20 files changed, 908 insertions, 0 deletions
diff --git a/3864/CH4/EX4.1/Ex4_1.sce b/3864/CH4/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..be8ad539f
--- /dev/null
+++ b/3864/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,47 @@
+clear
+//
+
+//Initilization of Variables
+
+L=5000 //mm //Length of Beam
+a=2000 //mm //Length of start of beam to Pt Load
+b=3000 //mm //Length of Pt load to end of beam
+A=150*250 //m**2 //Area of beam  
+b=150 //mm //Width of beam
+d=250 //mm //Depth of beam
+sigma=10//N/mm**2 //stress
+l=2000 //m //Load applied from one end
+
+//Calculations
+
+//Moment of Inertia
+I=1*12**-1*b*d**3 //m**4
+
+//Distance from N.A to end
+y_max=d*2**-1 //m
+
+//Section Modulus
+Z=1*6**-1*b*d**2 //mm**3
+
+//Moment Carrying Capacity
+M=sigma*Z //N-mm
+
+//Let w be the Intensity of the Load in N/m,then Max moment
+//M_max=w*L**2*8**-1 //N-mm
+//After substituting values and further simplifying we get
+//M_max=w*25*100*8**-1
+
+//EQuating it to moment carrying capacity,we get max intensity load
+w=M*(25*1000)**-1*8*10**-3
+
+//Part-2
+
+//Let P be the concentrated load,then max moment occurs under the load and its value
+//M1=P*a*b*L**-1 //N-mm
+
+//Equting it to moment carrying capacity we get
+P=M*1200**-1*10**-3 //N
+
+//Result
+printf("\n Max Intensity of u.d.l it can carry %0.3f  KN-m",w)
+printf("\n MAx concentrated Load P apllied at 2 m from one end is %0.3f  KN",P)
diff --git a/3864/CH4/EX4.10/Ex4_10.sce b/3864/CH4/EX4.10/Ex4_10.sce
new file mode 100644
index 000000000..d09c2efb0
--- /dev/null
+++ b/3864/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,53 @@
+clear
+//
+
+//Initilization of Variables
+H=10 //mm //Height
+A1=160*160 //mm**2 //area of square section at bottom
+L1=160 //mm //Length of square section at bottom
+b1=160 //mm //width of square section at bottom
+A2=80*80 //mm**2 //area of square section at top
+L2=80 //mm //Length of square section at top
+b2=80 //mm //Width of square section at top
+P=100 //N //Pull
+
+//Calculations
+
+//Consider a section at distance y from top.
+//Let the side of square bar be 'a'
+//a=L2+y*(H)**-1*(b1-b2)
+//After further simplifying we get
+//a=L2+8*y
+
+//Moment of Inertia
+//I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3
+//After further simplifying we get
+//I=a**4*12**-1
+
+//Section Modulus 
+//Z=a**4*(12*a*(2)**0.5)**-1
+//After further simplifying we get
+//Z=2**0.5*a**3*(12)**-1 //mm**3
+
+//Bending moment at this section=100*y N-mm
+//M=100*10**3*y //N-mm
+
+//But
+//M=sigma*Z
+//After sub values in above equation we get
+//sigma=M*Z**-1
+//After further simplifying we get
+//sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)
+
+//For Max stress df*(dy)**-1=0
+//After taking Derivative of above equation we get
+//df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)
+//After further simplifying we get
+y=80*16**-1 //m
+
+//Max stress at this level is
+sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1
