Modified the code

6 files changed, 28 insertions, 30 deletions

diff --git a/3776/CH10/EX10.1/Ex10_1.sce b/3776/CH10/EX10.1/Ex10_1.sce
index dfc409a95..63307cca9 100644
--- a/3776/CH10/EX10.1/Ex10_1.sce
+++ b/3776/CH10/EX10.1/Ex10_1.sce
@@ -1,10 +1,11 @@
 clear
-//Given 
+//Given
 dia = 400   //mm - The diameter of a pulley
-E = 2001     //Gpa - Youngs modulus
+E = 200    //GPa - Youngs modulus
 t = 0.6     //mm - The thickness of band
-c = t/2     //mm - The maximum stress is seen 
-//Caliculations
+c = t/2     //mm - The maximum stress is seen
+//calculations
 
 stress_max = E*c*(10**3)/(dia/2) //MPa - The maximum stress on the crossection occurs at the ends
 printf("\n The maximum bending stress developed in the saw  %0.3f MPa",stress_max)
+// answer varies from the book
diff --git a/3776/CH10/EX10.10/Ex10_10.sce b/3776/CH10/EX10.10/Ex10_10.sce
index b9e6eac7c..403def9d1 100644
--- a/3776/CH10/EX10.10/Ex10_10.sce
+++ b/3776/CH10/EX10.10/Ex10_10.sce
@@ -1,8 +1,8 @@
 clear
-k = 24.0*(10**12)    //N.mm2 Flexure rigidity
-E = 200.0            //Gpa - Youngs modulus of the string
+k = 24.0*(10**12)    //N.sq.mm Flexure rigidity
+E = 200.0            //GPa - Youngs modulus of the string
 l = 5000.0           //mm - The length of the string
-C_A = 300.0          //mm2 - crossection area 
+C_A = 300.0          //sq.mm - crossection area 
 P = 50.0             //KN - The force applies at the end 
 a = 2000.0           //mm - The distance C-F
 x = 1//X - let it be a variable X
diff --git a/3776/CH10/EX10.11/Ex10_11.sce b/3776/CH10/EX10.11/Ex10_11.sce
index a514a0eb8..7fd39419e 100644
--- a/3776/CH10/EX10.11/Ex10_11.sce
+++ b/3776/CH10/EX10.11/Ex10_11.sce
@@ -8,11 +8,11 @@ I_Z = 315       //in^4 - the moment of inertia wrt Z axis
 I_y = 8.13      //in^4 - the moment of inertia wrt Y axis
 o = 5           // degrees - the angle of acting force 
 P = 2000        //k the acting force 
-P_h = P*sin((%pi/180)*(o)) //k - The horizantal component of P
+P_h = P*sin((%pi/180)*(o)) //k - The horizontal component of P
 P_v = P*cos((%pi/180)*(o)) //k - The vertical  component of P
-e_h =  P_h*(L**3)/(3*E*I_y)       // the horizantal component of deflection 
+e_h =  P_h*(L**3)/(3*E*I_y)       // the horizontal component of deflection 
 e_v =  P_v*(L**3)/(3*E*I_Z )      // the vertical component of deflection
 e = ((e_h**2 + e_v**2)**0.5)
-printf("\n the horizantal component of deflection %0.3f in",e_h)
+printf("\n the horizontal component of deflection %0.3f in",e_h)
 printf("\n the vertical component of deflection %0.3f in",e_v)
 printf("\n the resultant deflection %0.3f in",e)
diff --git a/3776/CH10/EX10.13/Ex10_13.sce b/3776/CH10/EX10.13/Ex10_13.sce
index e2688eb46..75bc3aa0d 100644
--- a/3776/CH10/EX10.13/Ex10_13.sce
+++ b/3776/CH10/EX10.13/Ex10_13.sce
@@ -6,7 +6,7 @@ m = 15.3           // mass of the falling body
 h = 75.0           //mm - The height of the falling body 
 p = m*9.81         //N the force acted due to the body
 L = 1000.0         //mm The length of the cantilever
-E = 200  //Gpa The youngs modulus of the material used 
+E = 200  //GPa The youngs modulus of the material used 
 I = (l**4)/12 //mm - the moment of inertia 
 k = 300 //N/mm -the stiffness of the spring 
 //Rigid supports 
diff --git a/3776/CH10/EX10.15/Ex10_15.sce b/3776/CH10/EX10.15/Ex10_15.sce
index 72f3ed920..62eb98e8c 100644
--- a/3776/CH10/EX10.15/Ex10_15.sce
+++ b/3776/CH10/EX10.15/Ex10_15.sce
@@ -1,17 +1,14 @@
 clear
 //Given
-E = 30*(10**3) //ksi - The youngs modulus of the material 
+E = 30*(10**3) //ksi - The youngs modulus of the material
 stress_y = 40 //ksi - yield stress
-stress_max = 24.2 //ksi - the maximum stress
-l = 2          //in - The length of the crossection 
+stress_max = 24.4 //ksi - the maximum stress
+l = 2          //in - The length of the crossection
 b = 3       //in - the width of the crossection
-h = 3 //in - the depth of the crossection
-//lets check ultimate capacity for a 2 in deep section 
-M_ul = stress_y*b*(l**2)/4 //K-in the ultimate capacity 
-curvature = 2*stress_y/(E*(h/2) ) //in*-1 the curvature of the beam 
-curvature_max = stress_max/(E*(h/2)) //in*-1 The maximum curvature 
-printf("\n the ultimate capacity %0.3f k-in",M_ul)
-printf("\n the ultimate curvature %0.3f in *-1",curvature_max)
-printf("\n E given in equation is wrong")
-printf("\n Actual E in question is 30*10**3")
-
+h = 2 //in - the depth of the crossection
+//lets check ultimate capacity for a 2 in deep section
+M_ul = stress_max*b*(l**2)/4 //K-in the ultimate capacity
+curvature = 2*stress_y/(E*(h/2) ) //per inch the curvature of the beam
+curvature_max = stress_max/(E*(b/2)) //per inch The maximum curvature
+printf("\n the curvature in 11-in is %e per inch",curvature)
+printf("\n the ultimate curvature %e per inch",curvature_max)
diff --git a/3776/CH10/EX10.16/Ex10_16.sce b/3776/CH10/EX10.16/Ex10_16.sce
index 7cce6d1d2..d1e701290 100644
--- a/3776/CH10/EX10.16/Ex10_16.sce
+++ b/3776/CH10/EX10.16/Ex10_16.sce
@@ -1,12 +1,12 @@
 clear
-//Given 
-l_ad = 1600 //mm - The total length of the beam 
+//Given
+l_ad = 1600 //mm - The total length of the beam
 l_ab = 600  //mm - The length of AB
 l_bc = 600  //mm - The length of BC
-e_1 = 0.24  //mm - deflection 
+e_1 = 0.24  //mm - deflection
 e_2 = 0.48  //mm - deflection
-E = 35      //Gpa
-//Caliculation 
+E = 35      //GPa
+//calculation
 
 A_afe = -(l_ab+l_bc)*e_1*(10**-3)/(2*E)
 A_afe = -(l_ab)*e_2*(10**-3)/(4*E)
@@ -15,4 +15,4 @@ x_1 = 1200            //com from B
 x_2 = 800             //com from B
 y_b = A_afe*x_1 + A_afe*x_2 //mm The maximum deflection at tip B
 printf("\n The maximum deflection at tip B %0.2f mm",y_b)
-printf("\n The slope at the tip  B %0.2f radians",y_1_b)
+printf("\n The slope at the tip  B %e radians",y_1_b)