# Chapter7-Beams and Bendings¶

## Ex2-pg234¶

In [5]:
import math
import numpy
#calculate sf and Bm
import matplotlib
from matplotlib import pyplot
%matplotlib inline
#initialization of variables
s=3. #m
n=60.
p=50. #kg
## calculations
W=n*p
Rc=W*2./s
Rb=W-Rc
dx = 0.001;
x = numpy.linspace(0,s,3001)
n = round(s/dx +1);
n=int(n)
Sx=numpy.zeros(n)
Mx=numpy.zeros(n)
i=0;
for i in range (0,n):
Sx[i] = -Rb + Rc*x[i]**2./6.;
Mx[i] = Rb*x[i] - Rc*x[i]**3 /18.;

##Results
pyplot.plot(x,Sx);pyplot.title("Shear force diagram");pyplot.xlabel("X (in m)");pyplot.ylabel("Shear force (in kg)");
pyplot.show()
pyplot.plot(x,Mx);pyplot.title("Bending Moment diagram");pyplot.xlabel("X (in m)");pyplot.ylabel("Bending Moment (in kg-m)");
pyplot.show()


## Ex8-pg249¶

In [1]:
import math
##initialization of new variables
#centriod of the section and momets of inertia and product of inertia about xx and yy axes and direction of principal axes and value of maximum and minimum
b=10075 ##mm
h=10 ##mm
p1=7.5
p2=9
##part (a)
ybar=1*p1*0.5+1*p2*5.5
ybar=ybar/16.5
xbar=1*p1*0.5+1*p1*4.75
xbar=xbar/16.5
print('part (a)')
print'%s %.2f %s %.2f %s '%('\n Centroid coordinates (x,y) = ',xbar,''and '',ybar,'cm')

##part (b)
Ixx=p1*1**3/12.+p1*1*(3.23-0.5)**2+1*p2**3/12.+p2*1*(5.5-3.23)**2
Iyy=1*p1**3/12.+p1*1*(3.75-2.43)**2+p2*1**3/12.+p2*1*(2.43-0.5)**2
Ixy=p1*1.32*2.73+9*(-1.93)*(-2.27)
print('\n part (b)')
print'%s %.2f %s %.2f %s %.2f %s '%('\n Moment of Areas: \n Ixx = ',Ixx,' cm^4 \n Iyy = ',Iyy,' cm^4' and'Ixy=',Ixy,'cm^4')

##part (c)
alpha=0.5*math.atan(2*Ixy/(Iyy-Ixx))
alpha=alpha*180/math.pi
print('\n part (c)')
print('\n Direction of principal axes:')
print'%s %.2f %s'%('\n alpha = ',alpha,' degrees')

##part (d)
Iuu=(Ixx+Iyy)/2+math.sqrt((Iyy-Ixx)**2/4.+Ixy**2)
Ivv=(Ixx+Iyy)/2-math.sqrt((Iyy-Ixx)**2/4.+Ixy**2)
print('\n part (d)')
print'%s %.2f %s %.2f %s '%('\n Iuu = ',Iuu,' cm^4 \n Ivv = ',Ivv,' cm^4')

part (a)

Centroid coordinates (x,y) =  2.39  3.23 cm

part (b)

Moment of Areas:
Ixx =  163.65  cm^4
Iyy =  82.50 Ixy= 66.46 cm^4

part (c)

Direction of principal axes:

alpha =  -29.30  degrees

part (d)

Iuu =  200.94  cm^4
Ivv =  45.21  cm^4


## Ex10-pg256¶

In [2]:
import math
##initialization of variables
#determine bending stress and find the minimum radius
Ys=17000. ##kg/cm^2
E=2*10**6 ##kg/cm^2
d1=1. ##mm
d=1. ##cm
##calculations: 1 cm
R=E*d/(2*Ys)
M=Ys*math.pi*d**3/32.
## results
print'%s %.2f %s'%(' daimeter wire:',d,'cm')
print'%s %.2f %s'%('\n Minimum radius = ',R,' cm')
print'%s %.2f %s'%('\n Bending Moment = ',M,' kg-cm')
## calculations: 1 mm
R1=R/(d1*10.)
M1=M/(d1*1000.)
## results
print'%s %.2f %s'%('\n  daimeter wire:',d1,'mm')
print'%s %.2f %s'%('\n Minimum radius = ',R1,' cm')
print'%s %.2f %s'%('\n Bending Moment = ',M1,' kg-cm')

