*250**-1

#Now sigma_d+sigma_b=60*10**2

#P*300**-1+P*250**-1=6000

#After simplifying we get

#sigma_b-sigma_d=300

#Result
print"The maximum load column can carry",round(P_2,2),"N"

The maximum load column can carry 450000.0 N


## Problem 11.4,Page no.275¶

In [9]:
import math

#Initilization of variables

D=40 #cm #External diameter of column
d=30 #cm #Internal diameter of column
e=20 #cm #Eccentricity

#calculations

A=pi*4**-1*(D**2-d**2) #cm**2 #Area of the column
Z=pi*32**-1*((D**4-d**4)*D**-1) #cm**3 #Section modulus
M=P*10**3*e #N*cm #Moment

sigma_r_max=((P*10**3*A**-1)+(M*Z**-1)) #N/cm**2 #Max stress
sigma_r_min=((P*10**3*A**-1)-(M*Z**-1)) #N/cm**2 #Min stress

#Result
print"Max intensities of stress in the section is",round(sigma_r_max,2),"N/cm**2"
print"Min intensities of stress in the section is",round(sigma_r_min,2),"N/cm**2(tension)"

Max intensities of stress in the section is 971.3 N/cm**2
Min intensities of stress in the section is -425.63 N/cm**2(tension)


## Problem 11.5,Page no.277¶

In [27]:
import math

#Initilization of variables

b=4 #m #width of pier
d=3 #m #depth of pier
e_x=1 #m #distance from y axis
e_y=0.5 #m #distance from x axis

#Calculations

A=b*d #m**2 #Area of pier
I_x_x=b*d**3*12**-1 #m**4 #M.I about x-x axis
I_y_y=d*b**3*12**-1 #m**4 #M.I about y-y axis
M_x=P*e_y #KN*m #Moment about x-x axis
M_y=P*e_x #KN*m #Moment about y-y axis

x=2 #m #Distance between y-y axis and corners A and B
y=1.5 #m ##Distance between x-x axis and corners A and D

#Part-1
#Stress developed at each corner

sigma_A=P*A**-1+M_x*y*I_x_x**-1-M_y*x*I_y_y**-1 #KN/m**2 #stress at A
sigma_B=P*A**-1+M_x*y*I_x_x**-1+M_y*x*I_y_y**-1 #KN/m**2 #stress at B
sigma_C=P*A**-1-M_x*y*I_x_x**-1+M_y*x*I_y_y**-1 #KN/m**2 #stress at C
sigma_D=P*A**-1-M_x*y*I_x_x**-1-M_y*x*I_y_y**-1 #KN/m**2 #stress at D

#Part-2
#Let f be the additional load that should be placed at centre

#sigma_c=F*A**-1 #KN/m**2 #compressive stress

#For no tension in pier section, compressive stress is equal to tensile stress
sigma_c=10 #KN/m**2
F=sigma_c*A #KN

#Part-3

sigma_A_1=sigma_A+10 #stress at A
sigma_B_1=sigma_B+10 #stress at B
sigma_C_1=sigma_C+10 #stress at C
sigma_D_1=sigma_D+10 #stress at D

#Result
print"Stress at each corner are as follows:stress_A",round(sigma_A,2),"KN/m**2"
print"                                    :stress_B",round(sigma_B,2),"KN/m**2"
print"                                    :stress_C",round(sigma_C,2),"KN/m**2"
print"                                    :stress_D",round(sigma_D,2),"KN/m**2(tensile)"

print"Stresses at the corners with the  additional load in centre are as follows:Stress_A_1",round(sigma_A_1,2),"KN/m**2"
print"                                                                          :Stress_B_1",round(sigma_B_1,2),"KN/m**2"
print"                                                                          :Stress_C_1",round(sigma_C_1,2),"KN/m**2"
print"                                                                          :Stress_D_1",round(sigma_D_1,2),"KN/m**2"

Stress at each corner are as follows:stress_A 3.33 KN/m**2
:stress_B 23.33 KN/m**2
:stress_C 10.0 KN/m**2
:stress_D -10.0 KN/m**2(tensile)
Stresses at the corners with the  additional load in centre are as follows:Stress_A_1 13.33 KN/m**2
:Stress_B_1 33.33 KN/m**2
:Stress_C_1 20.0 KN/m**2
:Stress_D_1 0.0 KN/m**2


## Problem 11.11.6,Page no.278¶

In [5]:
import math

#Initilization of variables

#d=Diameter of rod
P=500 #KN
e=0.75 #cm #eccentricity

#calculation

#A=pi*d**2*4**-1 #cm**2 #Area of rod
#sigma_d=P*A**-1 #KN/cm**2 #stress due to direct load

#After substituting value and simplifying we get,
#sigma_d=2000*(pi*d**2)**-1 #KN/cm**2

M=P*e #Kn*cm #Moment

#Z=pi*d**3*32**-1 #cm**3 #section modulus
#sigma_b=M*Z**-1 #KN/cm**2 #Stress due to moment

#After substituting value and simplifying we get,
#sigma_b=12000*(pi*d**3)**-1 #KN/cm**2

#Max stress
#sigma=sigma_d+sigma_b

#After substituting value and simplifying we get,
#2000*(pi*d**2)**-1+12000*(pi*d**3)**-1=12.5

#After simplifying we get,
#d**3-53.05*d-318.3=0

#From Synthetic Division we get d**2+4.73*d-42.918
a=1
b=-4.73
c=-42.918

X=b**2-(4*a*c)

d_1=(-b+X**0.5)*(2*a)**-1
d_2=(-b-X**0.5)*(2*a)**-1

#Result
print"The minimum diameter of tie rod is",round(d_1,2),"cm"

The minimum diameter of tie rod is 9.33 cm