import math
#calculate the elongnation of the bar
P = 5.5 ;##Axial pull in tons
E = 13000 ;##modulus of elasticity tons/in^2
l = 120 ;##length in inches
A = math.pi/4. ;##Area of resisting section in^2
p = P/A ;##Intensity of stress in tons/in^2
e = p/E ;##strain
delta_l = l*e;##elongation of the bar in inches
print'%s %.4f %s'%('The elongation of the bar is',delta_l,'inch');
import math
#calculate the minimum diameter d of each stay bolt
s_p = 200.;##steam pressure in lb/in^2
l = 4.;##length in inches
b = 4.;##breadth in inches
p = 14000.;##permissible streaa in lb/in^2
P = s_p*l*b;##Pull on each bolt in lb-wt
A = P/p ;##necessary area of bolt-section
d = math.sqrt(4*A/math.pi) ;##minimum diameter in inches
print'%s %.2f %s'%('The minimum diameter d of each stay bolt =',d,'inch');
import math
#calculate the safe load in tons
D = 8.;##external diameter in inches
d = 6.;##internal diameter in inches
sigma = 36.;##ultimate stress in tons/in^2
n = 6.;##safety factor
A = 0.25*math.pi*(D**2 - d**2);##Area of section in in^2
P = sigma*A; ##crushing load for the column in tons
P_safe = P/n ;##safe load in tons
print'%s %.2f %s'%('Safe load =',P_safe,'tons');
##there is an error in the answer given in textbook.
import math
#calculate compressive stress of the punch
sigma = 20.;##ultimate sheat stress in tons/in^2
d = 1./2.;##diameter of the hole in inches
t = 3./8.;##thickness of the plate in inches
A = 0.25*math.pi*d**2;##area of the cross-section of the punch in^2
P = math.pi*d*t*sigma;##necessary force in tons
sigma_comp = P/A;##compressive stress on the punch
print'%s %.1f %s'%('The compressive stress of the punch =',sigma_comp,'tons/in^2');
##there is an error in the answer given in textbook.
import math
#calculate change in volume
b = 8.;##width in inches
t = 3./8.;##thickness in inches
l = 20.;##length in feets
P = 22.;##pull in tons
E = 13500.;##modulus of elasticity in tons/in^2
sigma = 0.3;##poisson/s ratio
A = b*t;##in in^2
V = l*A*12;##in cub.inch
p = P/A;##in tons/in^2
e = p/E;
delta_l = e*l*12;##stretch of the bar in inches
Lateral_strain = e*sigma ;##lateral strain
del_b = b*Lateral_strain;##in inches
del_t = t*Lateral_strain;##in inches
k = e*(1-2*sigma);##(del_V)/(V)
del_V = k*V;##change in volume in cub.inch
print'%s %.3f %s'%('The change in volume is',del_V,'cub.inch');
import math
#calculate contraction in diameter and change in volume and work done in streching the bar
d = 7./8.;##diameter of the bar in inches
l = 10.;##length in feets
P = 6.;##axial pull in tons
E = 13000.;##modulus of elsticity in tons/in^2
m = 4.;
A = 0.25*math.pi*d**2;##in in^2
V = 0.25*math.pi*d**2*l*12;##volume in cub.inches
p = P/A;##in tons/in^2
e = p/E;
del_l = e*l*12;##stretchof the bar in inches
Lateral_strain = e/m ;##lateral strain
del_d = Lateral_strain*d;##Contraction in diameter in inches
print'%s %.4f %s'%('The Contraction in diameter is',del_d,'inches');
k = e*(1-2/m);##(del_V)/(V)
del_V = k*V;##change in volume in cub.inch
print'%s %.4f %s'%('The change in volume is',del_V,'cub. inch');
W = 0.5*P*del_l;##work done in stretching the bar in in-ton
print'%s %.4f %s'%('The work done in stretching the bar is',W,'in-ton');
##there is an error in the answer given in textbook.
