From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 3761/CH4/EX4.1/Ex4_1.sce   |  63 ++++++++++++++++++++++++++++
 3761/CH4/EX4.10/Ex4_10.sce |  20 +++++++++
 3761/CH4/EX4.11/Ex4_11.sce |  71 +++++++++++++++++++++++++++++++
 3761/CH4/EX4.12/Ex4_12.sce |  50 ++++++++++++++++++++++
 3761/CH4/EX4.13/Ex4_13.sce |  50 ++++++++++++++++++++++
 3761/CH4/EX4.14/Ex4_14.sce |  33 +++++++++++++++
 3761/CH4/EX4.15/Ex4_15.sce | 102 +++++++++++++++++++++++++++++++++++++++++++++
 3761/CH4/EX4.16/Ex4_16.sce |  66 +++++++++++++++++++++++++++++
 3761/CH4/EX4.17/Ex4_17.sce |  54 ++++++++++++++++++++++++
 3761/CH4/EX4.2/Ex4_2.sce   |  40 ++++++++++++++++++
 3761/CH4/EX4.3/Ex4_3.sce   |  59 ++++++++++++++++++++++++++
 3761/CH4/EX4.4/Ex4_4.sce   |  62 +++++++++++++++++++++++++++
 3761/CH4/EX4.5/Ex4_5.sce   |  67 +++++++++++++++++++++++++++++
 3761/CH4/EX4.6/Ex4_6.sce   |  54 ++++++++++++++++++++++++
 3761/CH4/EX4.7/Ex4_7.sce   |  50 ++++++++++++++++++++++
 3761/CH4/EX4.8/Ex4_8.sce   |  61 +++++++++++++++++++++++++++
 3761/CH4/EX4.9/Ex4_9.sce   |  80 +++++++++++++++++++++++++++++++++++
 17 files changed, 982 insertions(+)
 create mode 100644 3761/CH4/EX4.1/Ex4_1.sce
 create mode 100644 3761/CH4/EX4.10/Ex4_10.sce
 create mode 100644 3761/CH4/EX4.11/Ex4_11.sce
 create mode 100644 3761/CH4/EX4.12/Ex4_12.sce
 create mode 100644 3761/CH4/EX4.13/Ex4_13.sce
 create mode 100644 3761/CH4/EX4.14/Ex4_14.sce
 create mode 100644 3761/CH4/EX4.15/Ex4_15.sce
 create mode 100644 3761/CH4/EX4.16/Ex4_16.sce
 create mode 100644 3761/CH4/EX4.17/Ex4_17.sce
 create mode 100644 3761/CH4/EX4.2/Ex4_2.sce
 create mode 100644 3761/CH4/EX4.3/Ex4_3.sce
 create mode 100644 3761/CH4/EX4.4/Ex4_4.sce
 create mode 100644 3761/CH4/EX4.5/Ex4_5.sce
 create mode 100644 3761/CH4/EX4.6/Ex4_6.sce
 create mode 100644 3761/CH4/EX4.7/Ex4_7.sce
 create mode 100644 3761/CH4/EX4.8/Ex4_8.sce
 create mode 100644 3761/CH4/EX4.9/Ex4_9.sce

(limited to '3761')

diff --git a/3761/CH4/EX4.1/Ex4_1.sce b/3761/CH4/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..aed818c3b
--- /dev/null
+++ b/3761/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,63 @@
+disp("EXAMPLE 4.1")
+disp("Material Properties","Applied Moment = 50kNm","Grade of Steel = Fe415","Grade of Concrete = M20","Ast = 4-25mm dia bars","d = 550mm","D = 600mm","b = 300mm","Given:")
+//disp("Given:")
+b=300
+d=550
+
+disp("sigmacbc= 7 MPa")
+m=280/(3*7)
+disp("modular ratio , m =" +string(m))
+// m= 280/(3*sigmacbc)
+
+
+fcr = 0.7 * sqrt(20)
+disp("modulus of rupture, fcr =" +string(fcr))
+// fcr = o.7 * sqrt(Fck)
+
+disp("Approximate Cracking Moment , assuming gross concrete section")
+b=300
+D=600
+Z= (b*D*D)/6
+disp(" in mm^3","Section Modulus Z=" + string(Z))
+//Cracking Moment
+Mcr= (fcr*Z)/(10^6)
+disp("in kNm","Cracking Moment =" +string(Mcr))
+
+disp("Transformed Section Properties")
+diabar=25
+Ast=(4*%pi*25*25)/4
+disp("in mm^2","Area of Tension Steel = " +string(Ast))
+disp("Transformed Area, At")
+disp("At = bD + (m-1)Ast")
+disp("Depth of neutral axis y")
+disp("At y = (bD)(D/2)+(m-1)Ast(d)")
+y=(((b*D*D)/2)+((Ast)*(m-1)*(d)))/(((b*D))+ ((m-1)*Ast))
+disp("in mm","Depth of neutral axis, y = " +string(y))
+yc=y
+yt= D-yc
+ys=d-yc
+
+disp("Distance from NA to extreme compression fibre, yc= " +string(yc))
+disp("Distance from NA to extreme tension fibre, yt=" +string(yt))
+disp("Distance from NA to reinforcing steel, ys=" +string(ys))              
+disp("Transformed Second Moment of Area")
+It = (b*yc^3/3)+(b*yt^3/3)+ ((m-1)*Ast*ys*ys)
+
+disp("Calculating Cracking Moment","4.1.a")
+Mcra = (fcr*It/(yt*10^6))
+disp("in kNm", "Cracking Moment=" +string(Mcra))
+
+disp("Stresses due to applied moment","4.1.b")
+
+M=50
+fc = M*yc*10^6/It
+disp("in MPa","Maximum Compressive Stress in Concrete, fc= " +string(fc))
+fct= (M*yt*10^6/It)
+disp("in MPa", "Maximum Tensile Stress in Concrete, fct=" +string(fct))
+
+if(fct<fcr)
+
+    disp("okay")
+
+fst= m*fc*ys/yc
+disp("Tensile Stress in Steel, fst=" +string(fst)) // Answer varies a little, calculation error in TB
