Chapter6-Strength of materials

Ex1-pg66

In [1]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
b=0.005##in
a=2.##tonf
p=10.##tonf
l=13500.##tonf/in^2
##CALCULATIONS
x=(p/a)*b##in
E=(l*b*1./2.)/a##in
##RESULTS
print'%s %.2f %s'%('the original length of bar = ',E,' in')

the original length of bar =  16.88  in

Ex2-pg67

In [3]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
p1=12000.##in
p2=0.0125##lbf/in
x=8.##in
w=14300.##in
r=0.122##in
##CALCULATIONS
M=(p1/p2)*(x/(math.pi/4.*1**2))##lbf/in^2
P=0.1*x/100.##in
S=w/(math.pi/4.*1**2)##lbf/in^2
P1=(r*100./x)##percent
##RESULTS
print'%s %.2f %s'%('the modulus of elasticity= ',M,' lbf/in^2')
print'%s %.2f %s'%('non-proportional elongation= ',S,' lbf/in^2')
print'%s %.2f %s'%('the percentage elongation= ',P1,' percent')

the modulus of elasticity=  9778479.70  lbf/in^2
non-proportional elongation=  18207.33  lbf/in^2
the percentage elongation=  1.52  percent

Ex3-pg69

In [4]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w=0.5##tonf/in^2
w1=7.##tonf/in^2
w2=10.##tonf/in^2
t=12.4##tonf/in^2
d1=1.5##in
d2=1.24##in
x=0.495##in
d3=3.02##in
##CALCULATIONS
Y=math.sqrt((d3/2.)**2-(d2/2.)**2)##in
S=(1/2.*t/(2.*Y*w))##tonf/in^2
##RESULTS
print'%s %.2f %s'%('the shear stress in fork end= ',S,' tonf/in^2')

the shear stress in fork end=  4.50  tonf/in^2

Ex4-pg70

In [5]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
g=2.##in
t=0.002##in
l=7500.##lbf
w=11000.##lbf
p=1./4.##in
##CALCULATIONS
W=1./2.*l*t##in lbf
P=t*(w/l)##in
S=w/p##lbf/in^2
E=S*g/P##lbf/in^2
R=(1./2.)*w*P##in lbf
##RESULTS
print'%s %.4f %s'%('The elongation at the elastic limit= ',P,' in')
print'%s %.2f %s'%('The stress at the elastic limit= ',S,' lbf/in^2')
print'%s %.2f %s'%('The modulus of elasticity E of the material is= ',E,' lbf/in^2')
print'%s %.2f %s'%('The resilience and modulus of elasticity= ',R,' in lbf')

The elongation at the elastic limit=  0.0029  in
The stress at the elastic limit=  44000.00  lbf/in^2
The modulus of elasticity E of the material is=  30000000.00  lbf/in^2
The resilience and modulus of elasticity=  16.13  in lbf

Ex5-pg71

In [6]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
v=4.##in
w=20.##tonf
d=10.##ft
m=13400.##tonf/in^2
q=2.##in
l=120.##in
##CALCULATIONS
Fmax=q*(w)/(math.pi/v*v**2)##tonf/in^2
M=Fmax*l/m##in
P=w*M##in tonf
##RESULTS
print'%s %.2f %s'%('The maximum instantneous stress= ',Fmax,' tonf/in^2')
print'%s %.2f %s'%('The maximum elongation is= ',M,' in')
print'%s %.2f %s'%('the strain energy stored= ',P,' in tonf')

The maximum instantneous stress=  3.18  tonf/in^2
The maximum elongation is=  0.03  in
the strain energy stored=  0.57  in tonf

Ex6-pg72

In [7]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
d=4.##in
p=2.##ft
d1=1./2.##in
e=13200.##tonf/in^2
f=9.51##tonf/in^2
k=0.0114##tonf/in^2
##CALCULATIONS
E=k*f##in tonf
F=(p/(math.pi/d*d**2))##tonf/in^2
##RESULTS
print'%s %.2f %s'%('the final stress after oscillation has died aways will load/area= ',F,' tonf/in^2')

the final stress after oscillation has died aways will load/area=  0.16  tonf/in^2