+
+//Result
+printf("\n Max Bending stress is Developed at %0.3f  m",y)
+printf("\n Value of Max Bending stress is %0.3f  N/mm**2",sigma)
diff --git a/3864/CH4/EX4.12/Ex4_12.sce b/3864/CH4/EX4.12/Ex4_12.sce
new file mode 100644
index 000000000..d1060e823
--- /dev/null
+++ b/3864/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,43 @@
+clear
+//
+
+//Initilization of Variables
+
+b=200 //mm //Width of timber 
+d=400 //mm //Depth of timber
+t=6 //mm //Thickness
+b2=200 //mm //width of steel plate
+t2=20 //mm //Thickness of steel plate
+M=40*10**6 //KN-mm //Moment
+//Let E_s*E_t**-1=X
+X=20 //Ratio of Modulus of steel to timber
+
+//Calculations
+
+//let y_bar be the Distance of centroidfrom bottom most fibre
+y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 //mm
+
+//Moment of Inertia
+I=1*12**-1*b*d**3+b*d*(b+t-(y_bar))**2+1*12**-1*t2*b2*t**3+b2*t2*t*((y_bar)-t*2**-1)**2
+
+
+//distance of the top fibre from N-A
+y_1=d+t-y_bar //mm
+
+//Distance of the junction of timber and steel From N-A
+y_2=y_bar-t //mm
+
+//Stress in Timber at the top
+Y=M*I**-1*y_1 //N/mm**2
+
+//Stress in the Timber at the junction point
+Z=M*I**-1*y_2
+
+//Coressponding stress in steel at the junction point
+Z2=X*Z //N/mm**2 
+
+//The stress in Extreme steel fibre 
+Z3=X*M*I**-1*y_bar
+
+//Result
+printf("\n Stress in Extreme steel Fibre %0.2f  N/mm**2",Z3)
diff --git a/3864/CH4/EX4.13/Ex4_13.sce b/3864/CH4/EX4.13/Ex4_13.sce
new file mode 100644
index 000000000..39953ccc4
--- /dev/null
+++ b/3864/CH4/EX4.13/Ex4_13.sce
@@ -0,0 +1,41 @@
+clear
+//
+
+//Initilization of Variables
+
+//Timber size
+b=150 //mm //Width
+d=300 //mm //Depth
+
+t=6 //mm //Thickness of steel plate
+l=6 //m //Span
+
+//E_s*E_t**-1=20 
+//m=E_s*E_t**-1
+m=20 
+sigma_timber=8 //N/mm**2  //Stress in timber
+sigma_steel=150 //N/mm**2 //Stress in steel plate
+
+//Let m*t=Y
+Y=m*t //mm
+L=(2*t+b)*m //mm //Width of flitched beam
+
+//Calculations
+
+//Due to  synnetry cenroid,the neutral axis is half the depth
+I=(1*12**-1*L*t**3+L*t*(b+t*2**-1)**2)*2+1*12**-1*(Y+b+Y)*d**3 //mm**4
+
+y_max1=150 //mm //For timber
+y_max2=156 //mm //For steel
+
+//stress in steel
+f_t1=1*m**-1*sigma_steel //N/mm**2
+
+//Moment of resistance
+M=f_t1*(I*y_max2**-1)
+
+//load
+w=8*M*(l**2)**-1*10**-6 //KN/m
+
+//Result
+printf("\n Load beam can carry is %0.2f  KN/m",w)
diff --git a/3864/CH4/EX4.14/Ex4_14.sce b/3864/CH4/EX4.14/Ex4_14.sce
new file mode 100644
index 000000000..dd7431bd4
--- /dev/null
+++ b/3864/CH4/EX4.14/Ex4_14.sce
@@ -0,0 +1,53 @@
+clear
+//
+
+//Initilization of Variables
+
+L=6000 //mm //Span of beam
+W=20*10**3 //N //Load
+sigma=8 //N/mm**2 //Stress
+b=200 //mm //Width of section
+d=300 //mm //Depth of section
+
+//Calculations
+
+//let x be the distance from left side of beam
+
+//Bending moment
+//M=W*2**-1*x //Nmm   .......(1)
+
+//But M=sigma*Z   ..........(2)
+
+//Equating equation 1 and 2 we get
+//W*2**-1*x=sigma*Z    ............(3)