diff --git a/3761/CH4/EX4.10/Ex4_10.sce b/3761/CH4/EX4.10/Ex4_10.sce
new file mode 100644
index 000000000..56bfc658e
--- /dev/null
+++ b/3761/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,20 @@
+disp("Example 4.10")
+disp("Grade of Steel,fy = Fe250","Grade of Concrete,fck = M20","D=600mm","d=550mm","b=300mm","Bars used = 4 - 25 dia")
+b=300
+d=550
+D=600
+fck=20
+Ast=%pi*4*25*25/4
+disp("mm^2",Ast,"Ast=")
+disp("For Fe415 Steel,")
+Es=2*10^5
+fy=250
+Est=0.87*fy/Es
+xumaxd=(0.0035/(0.0055+Est))
+disp(xumaxd,"xumax/d")
+xumax=xumaxd*d
+disp("mm",xumax,"xu,max=")
+disp("Assuming, xu</xu,max and applying the force equilibrium condition Cu=Tu")
+xu= (0.87*fy*Ast)/(0.362*fck*b)
+disp("mm",xu,"xu")
+disp("mm",xu,"xu<xu,max, therefore xu=")
diff --git a/3761/CH4/EX4.11/Ex4_11.sce b/3761/CH4/EX4.11/Ex4_11.sce
new file mode 100644
index 000000000..4bf023034
--- /dev/null
+++ b/3761/CH4/EX4.11/Ex4_11.sce
@@ -0,0 +1,71 @@
+disp("Example 4.11")
+disp("4.11.a (Referring to Example 4.9)")
+disp("Grade of Steel,fy = Fe415","Grade of Concrete,fck = M20","D=600mm","d=550mm","b=300mm","Bars used = 4 - 25 dia")
+b=300
+d=550
+D=600
+fck=20
+Ast=%pi*4*25*25/4
+disp("mm^2",Ast,"Ast=")
+disp("For Fe415 Steel,")
+Es=2*10^5
+fy=415
+Est=0.87*fy/Es
+//xumaxd=(0.0035/(0.0055+Est))
+//disp(xumaxd,"xumax/d")
+//xumax=xumaxd*d
+//disp("mm",xumax,"xu,max=")
+//disp("Assuming, xu</xu,max and applying the force equilibrium condition Cu=Tu")
+//xu= (0.87*fy*Ast)/(0.362*fck*b)
+xu=315
+disp("mm",xu,"xu")
+disp("Taking moments about the tension steel centroid")
+MuR=0.362*fck*b*xu*(d-0.416*xu)/10^6
+disp("kNm",MuR,"MuR=")
+disp("The value of MuR can also be calculated in terms of the steel tensile stress fst, whish is less than 0.87fy, as xu>xu,max. The value of fst obtained from last itteration is obtained as 349MPa, therefore, MuR=fst*Ast*(d-0.416*xu)")
+MuR1=fst*Ast*(d-0.416*xu)/10^6
+disp("kNm",MuR1,"Therefore, MuR=")
+disp("Alternative (using analysis aid)")
+pt=(100*Ast)/(b*d)
+disp(pt,"Referring Table A.2(a)-for M20 concrete and Fe415 steel for pt=")
+disp("MuR/bd^2 for pt,1.18  = 3.145 and for pt, 1.20 = 3.170, therefore, for M20 concrete and Fe415 steel and pt=1.19 MuR/bd^2=")
+MuR1bd2=(3.145+3.170)/2
+MuR1=MuR1bd2*b*d^2/10^6
+disp("kNm",MuR1,"MuR=")            
+
+disp("Example 4.11.b, (Refering Example 4.10")
+disp("Grade of Steel,fy = Fe250","Grade of Concrete,fck = M20","D=600mm","d=550mm","b=300mm","Bars used = 4 - 25 dia")
+b=300
+d=550
+D=600
+fck=20
+Ast=%pi*4*25*25/4
+disp("mm^2",Ast,"Ast=")
+disp("For Fe415 Steel,")
+Es=2*10^5
+fy=250
+Est=0.87*fy/Es
+xumaxd=(0.0035/(0.0055+Est))
+//disp(xumaxd,"xumax/d")
+//xumax=xumaxd*d
+//disp("mm",xumax,"xu,max=")
+//disp("Assuming, xu</xu,max and applying the force equilibrium condition Cu=Tu")
+xu= (0.87*fy*Ast)/(0.362*fck*b)
+disp("mm",xu,"xu")
+disp("mm",xu,"xu<xu,max, therefore xu=")
+disp("Taking moments about the tension steel centroid")
+MuR2=0.362*fck*b*xu*(d-0.416*xu)/10^6
+disp("kNm",MuR2,"MuR=")
+
+disp("Alternatively, as xu<xu,max, it follows that fst=0.87fy, and")
+MuR3=0.87*Ast*fy*(d-0.416*xu)/10^6
+disp("kNm",MuR3,"MuR=")
+disp("pt=1.190 as calculated above")
+disp("Referring to table A.2(a) for M20 concrete and Fe 250 Steel, for pt=1.180 Mu/bd^2= 2.188 and for pt= 1.20 it is = 2.219, therefore for pt=1.19, Mu/bd^2=")
+MuRbd2=(2.188+2.219)/2
+MuR4=MuRbd2*b*d^2/10^6
+disp("kNm",MuR4)
+
+
+
+
diff --git a/3761/CH4/EX4.12/Ex4_12.sce b/3761/CH4/EX4.12/Ex4_12.sce
new file mode 100644
index 000000000..76f25cff6
--- /dev/null
+++ b/3761/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,50 @@
+disp("Example 4.12")
+disp("fck=20MPa","fy=Fe250","Ast=3695mm^2","d=520mm","bw=250mm","Df=100mm","bf=850mm","Given:")
+bf=850
+Df=100
+bw=250
+d=520
+Ast=3695
+fy=250
+fck=20
+Es=2*10^5
+xumaxd=0.0035/(0.0055+0.87*(fy/Es))
+xumax=xumaxd*d
+disp("First assuming xu</Df and xu</xu,max and considering force equilibrium")
+disp("Cu=Tu=0.362*fck*bf*xu=0.87*fy*Ast")
+disp("Therefore,xu=")
+xu=(0.87*fy*Ast)/(0.362*bf*xu*fck)