 daimeter wire: 1.00 cm

Bending Moment =  1668.97  kg-cm

daimeter wire: 1.00 mm

Bending Moment =  1.67  kg-cm


## Ex11-pg258¶

In [3]:
#calculate the section modulus per meter width of plate
##initialization of variables
import math
t=0.5 ##cm
s=2. ##m
p1=7.5 ##cm
p2=10. ##cm
d=p2/2.
D=1650. ##kg/cm**2
## calculations
## ab
IxX=p1*t**3./12.+t*p1*d**2.
## bc
alpha=math.atan(3/4.)*57.3
Ixx=t*(p1+d)**3./12.
Iyy=0.
Ixy=0.
Iuu=Ixx*math.cos(alpha)**2+Iyy*math.sin(alpha)**2-Ixy*math.sin(2.*alpha)
Ixx=Iuu+IxX
IXX=Ixx*100./(2.*p1)
Z=IXX/(d+t/2.)
w=D*Z*8./(s**2*100.)
w=w/1000.
##Results
print'%s %.2f %s'%('w = ',w,' tonne/m')

w =  5.50  tonne/m


## Ex12-pg261¶

In [4]:
#find the total u.d load that can be supported
##initialization of variables
import math
wb=10. ##cm
wh=20. ##cm
sb=0.5 ##cm
sh=10. ##cm
s=6. ##m
fs=1650. ##kg/cm^2
fw=150. ##kg/cm^2
Es=2*10**6 ##kg/cm^2
Ew=12*10**4 ##kg/cm^2

##calculations
## Method 1
A=2*fs/(21.*Es)
aw=2.*fw/(20.*Ew)
a=min(A,aw)
ss=a*Ew*wh/2.
##Moment resistance of steel portion
F=(fs+1573.)/2.*sb*sh
k=sb/3*(fs+2*1573.)/(fs+1573.)
Ms=2*F*(10.5-k)
##Moment resistance of wooden portion
F=ss*wb*wb/2.
Mw=2*(F*(wb-wb/3.))
M=Ms+Mw
##Total udl supported
W=M*8./(s*100.)

##Results
print('Using method 1')
print'%s %.2f %s'%('\n W = ',W,' kg')

##Method 2
nE=Es/Ew
nf=fs/fw
Is=2*(0+sb*sh*10.25**2)
Iw=0.6*wh**3/12.
I=Is+Iw
W=fs*I*8./(s*100*10.5)

##Results
print('\n Using method 2')
print'%s %.2f %s'%('\n W = ',W,' kg')

Using method 1

W =  3040.91  kg

Using method 2

W =  3039.40  kg


## Ex13-pg264¶

In [5]:
#calculate stiffness ratio
##initialization of variables
import math
p=6. ##mm
Ixx=2375. ##cm**4
Es=2*10**6. ##kg/cm**2
EAl=0.667*10**6 ##kg/cm**2
d1=10.6 ##cm
d2=10. ##cm
## calculations
I1=2.*(0.+p/10*10*10.3**2)
I2=Ixx*EAl/Es
I=I1+I2
n=I/I2
## results
print'%s %.2f %s'%('stiffness ratio =  ',n,'')
n1=Es*d1/(d2*EAl)
print'%s %.2f %s'%('\n Stress ration = ',n1,'')

stiffness ratio =   2.61

Stress ration =  3.18


## Ex14-pg269¶

In [6]:
#calculate shape factor
##initilization of new variables
import math
wt=0.8 ##cm
ft=1.4 ##cm
w=10. ##cm
y=20. ##cm
## Sigma_y: yield stress is not given explicitly
k1=wt*(40.-2*ft)/2.
Zp=(14*19.3+k1*9.3)*2
If=2*(w*ft**3/12+w*ft*19.3**2)
Iw=wt*(40-2*ft)**3/12.
I=Iw+If
Z=I/y
sf=Zp/Z
##Results
print'%s %.2f %s'%('shape factor =  ',sf,'')

shape factor =   1.18


## Ex15-pg270¶

In [7]:
#calculate elastic core
##initilization of new variables
import math
wt=0.8 ##cm
ft=1.4 ##cm
w=10. ##cm
y=20. ##cm
T=750. ##T==750*sigma_y
## calculations
MpF=ft*w*(40-2*ft)
c1=((40.-2*ft)/2)**2-(T-MpF)/wt
c=math.sqrt(3*c1)
## results
print'%s %.2f %s'%('Elastic core of ',2*c,' cm depth is present')

Elastic core of  26.71  cm depth is present


## Ex17-pg274¶

In [8]:
#find stress at the inner side of the hook
##initialization of new variables
P=2000. ##kg
a=4. ##cm
b=1. ##cm
d=7. ##cm
r=3. ##cm
## calculations
A=(a+b)/2*d
xbar=(a+b*2.)*d/(r*(a+b))
rbar=r+xbar
I=b*d**3/12.+r*d**3/12.
Ixx=I-A*2.8**2
e=Ixx/(rbar*A)
f1=P*5.8*(xbar-0.62)/(A*0.62*r)
f2=P*5.8*(-d+2.18)/(A*0.62*(5.18+d-2.18))
str=P/A
Str_i=f1+str
Str_o=-f2-str
##Results
print'%s %.2f %s'%('stress at the inner side of the hook =',Str_i,' kg/cm^2 (tensile)')
print'%s %.2f %s'%('\n stress at the outer side of the hook = ',Str_o,' kg/cm^2 (compressive)')
## approximations involved in the text

stress at the inner side of the hook = 891.18  kg/cm^2 (tensile)

stress at the outer side of the hook =  401.03  kg/cm^2 (compressive)


