import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.1 Page No-41 \n');
F=20000.; ##[lb] Load applied to steel bar
L=6.; ##[in] Length of steel bar
d=1.; ##[in] Diameter of steel bar
A=math.pi*(d**2)/4.; ##[in^2] Area of cross section of steel bar
E=30.*10**6; ##[lb/in^2] Modulus of elasticity for AISI 1020 hot-rolled steel
Sy=30000.; ##[lb/in^2] Yield limit
S=F/A; ##[lb/in^2] Stress in bar
print'%s %.2f %s '%('\na. Stress in bar=',S,' lb/in^2.');
delta=F*L/(A*E); ##[in] Change in length of bar
print'%s %.2f %s '%('\nb. bar shorten by ',delta,' in.');
if Sy>S:
print'%s %.2f %s %.2f %s '%('\nc. The stress of ',S,' psi is less than Sy of ',Sy,' psi, so it will'and '\n return to its original size because the yield limit was not exceeded.');
else:
print('The bar will not return to its original length')
##Note: The deviation of answer from the answer given in the book is due to round off error.(In the book values are rounded while in scilab actual values are taken)
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.2 Page No.43\n');
b=2.; ##[in] Width of beam
h=2.; ##[in] Height of beam
I=(b*h**3)/12.; ##[in^4] Moment of inertia
F=3000.; ##[lb] Load applied to beam
L=36.; ##[in] Length of beam
c=1.; ##[in] Distance of outer most fiber from neutral axis
E=30*10**6; ##[lb/in^2] Modulus of elasticity
Sy=30000.; ##[lb/in^2] Yield strength
Su=55000.; ##[lb/in^2] Ultimate strength
SF=2.; ##[] Safety factor based on ultimate stress
M=F*L/4.; ##[lb*in] Bending moment
S=(M/I)*c; ##[lb/in^2] Bending stress
##Note-In the book I=1.33 in^4 is used instead of I=1.3333333 in^2
print'%s %.2f %s '%('\na. The maximum stress in beam is ',S,' lb/in^2');
delta=-F*L**3/(48.*E*I); ##[in] Maximum deflection in this beam
print'%s %.2f %s '%('\nb. The maximum deflection in this beam is ',delta,' in.');
if Sy>S:
print'%s %.2f %s %.2f %s '%('\nc. Yes, the stress of ',S,' lb/in^2'and ' is less than the yield of Sy=',Sy,' lb/in^2.');
else:
print'%s %.2f %s %.2f %s '%('\nc. No, the stress of ',S,' lb/in^2'and ' is greater than the yield of Sy=',Sy,' lb/in^2');
Sall=Su/SF; ##[lb/in^2] Allowable stress
if Sall>S:
print'%s %.2f %s '%('\nd. It is acceptable because allowable stress is greater than the acttual stress of ',S,' lb/in^2.');
else:
print'%s %.2f %s '%('\nd. Design is not acceptable because allowable stress is less than the actual stress of ',S,' lb/in^2.')
##Note: The deviation of answer from the answer given in the book is due to round off error.(In the book values are rounded while in scilab actual values are taken)
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.3 Page No.45\n');
Su=80.*10**3; ##[lb/in^2] Ultimate strength
d=0.5; ##[in] Diameter of pin
As=math.pi*d**2./4.; ##[in^2] Area of cross section of pin
F=20*10**3; ##[lb] Load acting
Ss=F/(2.*As); ##[lb/in^2] Shear stress
if 0.5*Su>=Ss:
print('Pin would not fail');
else:
if 0.6*Su>=Ss:
print('Pin would not fail');
else:
print('\n Actual stress is too high and the pin would fail.');
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.4 Page No.46\n');
hp=10.; ##[hp] Power transmitted
rpm=1750.; ##[rpm] Turning speed
d=0.5; ##[in] Diameter of shaft
L=12.; ##[in] Length of shaft
G=11.5*10**6 ##[lb/in^2] shear modulus of elasticity
Su=62000.; ##[lb/in^2]
T=63000.*hp/rpm; ##[in*lb] Torque transmitted
Z=math.pi*d**3/16.; ##[in^3] Polar section modulus
Ss=T/Z; ##[lb/in^2] Torsional shear stress
##Note- In the book Z=0.025 in^3 is used instead of Z=0.0245437 in^3
print'%s %.2f %s '%('\na. Stress in the shaft is ',Ss,' lb/in^2.')