+
+//Section Modulus 
+//Z=1*6*b*d**2   ...............(4)
+
+//Equating equation 3 and 4 we get
+//b*d**2=3*W*x*sigma**-1 .............(5)
+
+//Beam of uniform strength of constant depth
+//b=3*W*x*(sigma*d**2)    
+
+//When x=0
+b=0
+
+//When x=L*2**-1
+b2=3*W*L*(2*sigma*d**2)**-1 //mm
+
+//Beam with constant width of 200 mm
+
+//We have
+//d=(3*W*x*(sigma*d)**-1)**0.5
+//thus depth varies as (x)**0.5
+
+//when x=0
+d1=0
+
+//when x=L*2**-1
+d2=(3*W*L*(2*sigma*200)**-1)**0.5 //mm
+
+//Result
+printf("\n Cross section of rectangular beam is: %0.2f  mm",b2)
+printf("\n                                     : %0.2f  mm",d2)
diff --git a/3864/CH4/EX4.15/Ex4_15.sce b/3864/CH4/EX4.15/Ex4_15.sce
new file mode 100644
index 000000000..6991385fa
--- /dev/null
+++ b/3864/CH4/EX4.15/Ex4_15.sce
@@ -0,0 +1,33 @@
+clear
+//
+
+//Initilization of Variables
+
+L=800 //mm //Span
+n=5 //number of leaves
+b=60 //mm //Width
+t=10 //mm //thickness
+sigma=250 //N/mm**2 //Stress
+
+//Calculations
+
+//section Modulus
+Z=n*6**-1*b*t**2 //mm**3
+
+//from the relation
+//sigma*Z=M   ...................(1)
+//M=P*L*4**-1
+//sub values of M in equation 1 we get
+P=sigma*Z*4*L**-1*10**-3 //KN //Load
+
+//Length of Leaves
+L1=0.2*L //mm
+L2=0.4*L //mm
+L3=0.6*L //mm
+L4=0.8*L //mm
+L5=L //mm
+
+//Result
+printf("\n Max Load it can take is %0.2f  KN",P)
+printf("\n Length of leaves:L1 %0.2f  mm",L1)
+printf("\n                 :L2 %0.2f  mm",L2)
diff --git a/3864/CH4/EX4.17/Ex4_17.sce b/3864/CH4/EX4.17/Ex4_17.sce
new file mode 100644
index 000000000..0477cf682
--- /dev/null
+++ b/3864/CH4/EX4.17/Ex4_17.sce
@@ -0,0 +1,54 @@
+clear
+//
+//
+
+//Initilization of Variables
+
+F=40*10**3 //N //shear Force
+
+//I-section
+
+//Flanges
+b=80 //mm //Width of flange
+t=20 //mm //Thickness
+
+//Web
+d=200 //mm //Depth
+t2=20 //mm //Thickness
+
+//Flange-2
+b2=160 //mm //Width
+t3=20 //mm //Thickness
+
+D=240 //mm //Overall Depth
+
+//Calculations
+
+//Distance of N-A from Top Fibre 
+y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 //mm
+
+//Moment of Inertia
+I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 //mm**4
+
+//Shear stress bottom of flange
+sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 //N/mm**2
+
+//At same Level but in web
+sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 //N/mm**2
+
+//for shear stress at N.A
+X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 //mm**3
+sigma3=F*X*(t2*I)**-1 //N/mm**2
+
+//Shear stress at bottom of web
+
+X=b2*t3*((D-y)-t3*2**-1) //mm**3
+
+//Stress at bottom of web
+sigma4=F*X*(t2*I)**-1 //N/mm**2
+
+//Stress at Lower flange
+sigma5=F*X*(b2*I)**-1 //N/mm**2
+
+//Result
+printf("\n The Shear Force Diagram is the result")
diff --git a/3864/CH4/EX4.18/Ex4_18.sce b/3864/CH4/EX4.18/Ex4_18.sce
new file mode 100644
index 000000000..48f9ecb18
--- /dev/null
+++ b/3864/CH4/EX4.18/Ex4_18.sce
@@ -0,0 +1,36 @@
+clear
+//
+
+//Initilization of Variables
+