+disp(xu,"mm")
+disp("Since, xu>Df, the value is incorrect ")
+
+disp("Asxu>Df, the compression in web is given by: Cuw=0.362*fck*bw*xu")
+Cuw=0.362*fck*bw
+disp("*xu N",Cuw)
+disp("Assuming xu>/7/3*Df = 233.3, the commpression in the flange is given by: Cuf=0.447*fck*(bf-bw)*Df")
+Cuf=0.447*fck*(bf-bw)*Df
+disp("Cuf=")
+disp("N",Cuf)
+disp("Also assuming xu</xu,max = 276.1mm fst= 0.87*fy, and Tu=0.87*fy*Ast")
+Tu=0.87*fy*Ast
+disp("Tu=")
+disp("N",Tu)
+disp("Aplying the equilibrium conditions (Cuw+Cuf=Tu")
+xu=(Tu-Cuf)/Cuw
+disp("Since, xu<7/3*Df, hence this value of xu is also not correct")
+disp("As Df<xu<7/3*Df, the depth yf</Df of the equivalent concrete stress block is obtained as: yf=0.15*xu+0.65*Df = (0.15*xu+65)mm")
+disp("Cuf=Cuf*(yf/Df)= (804.6*xu+348660)N")
+disp("Cuw+Cuf=Tu")
+xu1=(Tu-(Cuf*(65/100)))/(Cuw+(Cuf*0.15/100))
+disp("xu=")
+disp("mm",xu1)
+disp("As, xu<xu,max; Hence the assumption fst=0.87*fy is OK")
+yf=(0.15*xu1)+65
+disp("mm",yf,"yf=")
+disp("Taking moments of Cuw and Cuf about the centroid of tension steel, MuR= Cuw*(d-0.416*xu)+Cuf*(d-yf/2)")
+Cuf=804.6*xu1+348660
+MuR=(Cuw*xu1*(d-0.416*xu1)+(Cuf*(d-yf/2)))/10^6
+disp("kNm",MuR,"MuR=")
+
+
+
+
diff --git a/3761/CH4/EX4.13/Ex4_13.sce b/3761/CH4/EX4.13/Ex4_13.sce
new file mode 100644
index 000000000..e9b3561e5
--- /dev/null
+++ b/3761/CH4/EX4.13/Ex4_13.sce
@@ -0,0 +1,50 @@
+disp("Example 4.13")
+disp("fck=20MPa","fy=Fe250","Ast=4926mm^2","d=520mm","bw=250mm","Df=100mm","bf=850mm","Given:")
+bf=850
+Df=100
+bw=250
+d=520
+Ast=4926
+fy=250
+fck=20
+Es=2*10^5
+xumaxd=0.0035/(0.0055+0.87*(fy/Es))
+xumax=xumaxd*d
+disp("mm",xumax,"xumax=")
+disp("First assuming xu</Df and xu</xu,max")
+disp("xu=(0.87*fy*Ast)/(0.362*fck*bf)")
+xu=(0.87*fy*Ast)/(0.362*fck*bf)
+disp("mm",xu,"xu=")
+disp("xu >Df, Hence this value of xu is not correct")
+disp("As xu>Df, Cuw = 0.362*fck*fy*bw*xu")
+Cuw=0.362*fck*fy
+disp("xu N", Cuw,"Cuw=")
+disp("ASssuming xu>/7/3*Df = 233.33mm, yf=Df=100mm and Cuf=0.447*fck*(bf-bw)*Df")
+Cuf=0.447*fck*(bf-bw)*Df
+disp("N",Cuf,"Cuf=")
+disp("Further assuming xu</xu,max = 276.1 mm, fst=0.87*fy, and")
+Tu=0.87*fy*Ast
+disp("N",Tu,"Tu=")
+disp("Applying the force equilibrium condition Cuw+Cuf=Tu")
+xu=(Tu-Cuf)/Cuw
+disp("mm",xu,"xu=")
+disp("which implies xu>7/3Df =233.3mm, but not xu</xu,max=276.1mm. The section is over-reinforcedas per the Code provisions")
+
+disp("Exact Solution (considering strain compatibility)")
+disp("mm",xu,"Corresponding to xu=")
+disp("Est=0.0035*(d/xu-1)")
+Est=0.0035*(d/xu-1)
+disp(Est,"Est=")
+strainyield=0.87*fy/Es
+disp("Est is greater than strain at yield for Fe250")
+disp(strainyield)
+disp("Hence the design steel stress isindeed fst=0.87*fy and the so calculated xu above, is the correct depth of the neutral axis")
+disp("Accordingly,MuR= Cuw*(d-0.416*xu)+Cuf*(d-Df/2)")
+MuR=(Cuw*xu*(d-0.416*xu)+Cuf*(d-Df/2))/10^6
+disp("kNm",MuR,"MuR=")
+
+disp("APPROXIMATE SOLUTION")
+disp("Limiting xu to xu,max=276.1 mm and taking moments of Cuw and Cuf about the centroid of the tension steel.(Note that, following the Code procedure, Df/d=100/520=0.192<0.2, yf=Df=100mm")
+xumax
+MuRl=((Cuw*xumax*(d-0.416*xumax))+(Cuf*(d-Df/2)))/10^6
+disp("kNm",MuRl,"MuR,lim=")
diff --git a/3761/CH4/EX4.14/Ex4_14.sce b/3761/CH4/EX4.14/Ex4_14.sce
new file mode 100644
index 000000000..75cbc9df6
--- /dev/null
+++ b/3761/CH4/EX4.14/Ex4_14.sce
@@ -0,0 +1,33 @@
+disp("Example 4.14")
+disp("Asc=2-25dia bars","Ast=3-36dia bars","M20 Grade of concrete and Fe250steel","dd = 50mm","d=550mm","b=300mm","Given:")
+b=300
+d=550
+dd=50
+Ast= %pi*36*36*3/4
+Asc=2*%pi*25*25/4
+m=13.33 //(280/(3*sigmacbc))
+Es=2*10^5
+fck=20
+fy=250
+xumaxd=0.0035/(0.0055+(0.87*fy)/Es)
+disp(xumaxd,"xumax/d for Fe250=")
+xumax=xumaxd*d
+disp("mm",xumax,"xu,max")
+disp("Assuming fsc=fst=0.87*fy,, and considering force equilibrium")
+disp("Cus+Cuc = Tu")
+Cuc=0.362*fck*b
+Cus=(0.87*fy-0.447*fck)*Asc
+Tu=0.87*fy*Ast
+xu=(Tu-Cus)/Cuc
+disp("mm",xu,"xu=")
+disp("xu<xu,max")
