# Chapter 6:Bending¶

### Example 6.10 Page No 271¶

In [3]:
#Given
Lab=12.0       #ft, length
f=120        #lb/ft , force
shear=240    #shear at x=0

#Calculation
a=f/Lab      #a=w/x
x=(shear*2/(a))**(1/2.0)
#Moment diagram
Mmax=shear*x-(0.5)*(a*x)*x*(1/3.0)*x

#Result
print"The Maximum Bending Moment is",round(Mmax,0),"lb-ft"
The Maximum Bending Moment is 1109.0 lb-ft

### Example 6.11 Page No:289¶

In [9]:
#Given:
b = 6.0         #inch, length
h = 12.0        #inch
sigma_max = 2.0 #ksi
c = b

#Part (a):
I = (1/12.0)*b*h**3
M1 = (sigma_max*I)/(c) #sigma_max = Mc/I Flexure Formula

F = (0.5*sigma_max*b*b)
c = (2/3.0)*(b) #distance between centroids of each volume.
d=2*c          #distance
M2 = F*d

#Display:
print"The internal moment M calculated using : "
print"a)The flexure formula = ",M1,"kip-inch or ",M1/12.0,"kip-ft"
print"b)The resultant of the stress distribution using the basic principles ",M2,"kip-inch or",M2/12.0,"kip-ft"
The internal moment M calculated using :
a)The flexure formula =  288.0 kip-inch or  24.0 kip-ft
b)The resultant of the stress distribution using the basic principles  288.0 kip-inch or 24.0 kip-ft

### Example 6.12 Page No:290¶

In [3]:
#Given:
udl = 5     #kN/m, force
l1 = 3        #m , lenght
l2 = 6        #m
t = 20/1000.0 #mm
yb = 0.15   #m

#Section Property:
I_bar1 = (1/12.0)*(0.25)*(t**3)
I_bar2 = (1/12.0)*(0.02)*(0.3**3)
I = 2*(I_bar1 + Ad2) + I_bar2

#Bending stress:
c = 0.15 + t
M= 22.5 #kNm
sigma_max = (M*c)/(I*1000)
sigma_B = (M*yb)/(I*1000)

#Display:
print"The absolute maximum bending stress is ",round(sigma_max,1),"MPa"
The absolute maximum bending stress is  12.7 MPa

### Example 6.13 Page No:291¶

In [35]:
#Given:
t1 = 15/1000.0 #m, thickness
t2 = 20/1000.0 #m
l = 250/1000.0 #m
b = 200/1000.0 #m
l_a = 2        #m
l_b = 1        #m

#Internal Moment:
y1 = b/2.0
y2 = t2/2.0
A = (2*t1*b)+(t2*l)
y_bar = ((2*y1*t1*b)+(y2*t2*l))/A
M = (P*l_a)+(1*y_bar)

#Section Property:
I1 = (1/12.0)*(l*t2**3) + (l*t2*(y_bar - y2)**2)
I2 = (1/12.0)*(t1*b**3) + (t1*b*(y1 - y_bar)**2)
I =I1+ 2*I2

#Maximum Bending Stress:
c = b - y_bar
sigma_max = (M*c)/(I*1000)

#Display:
print'The maximum bending stress at section a-a is',round(sigma_max,1),"MPa"
The maximum bending stress at section a-a is 16.2 MPa

### Example 6.14 Page No:292¶

In [39]:
#Given:
h = 30/1000.0    #m, height
M = 40           #Nm, moment
c1= h/2.0
rib_t = 5/1000.0  #m
rib_w = 10/1000.0 #m

#Without Ribs:
I1 = (1/12.0)*(b*h**3)
sigma_max1 = (M*c1)/(I1*10**6)

#With Ribs:
y1 = c1
y2 = h+(rib_t/2.0)
A1 = h*b
A2 = rib_t*rib_w
y_bar = ((y1*A1)+2*(y2*A2))/(A1 + 2*A2)

c2 = h+rib_t - y_bar
I2 = I1 + (b*h*(y_bar - y1)**2)
I3 = (1/12)*rib_w*rib_t**3 + (rib_w*rib_t*(y2 - y_bar)**2)
I = I2 + 2*I3
sigma_max2 = (M*c2)/(I*10**6)

if(sigma_max2>sigma_max1):
print"The maximum normal stress in the member without ribs",round(sigma_max1,2),"MPa"
print"The maximum normal stress in the member with ribs = ",round(sigma_max2,2),"MPa"
print"The ribs should be omitted."

else:
print"no"
The maximum normal stress in the member without ribs 4.44 MPa
The maximum normal stress in the member with ribs =  4.65 MPa
The ribs should be omitted.

