10 files changed, 152 insertions, 0 deletions
diff --git a/2744/CH6/EX6.1/Ex6_1.sce b/2744/CH6/EX6.1/Ex6_1.sce
new file mode 100755
index 000000000..ad6408598
--- /dev/null
+++ b/2744/CH6/EX6.1/Ex6_1.sce
@@ -0,0 +1,14 @@
+clear;
+clc;
+l = 5;// feet
+W = 150;// lb
+w = 120;// lb. per foot run
+l1 = 3;// feet
+b = 3;// inches
+d = 6;// inches
+E = 1.5*10^6;// lb/in^2
+I = (1/12)*b*d^3;// in^4
+y_B1 = (W*l^3)/(3*E*I);// feet
+y_B2 = (w*l1*l1^3)/(8*E*I) + (l-l1)*(w*l1*l1^2)/(6*E*I);// feet
+y_B = (12^3)*(y_B1+y_B2);// inches
+printf('The deflection at the free end = %.4f inches',y_B);
diff --git a/2744/CH6/EX6.10/Ex6_10.sce b/2744/CH6/EX6.10/Ex6_10.sce
new file mode 100755
index 000000000..332f01eed
--- /dev/null
+++ b/2744/CH6/EX6.10/Ex6_10.sce
@@ -0,0 +1,16 @@
+clear;
+clc;
+b = 18;// inches
+d = 6;// inches
+l = 16;// feet
+W = 2;// tons
+h = 1/2;// inches
+I_xx = 841.76;// in^4
+E = 13000;// tons/in^2
+P = W + sqrt(2*W*h*48*E*I_xx/(l*12)^3 + 2*W);// tons
+M_max = P*l*12/4;// ton-inches
+Z = 2*I_xx/b ;// in^3
+f = M_max/Z ;// tons/in^2
+delta = P*(l*12)^3 /(48*E*I_xx);// inches
+printf('The maximum instantaneous deflection delta = %.4f inches\n  and stress induced, f = %.3f tons/in^2',delta,f);
+//there is an error in the answer given in text book
diff --git a/2744/CH6/EX6.11/Ex6_11.sce b/2744/CH6/EX6.11/Ex6_11.sce
new file mode 100755
index 000000000..7b60364be
--- /dev/null
+++ b/2744/CH6/EX6.11/Ex6_11.sce
@@ -0,0 +1,18 @@
+clear;
+clc;
+l = 3;// feet
+b = 3;// inches
+t = 3/8;// inches
+W = 1500;// lb.
+f = 12;// tons/in^2
+E = 30*10^6;// tons/in^2
+M_max = W*l*12/4 ;// lb-inches
+M_r = f*(1/6)*b*t^2 *2240;// lb-inches
+n = M_max/M_r ;// no. of plates
+n = round(n+1);
+f = M_max/(n*(1/6)*b*t^2);// lb/in^2
+R = E/(2*f/t) ;// inches
+delta = (l*12)^2 /(8*R);// inches
+printf('Number of plates required, n = %d',n);
+printf('\n The central deflection, delta = %.4f inch.',delta);
+printf('\n The initial radius to which the plates must be bent, R = %.3f inches',R);
diff --git a/2744/CH6/EX6.3/Ex6_3.sce b/2744/CH6/EX6.3/Ex6_3.sce
new file mode 100755
index 000000000..4da2a37a0
--- /dev/null
+++ b/2744/CH6/EX6.3/Ex6_3.sce
@@ -0,0 +1,12 @@
+clear;
+clc;
+b = 4;// inches
+d = 9;// inches
+l = 12;// feet
+y_c = 1/4;// inches
+E = 1.5*10^6;// lb/in^2
+I = (1/12)*b*d^3;// in^4
+W = y_c*384*E*I/(5*12^3*l^3);// inches
+printf('Uniform distributed load, the beam should carry is, W = %d lb-wt',W);
+
+//there is an error in the answer given in text book
diff --git a/2744/CH6/EX6.4/Ex6_4.sce b/2744/CH6/EX6.4/Ex6_4.sce
new file mode 100755
index 000000000..cef3e2fad
--- /dev/null
+++ b/2744/CH6/EX6.4/Ex6_4.sce
@@ -0,0 +1,13 @@
+clear;
+clc;
+d = 6;// feet
+l = 60;// feet
+f = 15/2;// tons/in^2
+E = 13000;// tons/in^2
+k1 = 2*f/(12*d);// k1 = M_r/I 
+k2 = k1/(l*12/8);//k2 = W/I
+y_c = (5/384)*k2*l^3 *12^3 /E;// inches
+//If the giredr is of constant deapth and uniform strength, it bends to an arc of a circle of radius R 
+R = E*d*12/(2*f);// inches
+delta = (l*12)^2 /(8*R);// inches
+printf('The deflection for a uniformly distributed load on it is,delta = %.2f inches',delta);
diff --git a/2744/CH6/EX6.5/Ex6_5.sce b/2744/CH6/EX6.5/Ex6_5.sce
new file mode 100755
index 000000000..2a6ed40ca
--- /dev/null
+++ b/2744/CH6/EX6.5/Ex6_5.sce
@@ -0,0 +1,7 @@
+clear;
+clc;
+f = 8;//tons/in^2
+E = 12800;// tons/in^2
+k1 = 1/480;//central deflection = k = delta/l
+k2 = (5/24)*(f/E)/k1 ;//k2 = d/l =  deapth to span ratio
+printf('The ratio of deapth to span, d/l = %f ',k2);
diff --git a/2744/CH6/EX6.6/Ex6_6.sce b/2744/CH6/EX6.6/Ex6_6.sce
new file mode 100755
index 000000000..c3d159985
--- /dev/null
+++ b/2744/CH6/EX6.6/Ex6_6.sce
@@ -0,0 +1,18 @@
+clear;
+clc;
+w = 550;// lb. per foot run
+f = 1000;// lb/in^2
+l = 20;// feet
+d_limit = 15;// inches
+E = 1.5*10^6;// lb/in^2
+//central ddeflection
+delta = (1/2);// inches
+d = (5/24)*(f/E)*20*12/(1/(2*20*12));// inches
+M = w*l*l*12/8;// lb-inches
+b = M/(f*(1/6)*d^2);// inches
+printf('A section with d = %d inches, b = %d inches will do.',round(d),round(b));
+f1 = (1/(2*20*12))*(d_limit/(l*12))*E/(5/24);// lb/in^2
+b = M/(f1*(1/6)*d_limit^2);// inches
+printf('\n If the deapth of section is limited to %d inches, then \n f = %.1f lb/in^2\n b = %.1f inches',d_limit,f1,b);
+
+//tha answer is correct only, but it is approximated in the text book.
