14 files changed, 230 insertions, 0 deletions
diff --git a/2744/CH10/EX10.1/Ex10_1.sce b/2744/CH10/EX10.1/Ex10_1.sce
new file mode 100755
index 000000000..9c5d2753a
--- /dev/null
+++ b/2744/CH10/EX10.1/Ex10_1.sce
@@ -0,0 +1,7 @@
+clear;
+clc;
+d = 2;// feet
+p = 250;// lb/in^2
+f = 12000;// lb/in^2
+t_limit = p*d*12/(2*f) ;// inches
+printf('The necessary thickness of metal for seamless pipe is %.2f inches',t_limit);
diff --git a/2744/CH10/EX10.10/Ex10_10.sce b/2744/CH10/EX10.10/Ex10_10.sce
new file mode 100755
index 000000000..6ca3835e1
--- /dev/null
+++ b/2744/CH10/EX10.10/Ex10_10.sce
@@ -0,0 +1,27 @@
+clear;
+clc;
+d = 9;// inches
+p = 5000/2240;// lb/in^2
+f = 8;// tons/in^2
+PR = 0.3;// Poisson's ratio
+//(i) Maximum principal stress hypothesis:
+k_limit1 = sqrt((f + p)/(f - p));//k_limit = r1/r2
+r_limit1 = k_limit1*0.5*d;//inches
+printf('The outer radius in case(i), r2 = %.3f inches',r_limit1);
+//(ii) Maximum principal strain:
+k_limit2 = sqrt(((f/p - PR)+1)/(f/p - PR -1));
+r_limit2 = k_limit2*0.5*d;// inches
+printf('\n The outer radius in case(ii), r2 = %.3f inches',r_limit2);
+//(iii) Maximum shear stress:
+k_limit3 = sqrt(f/(2*p) /((f/(2*p)) - 1));
+r_limit3 = k_limit3*0.5*d;// inches
+printf('\n The outer radius in case(iii), r2 = %.3f inches',r_limit3);
+//(iv) Maximum strain energy
+K1 = (f^2 /p^2)/(2*((f^2 /(2*p^2)) - (1+PR)));
+K2 = K1^2;
+K3 = ((f^2 /(2*p^2)) - (1-PR))/((f^2 /(2*p^2)) - (1+PR));
+k_limit4 = sqrt(K1+sqrt(K2-K3));
+r_limit4 = k_limit4*0.5*d;// inches
+printf('\n The outer radius in case(iv), r2 = %.3f inches',r_limit4);
+
+//there are calculation errors in the answer given in text book
diff --git a/2744/CH10/EX10.11/Ex10_11.sce b/2744/CH10/EX10.11/Ex10_11.sce
new file mode 100755
index 000000000..b69489eb9
--- /dev/null
+++ b/2744/CH10/EX10.11/Ex10_11.sce
@@ -0,0 +1,29 @@
+clear;
+clc;
+r1 = 0.5*6;// inches
+r2 = 0.5*12;// inches
+r3 = 0.5*10;// inches
+p = 1500;// lb/in^2
+p_f = 12000;// lb/in^2
+//Initially, for the inner tube
+b = -p/((1/r1^2) - (1/r3^2));
+a = b/r1^2;
+f_3 = (b/r1^2) +a;// lb/in^2
+f_5 = (b/r3^2) +a;// lb/in^2
+//for the outer tube
+b1 = p/((1/r3^2)-(1/r2^2));
+a1 = b1/r2^2;
+f1_5 = (b1/r3^2)+a1;// lb/in^2
+f1_6 = (b1/r2^2)+a1;// lb/in^2
+//When the fluid pressure of 12000 lb/in^2, is admitted into the compound tube
+B = p_f/((1/r1^2)-(1/r2^2));
+A = B/(r2^2);
+f_3_ = (B/r1^2)+A;// lb/in^2
+f_5_ = (B/r3^2)+A;// lb/in^2
+f_6_ = (B/r2^2)+A;// lb/in^2
+printf('The hoop stresse are');
+printf('\n at x = 3 inches,x = 5 inches initially on inner tube are %.1f lb/in^2.., compressive,  %.1f lb/in^2..,compressive respectively',-f_3,-f_5);