+F=30*10**3 //N //Shear Force
+
+//Channel Section
+d=400 //mm //Depth of web 
+t=10 //mm //THickness of web
+t2=15 //mm //Thickness of flange
+b=100 //mm //Width of flange
+
+//Rectangular Welded section
+b2=80 //mm //Width
+d2=60 //mm //Depth
+
+//Calculations
+
+//Distance of Centroid From Top Fibre
+y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 //mm
+
+//Moment Of Inertia of the section about N-A
+I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2
+
+//Shear stress at level of weld
+sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 //N/mm**2
+
+//Max Shear Stress occurs at Neutral Axis
+X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1
+
+sigma_max=F*X*((b+t)*I)**-1
+
+//Result
+printf("\n Shear stress in the weld is %0.2f  N/mm**2",sigma)
+printf("\n Max shear stress is %0.2f  N/mm**2",sigma_max)
diff --git a/3864/CH4/EX4.19/Ex4_19.sce b/3864/CH4/EX4.19/Ex4_19.sce
new file mode 100644
index 000000000..319ec29a0
--- /dev/null
+++ b/3864/CH4/EX4.19/Ex4_19.sce
@@ -0,0 +1,30 @@
+clear
+//
+//
+
+//Initilization of Variables
+
+//Wooden Section
+b=300 //mm //Width
+d=300 //mm //Depth
+
+D=100 //mm //Diameter of Bore
+F=10*10**3 //N //Shear Force
+
+//Calculations
+
+//Moment Of Inertia Of Section
+I=1*12**-1*b*d**3-%pi*64**-1*D**4
+
+//Shear stress at crown of circle
+sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1
+
+//Let a*y_bar=X
+X=b*d*2**-1*d*4**-1-%pi*8**-1*D**2*4*D*2**-1*(3*%pi)**-1 //mm**3
+
+//Shear Stress at Neutral Axis
+sigma2=F*X*((b-D)*I)**-1 //N/mm**2
+
+//Result
+printf("\n Shearing Stress at Crown of Bore %0.3f  N/mm**2",sigma)
+printf("\n Shear Stress at Neutral Axis %0.3f  N/mm**2",sigma2)
diff --git a/3864/CH4/EX4.2/Ex4_2.sce b/3864/CH4/EX4.2/Ex4_2.sce
new file mode 100644
index 000000000..029c9dacc
--- /dev/null
+++ b/3864/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,35 @@
+clear
+//
+//
+
+//Initilization of Variables
+
+D=70 //mm //External Diameter
+t=8 //mm //Thickness of pipe
+L=2500 //mm //span 
+sigma=150 //N/mm**2 //stress
+
+//Calculations
+
+//Internal Diameter 
+d=D-2*t //mm
+
+//M.I Of Pipe
+I=%pi*64**-1*(D**4-d**4) //mm**4
+
+y_max=D*2**-1 //mm
+Z=I*(y_max)**-1 //mm**3
+
+//Moment Carrying capacity
+M=sigma*Z //N*mm
+
+//Max moment int the beam occurs at the mid-span and is equal to
+//m=P*L*4**-1
+
+//Equating Max moment to moment carrying capacity we get,
+//M=P*2.5*L*4**-1
+//After substituting and simplifying we get
+P=4*M*(L)**-1*10**-3 //N
+
+//Result
+printf("\n Max concentrated load that can be applied at the centre of span is %0.3f  KN",P)
diff --git a/3864/CH4/EX4.20/Ex4_20.sce b/3864/CH4/EX4.20/Ex4_20.sce
new file mode 100644
index 000000000..f4733fccf
--- /dev/null
+++ b/3864/CH4/EX4.20/Ex4_20.sce
@@ -0,0 +1,48 @@
+clear
+//
+
+//Initilization of Variables
+
+//flanges
+b=200 //mm //width
+t1=25 //mm //Thickness
+
+//web
+d=450 //mm //Depth 
+t2=20 //mm //thickness
+
+D=500 //mm //Total Depth of section
+
+//Calculations