diff --git a/2744/CH6/EX6.7/Ex6_7.sce b/2744/CH6/EX6.7/Ex6_7.sce
new file mode 100755
index 000000000..2ef11cde8
--- /dev/null
+++ b/2744/CH6/EX6.7/Ex6_7.sce
@@ -0,0 +1,18 @@
+clear;
+clc;
+l = 20;// feet
+b = 4;// feet
+W = 5;// tons
+d = 12;// inches
+h = 5;// inches
+I_xx = 220;// in^4
+E = 13000;// tons/in^2
+a = l-b;// feet
+//for maximum deflection
+x = sqrt((a^2 + 2*a*b)/3);// feet
+y_max = x*12^3 *((a^2 + 2*a*b) - x^2)/(6*E*I_xx);// inches
+//for deflection at the centre
+x1 = 0.5*l;// inches
+y_x1 = x1*12^3 *((a^2 + 2*a*b) - x1^2)/(6*E*I_xx);// inches
+printf('The position of maximum deflection occurs at x = %.2f feet\n The maximum deflection is, y_max = %.4f inches',x,y_max);
+printf('\n The deflection at the centre, y_%d = %.4f inches',x1,y_x1);
diff --git a/2744/CH6/EX6.8/Ex6_8.sce b/2744/CH6/EX6.8/Ex6_8.sce
new file mode 100755
index 000000000..8016c65ca
--- /dev/null
+++ b/2744/CH6/EX6.8/Ex6_8.sce
@@ -0,0 +1,20 @@
+clear;
+clc;
+d = 12;// inches
+h = 5;// inches
+l = 20;// feet
+E = 13000; //tons/in^2
+I_xx = 220;// in^4
+W = 4;// tons
+W1 = 3;// tons
+a = 15;// feet
+b = l-a;// feet
+a1 = 16;// feet 
+b1 = l-a1;// feet
+K1 = (-2*W1*b1*l)/(W1*b1-W*b);
+K2 = (W*b*a^2 + 2*a*W*b^2 + 2*W1*b1*l^2 - W1*b1*a1^2 -2*W1*a1*b1^2 +W1*b1*l^2)/(3*(W1*b1 - W*b));
+x = -0.5*K1 + sqrt(-K2 + 0.25*K1^2);// feet
+x1 = l-x;// feet
+y_max = W*b*x*1728*(a^2 +2*a*b -x^2)/(6*E*I_xx*l) + W1*b1*x1*1728*(a1^2 +2*a1*b1 -x1^2)/(6*E*I_xx*l);// inches
+printf('The position of the maximum deflection is, x = %.2f feet.',x);
+printf('\n And the maximum deflection is, y_max = %.4f inches.',y_max);
diff --git a/2744/CH6/EX6.9/Ex6_9.sce b/2744/CH6/EX6.9/Ex6_9.sce
new file mode 100755
index 000000000..d465d6c5e
--- /dev/null
+++ b/2744/CH6/EX6.9/Ex6_9.sce
@@ -0,0 +1,16 @@
+clear;
+clc;
+b = 18;// inches
+d = 7;// inches
+w1 = 1;// ton per foot run
+w2 = 3;// ton per foot run
+I_xx = 1149;// in^4
+E = 13000;// tons/in^2
+R_A = 0.5*b + (b/3);// tons
+R_B = 0.5*b + (2*b/3);// tons
+//integrating M = E*I*y'', to get E*I*y' and making y' = 0;, we get maximu deflection
+x = 9.18;// by trial and error method
+y_derivative = -R_A*0.5*x^3 + x^4 /6 +0.5*(2/3)*(1/b)*(1/4)*x^5 + 469.8;
+y = -R_A*0.5*x^3 /3 + x^4 /24 +0.5*(2/3)*(1/b)*(1/(4*5))*x^5 + 469.8*x;
+y_max = y;// inches
+printf('The position of maximum deflection from the end A, x = %.2f inches and \n  Maximum deflection, y_max = %.4f inches',x,y_max*12^3 /(E*I_xx));