16 files changed, 479 insertions, 0 deletions
diff --git a/3772/CH6/EX6.1/Ex6_1.sce b/3772/CH6/EX6.1/Ex6_1.sce
new file mode 100644
index 000000000..df498a429
--- /dev/null
+++ b/3772/CH6/EX6.1/Ex6_1.sce
@@ -0,0 +1,28 @@
+// Problem no 6.1,Page No.154
+
+clc;clear;
+close;
+
+b=0.12 //m //Width of beam
+d=0.2  //m //Depth of beam
+dell=0.005 //m //Deflection
+E=2*10**5*10**6 //N/m**2 
+L=2.5 //m //Length of beam
+
+//Calculations
+
+I=b*d**3*12**-1 //m**4 //M.I of rectangular section
+w=8*E*I*dell*(L**4)**-1 //N/m //U.d.l
+
+//Let slope at free end be theta
+theta=w*L**3*(6*E*I)**-1 //Radian 
+
+W=dell*3*E*I*(L**3)**-1*10**-3 //kN //Concentrated Load 
+
+theta_2=W*L**2*(2*E*I)**-1 //Slope at free end
+
+//Result
+printf("Uniformly distributed Load beam should carry is %.2f N/m",w)
+printf("\n Concentrated Load at free end is %.2f kN",W)
+
+//Answer is wrong in the textbook.
diff --git a/3772/CH6/EX6.10/Ex6_10.sce b/3772/CH6/EX6.10/Ex6_10.sce
new file mode 100644
index 000000000..83d0afa26
--- /dev/null
+++ b/3772/CH6/EX6.10/Ex6_10.sce
@@ -0,0 +1,45 @@
+// Problem no 6.10,Page No.161
+
+clc;clear;
+close;
+
+E=200*10**9 //Pa 
+I=20000*10**-8 //m**4
+
+//Calculations
+
+//Now Taking moment at B
+R_a=(1000*3*4.5+1000*2)*6**-1 //Reaction Force at pt A
+
+//On part BC u.d.l of 1KN/m is introduced both above and below 
+//consider section at distance x i.e X-X and considering moment at section X-X
+
+//M=15500*x*6**-1-1000*x**2*2**-1-1000(x-4)+1000*2**-1*(x-3)**2
+//EI*d**2y*d**x=-M=15500*x*6**-1-1000*x**2*2**-1-1000(x-4)+1000*2**-1*(x-3)**2
+
+//Now Integrating above Equation we get Equation of slope
+//EI*dy*dx**-1=-15500*x**2*12**-1+1000*x**3*6**-1+1000*(x-4)**2*2**-1+1000*6**-1*(x-3)**3+C_1
+
+//Now Integrating above Equation we get Equation of Deflection
+//EI*y=-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x+C_2
+
+//At x=0,deflection is zero,i.e y=0  C_2=0
+//At x=6,deflection is zero,i.e y=0  
+x=6
+C_1=-(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*(x-3)**4*24**-1)*x**-1 //Constant
+
+//Answer for constant C_1 is incorrect in Book
+
+//Now Deflection at C,put x=3 m
+x=3
+y_C=1*(E*I)**-1*(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x)*10**3
+
+//Now Deflection at D,put x=4 m
+x=4
+y_D=1*(E*I)**-1*(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x)*10**3
+
+//Answers for y_C & y_D are incorrect in book
+
+//Result
+printf("Deflection at pt C is %.2f mm",y_C)
+printf("\n Deflection at pt D is %.2f mm",y_D)
diff --git a/3772/CH6/EX6.11/Ex6_11.sce b/3772/CH6/EX6.11/Ex6_11.sce
new file mode 100644
index 000000000..c583a5429
--- /dev/null
+++ b/3772/CH6/EX6.11/Ex6_11.sce
@@ -0,0 +1,34 @@
+// Problem no 6.11,Page No.162
+
+clc;clear;
+close;
+
+L=2.5 //m //Length of beam
+L_1=1.5 //m //Length from Fixed end
+W=50*10**3 //N //Load
+
+//Calculations
+
+//Case-1
+y=W*L**3*3**-1 //Deflection of the cantilever at free end
+
+//Case-2
+//Deflection of cantilever at free end is