### Example 6.15 Page No:306¶

In [29]:
#Given:
M = 12     #kNm, moment
l_bc = 0.2 #m, length
l_be = 0.4 #m

#Internal Moment Components:
import math
My = (-4/5.0)*M
Mz = (3/5.0)*M

Iy = (1/12.0)*(l_be*l_bc**3)
Iz = (1/12.0)*(l_bc*l_be**3)

#Bending Stress:
sigma_B = (-Mz*1000*(l_be/2.0))/Iz + (My*1000*(-l_bc/2.0))/Iy
sigma_B = sigma_B/10.0**6
sigma_C = (-Mz*1000*(l_be/2.0))/Iz + (My*1000*(l_bc/2.0))/Iy
sigma_C = sigma_C/10.0**6
sigma_D = (-Mz*1000*(-l_be/2.0))/Iz + (My*1000*(l_bc/2.0))/Iy
sigma_D = sigma_D/10.0**6
sigma_E = (-Mz*1000*(-l_be/2.0))/Iz + (My*1000*(-l_bc/2.0))/Iy
sigma_E = sigma_E/10.0**6

#Orientation of Nuetral Axis:
z = (0.45)/(sigma_E + sigma_B)

#theta = -atan(4/3.0)
import math
tanA = (Iz/Iy)*(-4/3.0)
alpha = math.atan(tanA)
alpha = alpha*(180/math.pi)

#Display:
print"The normal stress at B",sigma_B,"MPa"
print"The normal stress at C ",sigma_C,"MPa"
print"The normal stress at D ",sigma_D,"MPa"
print"The normal stress at E ",sigma_E,"MPa"
print"The orientation of the nuetral axis ",round(alpha,1),"degree"

The normal stress at B 2.25 MPa
The normal stress at C  -4.95 MPa
The normal stress at D  -2.25 MPa
The normal stress at E  4.95 MPa
The orientation of the nuetral axis  -79.4 degree

### Example 6.16 Page No:308¶

In [34]:
#Given:
import math
M =20            #kN, moment
Iy = 0.96*10**-3 #m**4, moment of inertia
Iz = 7.54*10**-3 #m**4
theta = 57.1*(math.pi/180.0)  #angle

#Calculation
#Internal moment Components:
My = M*sin(theta)
Mz = M*cos(theta)
#Bending Stress:
y_p = -0.2 #y Coordinate of P
z_p = 0.35 #z Coordinate of P

theta1 = (math.pi/2)-(theta)
yp = -z_p*math.sin(theta1)+ y_p*math.cos(theta1)
zp = z_p*math.cos(theta1) + y_p*math.sin(theta1)

#Eq 6-17

sigma_p = ((Mz*-yp)/Iz) + ((My*zp)/Iy)
sigma_p = sigma_p/10.0**3
#Orientation of the Nuetral Axis:
alpha = math.atan((Iz/Iy)*math.tan(theta))
alpha = alpha*(180/math.pi)

#Display:
print"The maximum normal stress at point P is",round(sigma_p,1),"Mpa"
print"The orientation of the nuetral axis is",round(alpha,1),"degree"
The maximum normal stress at point P is 3.8 Mpa
The orientation of the nuetral axis is 85.3 degree

### Example 6.17 Page No:316¶

In [42]:
#Given:
M = 2       #kNm, moment
Ew = 12     #GPa, Pressure
Est = 200.0 #GPa
bw = 150/1000.0 #m
t = 20/1000.0 #m
rib = 9/1000.0#m

#Section Properties:
n = (Ew/Est)
bst = n*bw
y1 = t/2.0
A1 = t*bw
y2 = bw/2.0 + t
A2 = rib*bw

y_bar = (y1*A1 +y2*A2)/(A1+A2)
I1 = (1/12.0)*(bw)*(t**3) + A1*(y_bar - y1)**2
I2 = (1/12.0)*(rib)*(bw**3) + A2*(y2-y_bar)**2
Ina = I1+I2

#Normal Stress:
sigma_B = (M*(bw+t-y_bar))/(Ina*1000)
sigma_C = (M*(y_bar))/(Ina*1000)

#Normal Stress in the wood:
sigmaB = n*sigma_B

#Display:
print"The normal stress at point B = ",round(sigma_B,1),"MPa"
print"The normal stress at point C = ",round(sigma_C,1),"MPa"
print"The normal stress at point B in the wood =",round(sigmaB,2),"MPa"
The normal stress at point B =  28.6 MPa
The normal stress at point C =  7.8 MPa
The normal stress at point B in the wood = 1.71 MPa

### Example 6.18 Page No:308¶

In [15]:
#Given:
M = 60          #kip, moment
Est = 29*10**3  #GPa, stress
Econc = 3.6*10**3  #GPa
d = 25.0        #mm, diameter
w = 12      #inch, width
ht =16      #inch, height

#Section Properties:
import math
n = Est/Econc
Ast = 2*math.pi*r**2
A = n*Ast

#h**2+2.11h-33.7=0
a=1
b=2.11
c=-33.7
d=b**2-4*a*c
x1 = (-b+math.sqrt(b**2-4*a*c))/2*a
x2 = (-b-math.sqrt(b**2-4*a*c))/2*a
I = (1/12.0)*(w*x1**3) +w*x1*(x1/2.0)**2 + 12.65*(ht - x1)**2

#Normal Stress:
sigma_conc_max = (M*12*x1)/(I)
sigma_conc = (M*12*(ht-x1))/(I)
sigma_st = n*sigma_conc