+printf('\n at x = 5 inches,x = 6 inches initially on outer tube are %.1f lb/in^2.., tensile,  %.1f lb/in^2..,tensile respectively',f1_5,f1_6);
+printf('\n at x = 3 inches,x = 5 inches and x = 6 inches due to fluid pressure are %d lb/in^2.., tensile,  %d lb/in^2..,tensile, and %d lb/in^2..,tensile respectively',f_3_,f_5_,f_6_')
+printf('\n at x = 3 inches,x = 5 inches finally on inner tube are %.1f lb/in^2.., tensile,  %.1f lb/in^2..,tensile respectively',f_3_+f_3,f_5_+f_5);
+printf('\n at x = 5 inches,x = 6 inches finally on outer tube are %d lb/in^2.., tensile,  %d lb/in^2.., tensile respectively',f1_5+f_5_,f1_6+f_6_);
diff --git a/2744/CH10/EX10.12/Ex10_12.sce b/2744/CH10/EX10.12/Ex10_12.sce
new file mode 100755
index 000000000..f4c9fee1b
--- /dev/null
+++ b/2744/CH10/EX10.12/Ex10_12.sce
@@ -0,0 +1,11 @@
+clear;
+clc;
+p = 1500;// lb/in^2
+E = 30*10^6;// lb/in^2
+f1_5 = 8318;// lb/in^2
+f2_5 = 3187.5;// lb/in^2
+alpha = 0.0000062;// per F
+r3 = 6;// inches
+del_r3 = r3*(f1_5+f2_5)/E;// inches
+t = ((f1_5+f2_5)/E)/(alpha);// inches
+printf('The minimum temperature to which outer tube should be heated before it can be slipped on, t = %.2f F',t);
diff --git a/2744/CH10/EX10.13/Ex10_13.sce b/2744/CH10/EX10.13/Ex10_13.sce
new file mode 100755
index 000000000..034d3333d
--- /dev/null
+++ b/2744/CH10/EX10.13/Ex10_13.sce
@@ -0,0 +1,23 @@
+clear;
+clc;
+r1 = 0.5*9;// inches
+r2 = 0.5*3;// inches
+r3 = 0.5*6;// inches
+del_r3 = 0.5*0.003;// inches
+E = 13000;// tons/in^2
+k1 = r1/r3;
+k2 = r2/r3;
+a1 = (del_r3/r3)*E/((k1^2 +1)- (k2^2 +1)*(k1^2 -1)/(k2^2 -1));
+a = a1*(k1^2 -1)/(k2^2 -1);
+b1 = a1*r1^2;
+b = a*r2^2;
+p_ = (b/r3^2) -a;// tons/in^2
+// for the inner tube
+f_x1 = (b/r2^2) +a;// tons/in^2
+f_x2 = (b/r3^2) +a;// tons/in^2
+// for the outer tube
+f_x3 = (b1/r3^2) +a1;// tons/in^2
+f_x4 = (b1/r1^2) +a1;// tons/in^2
+printf('The hoop stresses are as under:');
+printf('\n For the inner tube, at x = 1/5 inches, f = %.2f tons/in^2., compressive\n   at x = 3 inches, f = %.2f tons/in^2.,compressive',-f_x1,-f_x2);
+printf('\n For the outer tube, at x = 3 inches, f = %.2f tons/in^2., tensile\n   at x = 4.5 inches, f = %.2f tons/in^2.,tensile',f_x3,f_x4);
diff --git a/2744/CH10/EX10.14/Ex10_14.sce b/2744/CH10/EX10.14/Ex10_14.sce
new file mode 100755
index 000000000..d54026b1a
--- /dev/null
+++ b/2744/CH10/EX10.14/Ex10_14.sce
@@ -0,0 +1,13 @@
+clear;
+clc;
+r1 = 0.5*5;// inches
+p = 5000;// lb/in^2
+f = 5;// tons/in^2
+b = (f + p/2240)/((1/r1^3) + (2/r1^3));
+a = f - (b/r1^3);
+//external diameter