J=math.pi*d**4/32.; ##[in^4] Polar moment of inertia
theta=T*L/(J*G); ##[radians]
##Note- In the book J=0.0061 in^4 is used instead of J=0.0061359 in^4
print'%s %.2f %s '%('\nb. The angular deflection of shaft would be ',theta,' radians');
SF=3; ##[] Safety factor based on ultimate strength
Zd=T/(0.5*Su/SF); ##[in^3] Polar section modulus required for SF=3
Dd=(16.*Zd/math.pi)**(1/3.); ##[in] Diameter of shaft required Z=%pi*d^3/16
print'%s %.2f %s '%('\nc. Diameter of shaft required is ',Dd,' in.');
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.5 Page No.53\n');
L=30.; ##[in] Length of link
d=5/8.; ##[in] Diameter of link
I=math.pi*d**4/64.; ##[in^4] Moment of inertia
A=math.pi*d**2/4.; ##[in^2] Area of cross section
E=30*10**6; ##[lb/in^2] Modulus of elasticity
r=math.sqrt(I/A); ##[in] Radius of gyration
print'%s %.2f %s '%('\n The radius of gyration ',r,' in.');
K=1; ##[] End support condition factor
Le=K*L; ##[in] Effective length
print'%s %.2f %s '%('\n Effective length is ',Le,' in');
SR=Le/r; ##[] Slenderness ratio
print'%s %.2f %s '%('\n Slenderness ratio is ',SR,'.')
Sy=42000.; ##[lb/in^2] Yield strength
Cc=math.sqrt(2*math.pi**2*E/Sy); ##[] Column constant
print'%s %.2f %s '%('The column constant is',Cc,'');
if SR>Cc:
print('\n Slenderness ratio is greater than column constant, so use the euler formula')
I=math.pi*d**4/64.; ##[in^4] Moment of inertia
print'%s %.2f %s '%('\n The moment of inertia is ',I,' in^4');
Pc=math.pi**2*E*I/Le**2; ##[lb] Critical force
##Note- In the book I=0.0075 in^4 is used instead of I=0.0074901 in^4
print'%s %.2f %s '%('\n The critical force is ',Pc,' lb.');
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.6 Page No.55\n');
L=60.; ##[in] Length of column
Sy=36000.; ##[lb/in^2] Yield strength
SF=2.; ##[]Safty factor
E=30*10**6; ##[lb/in^2] Modulus of elasticity
A=2.26; ##[in^2] Area of cross section (Appendix 5.4)
I=0.764; ##[in^4] Moment of inertia (Appendix 5.4)
r=math.sqrt(I/A); ##[in] Radius of gyration
K=0.65; ##[] End support condition factor from Figure 3.8
Le=K*L; ##[in] Effective length
print'%s %.2f %s '%('\n The effective length is ',Le,' in.');
SR=Le/r; ##[] Slenderness ratio
print'%s %.2f %s '%('\n The slenderness ratio is ',SR,'');
Cc=math.sqrt(2*math.pi**2*E/Sy); ##[] Column constant
print'%s %.2f %s '%('\n The column constant is ',Cc,'');
if Cc>SR :
print('\n The column constant is greater than slenderness ratio, so use the Johnson formula.');
F=(A*Sy/SF)*(1.-Sy*SR**2/(4.*math.pi**2*E));
print'%s %.2f %s '%('\n The acceptable load for a safty factor of 2 is ',F,' lb.');