+
+//Moment Of Inertia of the section about N-A
+I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 //mm**4
+
+//Consider an element in the web at distance y from y from N-A
+//Depth of web section=225-y
+
+//C.G From N-A
+//y2=y+(((D*2**-1-t)-y)*2**-1)
+
+//ay_bar for section at y
+//Let ay_bar be X
+//X=X1 be of Flange + X2 be of web above y
+//X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1
+//After Sub values and Further simplifying we get
+//X=1187500+10*(225**2-y**2)
+
+//Shear stress at y
+//sigma_y=F*(X)*(t2*I)**-1
+
+//Shear Force resisted by the Element
+//F1=F*X*t2*dy*(t2*I)**-1
+
+//Shear stress resisted by web 
+//sigma=2*F*I**-1*(X)*dy
+
+//After Integrating above equation and further simplifying we get
+//sigma=0.9578*F
+
+sigma=0.9578*100
+
+//Result
diff --git a/3864/CH4/EX4.21/Ex4_21.sce b/3864/CH4/EX4.21/Ex4_21.sce
new file mode 100644
index 000000000..d8a38b8be
--- /dev/null
+++ b/3864/CH4/EX4.21/Ex4_21.sce
@@ -0,0 +1,40 @@
+clear
+//
+
+//Initilization of Variables
+
+//Wooden Beam
+
+b=150 //mm //width
+d=250 //mm //Depth
+
+L=5000 //mm //span
+m=11.2 //N/mm**2 //Max Bending stress
+sigma=0.7 //N/mm**2 //Max shear stress
+
+//Calculations
+
+//Let 'a' be the distance from left support
+//Max shear force
+//F=R_A=W*(L-a)*L**-1 
+
+//Max Moment
+//M=W*(L-a)*a*L**-1
+
+//But M=sigma*Z
+//W*(L-a)*a*L**-1=m*1*6**-1*b*d**2   .....................(1)
+
+//In Rectangular Section MAx stress is 1.5 times Avg shear stress
+F=sigma*b*d*1.5**-1
+
+//W*(L-a)*L**-1=F                     .....................(2)
+
+//Dividing Equation 1 nad 2 we get
+a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1
+
+//Sub above value in equation 2 we get
+W=(L-a)**-1*L*F*10**-3 //KN  
+
+//Result
+printf("\n Load is %0.2f  KN",W)
+printf("\n Distance from Left support is %0.2f  mm",a)
diff --git a/3864/CH4/EX4.22/Ex4_22.sce b/3864/CH4/EX4.22/Ex4_22.sce
new file mode 100644
index 000000000..d25a86e08
--- /dev/null
+++ b/3864/CH4/EX4.22/Ex4_22.sce
@@ -0,0 +1,38 @@
+clear
+//
+
+//Initilization of Variables
+
+L=1000 //mm //span
+
+//Rectangular Section
+
+b=200 //mm //width
+d=400 //mm //depth
+
+sigma=1.5 //N/mm**2 //Shear stress
+
+//Calculations
+
+//Let AB be the cantilever beam subjected to load W KN at free end
+
+//MAx shear Force
+//F=W*10**3 //KN
+
+//Since Max shear stress in Rectangular section
+//sigma_max=1.5*F*A**-1 
+//After sub values and further simplifyng we get
+W=1.5*b*d*(1.5*1000)**-1 //KN
+
+//Moment at fixwed end
+M=W*1 //KN-m
+y_max=d*2**-1 //mm
+
+//M.I
+I=1*12**-1*b*d**3 //mm**3
+
+//MAx Stress
+sigma_max=M*10**6*I**-1*y_max
+
+//Result
+printf("\n Concentrated Load is %0.2f  N/mm**2",sigma_max)
diff --git a/3864/CH4/EX4.24/Ex4_24.sce b/3864/CH4/EX4.24/Ex4_24.sce
new file mode 100644
index 000000000..5b2e0cad5
--- /dev/null
+++ b/3864/CH4/EX4.24/Ex4_24.sce
@@ -0,0 +1,40 @@
+clear
+//
+
+//Initilization of Variables
+
+L=4000 //mm //span
+
+//Rectangular Cross-section