+//y_1=W_1*L**3*3**-1+W_1*L_1**3*3**-1+W_1*L_1**3*3**-1*(L-L_1)
+//After substituting values in above equation and simplifying further we get
+
+//y_1=22.375*W_1*3**-1
+
+W_1=y*3*22.375**-1*10**-3 //Magnitude of equal Loads
+M_1=W*L*10**-3
+M_2=W_1*L+W_1*L_1
+
+//Let M_1=sigma_1*z  and M_2=sigma_2*z
+//Dividing above two equations we get
+
+//Let X=sigma_1*sigma_2**-1
+X=M_2*M_1**-1*100
+
+//Result
+printf("Magnitude of equal Loads is  %.2f kN",W_1)
+printf("\n Max Bending stress is  %.2f %%",X)
diff --git a/3772/CH6/EX6.12/Ex6_12.sce b/3772/CH6/EX6.12/Ex6_12.sce
new file mode 100644
index 000000000..a5cbe7c4b
--- /dev/null
+++ b/3772/CH6/EX6.12/Ex6_12.sce
@@ -0,0 +1,30 @@
+// Problem no 6.12,Page No.163
+
+clc;clear;
+close;
+
+L=4 //m //Length of Beam
+
+//calculations
+
+//Consider a section at distance x from A and B.M at this section is
+//M=P*(3-x)-10*x**2+90*x-195 
+
+//Now //EI*d**2*y*d**2*x=-P*(3-x)+10*x**2-90*x+195
+
+//On Integrating above equation we get 
+//E*I*dy*dx**-1=-P*(3*x-x**2*2**-1)+10*x**3*2**-1-45*x**2+195*x+C_1
+
+//Again On Integrating above equation we get 
+//E*I*y=-P*(3*x**2*2**-1-x**3*6**-1)+10*x**4*12**-1-15*x**3+195*x**2*2**-1+C_1*x+C_2
+
+//But at x=0,dy*dx**-1=0 we get ,C_1=0
+//        x=0,y=0     we get    ,C_2=0
+//At x=3 m,y=0
+x=3
+C_1=0
+C_2=0
+P=(10*x**4*12**-1-15*x**3+195*x**2*2**-1+C_1*x+C_2)*(3*x**2*2**-1-x**3*6**-1)**-1
+
+//Result
+printf("Load taken by prop is %.2f",P);printf(" KN")
diff --git a/3772/CH6/EX6.13/Ex6_13.sce b/3772/CH6/EX6.13/Ex6_13.sce
new file mode 100644
index 000000000..fcaa46c63
--- /dev/null
+++ b/3772/CH6/EX6.13/Ex6_13.sce
@@ -0,0 +1,51 @@
+// Problem no 6.13,Page No.163
+
+clc;clear;
+close;
+
+L=6 //m //Span of Beam
+sigma=100*10**6 //Pa //Bending stress
+E=210*10**9
+y=0.45 //m //Depth
+
+//Calculations
+
+//Taking moment at B
+R_a=20*6*3+6*40*2*2**-1
+
+//At a section x from A the rate of Loading=20+2*3**-1*x //KN/m
+//S.F=100-20*x-x**2*3**-1
+//M=100*x-10*x**2-x**3*9**-1 
+
+//Thus B.M will be max where S.F is zero,we get equation as
+//x**2+60*x-300=0
+a=1
+b=60
+c=-300
+
+X=b**2-4*a*c
+x_1=(-b+X**0.5)*(2*a)**-1
+x_2=(-b-X**0.5)*(2*a)**-1
+
+x=4.641
+M=100*x-10*x**2-x**3*9**-1 //KN*m //Max bending moment
+I=M*sigma**-1*y*1000*2**-1 //m**4 //M.I
+
+//E*I*d**2*y*(d*x**2)**-1=-100*x+10*x**2+x**3*9**-1
+
+//AFter Integrating above EquATION WE get
+//E*I*dy*(dx)**-1=-50*x**2+10*3**-1*x**3+x**4*36**-1+C_1
+//Again Integrating above EquATION WE get
+//E*I*y=-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1+C_1*x+C_2
+
+//At x=0,y=0  ,C_2=0
+//At x=6,y=0 
+x=6
+C_2=0
+C_1=-(-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1)*x**-1
+
+x=3 //m
+y=1*(E*I)**-1*(-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1+C_1*x+C_2)*1000*100
+
+//Result
+printf("The central Deflection is %.2f",y);printf(" cm")
diff --git a/3772/CH6/EX6.14/Ex6_14.sce b/3772/CH6/EX6.14/Ex6_14.sce
new file mode 100644
index 000000000..eaba57222
--- /dev/null
+++ b/3772/CH6/EX6.14/Ex6_14.sce
