# Chapter12-Shafts and springs in torsion¶

## Ex1-pg450¶

In [1]:
import math
#calculate The maximum intensity of shear stress induced and The angle of twist in degree
d = 3.;##inches
HP = 120.;##horse power
RPM = 180.;
l = 25.;##feet
N = 12.*10**6.;## lb/in**2
T = 33000.*HP/(2.*math.pi*RPM);##  lb-feet
f_s = 16*T*12/(math.pi*d**3.);## lb/in**2
print'%s %.d %s'%('The maximum intensity of shear stress induced is f_s =',f_s,' lb/in**2');
print'%s %.2f %s'%('The angle of twist in degrees is theta = ',theta*180/math.pi,'');

##there is a minute error in the answer given in textbook.

The maximum intensity of shear stress induced is f_s = 7925  lb/in**2
The angle of twist in degrees is theta =  7.57


## Ex2-pg451¶

In [2]:
import math
#calculate H.P transmitted is
D = 2.;## inches
N = 150.;## RPM
f_s = 9000.;## lb/in^2
M_r = f_s*(math.pi/16.)*D**3.;## lb-inches
HP = M_r*2.*math.pi*N/(12.*33000.);##
print'%s %.2f %s'%('H.P transmitted is',HP,'');

H.P transmitted is 33.65


## Ex3-pg451¶

In [3]:
import math
#calculate Suitable diameter
HP = 80.;
N = 200.;## RPM
m = 30/100.;
f = 12000.;## lb/in^2
T = HP*33000./(2.*math.pi*N);## lb-feet
T_max = (1+m)*T;## lb-feet
D = (T_max*12.*16./(math.pi*f))**(1/3.);## inches
print'%s %.3f %s'%('Suitable diameter is D =',D,'inches');

##the answer is approximated in the textbook.

Suitable diameter is D = 2.405 inches


## Ex4-pg451¶

In [3]:
import math
#check whether it is satisfactory or not for given st A shaft',round(D),'inches external diameter and',round(d),' inches internal diameter
HP = 750.;
N = 90.;## RPM
m = 40/100.;
f = 12000.;## lb/in^2
t = 1.;## inch
T = HP*33000./(2.*math.pi*N);## lb-inches
T_max = (1.+m)*T;## lb-inches
##On solving (4*t)D^3 - (6*t^2)D^2 +(4*t^3 -(16*M/f*%pi))D -t^4 = 0, we get D
D = 7.6;##inches
d = D - 2.;##inches
print'%s %.d %s %d %s '%('A shaft',round(D),'inches external diameter and',round(d),' inches internal diameter will be satisfactory.');

A shaft 8 inches external diameter and 6  inches internal diameter will be satisfactory.


## Ex5-pg452¶

In [4]:
import math
#calculate'D = ',round(D1),' inches will be suitable for the shaft
RPM = 180.;## RPM
HP = 130.;
f = 9000.;## lb/in^2
alpha = 1.;##degree
l = 10.;## feet
N = 6000.;## tons/in^2
T = 33000.*HP/(2*math.pi*RPM);## lb-feet
D1 = (16*T*12/(f*math.pi))**(1/3);## inches
D2 = (T*12*l*12*32*alpha*180/(math.pi*N*math.pi*2240.))**(1/4.);## inches
if D1 > D2 :
print'%s %.d %s'%('D = ',round(D1),' inches will be suitable for the shaft');
else:
print'%s %.d %s'%('D =',round(D2),'inches will be suitable for the shaft');