+b=100 //mm //Width
+d=200 //mm //Thickness
+
+F_per=10 //N/mm**2 //Max Bending stress
+q_max=0.6 //N/mm**2 //Shear stress
+
+//Calculations
+
+//If the Load W is in KN/m
+
+//Max shear Force
+//F=w*l*2**-1 //KN
+//After substituting values and further simplifying we get
+//M=2*w //KN-m
+
+//Max Load from Consideration of moment
+//M=1*6**-1*b*d**2*F_per
+//After substituting values and further simplifying we get
+w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 //KN/m
+
+//Max Load from Consideration of shear stress
+//q_max=1.5*F*(b*d)**-1 //N
+//After substituting values and further simplifying we get
+F=q_max*(1.5)*b*d //N
+
+//If w is Max Load in KN/m,then
+//2*w*1000=8000
+//After Rearranging and Further simplifying we get
+w2=8000*(2*1000)**-1 //KN/m
+
+//Result
+printf("\n Uniformly Distributed Load Beam can carry is %0.2f  KN/m",w)
diff --git a/3864/CH4/EX4.4/Ex4_4.sce b/3864/CH4/EX4.4/Ex4_4.sce
new file mode 100644
index 000000000..40556ec38
--- /dev/null
+++ b/3864/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,52 @@
+clear
+//
+
+//Initilization of Variables
+
+//Flange (Top)
+b1=80 //mm //Width 
+t1=40 //mm //Thickness
+
+//Flange (Bottom)
+b2=160 //mm //width
+t2=40 //mm //Thickness
+
+//web
+d=120 //mm //Depth
+t3=20 //mm //Thickness
+
+D=200 //mm //Overall Depth
+sigma1=30 //N/mm**2 //Tensile stress
+sigma2=90 //N/mm**2 //Compressive stress
+L=6000 //mm //Span
+
+//Calculations
+
+//Distance of centroid from bottom fibre
+y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 //mm
+
+//Moment of Inertia
+I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-(y_bar))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-(y_bar))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-(y_bar))**2
+
+
+//Extreme fibre distance of top and bottom fibres are y_t and y_c respectively
+
+y_t=y_bar //mm
+y_c=D-y_bar //mm
+
+//Moment carrying capacity considering Tensile strength 
+M1=sigma1*I*y_t**-1*10**-6  //KN-m
+
+//Moment carrying capacity considering compressive strength 
+M2=sigma2*I*y_c**-1*10**-6  //KN-m
+
+//Max Bending moment in simply supported beam 6 m due to u.d.l
+//M_max=w*L*10**-3*8**-1
+//After simplifying further we get
+//M_max=4.5*w
+
+//Now Equating it to Moment carrying capacity, we get load carrying capacity
+w=M1*4.5**-1 //KN/m
+
+//Result
+printf("\n Max Uniformly Distributed Load is %0.3f  KN/m",w)
diff --git a/3864/CH4/EX4.5/Ex4_5.sce b/3864/CH4/EX4.5/Ex4_5.sce
new file mode 100644
index 000000000..0b6583da4
--- /dev/null
+++ b/3864/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,49 @@
+clear
+//
+//
+
+//Initilization of Variables
+
+//Flanges
+b=200 //mm //Width
+t=25 //mm //Thickness 
+
+D1=500 //mm //Overall Depth
+t2=20 //mm //Thickness of web
+
+d=450 //mm //Depth of web
+
+//Calculations
+
+//Consider,Element of Thickness "y" at Distance "dy" from N.A 
+//Let Bending stress "sigma_max"
+
+//Stress on the element 
+//sigma=y*(D*2**-1)*sigma_max   ..............(1)