+r = (2*b/a)^(1/3);// inches
+t = r - r1;// inches
+printf('The thickness of the shell required, t = %.3f inches',t);
+
+//the answer is approximated in the text book
diff --git a/2744/CH10/EX10.2/Ex10_2.sce b/2744/CH10/EX10.2/Ex10_2.sce
new file mode 100755
index 000000000..1f7d8190a
--- /dev/null
+++ b/2744/CH10/EX10.2/Ex10_2.sce
@@ -0,0 +1,18 @@
+clear;
+clc;
+l = 8;//feet
+d = 3;// feet
+t = 1/2;// inches
+p = 200;// lb/in^2
+E = 30*10^6;// lb/in^2
+PR = 0.3;// poisson's ratio
+f1 = p*d*12/(2*t);// lb/in^2
+f2 = p*d*12/(4*t);// lb/in^2
+f_s = 0.5*(f1-f2);// lb/in^2
+e1 = (f1/E)-(PR*f2/E);// lb/in^2
+e2 = (f2/E)-(PR*f1/E);// lb/in^2
+del_d = e1*d*12;// inches
+del_l = e2*l*12;// inches
+del_V = (e2+2*e1)*0.25*%pi*(12*d)^2 * l*12;// cub. inches
+printf('Maximum intensity of shear stress induced = %d lb/in^2',f_s);
+printf('\n del_d = %.6f inches\n del_l = %.6f inches\n del_V = %.1f cub. inches',del_d,del_l,del_V);
diff --git a/2744/CH10/EX10.3/Ex10_3.sce b/2744/CH10/EX10.3/Ex10_3.sce
new file mode 100755
index 000000000..177d7a6a4
--- /dev/null
+++ b/2744/CH10/EX10.3/Ex10_3.sce
@@ -0,0 +1,12 @@
+clear;
+clc;
+d = 30;// inches
+H = 300;// feet
+w = 62.5;
+f = 2800;
+//intensity of water pressur
+p = w*H/144;// lb/in^2
+t_limit = p*d/(2*f);// inches
+printf('Thickness of metal required is %.4f inches',t_limit);
+
+//the answer is correct only, but it is approximated in the text book.
diff --git a/2744/CH10/EX10.4/Ex10_4.sce b/2744/CH10/EX10.4/Ex10_4.sce
new file mode 100755
index 000000000..98bdd536a
--- /dev/null
+++ b/2744/CH10/EX10.4/Ex10_4.sce
@@ -0,0 +1,14 @@
+clear;
+clc;
+d = 78;// inches
+t = 3/4;// inches
+n1 = 70/100;// efficiency of the longitudinal riveted joint
+f = 6;// tons/in^2
+n2 = 60/100;// efficiency of the circumferential riveted joint
+p = f*2240/(d/(2*t*n1));//lb/in^2
+p = round(p-1);
+f1 = p*d/(2*t);// lb/in^2
+f2 = p*d/(4*t*n2);// lb/in^2
+printf('The permissible steam pressure, p = %d lb/in^2',p);
+printf('\n The circumferential stress, f1 = %d lb/in^2 = %.2f tons/in^2',f1,f1/2240);
+printf('\n The longitudinal stress, f2 = %d lb/in^2 = %.2f tons/in^2',f2,f2/2240);
diff --git a/2744/CH10/EX10.5/Ex10_5.sce b/2744/CH10/EX10.5/Ex10_5.sce
new file mode 100755
index 000000000..a1b9849f6
--- /dev/null
+++ b/2744/CH10/EX10.5/Ex10_5.sce
@@ -0,0 +1,10 @@
+clear;
+clc;
+d = 4;// feet
+p = 200;// lb/in^2
+f = 15000;// lb/in^2
+n = 0.7;// efficiency
+t_limit = p*d*12/(4*f*n);// inches
+printf('The thickness of the plate required = %.2f inches',t_limit);
+
+//the answer is correct only, but it is approximated in the text book.