D = 4 inches will be suitable for the shaft


## Ex6-pg453¶

In [5]:
import math
#calculate 'The saving in weight per foot run and percentage saving
HP = 3000.;
RPM = 60.;
f = 12000.;##lb/in^2
rho = 480.;##lb. per sq. foot
k = 3./4.;## k = d/D
T = HP*33000.*12./(2.*math.pi*RPM);## lb-inches
D1 = (T*16/(f*math.pi))**(1/3.);##inches
D2 = (T/((1.+k**2.)*(1.-k**2.)*math.pi*f/16.))**(1/3.);##inches
d = k*D2;## inches
w1 = 0.25*math.pi*D1**2 *rho/144. ;## lb-wt
w2 = 0.25*math.pi*(D2+d)*(D2-d)*rho/144. ;## lb-wt
w = w1-w2;## lb-wt
n = (w/w1)*100.;
print'%s %.d %s'%('The saving in weight per foot run is w =',w,' lb-wt');
print'%s %.2f %s'%('Percentage saving is ',n,'');

##there is a minute error in the answer given in textbook.

The saving in weight per foot run is w = 138  lb-wt
Percentage saving is  43.62


## Ex7-pg454¶

In [7]:
import math
#calculate T1 and T2
l1 = 3.;## feet
d1 = 1.;## feet
l2 = 9.;## feet
M = 200.;## lb-wt
l = 9.;## inches
N = 12.*10**6;## lb/in^2
k = l2/l1;
T1 = M/(1.+k);## lb-feet
T2 = k*T1;## lb-feet
f_s = T2*12./(math.pi/16.);## lb/in^2
print'%s %.d %s %.d %s '%('T1 =',T1,'lb-feet, T2 = ',T2,' lb-feet');
print'%s %.d %s'%('f_s = ',f_s,' lb/in^2');
print'%s %.4f %s %.3f %s'%(' theta = ',theta,' radian ,  theta    = ',theta*180/math.pi,' degrees');
##there is a minute error in the answer given in twxtbook

T1 = 50 lb-feet, T2 =  150  lb-feet
f_s =  9167  lb/in^2
theta =  0.0138  radian ,  theta    =  0.788  degrees


## Ex8-pg456¶

In [8]:
import math
#calculate forces
D = 5.;## inches
HP = 120.;
RPM = 150.;
b = 5.;## inches
h = 1.;##inch
n = 6.;## no. of bolts
d = 3/4.;## inches
T = HP*33000.*12./(2.*math.pi*RPM);## lb-inches
f_s = T*16./(math.pi*27.);
f_k = T/(b*h*2.*d);
f_b = T/(n*0.25*math.pi*d**2 * b);## lb-inches
print'%s %.d %s %d %s %d %s'%('f_s =',f_s,'lb/in^2, f_k =',f_k,'lb/in^2, f_b =',f_b,' lb/in^2');

##there are errors given in the answers given in the textbook

f_s = 9510 lb/in^2, f_k = 6722 lb/in^2, f_b = 3804  lb/in^2


## Ex9-pg460¶

In [9]:
import math
#calculate 'Maximum strain  energys
d = 4.;##inches
T = 30.;## ton-inches
M = 20.;##ton-inches
m = 1/0.3;
f_s = 16.*T/(math.pi*d**3);## tons/in^2
f_b = 32*M/(math.pi*d**3);## tons/in^2
theta1 = theta*180./math.pi;
theta2 = theta1+90.;
f1 = 0.5*f_b + math.sqrt(f_s**2 + 0.25*f_b**2);## tons/in^2
f2 = 0.5*f_b - math.sqrt(f_s**2 + 0.25*f_b**2);## tons/in^2
Ee = f1 - (f2/m);## tons/in^2
f = math.sqrt(f1**2 + f2**2 - 2*f1*f2/m);## tons/in^2
print'%s %.3f %s'%('Maximum strain is Ee = ',Ee,' tons/in^2');
print'%s %.3f %s'%(' Maximum strain energy is f = ',f,' tons/in^2');

Maximum strain is Ee =  4.844  tons/in^2
Maximum strain energy is f =  4.995  tons/in^2