@@ -0,0 +1,42 @@
+// Problem no 6.14,Page No.168
+
+clc;clear;
+close;
+
+L=10 //m //Lenght of cantilever beam
+P_1=20*10**3 //N //Load at free end
+P_2=20*10**3 //N //Load at middle of beam
+E=200*10**9 //Pa 
+I=20000*10**-8 //m**4
+
+//Calculations
+
+//Taking moment at pt B we get
+R_a=20*5*10**-1 //Force at pt A
+
+//Now B.M at b=0,at C=-100,at A=-300 KN*m
+
+//Now Area of B.M 
+A_1=2**-1*5*100 //KN*m**2
+A_2=5*100       //KN*m**2
+A_3=2**-1*5*200 //KN*m**2
+
+//Total Area of B.M diagram is given by A
+A=A_1+A_2+A_3
+
+theta=A*10**3*(E*I)**-1 //radian
+
+x_1=2*3**-1*5
+x_2=3*2**-1*5
+x_3=5*3**-1*5
+M_1=A_1*x_1
+M_2=A_2*x_2
+M_3=A_3*x_3
+
+M=M_1+M_2+M_3 //Total moments of B.M about B
+
+y_B=M*10**3*(E*I)**-1 //Deflection a tfree end
+
+//REsult
+printf("Slope of cantilever at free end is %.2f",theta);printf(" radian")
+printf("\n Deflection of cantilever at free end is %.2f",y_B);printf(" m")
diff --git a/3772/CH6/EX6.15/Ex6_15.sce b/3772/CH6/EX6.15/Ex6_15.sce
new file mode 100644
index 000000000..3172da236
--- /dev/null
+++ b/3772/CH6/EX6.15/Ex6_15.sce
@@ -0,0 +1,14 @@
+// Problem no.6.15,Page no.169
+
+clc;clear;
+close;
+
+//Calculations
+
+//Slope at A is Zero and deflection at C is zero According to Mohr's second theorem
+//Let A_1*x_1=Y
+Y=1*30**-1*80*4*(3*4**-1*4+2)
+P=200*27**-1 //Reaction at ens D
+
+//Result
+printf("The reaction at end C is %.2f",P);printf(" KN")
diff --git a/3772/CH6/EX6.16/Ex6_16.sce b/3772/CH6/EX6.16/Ex6_16.sce
new file mode 100644
index 000000000..d11eb4225
--- /dev/null
+++ b/3772/CH6/EX6.16/Ex6_16.sce
@@ -0,0 +1,27 @@
+// Problem no 6.16,Page No.171
+
+clc;clear;
+close;
+
+E=200*10**9 //Pa
+I=2500*10**-8 //m**4
+
+//Calculations
+
+//Taking moment about A we get
+R_a=(30*5+30*1)*6**-1 //Reaction at pt A
+R_b=60-R_a //Reaction at pt B
+
+M_c=30*1 //B.M at C
+M_d=30*1 //B.M at D
+M_a=0 //B.M at a
+M_b=0 //B.M at b
+
+//For conjugate beam taking moment about B_dash
+R_a_dash=(30*2**-1*(5+1*3**-1)+30*4*3+30*2*3**-1*2**-1)*6**-1
+R_b_dash=150-R_a_dash
+
+y_e=1*(E*I)**-1*(R_a_dash*3-30*2*1-2**-1*1*30*(2+1*3**-1))*1000
+
+//Result
+printf("Deflection at the centre is %.2f",y_e);printf(" m")
diff --git a/3772/CH6/EX6.2/Ex6_2.sce b/3772/CH6/EX6.2/Ex6_2.sce
new file mode 100644
index 000000000..3e26a3c29
--- /dev/null
+++ b/3772/CH6/EX6.2/Ex6_2.sce
@@ -0,0 +1,24 @@
+// Problem no 6.2,Page No.155
+
+clc;clear;
+close;
+
+L=6 //m //Length of beam
+y_b=1.5*10**-2 //m //Deflection
+E=2*10**7*10**4
+sigma=10*10**3*10**4
+//d=2*b
+
+//Calculations
+
+//Let w*I**-1=X    //From Deflection at the free end Equation 
+X=y_b*8*E*(L**4)**-1*10**-3    //Equation 1
+
+//Let w*b*I**-1=Y     //From Max bending stress at the extreme fibre From N.A
+Y=sigma*2*(L**2)**-1           //Equation 2
+                                
+b=Y*X**-1  //width of beam //mm
+d=2*b      //depth of beam //mm
+                                
+//Result
+printf("The Dimension of Beam are:\n\t\t\t  b=%.2f mm (width)\n\t\t\t  d=%.2f mm (depth)",b,d)