+disp("Hence the assumption fst=0.87*fy is justified")
+disp("Also Esc = 0.0035*(1-dd/xu)")
+Esc=0.0035*(1-dd/xu)
+Ey=0.87*fy/(Es)
+disp(Esc,"Esc=")
+disp(Ey,"Ey=")
+disp("Therefore, fsc=0.87*fy is also justified ")
+disp("Ultimate Moment of Resistance")
+MuR=(Cuc*xu*(d-0.416*xu)+Cus*(d-dd))/10^6
+disp("kNm",MuR,"MuR=")
diff --git a/3761/CH4/EX4.15/Ex4_15.sce b/3761/CH4/EX4.15/Ex4_15.sce
new file mode 100644
index 000000000..c98598035
--- /dev/null
+++ b/3761/CH4/EX4.15/Ex4_15.sce
@@ -0,0 +1,102 @@
+disp("Example 4.15")
+disp("Asc=2-25dia bars","Ast=3-36dia bars","M20 Grade of concrete and Fe250steel","dd = 50mm","d=550mm","b=300mm","Given:")
+b=300
+d=550
+dd=50
+Ast= %pi*36*36*3/4
+Asc=2*%pi*25*25/4
+m=13.33 //(280/(3*sigmacbc))
+Es=2*10^5
+fck=20
+fy=415
+xumaxd=0.0035/(0.0055+(0.87*fy)/Es)
+disp(xumaxd,"xumax/d for Fe250=")
+xumax=xumaxd*d
+disp("mm",xumax,"xu,max")
+disp("Assuming fsc=fst=0.87*fy,, and considering force equilibrium")
+disp("Cus+Cuc = Tu")
+Cuc=0.362*fck*b
+Cus=(0.87*fy-0.447*fck)*Asc
+Tu=0.87*fy*Ast
+xu=(Tu-Cus)/Cuc
+disp("mm",xu,"xu=")
+disp("xu>xu,max, hence the section is over-reinforced")
+disp("Exact Solution considering strain compatibility")
+disp("Applying Eq. 4.81: xu = fst*Ast - (fsc-0.447*fck)*Asc/(0.362*fck*b)")
+disp("Therefore,xu=(3054*fst - 982*fsc+8779)/2172")
+
+disp("First Cycle")
+disp("1. xu lies within the two limits above; 263.5 mm < xu < 348.5mm")
+disp("2. xu = (xu,max+xu)/2")
+xu1=(xumax+xu)/2
+disp("mm",xu1,"xu=")
+disp("3.Esc = 00035*(1-dd/xu1)")
+Esc = 0.0035*(1-dd/xu1)
+disp(Esc,"Esc=")
+disp("4.Est = 0.0035*(d/xu1-1)")
+Est = 0.0035*(d/xu1-1)
+disp(Est,"Est=")
+disp("for Esc= 0.00380 fsc = 360.9 and for Esc = 0.00276 fsc=351.8")
+fst1=351.8
+fst2=360.9
+fsc=fst1+((fst2-fst1)*((Esc*10^5-276)/(380-276)))
+disp("MPa",fsc,"fsc=")
+fst=fst1+((fst2-fst1)*((Est*10^5-276)/(380-276)))
+disp("MPa",fst,"fst=")
+disp("Therefore, xu = ")
+xu2=(3054*fst - 982*fsc+8779)/2172
+disp("mm",xu2,"xu=")
+
+disp("Second Cycle")
+disp("Assume xu= ")
+xu3=(xu2+xu1)/2
+disp("mm",xu3,"xu=")
+Esc = 0.0035*(1-dd/xu3)
+disp(Esc,"Esc=")
+Est1=0.0035*(d/xu3-1)
+disp(Est1,"Est=")
+disp("for Esc= 0.00380 fsc = 360.9 and for Esc = 0.00276 fsc=351.8")
+fst1=351.8
+fst2=360.9
+fsc=fst1+((fst2-fst1)*((Esc*10^5-276)/(380-276)))
+disp("MPa",fsc,"fsc=")
+disp("For strain, 0.00276 fst = 351.8 and for strain 0.00241 fst=342.8  From table 3.2")
+fst4=351.8
+fst3=342.8
+fst11=(fst3+(fst4-fst3)*((Est1*10^5-241)/(276-241)))
+disp("MPa",fst11,"fst1=")
+xu4=(3054*fst11- 982*fsc+8779)/2172
+disp("mm",xu4,"xu=")
+
+disp("Third Cycle")
+disp("1.Assume xu=")
+xu5=(xu3+xu4)/2
+disp("mm",xu5,"xu=")
+Esc = 0.0035*(1-dd/xu5)
+disp(Esc,"Esc=")
+Est2=0.0035*(d/xu5-1)
+disp(Est2, "Est=")
+disp("for Esc= 0.00380 fsc = 360.9 and for Esc = 0.00276 fsc=351.8")
+fst1=351.8
+fst2=360.9
+fsc=fst1+((fst2-fst1)*((Esc*10^5-276)/(380-276)))
+disp("For strain, 0.00276 fst = 351.8 and for strain 0.00241 fst=342.8  From table 3.2")
+fst12=342.8
+disp("MPa",fst12,"fst2=")
+xu6=(3054*fst12- 982*fsc+8779)/2172
+disp("mm",xu6,"xu,final=")
+Cuc=0.362*fck*b
+Cus=(fsc-0.447*fck)*Asc
+MuR=(Cuc*xu6*(d-0.416*xu6)+Cus*(d-dd))/10^6
+disp("kNm",MuR,"MuR,final=")
+
+disp("Approximate Solution")
+disp("As an approximate and conservative estimate limiting xu to xu,max=263.5mm,")
+Esc=0.0035*(1-dd/xumax)
+disp(Esc,"Esc=")
+fsc=352.5
+disp("MPa",fsc,"fsc=")
+disp("This value is alternatively obtainable from Table 4.5 for dd/d=0.09 and Fe415")
+disp("Accordingly, limiting the ultimate moment of resistance MuR to the limiting moment Mu,lim")
+Mulim=(0.362*fck*b*xumax*(d-0.416*xumax)+(fsc-0.447*fck)*Asc*(d-dd))/10^6
+disp("kNm",Mulim,"Mu,lim=")
diff --git a/3761/CH4/EX4.16/Ex4_16.sce b/3761/CH4/EX4.16/Ex4_16.sce
new file mode 100644
index 000000000..f817fdc2e
--- /dev/null
+++ b/3761/CH4/EX4.16/Ex4_16.sce
@@ -0,0 +1,66 @@
+disp("Example 4.16")
+disp("Ast= 4-25dia bars","Asc= 2-25 dia bars","fck=20MPa","fy=415MPa","dd=45mm","d=655mm","b=300mm","Given:")
+disp("xu,max/d=0.479")
+Es=2*10^5
+dd=45
+d=655
+b=300