## Ex10-pg461¶

In [6]:
import math
#calculate diameter of the shaft in all  different cases
HP = 80.;
RPM = 120.;
b = 10.;## feet
h = 3.;## feet
F = 8000.;## lb-wt
m = 4.;
T = HP*33000.*12./(2.*math.pi*RPM*2240.);## ton-inches
M = F*h*(b-h)*12./(b*2240.);## ton-inches
##(i) The major principal stress f1 is given by
f1 = 6;## tons/in^2
d1 = ((M+math.sqrt(M**2 + T**2))*16/(math.pi*f1))**(1/3.);## inches

##(ii) If f_s_dash is the maximum intensity of shear stress
f_s_dash = 3;## tons/in^2
d2 = (math.sqrt(M**2 + T**2) * 16./(math.pi*f_s_dash))**(1/3.);## inches

##(iii) If e is the major principal strain
Ee = 6.;## tons/in^2
d3 = (((1.-(1./m))*M + (1.+(1./m))*math.sqrt(M**2 + T**2))*16./(math.pi*Ee))**(1/3.);## inches

##(iv) If f is the direct stress which, acting alone will produce the same maximum strain energy
f = 6.;## tons/in^2
d4 = ((math.sqrt(4*M**2 + 2.*(m+1.)*(T**2)/m))*16./(math.pi*f))**(1/3.);## inches
print'%s %.3f %s %.3f %s %.3f %s %.3f %s'%('The diameter of the shaft in different cases will be, (i) d = ',d1,' inches,(ii) d = ',d2,' inches  , (iii) d = ',d3,' inches  ,(iv) d = ',d4,'inches');
##there are round-off errors in the answers given in textbook.

The diameter of the shaft in different cases will be, (i) d =  5.365  inches,(ii) d =  5.384  inches  , (iii) d =  5.370  inches  ,(iv) d =  5.370 inches


## Ex11-pg463¶

In [7]:
import math
#calculate The maximum principal stress and The maximum shear intensity and maximum strain
D = 12.;## inches
d = 6.;## inches
HP = 2400.;
RPM = 80.;
M = 40.;## ton-feet
P = 25.;## tons
PR = 0.3;##poisson's ratio
A = 0.25*math.pi*(D**2 - d**2);## in**2
Z = (math.pi/32)*(D**4 - d**4)/D;## in**3
J = (math.pi/16)*(D**4 - d**4)/D;## in**3
p_0 = P/A ;## ton/in**2
p_b = M*12/Z ;## tons/in**2
f_b = p_0 + p_b;##tons/in**2
f_s = HP*33000*12/(2*math.pi*RPM*2240*J);## tons/in**2
theta1 = theta*180/math.pi;## degrees
theta2 = theta1+90;##degrees
f_1 = 0.5*f_b + math.sqrt(f_s**2 + 0.25*f_b**2);##tons/in**2
f_2 = 0.5*f_b - math.sqrt(f_s**2 + 0.25*f_b**2);##tons/in**2
f = math.sqrt(0.25*f_b**2 + f_s**2);## tons/in**2
Ee = f_1 - PR*f_2;## tons/in**2
print'%s %.2f  %s %.2f %s'%('The maximum principal stresse are f_1 = ',f_1,' tons/in**2.,compressive ,  f_2 = ',-f_2,' tons/in**2., tensile');
print'%s %.1f %s %.1f %s'%('theta1 = ',theta1,' degrees,theta2 =',theta2,'degrees');
print'%s %.1f %s '%(' The maximum shear intensity = ',f,'tons/in**2');
print'%s %.3f %s'%(' Maximum strain is, Ee =',Ee,'tons/in**2');

##there are minute errors in the answers given in textbook.

