Modified the code

7 files changed, 47 insertions, 38 deletions

diff --git a/3776/CH2/EX2.11/Ex2_11.sce b/3776/CH2/EX2.11/Ex2_11.sce
index c4e15f631..4f9a017ec 100644
--- a/3776/CH2/EX2.11/Ex2_11.sce
+++ b/3776/CH2/EX2.11/Ex2_11.sce
@@ -1,21 +1,26 @@
 clear
-mass = 4 //kg 
+mass = 4 //kg
 dist = 1 //mt freely falling distance
 l = 1500 //mm length of rod
 d = 15 //mm diameter
 l_ab = 6.71 //inch
 l_bc = 8.29 //inch
-E =  200  //GPA youngs modulus 
+E =  200  //GPA youngs modulus
 k = 4.5 // N/mm stiffness costant
 F =  mass*9.81// The force applying
-Area =  3.14*(d**2)/4  
-// Two cases 
-//youngs modulus 
+Area =  3.14*(d**2)/4
+// Two cases
+//youngs modulus
 e_y = F*l/(Area*E*(10**3))
 // stiffness
-e_f = F/k 
+e_f = F/k
 //total
 e = e_y +e_f
 k = 1+(2/(e*(10**-3)))
 stress_max_1 = F*(1+(k**0.5))/Area
 printf("\n The maximum stress is: %0.3f MPa",stress_max_1)
+
+//for the rod without washer
+k2 = 1+(2/(e_y*(10**-3)))
+stress_max_2 = F*(1+(k2**0.5))/Area
+printf("\n The maximum stress for the rod without washer is: %0.3f MPa",stress_max_2)
diff --git a/3776/CH2/EX2.12/Ex2_12.sce b/3776/CH2/EX2.12/Ex2_12.sce
index b4a566a08..fb8e4ffb6 100644
--- a/3776/CH2/EX2.12/Ex2_12.sce
+++ b/3776/CH2/EX2.12/Ex2_12.sce
@@ -1,9 +1,10 @@
 clear
-flex_a = 1//f
-flex_b = 2//f
+flex_a = 1   //f
+flex_b = 2    //f
 //removing lower support and solving FBD
 e = -2 -(2+1)//fp
 //e_1 = (2+1+1)*R
 //e_1 = -e Making the elongations zero since the both ends are fixed
-R = e/(2+1+1.0) //P
+R = -e/(2+1+1.0) //P
+//since sum of forces are 0
 printf("\n The reactions at bottom is %0.3f P",R)
diff --git a/3776/CH2/EX2.19/Ex2_19.sce b/3776/CH2/EX2.19/Ex2_19.sce
index 2f56f7092..cf1308495 100644
--- a/3776/CH2/EX2.19/Ex2_19.sce
+++ b/3776/CH2/EX2.19/Ex2_19.sce
@@ -3,8 +3,8 @@ clear
 l = 30 //in - The length of the rod
 p_1 = 80 //kips - The Force on the end
 p_2  = 125 //kips - The force on the other end
-A_s = 0.5 //in2 - The crossection of the steel rod
-A_a = 0.5 //in2 - The crossection of the aluminium 
+A_s = 0.5 //sq.in - The crossection of the steel rod
+A_a = 0.5 //sq.in - The crossection of the aluminium 
 E_a = 10*(10**6) //psi - The youngs modulus of the aluminium 
 E_s = 30*(10**6) //psi - The youngs modulus of the steel
 //Internally stastically indeterminant 
diff --git a/3776/CH2/EX2.2/Ex2_2.sce b/3776/CH2/EX2.2/Ex2_2.sce
index 9e6317279..88d9c4ad6 100644
--- a/3776/CH2/EX2.2/Ex2_2.sce
+++ b/3776/CH2/EX2.2/Ex2_2.sce
@@ -5,9 +5,9 @@ l_cd = 1500 //mm - length of rod cd
 p_ob = 100  //kN - Force in rods 
 p_bc = -150 //KN
 p_cd = 50   //KN 
-A_ob = 1000 //mm2 - Area of rod ob
-A_bc = 2000 //mm2 - Area of rod bc 
-A_cd = 1000 //mm2 - Area of rod cd
+A_ob = 1000 //sq.mm - Area of rod ob
+A_bc = 2000 //sq.mm - Area of rod bc 
+A_cd = 1000 //sq.mm - Area of rod cd
 E = 200.0 //GPA 
 // the total deflection is algebraic sums of `deflection in each module 
 e_1 = p_ob*l_ob/(A_ob*E)
diff --git a/3776/CH2/EX2.4/Ex2_4.sce b/3776/CH2/EX2.4/Ex2_4.sce
index 76ea23bb6..2a8f3fa43 100644
--- a/3776/CH2/EX2.4/Ex2_4.sce
+++ b/3776/CH2/EX2.4/Ex2_4.sce
@@ -1,21 +1,23 @@
 clear
-p_app = 3 //kips - applied force 
-P_A = 2.23 //kips  
+p_app = 3 //kips - applied force
+P_A = 2.23 //kips
 p_B = -2.83 //kips - compressive force
 l_ab = 6.71 //inch
 l_bc = 8.29 //inch
 s_ab = 17.8 //ksi - tensile stress
 s_bc = -12.9 //ksi - compressive stress
-E = 10.6 * (10**3) //ksi -youngs modulus 
+E = 10.6 * (10**3) //ksi -youngs modulus
 e_ab = s_ab*l_ab/E //elongation
 