#Display:
print"The normal stress in each steel reinforcing rod = ",round(sigma_st,1),"ksi"
print"The maximum normal stress in the concrete = ",round(sigma_conc_max,2),"ksi"
The normal stress in each steel reinforcing rod =  31.9 ksi
The maximum normal stress in the concrete =  1.72 ksi

### Example 6.19 Page No:324¶

In [46]:
#Given:
ri = 200/1000.0 #m
r1 = 250/1000.0 #m
ro = 280/1000.0 #m
M = 4 #kNm
a = 0.05 #m
h = 0.03 #m

#Section Properties:
import math
A1 = a**2
A2 =  (0.5*a*h)
A = A1+A2
r_avg1 = (r1+ri)/2.0
r_avg2 = r1+(h/3.0)
r_bar =((r_avg1*A1)+(r_avg2*A2))/A

int_dA_r1 = a*math.log(r1/ri)
int_dA_r2 = (a*ro*math.log(ro/r1))/(ro-r1) - a
R = (A)/(int_dA_r1+ int_dA_r2)
k= r_bar - R

#Normal Stress:
sigma_B = (-M*(R-ri))/(A*ri*k*1000)
sigma_A = (-M*(R-ro))/(A*ro*k*1000)
sigma = max(abs(sigma_B),abs(sigma_A))

#Display:
print"The maximum normal stress in the bar =",round(sigma,0),"MPa"
The maximum normal stress in the bar = 129.0 MPa

### Example 6.20 Page No:328¶

In [50]:
#Given:
M = 5         #kNm, moment
sigma_y = 500 #MPa, stress
h = 80.0      #mm, height
w = 120        #mm, width
r_h = r/h
w_h = w/h
k = 1.45
c = h/(2000.0)
t = 20/1000.0 #m

#Calculations:
I = (1/12.0)*(t)*(h/1000.0)**3
sigma_max = (k*M*c)/(I*1000)

#Display:
print"The maximum normal stress in the steel =",round(sigma_max,0),"Mpa"

The maximum normal stress in the steel = 340.0 Mpa

### Example 6.21 Page No:341¶

In [60]:
#Given:
sigma_y = 36 #ksi, stress
t = 0.5       #inch, thickness
w = 8       #inch, width
h = 9         #inch, height

#Maximum Elastic Moment:
yy = (h+t)/2.0
I1 = (1/12.0)*(w*t**3) + (w*t*yy**2)
I = (1/12.0)*(t*h**3) + 2*(I1)
c = 5.0 #mm

My = (sigma_y*I)/(c) #Flexure Formula
#Plastic Moment:
C1= sigma_y*t*(h/2.0)
C2= sigma_y*t*(w)
Mp = (2*2.25*C1) + (2*yy*C2)

#Shape Factor:
k = Mp/My

#Display:
print"The shape factor for the beam = ",round(k,2)
The shape factor for the beam =  1.14

### Example 6.22 Page no 342¶

In [62]:
#Given:
sigma_y = 250 #MPa, bending stress
t = 15/1000.0 #m, thickness
w = 100/1000.0 #m, width
h = 120/1000.0 #m, height
c = 10/1000.0 #m

#Calculations:
d = ((sigma_y*t*w)+(sigma_y*t*h))/(sigma_y*t*2)
T = sigma_y*t*d*10**3
C1 = sigma_y*t*c*10**3
C2 = sigma_y*t*w*10**3
Mp = (T*d/2.0)+(C1*c/2.0)+(C2*(c+t/2.0))

#Display:
print"The plastic moment that can be resisted by the beam = ",round(Mp,1),"MPa"
The plastic moment that can be resisted by the beam =  29.4 MPa

### Example 6.23 Page no 343¶

In [65]:
#Given:
sigma_y = 36   #ksi, bending stress
t = 12.5       #mm, thickness
w = 8          #inch, width
h = 9        #inch, height
c = (h/2.0)+t
I = 211        #inch**4, moment of inertia
Mp = 1732.5   #kip

#Calculations:
sigma_allow = (Mp*c)/(I)
y = (sigma_y*c)/(sigma_allow)

#Display:
print"The point of zero normal stress = ",round(y,2),"inch"

The point of zero normal stress =  4.38 inch

### Example 6.24,Page no :344¶

In [71]:
#Given:
ep1 = 0.01      #Strain at top
ep2 = 0.05      #strain at bottom
sig1 = 150      #kip
sig2 = 150     #N/mm**2
sig3 = 40      #N/mm**2
y = 0.3         #in
h = 3           #in
w = 2           #in

#Calculations:
yy = (h/2.0)-y
T1 = (1/2.0)*(sig3*yy*w)
y1 = y +(2/3.0)*(yy)
T2 = yy*sig1*w
y2 = y+(0.5*yy)
T3 = (0.5*y*sig1*w)
y3 = (2/3.0)*(y)
M = 2*(T1*y1 + T2*y2 + T3*y3)

#Display:
print"The bending moment applied that will cause a strain of ",round(M,0),"kip in"
The bending moment applied that will cause a strain of  772.0 kip in