Ex7-pg73

In [8]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
h=3.##in
s=10.2##tonf/in^2
v=0.006##in
d=0.5##in
d1=0.75##in
w=20.##lbf
q=v/8.##tonf/in^2
x=0.029##in
##CALCULATIONS
M=s/q##tonf/in^2
E=M*(x)/(h*12.)##tonf/in^2
##RESULTS
print'%s %.2f %s'%('the corresponding stress= ',E,' tonf/in^2')

the corresponding stress=  10.96  tonf/in^2

Ex8-pg73

In [11]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
e=30.*10**2##lbf/in^2
b=15.##in
t=50.##percent
p=1.5##in
v=6.##in
h=2240.##lbf
I=0.0038##in
##CALCULATIONS
W=1/2.*v*I##in tonf
w1=W*p##in tonf
T=math.sqrt((v**2*h)/(2.*math.pi/4.*e))/((b)/(p)**2/(1)**2)*10.##in
##RESULTS
print'%s %.2f %s'%('the total energy in the bar= ',T,' in')

the total energy in the bar=  6.21  in

Ex9-pg74

In [12]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables

E=13400.##tonf/in^2
E1=5600.##tonf/in^2
h=7.##tonf/in^2
h1=3.5##tonf/in^2
w=1.5##ij
l=5.##tonf
A=math.pi/4.*1**2##in^2
A1=math.pi/4.*(w**2-1**2)##in^2
s=1.91##tonf
t=0.787##in
pg=1.72##tonf
##CALCULATIONS\
m=h*t##tonf
p=m/s##tonf
g=p/A1##tonf/in^2
m1=m+p##tonf
S=pg/A1##tonf/in^2
Ps=pg*s##tonf
S1=Ps/t##tonf/in^2
##RESULTS
print'%s %.2f %s'%('the stress in the steel= ',S1,' tonf/in^2')

the stress in the steel=  4.17  tonf/in^2

Ex10-pg75

In [13]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables

E=2.*10**6##lbf/in^2
s=600.##lbf/in^2
w=12.##in
l=80.##tonf
w1=4.##ft
E1=30.*10**6##lbf/in^2
h=2240.##in
s2=10.9##in^2
F=9000.##lbf/in^2
##CALCULATIONS
L=(F*w1*w/E1)##in
##RESULTS
print'%s %.2f %s'%('the column shortens by= ',L,' in')

the column shortens by=  0.01  in

Ex11-pg76

In [14]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
E1=30.*10**6##lbf/in^2
E2=15.*10**6##lbf/in^2
alf=6.4*10**-6##degF-1
alf1=9.5*10**-6##degF-1
t=170.##deg
t1=50.##deg
w=5.##tonf
ec=0.000248##lbf/in^2
es=0.000124##lbf/in^2
h=2240.##in
##CALCULATIONS
e=(alf1-alf)*(t-t1)##in
Ec=E2*ec##lbf/in^2
Es=E1*es##lbf/in^2
F=E1/E2##fc
S=w*h/(2.*1.+1.)##lbf/in^2
S1=S*2.##lbf/in^2
R=-Es+S##lbf/in^2
R1=Es+S1##lbf/in^2
##RESULTS
print'%s %.2f %s'%('The final stress in the steel and applied to the compound = ',R1,' lbf/in^2')

The final stress in the steel and applied to the compound =  11186.67  lbf/in^2

Ex12-pg77

In [15]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
a=1/16.##ft/s
h=100.##lbf/in^2
w=10.##lbf/in^2
q=2.##in
b=math.pi/4.*(3./16.)**2##in^2
p=5.##inch valu per 12.7
##CALCULATIONS
H=(h*w)/(q*a)##lbf/in^2
F=H*1.*a##lbf
A=H/2.##lbf/in^2
R=(b)/(F/A)*5.14*4.##per inch
F1=A*1.*a##lbf
m=(b)/(F1/A)*5.14##per inch
##RESULTS
print'%s %.2f %s'%('the force per inch of circumferential seam= ',m,' per in')

the force per inch of circumferential seam=  2.27  per in

Ex13-pg77

In [16]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables

p=14.7##lbf/in^2
w=15000.##lbf/in^2
p1=190.##lbf/in^2
q=0.35##percent
q1=0.75##percent
w1=2.##ft
q2=36.##tonf/in^2
f=6.##in
r1=3/8.##in
p2=4.##in
h=2240.##in
##CALCULATIONS
A=w*q##lbf/in^2
E=w*q1##lbf/in^2
M1=(p2*A*(1/2.)/(p1-p))##in
M2=(w1*E*(1/2.)/(p1-p))##in
M3=p2*r1*((q2*h)/f)/(w1*12.)##lbf/in^2 gauge
##RESULTS
print'%s %.2f %s'%('the Maximum possible diameter of cylinder = ',M2,' in')
print'%s %.2f %s'%('the Maximum allowable pressure= ',M3,' lbf/in^2 gauge')

the Maximum possible diameter of cylinder =  64.18  in
the Maximum allowable pressure=  840.00  lbf/in^2 gauge

Ex14-pg81

In [17]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables

w=450.##lbf/in^2
m=3000.##lbf/in^2
g=32.2##lbf/in^2
h=144.##in
##CALCULATIONS
M=math.sqrt(g*m*h/w)##ft/f
##RESULTS
print'%s %.2f %s'%('the maximum rim speed of flywheel= ',M,' ft/f')

the maximum rim speed of flywheel=  175.82  ft/f