diff --git a/3772/CH6/EX6.3/Ex6_3.sce b/3772/CH6/EX6.3/Ex6_3.sce
new file mode 100644
index 000000000..865c03e6b
--- /dev/null
+++ b/3772/CH6/EX6.3/Ex6_3.sce
@@ -0,0 +1,24 @@
+// Problem no 6.3,Page No.156
+
+clc;clear;
+close;
+
+L=3 //m //Length of beam
+L_1=1.2 //m //Distance from fixed end
+d=0.25 //m //Depth of beam
+w=15*10**3 //N //U.d.L
+W=40*10**3 //N //Point Load
+E=2*10*10**4 //N/m**2 
+I=13500*10**-4 //M.I
+
+//Calculations
+
+y_b=W*L_1**3*(3*E*I)**-1+W*L_1**2*(2*E*I)**-1*(L-L_1)+w*L**4*(8*E*I)**-1 //Deflection at free end
+
+M=W*L_1+w*L*L*2**-1 //Max Bending moment at the fixed end A //Nm
+y=d*2**-1
+sigma_max=M*y*I**-1  //N/cm**2 //Max Bending stress at extreme fibre 
+
+//Result
+printf("Deflection at the free end is %.4f cm",y_b)
+printf("\n Max stress due to bending is %.2f N/cm^2",sigma_max)
diff --git a/3772/CH6/EX6.4/Ex6_4.sce b/3772/CH6/EX6.4/Ex6_4.sce
new file mode 100644
index 000000000..6fb9ceb0d
--- /dev/null
+++ b/3772/CH6/EX6.4/Ex6_4.sce
@@ -0,0 +1,20 @@
+// Problem no 6.4,Page No.156
+
+clc;clear;
+close;
+
+M=100*10**3 //N //Moment
+L=3 //m //Length
+d=0.15 //m //Width
+b=0.1 //m //width
+E=2.1*10**7*10**4 //N/cm**2 
+
+//Calculations
+
+I=b*d**3*12**-1 //cm**4 //M.I of beam section
+B_1=M*L*(E*I)**-1 //radian //Slope at B
+B_2=M*L**2*(2*E*I)**-1*10**2 //cm //Deflection at point B
+
+//Result
+printf("The slope at Point B is %.2f radian",B_1)
+printf("\n The Deflection at point B is %.2f cm",B_2)
diff --git a/3772/CH6/EX6.5/Ex6_5.sce b/3772/CH6/EX6.5/Ex6_5.sce
new file mode 100644
index 000000000..3d7e0a958
--- /dev/null
+++ b/3772/CH6/EX6.5/Ex6_5.sce
@@ -0,0 +1,26 @@
+// Problem no 6.5,Page No.157
+
+clc;clear;
+close;
+
+b=0.1 //m //width 
+d=0.2 //m //depth
+L=2 //m //Length of beam
+L_1=1 //m //Length from free end
+E=210*10**9 
+W=1*10**3 //N //Concentrated Load
+w=2*10**3 //N/m
+
+//Calculations
+
+I=b*d**3*12**-1 //m**4 //M.I of the beam section
+
+//Slope at free end
+theta=W*L**2*(2*E*I)**-1+w*L**3*(6*E*I)**-1-w*(L-L_1)**3*(6*E*I)**-1 
+
+//Deflection at free end
+y_b=(W*L**3*(3*E*I)**-1+w*L**4*(8*E*I)**-1-w*(L-L_1)**4*(8*E*I)**-1-w*(L-L_1)**3*L_1*(6*E*I)**-1)*10**3 
+
+//Result
+printf("Slope at free end is %.5f radian",theta)
+printf("\n Deflection at free end is %.2f mm",y_b)
diff --git a/3772/CH6/EX6.6/Ex6_6.sce b/3772/CH6/EX6.6/Ex6_6.sce
new file mode 100644
index 000000000..1f80132f7
--- /dev/null
+++ b/3772/CH6/EX6.6/Ex6_6.sce
@@ -0,0 +1,27 @@
+// Problem no 6.6,Page No.158
+
+clc;clear;
+close;
+
+L=10 //m //span of beam
+W=10*10**3 //N //Point Load
+a=6 //m //Distance from left end of beam to point Load
+b=4 //m ////Distance from right end of beam to point Load
+E=210*10**9 
+I=10**-4 //m //M.I of beam
+
+//Calculation 
+
+//slope at left end is given by
+theta_A=W*b*(L**2-b**2)*(6*E*I*L)**-1 //radian
+
+//Deflection under Load is
+y_c=W*a*b*(L**2-a**2-b**2)*(6*E*I*L)**-1*10**3 //mm
+
+//Maximum Deflection of the beam is
+y_max=W*a*(L**2-a**2)**1.5*(15.588457*E*I*L)**-1*10**3 //mm