## Ex20-pg281¶

In [9]:
#  find Ratio of maximum bending stress in the stem and flange at various point
##initialization of new variables
import math
t=1. ##cm
a=40. ##cm
A=236.
## calculations
ybar=a*t*0.5+(50-1)*4*0.5/(a*t+(50-1)*4)
y1bar=1.25*a-ybar
IAA=a*t**3/3.+(50-1)**3*4/12.+(50-1)*4*25.5**2
Io=IAA-A*ybar**2.
##part (1)
r=y1bar/ybar
## results
print('Ratio of maximum bending stress in the stem and flange')
print'%s %.2f %s'%('\n Ratio = ',r,'')
##part(2)
## calculations
r=(2/3.*388*29.56)-(2/3.*160*20.44)-(228*20.44)
r=r/(2*2/3.*388*29.56)
## results
print('\n Ratio of S.F in flange to total S.F')
print'%s %.2f %s'%('\n Ratio = ',r*100,' percent')
## part (3)
## calculations
r=359.*200./Io
## results
print('\n Ratio of maximum shear stress in the flange to average sher stress in the stem')
print'%s %.2f %s'%('\n Ratio =  ',r,'')
##part (4)
## calculations
s=10 ##m
r=r/0.922
sigma=1650 ##kg/cm**2
shear=945 ##kg/cm**2
wsh=2*200*shear/(r*s)
wsi=8*Io*sigma/(s**2*10*29.56)
w=min(wsh,wsi)
## results
print'%s %.2f %s'%('\n Maximum u.d.l. = ',w,' kg/m ')

print('wrong moment of Inertia (Io) in the text and hence part (3) and part (4) are wrong')

Ratio of maximum bending stress in the stem and flange

Ratio =  1.45

Ratio of S.F in flange to total S.F

Ratio =  5.27  percent

Ratio of maximum shear stress in the flange to average sher stress in the stem

Ratio =   1.05

Maximum u.d.l. =  30507.35  kg/m
wrong moment of Inertia (Io) in the text and hence part (3) and part (4) are wrong


## Ex21-pg284¶

In [10]:
#find the design force two cases
##initialization of new variables
import math
a=30. ##cm
t=2.5 ##cm
S=15. ##cm
s=5. ##Tonne
## calculations
I=a*a**3-25*25**3
I=I/12.
tau_zx=s*1000.*27.5*t*25/(4*35000.*t)
FA=S*t*tau_zx
tau_xy=s*1000.*a*t*27.5/(4.*35000.*t)
FB=tau_xy*t*S
##Results
print'%s %.2f %s'%('case A \n F = ',FA,' kg')
print'%s %.2f %s'%('\n case B \n F= ',FB,' kg')

case A
F =  920.76  kg

case B
F=  1104.91  kg


## Ex23-pg288¶

In [11]:
#calculate shear centre
##initialization of variables
import math
h=40. ##cm
b=10. ##cm
t1=1.4 ##cm
t2=0.8 ##cm
Ixx=13989.5 ##cm^4
##calculations
e=b**2*h**2*t1/(4.*Ixx)*(1-t1/h-t1/b+t1**2/(b*h))*(1-t1/h)
##Results
print'%s %.2f %s'%('Shear center: \n e = ',e,' cm')

Shear center:
e =  3.21  cm


## Ex33-pg303¶

In [12]:
#calculate deflection
##initialization of new variables
import math
L=50. ##cm
k=15. ##cm
I=200. ##cm^4
II=40. ##cm^4
d=30. ##cm
Pd=40. ##cm
E=0.6667*10**6 ##kg/cm^2
##calculations
delta=(100.*10./2*16.33+L*d*35+L*k/2.*25+d*k/2.*45)
delta1=delta/E
##Results
print'%s %.2f %s'%('deflection = ',delta1*10**1,'mm')

deflection =  1.20 mm


## Ex39-pg312¶

In [19]:
#design suitable I design section
W=1
L=6
f=1650
Z=305.9
w=31.2
z=8
t=0.65
d=22.5
M=4.5*10**5
##part(i)
#maximum B.M
BM= W*L**2/(z)
SF=W*L/2
print'%s %.2f %s %.2f %s '%('Bm=',Bm,''and '',SF,'')
#part(ii)
Z=M/f
#part(iii)
#for ISMB 225 having weight 31.2
Z1=305.9
print'%s %.2f %s'%('z=',Z1,'')

Bm= 4.00  3.00