The maximum principal stresse are f_1 =  4.78   tons/in**2.,compressive ,  f_2 =  1.47  tons/in**2., tensile
theta1 =  29.0  degrees,theta2 = 119.0 degrees
The maximum shear intensity =  3.1 tons/in**2
Maximum strain is, Ee = 5.226 tons/in**2


## Ex12-pg466¶

In [8]:
import math
#calculate 'The size of the shaft and angle of the twist in the shaft for a length
RPM = 180.;
P = 10.;## tons
v = 25.;## feet per minute
n = 64/100.;## efficiency of the crane
f = 5500.;## lb/in^2
l = 10.;## feet
N = 12*10**6.;## lb/in^2
W = P*v*2240./n;## ft-lbs
T = W*12./(2.*math.pi*RPM);## lb-inches
s = (T/(0.208*f))**(1/3.);## inches
theta = 7.11*T*l*12.*180./(math.pi*N*s**4.);## degrees
print'%s %.3f %s'%('The size of the shaft is s = ',s,' inches');
print'%s %.1f %s %.2f %s'%('The angle of the twist in the shaft for a length of ',l,' feet, theta =',theta,' degrees');
##there is a round-off error in the answer given in textbook.

The size of the shaft is s =  2.010  inches
The angle of the twist in the shaft for a length of  10.0  feet, theta = 2.32  degrees


## Ex13-pg469¶

In [9]:
import math
#calculate Shear stress induced  and  Shear stress induced  and  Energy stored
d = 3/8.;## inches
n = 12.;##no. of complete turns
D = 4.;## inches
W = 50.;## lb-wt
N = 12*10**6;## lb/in^2
T = W*0.5*D;## lb-inches
f_s = T*16./(math.pi*d*83);##lb/in^2
delta = 64.*W*(D**3 /8.)*n/(N*d**4.);## inches
E = 0.5*W*delta;## inch-lbs
print'%s %.d %s'%('Shear stress induced is f_s =',f_s,' lb/in^2');
print'%s %.3f %s'%(' Deflection under the pull is delta = ',delta,' inches');
print'%s %.3f %s'%(' Energy stored = ',E,' lb-inches');

##there is a minute error in the answer given in textbook.

Shear stress induced is f_s = 16  lb/in^2
Deflection under the pull is delta =  1.295  inches
Energy stored =  32.363  lb-inches


## Ex14-pg469¶

In [10]:
import math
#calculate The number of springs required
W = 2.;## tons
v = 4.;## miles per hour
n = 18.;## no. of coils
delta = 9.;## inches
N = 6000.;## tons/in^2
d = 1.;## inch
D = 8.;## inches
KE = 12.*(W*(v*44./30.)**2.)/(2.*32.);## inch-tons
P = (delta*N*d**4)/(64.*n*(0.5*D)**3);## tons
E = 0.5*P*delta;## inch-tons
m = KE/E ;## no. of springs required
print'%s %.d %s'%('The number of springs required m =',round(m),'');

The number of springs required m = 4


## Ex15-470¶

In [11]:
import math
#calculate height of drop
W = 5.;## cwt
n = 18.;## no. of coils
delta = 9.;## inches
d = 1.;## inch
D = 8.;## inches
N = 6000.;## tons/in^2
P = (delta*N*d**4.)/(64.*n*(0.5*D)**3);## tons
h = (0.5*P*delta*20./W)-delta;## inches
print'%s %.3f %s'%('The height of drop h = ',h,' inches');

The height of drop h =  4.184  inches


## Ex16-pg471¶

In [12]:
import math
#calculate  Maximum possible axial load and  Deflection, delta
s = 1/4.;## inch
n = 12.;## no. of coils
D = 3.;## inches
f_s = 45000.;## lb/in^2
N = 12*10**6;## lb/in^2
T = 0.208*f_s*s**3;## lb-inches
W = T/(0.5*D);## lb-wt
theta = 7.11*T*math.pi*D*12./(N*s**4);##rdaians
delta = 0.5*D*theta;## inches
print'%s %.1f %s'%('Maximum possible axial load is W = ',W,' lb-wt');
print'%s %.3f %s'%(' Deflection, delta = ',delta,' inches');

##there is a minute error in the answer given in textbook.