+fy=415
+fck=20
+Ast=%pi*25*25
+Asc=%pi*25*25*2/4
+xumaxd=0.0035/(0.0055+(0.87*fy/Es))
+xumax=xumaxd*d
+disp("mm",xumax,"xu,max=")
+disp("Assuming for a first approximation fsc=fst=0.87*fy")
+Cuc=0.362*fck*b
+disp("xu N",Cuc,"Cuc=")
+Cus=(0.87*fy-0.447*fck)*Asc
+disp("N",Cus,"Cus=")
+Tu=0.87*fy*Ast
+disp("N",Tu,"Tu=")
+disp("Considering force equilibrium:Cuc+Cus = Tu")
+xu=(Tu-Cus)/Cuc
+disp("mm",xu,"xu=")
+disp("xu<xu,max, therefore, the assumption fst=0.87*fy is justified")
+disp("Further Esc= 0.0035*(1-dd/xu)")
+Esc= 0.0035*(1-dd/xu)
+disp(Esc,"Esc=")
+disp("For Fe415, Ey=0.87*fy/Es +0.002")
+Ey=0.87*fy/Es +0.002
+disp(Ey,"Ey=")
+disp("As, Esc<Ey the assumption fsc=0.87*fy is not justified whereby the calculated value of Cus and henec of xy =167.3mm is alos not correct. the correct value has to be obtained iteratively using strain compatibility")
+disp("FIRST CYCLE")
+disp("Assuming Esc=0.00256")
+//according to the table 3.2, interpolating the values
+//for Esc=0.00241 fst=342.8
+//for Esc=0.00276 fst = 351.8
+fst1=342.8
+fst2=351.8
+fsc=fst1+((fst2-fst1)*((Esc*10^5-241)/(276-241)))
+disp("MPa",fsc,"fsc=")
+Cus=(fsc-0.447*fck)*Asc
+disp("N",Cus,"Cus=")
+xu=(Tu-Cus)/Cuc
+disp("mm",xu,"xu=")
+Esc1= 0.0035*(1-dd/xu)
+disp(Esc1,"Esc=")
+
+disp("SECOND CYCLE")
+disp("Assuming Esc=0.00259")
+//according to the table 3.2, interpolating the values
+//for Esc=0.00241 fst=342.8
+//for Esc=0.00276 fst = 351.8
+fst1=342.8
+fst2=351.8
+fsc1=fst1+((fst2-fst1)*((Esc1*10^5-241)/(276-241)))
+disp("MPa",fsc1,"fsc=")
+Cus1=(fsc1-0.447*fck)*Asc
+disp("N",Cus1,"Cus=")
+xu1=(Tu-Cus1)/Cuc
+disp("mm",xu1,"xu=")
+
+disp("Taking xu=173.4mm")
+MuR=(Cuc*xu1*(d-0.416*xu1)+Cus1*(d-dd))/10^6
+disp("kNm",MuR,"MuR=")
+
diff --git a/3761/CH4/EX4.17/Ex4_17.sce b/3761/CH4/EX4.17/Ex4_17.sce
new file mode 100644
index 000000000..8221f051a
--- /dev/null
+++ b/3761/CH4/EX4.17/Ex4_17.sce
@@ -0,0 +1,54 @@
+disp("Example 4.17")
+disp("fck=20MPa","fy=415MPa","d=125mm","Ast=10-mmdia bars at 200mm spacing")
+Ast=(%pi*10*10/4)*1000/200
+fck=20
+fy=415
+d=125
+disp("mm^2/m",Ast,"Ast=")
+disp("a. Analysis at working loads")
+disp("For M20 concrete,sigmacbc=7.0 MPa and m =13.33")
+disp("For Fe415 Steel,sigmast=230 MPa and kb=280/(280+3sigmast)")
+sigmacbc=7
+sigmast=230
+m=13.33 //for M20 concrete
+b=1000
+mAst=m*Ast
+kb=280/(280+(3*sigmast))
+disp(kb,"kb=")
+disp("The neutral axis depth kd is obtained by considering moments of areas in the transformed-cracked section, and considering b=1000mm")
+disp("b*(kd)^2/2 = m*Ast*(d-kd)")
+a=b/2
+b1=mAst
+c=-mAst*d
+D=b1*b1-4*a*c
+kd1=(-b1+sqrt(D))/(2*a)
+disp("mm",kd1,"kd=")
+kbd=kb*d
+disp("mm",kbd,"kbd=")
+disp("kd<kbd, therefore the section is under-reinforced")
+Mall=((sigmast*Ast*(d-kd1/3))/10^6)
+disp("kNm/m",Mall,"M,all = ")
+
+disp("Analaysis at ultimate limit state")
+disp("Fe 415 steel, xu,max/d = 0.479")
+//using formula 0.0035/(0.0055+0.87*fy)
+xumax=0.479*d
+disp("mm",xumax,"xu,max=")
+disp("Assuming xu</xu,max and considering Cu=Tu")
+xu=(0.87*fy*Ast)/(0.362*fck*b)
+disp("mm",xu,"xu=")
+disp("xu</xu,max, therefore ok")
+disp("Accordingly,")
+MuR=((0.362*fck*b*xu*(d-0.416*xu))/10^6)
+disp("kNm/m",MuR,"MuR=")
+
+disp("Alternatively")
+pt=(100*Ast)/(b*d)
+disp(pt,"pt=")
+
+//Eq. pt,lim = 41.61*(fck/fy)*(xu,max/d)
+ptlim = 41.61*(fck/fy)*(xumax/d)
+disp(ptlim,"pt,lim=")
+disp("pt<pt,lim therefore using Eq. MuR/bd^2 = 0.87*fy*pt*(1-(fy/fck)*pt)")
+MuR = (0.87*fy*(pt/100)*(1-(fy/fck)*(pt/100))*b*d*d)/10^6
+disp("kNm/m",MuR,"MuR=")
diff --git a/3761/CH4/EX4.2/Ex4_2.sce b/3761/CH4/EX4.2/Ex4_2.sce
new file mode 100644
index 000000000..386a1b5e7
--- /dev/null
+++ b/3761/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,40 @@
+disp("Example 4.2")
+disp("Given Moment value = 140kNm")
+disp("Grade of Steel = Fe415","Grade of Concrete = M20","D=600mm","d=550mm","b=300mm","Bars used = 4 - 25 dia")
+disp("Transformed Section Properties")
+disp("Modula ratio m = 13.33 for M20")
+disp("Transformed Steel Area = m*Ast")
+m=13.33
+d=550
+b=300
+D=600
+M=140