 e_bc = s_bc*l_bc/E //contraction
-x = -e_bc/e_ab //the Ratio of cosines of the deflected angles 
-// t_1 and t_2 be deflected angles 
+x = -e_bc/e_ab //the Ratio of cosines of the deflected angles
+// t_1 and t_2 be deflected angles
 //t_2 = 180-45-26.6-t_1 the sum of angles is 360
 //applying cos on both sides
 t_1=atand(1.29)
 
 e = e_ab/cosd((t_1)) //inch
-k  = p_app/e // kips/in vertical stiffness of the combination
+e_t = e*cosd(11.2)
+k  = p_app/e_t // kips/in vertical stiffness of the combination
 printf("\n The vertical stiffness of the combination is %0.3f kips/inch",k) //answer in textbook is 167
+// answer varies due to rounding off errors
diff --git a/3776/CH2/EX2.6/Ex2_6.sce b/3776/CH2/EX2.6/Ex2_6.sce
index 844997aec..a5dcb9d0f 100644
--- a/3776/CH2/EX2.6/Ex2_6.sce
+++ b/3776/CH2/EX2.6/Ex2_6.sce
@@ -1,15 +1,15 @@
 clear
-dia = 50 //mm - diameter of aluminium 
-p = 100 // KN - instant force applid
-dia_c = 0.1215 //mm- change in diameter 
+dia = 50 //mm - diameter of aluminium
+p = 100 // KN - instant force applied
+dia_c = 0.01215 //mm- change in diameter
 l_c = 0.219 //mm - change in length
-l = 300 //mm - length 
-strain_dia = dia_c/dia // lateral strain 
-strain_l = l_c/l //longitudinal strain 
-po = strain_dia/strain_l // poission ratio 
-area = 3.14*dia*dia/4 //mm2 area
-E = p*l/(area*l_c) //N/mm2 youngs modulus 
-printf("\n The lateral strain is: %0.3f no units",strain_dia)
-printf("\n The longitudinal strain is: %0.3f no units",strain_l)
-printf("\n The poissions ratio is: %0.3f no units",po)
-printf("\n Youngs modulus: %0.2f N/sq.mm",E)
+l = 300 //mm - length
+strain_dia = -dia_c/dia // lateral strain
+strain_l = -l_c/l //longitudinal strain
+po = strain_dia/strain_l // Poisson ratio
+area = 3.14*dia*dia/4 //sq.mm area
+E = p*l/(area*l_c) //N/sq.mm youngs modulus
+printf("\n The lateral strain is: %e mm/mm",strain_dia)
+printf("\n The longitudinal strain is: %e no units",strain_l)
+printf("\n The Poissons ratio is: %0.3f no units",po)
+printf("\n Youngs modulus: %0.2f GPa",E)
diff --git a/3776/CH2/EX2.7/Ex2_7.sce b/3776/CH2/EX2.7/Ex2_7.sce
index c452c6448..0bfdaefa3 100644
--- a/3776/CH2/EX2.7/Ex2_7.sce
+++ b/3776/CH2/EX2.7/Ex2_7.sce
@@ -1,15 +1,16 @@
 clear
 T = 12.9*(10**-6) ///F
-t = 100.00 // F 
+t = 100.00 // F
 l_ab = 6.71 //inch
 l_bc = 8.49 //inch
-e_ab = T*t*l_ab //in-elongation 
+e_ab = T*t*l_ab //in-elongation
 e_bc = T*t*l_bc //in-elongation
-k = e_ab/e_bc // ratio of cosines of deflected angles 
-// t_1 and t_2 be deflected angles 
-//t_2 = 180-45-26.6-t_1 the sum of angles is 360
+k = e_ab/e_bc // ratio of cosines of deflected angles
+// t_1 and t_2 be deflected angles
+//t_2 = 45+26.6-t_1 the sum of angles is 360
 //applying cos on both sides
 t_1 = atand(0.5)
 //
 e = e_bc/cosd(t_1)
 printf("\n The displacement in point B is : %e in",e )
+printf("\n It forms an angle of %f degrees with vertical",45-t_1 )