Maximum possible axial load is W =  97.5  lb-wt
Deflection, delta =  3.763  inches


## Ex17-pg472¶

In [13]:
import math
#calculate The bending stresses
d = 3/8.;## inches
n = 12.;##no. of complete turns
D = 4.;## inches
W = 50.;## lb-wt
N = 12.*10**6;## lb/in^2
E = 30.*10**6;## lb/in^2
M = 75.;## lb-inches
I = (math.pi/64.)*d**4;## in^4
Z = 2.*I/d;## in^3
f = M/Z ;## lb/in^2
n_ = (phi/(2*math.pi)) + n;## increase in no. of turns
print'%s %.d %s'%('The bending stress is f = ',f,' lb/in^2');
print'%s %.3f %s'%(' n_new = ',n_,' turns');

##there are minute errors in the answers given in textbook.

The bending stress is f =  14486  lb/in^2
n_new =  12.062  turns


## Ex18-pg476¶

In [14]:
import math
#calculate The maximum intensity of shear stress and 'Deflection
d = 3/8.;## inches
n = 12.;##no. of complete turns
D = 4.;## inches
W = 50.;## lb-wt
N = 12.*10**6.;## lb/in^2
alpha = 15*math.pi/180.;## degrees
E = 30.*10**6.;## lb/in^2
T = W*0.5*D*math.cos(alpha);## lb-inches
M = W*0.5*D*math.sin(alpha);## lb-inches
J = math.pi*d**4 /32.;## in^4
I = math.pi*d**4 /64.;## in^4
delta = 64.*W*((D/2.)**3.)*n/math.cos(alpha)*((math.cos(alpha)**2.)/N + (2.*math.sin(alpha)**2.)/E)/d**4. ;## inches
f = 32.*W*0.5*D*math.sin(alpha)/(math.pi*d**3) ;## lb/in^2
f_s = T*16./(math.pi*d**3);## lb/in^2
f_1 = 0.5*f + math.sqrt(f_s**2. + 0.25*f**2.);## lb/in^2
f_2 = 0.5*f - math.sqrt(f_s**2. + 0.25*f**2.);## lb/in^2
f_s_dash = math.sqrt(f_s**2. + 0.25*f**2.);## lb/in^2
print'%s %.3f %s'%('Deflection, delta = ',delta,' inches');
print'%s %.d %s %.d %s'%(' f = ',f,' lb/in^2, f_s = ',f_s,' lb/in^2');
print'%s %.d %s'%('The maximum intensity of shear stress =',f_s_dash,' lb/in^2');

##there are calculation errors in the answers given in textbook

Deflection, delta =  1.322  inches
f =  4999  lb/in^2, f_s =  9328  lb/in^2
The maximum intensity of shear stress = 9657  lb/in^2


## Ex19-pg477¶

In [15]:
import math
#calculate 'Angle of rotation and 'Angle of rotation
d = 3/8.;## inches
n = 12.;##no. of complete turns
D = 4.;## inches
M = 75.;## lb-inches
N = 12.*10**6.;## lb/in**2
alpha = 15*math.pi/180.;## degrees
E = 30.*10.**6.;## lb/in**2
phi_dash = (64./d**4.)*M*0.5*D*n/math.cos(alpha)*((2.*(math.cos(alpha))**2.)/E + ((math.sin(alpha))**2)/N);## radians
DELTA = 64.*M*((0.5*D)**2.)*n*math.sin(alpha)*((1./N) - (2./E))/d**4.;## inches
print'%s %.2f %s %.3f %s'%('Angle of rotation phi_dash = ',phi_dash,' radians or ',phi_dash*180/math.pi,' degrees');
print'%s %.4f %s '%(' The axial deflection = ',DELTA,' inches');

Angle of rotation phi_dash =  0.41  radians or  23.422  degrees
The axial deflection =  0.0503  inches