+
+//Result
+printf("slope at left end is %.5f radian",theta_A)
+printf("\n Deflection under Load is %.2f mm",y_c)
+printf("\n #Maximum Deflection of the beam is %.2f mm",y_max)
diff --git a/3772/CH6/EX6.7/Ex6_7.sce b/3772/CH6/EX6.7/Ex6_7.sce
new file mode 100644
index 000000000..4e53cbaa6
--- /dev/null
+++ b/3772/CH6/EX6.7/Ex6_7.sce
@@ -0,0 +1,24 @@
+// Problem no 6.7,Page No.158
+
+clc;clear;
+close;
+
+L=5 //m //Length of beam
+w=40*10**3 //N //U.d.L 
+y_max=0.01 //Deflection
+sigma_s=7*10**6 //Bending stress
+E=10.5*10**9
+
+//Calculation
+
+M=w*L*8**-1 //N*m //Max Bending moment 
+
+//From equation of max deflection
+I=5*w*L**3*(y_max*384*E)**-1 //m**4 
+
+d=sigma_s*2*I*M**-1*10**2 //cm
+b=12*I*((d*10**-2)**3)**-1*10**2 //cm //Breadth
+
+//Result
+printf("Minimum value of breadth is %.2f cm",b)
+printf("\n Minimum value of Depth is %.2f cm",d)
diff --git a/3772/CH6/EX6.8/Ex6_8.sce b/3772/CH6/EX6.8/Ex6_8.sce
new file mode 100644
index 000000000..337ee732e
--- /dev/null
+++ b/3772/CH6/EX6.8/Ex6_8.sce
@@ -0,0 +1,21 @@
+// Problem no 6.8,Page No.159
+
+clc;clear;
+close;
+
+L=6 //m //Length of beam
+d=0.15 //m //diameter
+y_max=1.035*10**-2 //m //Deflection
+E=210*10**9 
+
+//Calculations
+
+I=%pi*64**-1*d**4 //M.I of Beam
+W=y_max*48*E*(L**3)**-1 //Point Load
+theta_A=3*y_max*L**-1
+theta_B=-theta_A
+
+//Result
+printf("The Heaviest central Point Load placed is %.2f N",W)
+printf("\n Slope at supports are:theta_A = %.5f radian",theta_A)
+printf("\n                      :theta_B = %.5f radian",theta_B)
diff --git a/3772/CH6/EX6.9/Ex6_9.sce b/3772/CH6/EX6.9/Ex6_9.sce
new file mode 100644
index 000000000..9e6f934b6
--- /dev/null
+++ b/3772/CH6/EX6.9/Ex6_9.sce
@@ -0,0 +1,42 @@
+// Problem no 6.9,Page No.160
+
+clc;clear;
+close;
+
+L=14 //m //Lenth of steel girder
+E=210*10**9 //modulus of Elasticity of steel
+I=16*10**4*10**-8 //M.I of girder section
+
+//Calculations
+
+//R_a+R_b=200  //R_a & R_b are the Reactions at supports A & B respectively
+
+//After taking moment at B We get
+R_a=(120*11+80*4.5)*14**-1 //KN
+R_b=200-R_a
+
+//After considering section at X-X at a distance x from left end A and taking B.M at X-X
+//M=120*x-120(x-3)-80*(x-9.5)
+
+//After Integrating twice we get
+//EI*dy*dx**-1=-60*x**2*+60(x-3)**2+40(x-9.5)**2+C_1 //slope 
+
+//Again on Integrating we get
+//EI*y=-20*x**3+20(x-3)**3+40*3**-1*(x-9.5)**3+C_1*x+C_2 //Deflection
+
+//At A deflection is zero,i.e at x=0,y=0 
+//At B deflection is zero,i.e at x=14,y=0 So C_2=0
+
+C_1=-(-20*(14)**3+20*(11)**3+40*3**-1*(14-9.5)**3)*14**-1 //constant 
+
+//Now Deflection at D i.e at x=3 m
+x=3
+y_D=1*(E*I)**-1*(-20*x**3+20*(x-3)**3+C_1*x)*10**3
+
+//Now Deflection at D i.e at x=9.5 m
+x=9.5
+y_C=1*(E*I)**-1*(-20*x**3+20*(x-3)**3+40*3**-1*(x-9.5)**3+C_1*x)*10**3
+
+//Result
+printf("Deflection under points of two Loads are i.e: at pt D = %.4f m",y_D)
+printf("\n                                             : at pt C = %.4f m",y_C)