import math
#calculate total change in length of the bar and energy stored in bar
L = 24.;##length of the bar in ft
d1 = 9./8.;##diameter of the bar in inches
l1 = 6.;##in ft
d2 = 1.;##in inches
l2 = 12.;##in ft
d3 = 5./4.;##in inches
l3 = L-l1-l2;##in ft
P = 10000.;##axial compression in lb-wt
E = 28.*10**6;##modulus of elasticity in lb/in^2
A1 = 0.25*math.pi*d1**2;##in in^2
A2 = 0.25*math.pi*d2**2;##in in^2
A3 = 0.25*math.pi*d3**2;##in in^2
p1 = P/A1 ;##in lb/in^2
e1 = p1/E;
p2 = P/A2 ;##in lb/in^2
e2 = p2/E;
p3 = P/A3 ;##in lb/in^2
e3 = p3/E;
del_l1 = e1*l1*12;##in inches
del_l2 = e2*l2*12;##in inches
del_l3 = e3*l3*12;##in inches
del_l = del_l1+del_l2+del_l3;##total change in length in ft
W = 0.5*P*del_l/12;##energy stored in the bar in ft-lbs
print'%s %.3f %s'%('Total change in length of the bar is',del_l,'inches');
print'%s %.f %s'%('The energy stored in the bar is',W,'ft-lbs');
##there is an error in the answer given in textbook.
import math
#calculate change in lenght of the rod
P = 1200.;##axial pull in lb-wt
d1 = 1.;##diameter of one end in inches
d2 = 0.5;##diameter of other end in inches
l = 10.;##length of the rod in inches
E = 14.*10**6;##modulus of elsticity in lb/in^2
del_l = 4*P*l/(math.pi*E*d1*d2);##change in length in inches
print'%s %.4f %s'%('The change in length of the rod is',del_l,'inches');
import math
#calculate the strain and extension of the bar and the work done in streching
d = 1.;##diameter of the steel bar in inches
l = 12.;##length of the steel bar in inches
d1 = 3./2.;##external diameter in inches
d2 = 1.;##internal diameter in inches
P = 5.;##axial pull in tons
E_s = 30.*10**6;##modulus of elasticity of steel in lb/in^2
E_b = 14.*10**6;##modulus of elasticity of brass in lb/in^2
A_s = 0.25*math.pi*d**2;##area of the steel section in in^2
A_b = 0.25*math.pi*(d1**2-d2**2);##area of the brass section in in^2
P_b = (P/((E_s/E_b)*A_s+A_b))*A_b;##load resisted by the brass tube in tons
P_s = P-P_b;##bal;ance load resisted by the steel tube
e = (P_b/A_b)*2240./E_b ;##strain
print'%s %.4f %s'%('The strain e =',e,'');
del_l = e*l ;##extension of the bar in inches
print'%s %.4f %s'%('The extension of the bar =',del_l,'inches');
W = 0.5*P*del_l;##work done in stretching in inch-ton
print'%s %.3f %s'%('The work done in stretching is',W,'inch-ton');
import math
#calculate safe central load and the reinforcing bar
a = 12.;##length of each side in inches
d = 9./8.;##diameter of each reinforced bar in inches
r = 3.;##distance of centre from the edges in inches
p_c = 600.;##in lb/in^2
n = 18.;##modular ration E_s/E_c
A_s = 4.*0.25*math.pi*d**2;##in in^2
A_c = a**2 - A_s;##in in^2
p_s = n*p_c;##in lb/in^2
P = p_s*A_s+p_c*A_c;##safe central load in lb-wt
print'%s %.f %s'%('Safe central load =',P,'lb-wt');
print'%s %.d %s'%('Of this, the reinforcing bars carry',p_s*A_s,'lb-wt',);
##there is an error in the answer given in textbook.