+
+//Area of Element
+//A=b*dy       .................................(2)
+
+//Force on Element 
+//F=y*250**-1*sigma_max*b*dy
+
+//Let M be the Moment of resistance
+//M=y*250**-1*sigma_max*b*dy*y
+
+//Moment of Resistance of top flange after simplification we gget
+//M.R=2258333.3*f
+
+//M.I of I section
+I=1*12**-1*(b*D1**3-180*d**3)*10**-8
+
+//Moment acting on section 
+//After simplifying we get
+//M=2865833.3*f
+
+//Percentage moment resistance
+M1=2258333.3*2865833.3**-1*100
+
+//Percentage moment resisted by web
+M2=100-M1
+
+//Result
diff --git a/3864/CH4/EX4.6/Ex4_6.sce b/3864/CH4/EX4.6/Ex4_6.sce
new file mode 100644
index 000000000..d6659cef7
--- /dev/null
+++ b/3864/CH4/EX4.6/Ex4_6.sce
@@ -0,0 +1,63 @@
+clear
+//
+//
+
+//Initilization of Variables
+
+//Flanges
+b1=200 //mm //Width
+t1=10 //mm //Thickness
+
+//Web
+d=380 //mm //Depth 
+t2=8 //mm //Thickness
+
+D=400 //mm //Overall Depth
+sigma=150 //N/mm**2
+
+//Calculations
+
+//Area
+A=b1*t1+d*t2+b1*t1 //mm**2
+
+//Moment of Inertia
+I=1*12**-1*(b1*D**3-(b1-t2)*d**3)
+
+//Bending Moment
+M=sigma*I*(D*2**-1)**-1
+
+//Square Section
+
+//Let 'a' be the side
+a=A**0.5
+
+//Moment of Resistance of this section
+M1=1*6**-1*a*a**2*sigma
+
+X=M*M1**-1
+
+//Rectangular section
+//Let 'a' be the side and depth be 2*a
+
+a=(A*2**-1)**0.5
+
+//Moment of Rectangular secction
+M2=1*6**-1*a*(2*a)**2*sigma
+
+X2=M*M2**-1
+
+//Circular section
+//A=%pi*d1**2*4**-1
+
+d1=(A*4*%pi**-1)**0.5
+
+//Moment of circular section
+M3=%pi*32**-1*d1**3*sigma
+
+X3=M*M3**-1
+
+//Result
+printf("\n Moment of resistance of beam section %0.2f  mm",M)
+printf("\n Moment of resistance of square section %0.2f  mm",X)
+printf("\n Moment of resistance of rectangular section %0.2f  mm",X2)
+printf("\n Moment of resistance of circular section %0.2f  mm",X3)
diff --git a/3864/CH4/EX4.7/Ex4_7.sce b/3864/CH4/EX4.7/Ex4_7.sce
new file mode 100644
index 000000000..e50fc587f
--- /dev/null
+++ b/3864/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,40 @@
+clear
+//
+
+//Initilization of Variables
+
+F=12 //KN //Force at End of beam
+L=2 //m //span
+
+//Square section 
+b=200 //mm //Width and depth of beam
+d=200
+
+//Rectangular section
+b1=150 //mm //Width
+d1=300 //mm //Depth
+	
+//Calculations
+
+//Max bending Moment
+M=F*L*10**6 //N-mm
+
+//M=sigma*b*d**2
+sigma=M*6*(b*d**2)**-1 //N/mm**2
+
+//Let W be the central concentrated Load in simply supported beam of span L1=3 m
+//MAx Moment
+//M1=W*L1*4**-1
+//After Further simplifying we get
+//M1=0.75*10**6 //N-mm
+
+//The section has  a moment of resistance
+M1=sigma*1*6**-1*b1*d1**2
+
+//Equating it to moment of resistance we get max load W
+//0.75*10**6*W=M1
+//After Further simplifying we get
+W=M1*(0.75*10**6)**-1
+
+//Result
+printf("\n Minimum Concentrated Load required to brek the beam %0.2f  KN",W)
diff --git a/3864/CH4/EX4.8/Ex4_8.sce b/3864/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..73bf0efb6