+Ast=%pi*25*25*4/4 // Value of Ast is varying a bit b'cos of value of pi considered
+mAst=m*Ast
+disp(mAst) // and hence value of mAst is varying a little
+disp("Equating Moments")
+disp("b*k*d*d/2 = mAst*(d-k*d)")
+b1=(2*mAst/b)
+c=(-mAst*2*d/b)
+a=1
+kd1=((-b1+sqrt((b1*b1)-(4*a*c)))/(2*a))
+kd2=((-b1-sqrt((b1*b1)-(4*a*c)))/(2*a))
+disp("mm",kd1,"therefore, kd=")
+k=kd1/d
+disp("mm",k,"k=")
+disp("Lever arm, jd = d(1-k/3)")
+j=(1-k/3)*d
+disp("mm",j,"jd=")
+
+//Calculating Stresses
+
+disp("Maximum Concrete Stresses")
+disp("Taking moments about the tension steel centroid,")
+disp("M=0.5*fc*b*kd*jd")
+fc=((140*10^6)/(0.5*300*234.6*471.8))
+disp("in MPa",fc,"fc=")
+disp("Tensile Stress in Steel")
+disp("Taking moments about the line of action of C")
+fst=((M*10^6)/(Ast*j))
+disp("MPa",fst,"fst=")
+
diff --git a/3761/CH4/EX4.3/Ex4_3.sce b/3761/CH4/EX4.3/Ex4_3.sce
new file mode 100644
index 000000000..10aca254c
--- /dev/null
+++ b/3761/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,59 @@
+disp("Example 4.3")
+b=300
+d=550
+D=600
+dia=25
+Ast=%pi*dia*dia*4/4
+sigmacbc=7
+sigmast=230
+m= 280/(3*sigmacbc)
+
+disp("Grade of Steel = Fe415","Grade of Concrete = M20","D=600mm","d=550mm","b=300mm","Bars used = 4 - 25 dia")
+disp("Transformed Section Properties")
+disp("Modula ratio m = 13.33 for M20")
+disp("mm^2",Ast,"Area of Steel=")
+disp(m,"m=")
+
+mAst=m*Ast
+//disp(mAst)
+//disp("Equating Moments")
+//disp("b*k*d*d/2 = mAst*(d-k*d)")
+b1=(2*mAst/b)
+c=(-mAst*2*d/b)
+a=1
+kd1=((-b1+sqrt((b1*b1)-(4*a*c)))/(2*a))
+kd2=((-b1-sqrt((b1*b1)-(4*a*c)))/(2*a))
+//disp("mm",kd1,"therefore, kd=")
+k=kd1/d
+disp("mm",k,"Transformed Section Properties, k=")
+disp("Neutral Axis depth factor kb")
+disp("kb=280/(280+(3*sigmast)")
+kb=280/(280+(3*sigmast))
+disp(kb,"kb=")
+disp("Calculating for stresses")
+disp("As k>kb, the section is over reinforced (WSM method)")
+disp("therefore, concrete stress controls, fc=sigmacbc= 7MPa")
+disp("Applying Tension force=Compressive force")
+disp("fst*Ast = 0.5*sigmacbc*b*kd")
+fst= 0.5*sigmacbc*b*kd1/Ast
+disp("MPa",fst,"fst=")
+disp("Alternatively, considering the linear stress distribution")
+fc=7
+fst1=(m*fc*(1-k))/k
+disp("MPa",fst1,"fst=")
+
+disp("Calculating Allowable Bending Moment")
+disp("Taking moments of forces about the tension steel considered")
+disp("Mall= (0.5*sigmacbc*b*kd)*(d-(kd/3)")
+Mall= (0.5*sigmacbc*b*kd1)*(d-(kd1/3))
+Mall1=Mall/10^6
+disp("kN-m",Mall1,"Mall=")
+disp("Alternatively, using the analysis aids given in TABLE A.1(a) ")
+pt=1.190
+disp("Muall = 1.28*bd^2")
+Muall=1.28*b*d*d/10^6
+disp("kNm",Muall,"Muall=")
+
+
+
+
diff --git a/3761/CH4/EX4.4/Ex4_4.sce b/3761/CH4/EX4.4/Ex4_4.sce
new file mode 100644
index 000000000..025fbc195
--- /dev/null
+++ b/3761/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,62 @@
+L=6
+Df=100
+bw=250
+b=1000
+D=600
+d=520
+M=200
+disp("Example 4.4")
+disp("Service Load Moment = 200kNm","Ast=6-28mmdia bar","Effective depth d=520 mm","Depth D=600mm","b=1000mm","bw=250mm","Depth of Flange Df=100mm","Span of Beam L=6m","Given Data:")
+disp("Grade of Steel =Fe250","Grade of Concrete=M20")
+disp("Verifying for the effective flange width b")
+disp("Refering IS456:2000, Clause 23.1.2 c or Eq 4.30(b) from TB")
+disp("For T-beams, bf = (lo/((lo/b)+4)+bw")
+disp("bw=250mm","lo=6000mm")
+lo=6000
+bf=((lo)/((lo/b)+4)+bw)
+disp("mm",bf,"bf=")
+
+if(bf<b)
+       disp(bf,"Since, bf<b,b=")
+
+sigmacbc=7
+m=280/(3*sigmacbc)
+disp(m,"m=")
+dia=28
+Ast= %pi*dia*dia*6/4
+mAst=m*Ast
+disp("Assuming kd</Df, and equating moments of compression and transformed tension areas about the neutral axis,")
+disp("bf*(kd)^2/2 = m*Ast*(d-kd)")
+b1=(2*mAst/bf)
+c=(-mAst*2*d/bf)
+a=1
+kd1=((-b1+sqrt((b1*b1)-(4*a*c)))/(2*a))
+kd2=((-b1-sqrt((b1*b1)-(4*a*c)))/(2*a))
+disp("mm",kd1,"therefore, kd=")
+disp("kd1>Df, the assumption kd</Df is incorrect")
+disp(" For kd>Df, neutral axis located in the web")
+disp("Using Equation 4.31 of TB")
+disp("(bf-bw)*Df*(kd-Df/2)+ bw*(kd)^2/2 = mAst*(d-kd)")