import math
#calculate P_sand p_b and x
l = 8.;##length in feet
d = 0.5;##diameter in inches
r = 30.;##distance between two rods in inches
P = 2000.;##load in lb-wt
E_s = 30.*10**6;##modulus of elsticity of steel rod
E_b = 16.*10**6;##modulus of elsticity of brass rod
A_s = 0.25*math.pi*d**2;##section area in in**2
p_b = P/(A_s*(1+(E_s/E_b)));
p_s = (P/A_s) - p_b ;
P_b = A_s*p_b;
P_s = A_s*p_s;
print'%s %.1f %s'%('P_s =',P_s,'lb/in**2')
print'%s %.1f %s'%('and P_b =',P_b,'lb/in**2');
x = r*P_b/P ;##
print'%s %.2f %s'%('x =',x,'inches');
import math
#calculate tensile strain imposed by end grips and p and P
alpha = 0.0000062 ;##co-efficient of the expansion in "per F"
t = 100.;##in F
d = 3/4.;##in inches
D = 0.02;##in inches
l = 15.;##in ft
E = 13000.;##in tons/in**2
e = alpha*t - (D/(l*12));
p = E*e;##in tons/in**2
A = 0.25*math.pi*d**2;##in in**2
P = p*A ;##in tons
print'%s %.4f %s'%('Tensile strain imposed by end-grips,e =',e,'');
print'%s %.2f %s'%('p =',p,'tons/in**2')
print'%s %.4f %s'%('P =',P,'tons');
import math
#calculate the stresses induced in each metal and p_s
d = 1.;##diameter of steel bar in inches
d1 = 3./2.;##external diameter of brass tube in inches
d2 = 1.;##internal diameter of brass tube in inches
t = 100.;##in F
alpha_s = 0.0000062;##alpha of steel in "per F"
alpha_b = 0.000010;##alpha of brass in "per F"
E_s = 30.*10**6;##in lb/in^2
E_b = 14.*10**6;##in lb/in^2
A_s = 0.25*math.pi*d**2;##section area of steel bar in in^2
A_b = 0.25*math.pi*(d1**2-d2**2);##section area of brass tube in in^2
p_b = t*(alpha_b-alpha_s)*E_s/((A_b/A_s)+(E_s/E_b));
p_s = (A_b/A_s)*p_b;
print'%s %.d %s'%('The stresses induced in each metal are, p_b =',p_b,'lb/in^2')
print'%s %.d %s'%('p_s =',p_s,'lb/in^2');
import math
#calculate the least temperature the tube must be heated
D = 4.;##diameter of the wheel in ft
p = 6. ;##hoop stress in tons/in^2
alpha = 0.0000062;##in "per F"
E = 13000.;##in tons/in^2
d = (1./(1.+(p/E)))*D*12.;##internal diameter in inches
t = (D*12.-d)/(d*alpha);
print'%s %.1f %s'%('The least temperature the tube must be heated is, t =',t,'F');
import math
#calculate Resultant stress intensity and normal stress intensity and tangential stress intensity and The maximum possible shear on any plane and these planes are inclined at',angle,'degrees to the normal section.
p = 8.;##normal stress intensity in tons/in^2
theta = 35.*math.pi/180.;##inclination of the section in degrees
P = p*math.cos(theta);##resultant stress intensity in tons/in^2
p_n = P*math.cos(theta);##normal stress intensity in tons/in^2
p_t = P*math.sin(theta);##tangential stress intensity in tons/in^2
p_max = 0.5*p;##maximum possible shear in tons/in^2
angle = 45.;##inclination of these planes in degrees
print'%s %.2f %s'%('Resultant stress intensity =',P,'tons/in^2');
print'%s %.2f %s'%('normal stress intensity =',p_n,'tons/in^2');
print'%s %.2f %s'%('tangential stress intensity =',p_t,'tons/in^2');
print'%s %.d %s'%('The maximum possible shear on any plane is',p_max,'tons/in^2');
print'%s %.d %s'%('and these planes are inclined at',angle,'degrees to the normal section.');
import math
#calculate poisson ratio and E,N,K,
d = 9./8.;##diameter of the steel bar in inches
P = 6.;##tensile load in tons
del_l = 0.0036 ;##extension of length inches
l = 8.;##gauge length in inches
del_d = 0.00015;##change in diameter in inches
A = 0.25*math.pi*d**2;##section area in in^2
p = P/A;##stress in tons/in^2
e = del_l/l;##strain
E = p/e;##modulus of elasticity in tons/in^2
LS = del_d/d;##lateral strain
PR = LS/e;##poisson's ratio
N = E/(2.*(1.+PR));##rigidity modulus in tons/in^2
K = E/(3.*(1.-2.*PR));##bulk modulus in tons/in^2
print'%s %.4f %s'%('Poisson ratio 1/m =',PR,'');
print'%s %.d %s'%('E =',E,'tons/in^2');
print'%s %.d %s'%('N =',N,'tons/in^2');
print'%s %.d %s'%('K =',K,'tons/in^2');
##there is an error in the answer given in textbook.
import math
#calculate Poisson ratio and E
N = 2640.;##rigidity modulus in tons/in^2
d = 3./8.;##diameter of the rod in inches
P = 1./2.;##axial pull in tons
del_d = 0.000078;##change in diameter in inches
A = 0.25*math.pi*d**2;##section area in in^2
p = P/A ;##stress tons/in^2
LS = del_d/d;##lateral strain
m = p/(LS*2.*N) - 1.;
E = 2.*N*(1. + 1./m);##modulus of elasticity in ton/in^2
PR = 1./m;##poisson's ratio
print'%s %.3f %s'%('Poisson ratio 1/m =',PR,'');
print'%s %.d %s'%('E =',E,'ton/in^2');
##there is an error in the answer given in textbook.