Chapter7-Impact and Energy Analysis

Ex1-pg137

In [1]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-7.1 Page No.137\n');

D=2.;                  ##[in] Diameter of bar
W=500.;                ##[lb] Weight
h=1.;                  ##[in] Height from which the weight falls
A=math.pi*D**2/4.;          ##[in^2] Area of cross section of bar
L=10.;                 ##[in] Length of bar
E=30*10**6;            ##[lb/in^2] Modulus of elasticity

S=(W/A)+(W/A)*(1.+(2.*h*E*A/(L*W)))**(0.5);  ##[lb/in^2] Stress in the bar

print'%s %.2f %s '%('\n Stress in the bar is ',S,' lb/in^2.');

Delta=S*L/E;          ##[in] Deflection

print'%s %.2f %s '%('\n Deflecton in the bar is ',Delta,' in.');
MACHINE DESIGN 
 Timothy H. Wentzell, P.E. 
 EXAMPLE-7.1 Page No.137


 Stress in the bar is  31061.50  lb/in^2. 

 Deflecton in the bar is  0.01  in. 

Ex2-pg139

In [2]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-7.2 Page No.139\n');

W=2000.;             ##[lb] Weight of automobile
L=36.;               ##[in] Length of stop
D=2.;                ##[in] Diameter of steel bar
V=5.*5280.*12./3600.;   ##[in/s] Velocity of automobile

A=math.pi*D**2/4.;        ##[in^2] Area of cross section of bar
E=30.*10**6;          ##[lb/in^2] Modulus of elasticity

k=A*E/L;            ##[lb/in] Stiffness of the bar
g=386.;              ##[in/s^2] Acceleration due to gravity

Delta=math.sqrt(2./k*W*(V**2./(2.*g)+0.));  ##[in] Deflection

print'%s %.2f %s '%('\n The deflection in the bar is ',Delta,' in.');

S=Delta*E/L;        ##[in] Stress in the bar

##Note-In the book Delta=0.124 is used instead of Delta=0.123800

print'%s %.2f %s '%('\n The stress in the bar is ',S,' lb/in^2.');
MACHINE DESIGN 
 Timothy H. Wentzell, P.E. 
 EXAMPLE-7.2 Page No.139


 The deflection in the bar is  0.12  in. 

 The stress in the bar is  103166.44  lb/in^2. 

Ex3-pg141

In [3]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-7.3 Page No.141\n');

W=3000.;                ##[lb] Weight of automobile
L=40*12;               ##[in] Length of the beam
I=64.2;                ##[in^4] Moment of inertia of the beam
Sy=48000.;              ##[lb/in^2] Yield strength of the beam
c=8/2.;                 ##[in] Distance from the outermost fiber to neutral axis
E=30.*10**6;             ##[lb/in^2] Modulus of elasticity
g=32.2;                ##[ft/s^2] Acceleration due to gravity

M=I*Sy/c;              ##[lb*in] Moment at which beam will yield
F=4.*M/L;               ##[lb] Force at which beam will yield

Delta=F*L**3/(48.*E*I);  ##[in] Deflection
KE=F*Delta/2.;          ##[lb*in] Kinetic energy

V=math.sqrt(2.*g*KE/W);      ##[in/s] Velocity
V=V/5280.*3600.;         ##[miles/hr] Velocity

print'%s %.2f %s '%('\n At ',V,' miles/hr velocity the beam will yield.');
MACHINE DESIGN 
 Timothy H. Wentzell, P.E. 
 EXAMPLE-7.3 Page No.141


 At  15.68  miles/hr velocity the beam will yield. 

Ex4-pg143

In [9]:
import math
import numpy

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-7.4 Page No.143\n');

D=3/4.;                ##[in] Diameter of the bolt
At=0.334;             ##[in^2] Area of thread
As=math.pi*D**2/4.;         ##[in^2] Area of shank

##Note-In the book As=0.442 in^2 is used instead of As=0.4417865 in.

E=30*10**6;            ##[lb/in^2] Modulus of elasticity
Lt=2.;                 ##[in] Length of the thread
Ls=6.;                 ##[in] Length of the shank
h=0.03;               ##[in] Height from which the weight falls
W=500.;                ##[lb] Falling load

Kt=At*E/Lt;           ##[lb/in] Stiffness of threaded portion
Ks=As*E/Ls;           ##[lb/in] Stiffness of shank

K=Kt*Ks/(Kt+Ks);      ##[lb/in] Overall stiffness

Delta=(W/K)+(W/K)*math.sqrt(1.+2.*h*K/W); ##[in] Deflection

A=numpy.matrix([[Ls/E, Lt/E], [0.442, -0.334]]);
b=numpy.matrix([[Delta] ,[0]]);
S=A/b


Ln=8.;               ##[in] Length when shank has same area as threads
Kn=At*E/Ln;         ##[lb/in] Stiffness
Deltan=(W/Kn)+(W/Kn)*math.sqrt(1.+2.*h*Kn/W);  ##[in] Deflection
S=Deltan*E/Ln;      ##[ln/in^2] Stress

print'%s %.2f %s %.2f %s '%('\n If shank has the same area as threads then stress is ',S,' lb/in^2 and deflection is ',Deltan,' in.');
MACHINE DESIGN 
 Timothy H. Wentzell, P.E. 
 EXAMPLE-7.4 Page No.143


 If shank has the same area as threads then stress is  19910.79  lb/in^2 and deflection is  0.01  in.