+a=bw/2
+B=(bf*Df-bw*Df+mAst)
+c=(bw*Df*Df/2 - bf*Df*Df/2-mAst*d)
+
+Dis=(B*B)-(4*a*c)
+kd1=((-B+sqrt(Dis))/(2*a))
+disp("mm",kd1,"kd=")
+disp("Relating the compressive stress fc1 at the flange bottom to fc,")
+disp("fc1=fc*((kd-Df)/kd)")
+fact=((kd1-Df)/kd1)
+disp("*fc",fact,"fc1=")
+disp("Compressive Force C= 0.5*fc*bf*(kd)-0.5*fc1*(bf-bw)*(kd-Df)")
+disp("Taking moments of forces about the tension steel centriod, using equation 4.34")
+fc=((M*10^6)/((0.5*bf*(kd1)*(d-kd1/3)-(0.5*fact*(bf-bw)*(kd1-Df))*(d-Df-((kd1-Df)/3)))))
+disp("MPa",fc,"Therefore, on solving we get, fc=")
+disp("Applying C= T")
+fst= (0.5*fc*bf*(kd1)-0.5*fact*fc*(bf-bw)*(kd1-Df))/Ast
+disp("MPa",fst,"Therefore, fst = ")
+disp("From the stress distribution diagram: fst = m*fc*((d-kd)/kd)")
+fst1=m*fc*((d-kd1)/kd1)
+disp("MPa",fst1, "As before")
+
+
diff --git a/3761/CH4/EX4.5/Ex4_5.sce b/3761/CH4/EX4.5/Ex4_5.sce
new file mode 100644
index 000000000..27366a9e7
--- /dev/null
+++ b/3761/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,67 @@
+disp("Example 4.5")
+disp("Calculating allowable moment capacity for beam of section as in 4.4")
+L=6
+Df=100
+bw=250
+b=1000
+D=600
+d=520
+M=200
+disp("Example 4.4")
+disp("Service Load Moment = 200kNm","Ast=6-28mmdia bar","Effective depth d=520 mm","Depth D=600mm","b=1000mm","bw=250mm","Depth of Flange Df=100mm","Span of Beam L=6m","Given Data:")
+disp("Grade of Steel =Fe250","Grade of Concrete=M20")
+disp("Verifying for the effective flange width b")
+disp("Refering IS456:2000, Clause 23.1.2 c or Eq 4.30(b) from TB")
+disp("For T-beams, bf = (lo/((lo/b)+4)+bw")
+disp("bw=250mm","lo=6000mm")
+lo=6000
+bf=((lo)/((lo/b)+4)+bw)
+disp("mm",bf,"bf=")
+
+if(bf<b)
+       disp(bf,"Since, bf<b,b=")
+
+sigmacbc=7
+m=280/(3*sigmacbc)
+disp(m,"m=")
+dia=28
+Ast= %pi*dia*dia*6/4
+mAst=m*Ast
+//disp("Assuming kd</Df, and equating moments of compression and transformed tension areas about the neutral axis,")
+//disp("bf*(kd)^2/2 = m*Ast*(d-kd)")
+b1=(2*mAst/bf)
+c=(-mAst*2*d/bf)
+a=1
+kd1=((-b1+sqrt((b1*b1)-(4*a*c)))/(2*a))
+kd2=((-b1-sqrt((b1*b1)-(4*a*c)))/(2*a))
+//disp("mm",kd1,"therefore, kd=")
+//disp("kd1>Df, the assumption kd</Df is incorrect")
+disp(" For kd>Df, neutral axis located in the web")
+disp("Using Equation 4.31 of TB")
+disp("(bf-bw)*Df*(kd-Df/2)+ bw*(kd)^2/2 = mAst*(d-kd)")
+a=bw/2
+B=(bf*Df-bw*Df+mAst)
+c=(bw*Df*Df/2 - bf*Df*Df/2-mAst*d)
+
+Dis=(B*B)-(4*a*c)
+kd1=((-B+sqrt(Dis))/(2*a))
+disp("mm",kd1,"kd=")
+disp("The neutral axis depth factor, k")
+k=kd1/d
+disp(k,"k=")
+disp("For a balanced section as per Eq 4.23, kb")
+sigmast=130
+kb=280/(280+3*sigmast)
+disp(kb,"kb=")
+disp("As k<kb the section is under-reinforced section(WSM)")
+disp("fst=sgimgast=130MPa (For Fe 250 Steel, dia>20mm")
+fst=sigmast
+fc=(kd1/(d-kd1))*(fst/m)
+fc1=0.526*fc //As derived in previous example 4.4
+disp("MPa",fc,"fc=")
+disp("MPa",fc1,"fc1=")
+disp("Substituting in Eq 4.34")
+fact=0.526
+M=(fc*((0.5*bf*(kd1)*(d-kd1/3)-(0.5*fact*(bf-bw)*(kd1-Df))*(d-Df-((kd1-Df)/3)))))/10^6
+disp("Therefore, Moment carrying capacity Mall=")
+disp("kNm", M , "Mall=")
diff --git a/3761/CH4/EX4.6/Ex4_6.sce b/3761/CH4/EX4.6/Ex4_6.sce
new file mode 100644
index 000000000..b2a351109
--- /dev/null
+++ b/3761/CH4/EX4.6/Ex4_6.sce
@@ -0,0 +1,54 @@
+disp("Example 4.6")
+disp("Asc=2-25dia bars","Ast=3-36dia bars","Service Load Moment = 175kNm","M20 Grade of concrete and Fe250steel","sigmast=130MPa","Sigmacbc=7 MPa","dd = 50mm","d=550mm","b=300mm","Given:")
+b=300
+d=550
+dd=50
+sigmacbc=7
+sigmast=130
+M=175
+Ast= %pi*36*36*3/4
+Asc=2*%pi*25*25/4
+disp("Transformed Section Properties")
+m=13.33 //(280/(3*sigmacbc))
+mAst=m*Ast
+CSA=(1.5*m-1)*Asc
+disp("mm^2",mAst,"Transformed tension steel area=")
+disp("mm^2",CSA,"Transformed Compression Steel Area=")
+
+disp("Neutral Axis Depth")
+disp("Considering moments of areas about the neutral axis,")
+disp("b*kd^2/2 + CSA*(kd-dd) = mAst*(d-kd)")
+a=b/2
+b1=(CSA+mAst)
+c=-CSA*dd-mAst*d
+D=b1*b1-4*a*c
+kd1=(-b1+sqrt(D))/(2*a)