--- /dev/null
+++ b/3864/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,65 @@
+clear
+//
+
+//Initilization of Variables
+
+L=3 //m //span
+sigma_t=35 //N/mm**2 //Permissible stress in tension
+sigma_c=90 //N/mm**2 //Permissible stress in compression
+
+//Flanges
+t=30 //mm //Thickness
+d=250 //mm //Depth
+
+//Web
+t2=25 //mm //Thickness
+b=600 //mm //Width
+
+//Calculations
+
+//Let y_bar be the Distance of N.A from Extreme Fibres
+y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1
+
+//Moment of Inertia
+I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2
+
+//Part-1
+
+//If web is in Tension
+y_t=y_bar //mm
+y_c=d-y_bar //mm
+
+//Moment carrying caryying capacity From consideration of tensile stress
+M=sigma_t*I*(y_bar)**-1 //N-mm
+
+//Moment carrying caryying capacity From consideration of compressive stress
+M1=sigma_c*I*(y_c)**-1 //N-mm
+
+//If w KN/m is u.d.l in beam,Max bending moment
+//M=wl**2*8**-1
+//After further simplifyng we get
+//M=1.125*w*10**6 N-mm
+w=M*(1.125*10**6)**-1 //KN
+
+//Part-2
+
+//If web is in compression
+y_t2=178.299 //mm
+y_c2=71.71 //mm 
+
+//Moment carrying caryying capacity From consideration of tensile stress
+M2=sigma_t*I*(y_t2)**-1 //N-mm
+
+//Moment carrying caryying capacity From consideration of compressive stress
+M3=sigma_c*I*(y_c2)**-1 //N-mm
+
+//Moment of resistance is M2
+
+//Equating it to bending moment we get
+//M2=1.125*10**6*w2
+//After further simplifyng we get
+w2=M2*(1.125*10**6)**-1
+
+//Result
+printf("\n Uniformly Distributed Load carrying capacity if:web is in Tension %0.2f  KN",w)
+printf("\n                                                :web is in compression %0.3f  KN",w2)
diff --git a/3864/CH4/EX4.9/Ex4_9.sce b/3864/CH4/EX4.9/Ex4_9.sce
new file mode 100644
index 000000000..93537e581
--- /dev/null
+++ b/3864/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,48 @@
+clear
+//
+//
+
+//Initilization of Variables
+
+b1=200 //mm //Width at base
+b2=100 //mm //Width at top
+
+L=8 //m Length
+P=500 //N //Load
+
+//Calculations
+
+//Consider a section at y metres from top
+
+//At this section diameter d is
+//d=b2+y*L**-1*(b1-b2)
+//After Further simplifying we get
+//d=b2+12.5*y //mm
+
+//Moment of Inertia
+//I=%pi*64**-1*d**4
+
+//Section Modulus 
+//Z=%pi*32**-1*(b1+12.5*y)**3
+
+//Moment 
+//M=5*10**5*y //N-mm
+
+//Let sigma be the fibre stress at this section then
+//M=sigma*Z
+//After sub values in above equation and further simplifying we get
+//sigma=5*10**5*32*%pi**-1*y*((b2+12.5*y)**3)**-1
+
+//For sigma to be Max,d(sigma)*(dy)**-1=0
+//16*10**6*%pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)
+//After Further simplifying we get
+//b2+12.5*y=37.5*y
+//After Further simplifying we get
+y=b2*25**-1 //m
+
+//Stress at this section
+sigma=5*10**5*32*%pi**-1*y*((b2+12.5*y)**3)**-1
+
+//Result
+printf("\n Stress at Extreme Fibre is max %0.2f  m",y)
+printf("\n Max stress is %0.2f  N/mm**2",sigma)