diff --git a/2744/CH10/EX10.6/Ex10_6.sce b/2744/CH10/EX10.6/Ex10_6.sce
new file mode 100755
index 000000000..492f1f8ae
--- /dev/null
+++ b/2744/CH10/EX10.6/Ex10_6.sce
@@ -0,0 +1,14 @@
+clear;
+clc;
+d = 3;// feet
+t = 1/4;// inches
+del_V = 9;// cub. inches
+E = 30*10^6;// lb/in^2
+PR = 0.3;// poisson's ratio
+V = (%pi/6)*(12*d)^3;// in^3
+k = del_V/V;
+f = k*E/(3*(1-PR));// lb/in^2
+p = 4*f*t/(12*d);// lb/in^2
+printf('The pressure exerted by fluid on the shell, p = %d lb/in^2',p);
+
+//there is a minute calculation error in the answer given in text book
diff --git a/2744/CH10/EX10.7/Ex10_7.sce b/2744/CH10/EX10.7/Ex10_7.sce
new file mode 100755
index 000000000..8a70754af
--- /dev/null
+++ b/2744/CH10/EX10.7/Ex10_7.sce
@@ -0,0 +1,23 @@
+clear;
+clc;
+d = 12;// feet
+t = 1/2;// inches
+d1 = 1/4;// inches
+p = 500;// lb/in^2
+E_c = 6000;// tons/in^2
+PR = 0.3;// Poisson's ratio
+E_s = 13000;// tons/in^2
+f_t = 8000;// lb/in^2
+l = 8;
+P_c = l*(%pi/64)*f_t;// lb-wt
+f_c = P_c/(2*t);// lb/in^2
+//bursting force per inch unit length
+f_b = p*d;// lb-wt
+f_p = (f_b + (l*0.049*PR*p*d/(4*t))*(E_s/E_c))/(1 + (l*0.049)*E_s/E_c);// lb/in^2 
+f_w = (f_p - PR*p*d/(4*t))*E_s/E_c;// lb/in^2
+printf('                  Pipe                         Steel-wire');
+printf('\n Initually,    %d lb/in^2.,compr.          %d lb/in^2., tensile',f_c,f_t);
+printf('\n Due to p,     %d lb/in^2.,tensile.        %d lb/in^2., tensile',f_p,f_w);
+printf('\n Finally,      %d lb/in^2.,tensile.        %d lb/in^2., tensile',f_p-f_c,f_w+f_t);
+
+//there is a calculation error in the answer given in text book
diff --git a/2744/CH10/EX10.8/Ex10_8.sce b/2744/CH10/EX10.8/Ex10_8.sce
new file mode 100755
index 000000000..f665b3e64
--- /dev/null
+++ b/2744/CH10/EX10.8/Ex10_8.sce
@@ -0,0 +1,14 @@
+clear;
+clc;
+d = 12;// inches
+t = 3;// inches
+p_x1 = 900;// lb/in^2
+x1 = 0.5*d;// inches
+p_x2 = 0;
+x2 = 0.5*d+t;// inches
+//from Lame's formulae
+b = (p_x1-p_x2)/((1/x1^2)-(1/x2^2));
+a = (b/x1^2)- p_x1;
+f_x1 = (b/x1^2)+a;// lb/in^2
+f_x2 = (b/x2^2)+a;// lb/in^2
+printf('The maximum and minimum intensities of circumferential stresses are: f_6 = %d lb/in^2.,tensile\n   f_9 = %d lb/in^2., tensile',f_x1,f_x2);
diff --git a/2744/CH10/EX10.9/Ex10_9.sce b/2744/CH10/EX10.9/Ex10_9.sce
new file mode 100755
index 000000000..f11f7dc79
--- /dev/null
+++ b/2744/CH10/EX10.9/Ex10_9.sce
@@ -0,0 +1,15 @@
+clear;
+clc;
+d = 5;// inches
+p = 3;// tons/in^2
+f = 8;// tons/in^2
+x = 0.5*d;// inches
+b = (p+f)/(2/x^2);
+a = f-(b/x^2); 
+r = sqrt(b/a);//outer radius
+t = r-0.5*d;//thickness
+D = 2*t+d;//outer diameter
+printf('The thickness of metal necessary, t = %.3f inches',t);
+printf('\n the outer diameter will be, D = %.1f inches',D);
+
+//the answer is correct, but it is approximated in the text book.