+disp("mm",kd1,"Solving kd=")
+disp("Stresses due to M=175kNm")
+disp("Considering the linear stress distribution")
+disp("fcsc= fc*(kd-dd/kd)")
+Ccfact=0.5*b*kd1
+Csfact=CSA*((kd1-dd)/kd1)
+fc=(M*10^6)/(Ccfact*(d-kd1/3)+Csfact*(d-dd))
+disp("MPa",fc,"fc=")
+disp("Compressive Stress in Steel,fsc")
+fcsc= fc*((kd1-dd)/kd1)
+fsc=1.5*m*fcsc
+disp("MPa",fsc,"fsc=")
+disp("Tensile Stress in Steel,fst")
+fst=m*fc*((d-kd1)/kd1)
+disp("MPa",fst,"fst=")
+fst=(fc*(Ccfact+Csfact)/Ast)
+disp("MPa",fst,"Alternatively, Cc+Cs = T --> fst=")
+disp("Allowable Bending Moment")
+disp("For a balanced (WSM) section, kb= 280/(280+3*sigmast)")
+kb=280/(280+(3*sigmast))
+disp(kb,"kb=")
+disp("For the given section, k =kd/d")
+k=kd1/d
+disp(k,"k=")
+disp("Here, k >kb")
+disp("Hence the section is over reinforced(WSM)")
+disp("whereby fc =sigmacbc =7 MPa")
+Mall=fc*(Ccfact*(d-kd1/3)+Csfact*(d-dd))/10^6
+disp(Mall)
diff --git a/3761/CH4/EX4.7/Ex4_7.sce b/3761/CH4/EX4.7/Ex4_7.sce
new file mode 100644
index 000000000..b3c723353
--- /dev/null
+++ b/3761/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,50 @@
+disp("Example 4.7")
+disp("M20 Grade of concrete and Fe250steel","sigmast=130MPa","Sigmacbc=7 MPa","dd = 50mm","d=550mm","b=300mm","Given:")
+b=300
+d=550
+dd=50 //Mentioning the top cover d' as dd throughout the example
+sigmacbc=7
+sigmast=130
+disp("Transformed Section Properties")
+m=13.33 //(280/(3*sigmacbc))
+disp("Neutral Axis Depth, here in this case k=kb, corresponding to balanced section")
+disp("For a balanced (WSM) section, kb= 280/(280+3*sigmast)")
+kb=280/(280+(3*sigmast))
+disp(kb,"kb=")
+kbd=kb*d
+disp("mm",kbd,"kb d=")
+disp("Considering moments of areas about the neutral axis,")
+disp("b*kb d^2/2 + CSA*(kb d-dd) = mAst*(d-kb d)")
+disp("On replacing values stated above we get equation as : Ast = (0.8Asc + 1856)mm^2")
+disp("Asc is to be determined using equation 4.39 as stated")
+disp("M= Cc(d-kb d/3)+Cs(d-dd)")
+disp("Cc=0.5*fc*b*(kb.d)")
+disp("Cs=(1.5*m-1)*Asc*fc*((kd-dd)/kd)")
+M=175*10^6
+fc=7
+Cc=0.5*fc*b*(kbd)
+Cs1=(1.5*m-1)*fc*((kbd-dd)/kbd)
+Asc=(M-(Cc*(d-kbd/3)))/(Cs1*(d-dd))
+disp("mm^2",Asc,"Therefore, Asc=")
+Ast=(0.8*Asc+1856)
+disp("mm^2", Ast,"Therefore, Ast=")
+
+disp("Alternate Solution to Example 4.7")
+disp("Calculating Mwb=0.5*kb*(1-kb/3)*sigmacbc*b*d*d")
+Mwb=0.5*kb*(1-kb/3)*sigmacbc*b*d*d/10^6
+disp("kNm",Mwb,"Mwb=")
+disp("Calculating Ast1=Mwb/(sigmast*d*(1-kb/3)")
+Ast1=(Mwb*10^6)/(sigmast*d*(1-kb/3))
+disp("mm^2",Ast1,"Ast1=")
+disp("Calculating Ast2, Ast2= (M-Mwb)/(sigmast*(d-dd))")
+Ast2= (M-Mwb*10^6)/(sigmast*(d-dd))
+disp("mm^2",Ast2,"Ast2=")
+disp("Therefore, Ast= Ast1+Ast2")
+Astf=Ast1+Ast2
+disp("mm^2",Astf,"Therefore, final Ast=")
+disp("Calculating fcsc=sigmacbc*(1-dd/kbd)")
+fcsc=sigmacbc*(1-dd/kbd)
+disp(fcsc)
+disp("Calculating Asc=(M-Mwb)/(1.5*m-1)*fcsc*(d-dd)")
+Asc=(M-Mwb*10^6)/((1.5*m-1)*fcsc*(d-dd))
+disp("mm^2",Asc,"Asc=")
diff --git a/3761/CH4/EX4.8/Ex4_8.sce b/3761/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..4461d96f5
--- /dev/null
+++ b/3761/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,61 @@
+disp("Example 4.8")
+disp("Asc=3-28dia bars","Fe250","dd=50mm","Ast=6-28mmdia bar","Effective depth d=520 mm","Depth D=600mm","b=1000mm","bw=250mm","Depth of Flange Df=100mm","Span of Beam L=6m","Given Data:")
+Asc=3*%pi*28*28/4
+L=6
+Df=100
+bw=250
+b=1000
+D=600
+d=520
+dd=50
+disp("Grade of Steel =Fe250","Grade of Concrete=M20")
+disp("Verifying for the effective flange width b")
+disp("Refering IS456:2000, Clause 23.1.2 c or Eq 4.30(b) from TB")
+disp("For T-beams, bf = (lo/((lo/b)+4)+bw")
+disp("bw=250mm","lo=6000mm")
+lo=6000
+bf=((lo)/((lo/b)+4)+bw)
+disp("mm",bf,"bf=")
+
+if(bf<b)
+       disp(bf,"Since, bf<b,b=")
+
+sigmast=130
+sigmacbc=7
+m=280/(3*sigmacbc)
+disp(m,"m=")
+dia=28
+Ast= %pi*dia*dia*6/4
+mAst=m*Ast
+disp("mm^2",mAst,"Transformed Steel Area=mAst")
+CSA=(1.5*m-1)*Asc
+disp(CSA)
+disp("Assuming first kd<Df and kd>dd, and solving Eq4.35 with b=bf")
+a=bf/2
+b1=CSA+mAst
+c=(-CSA*dd)-(mAst*d)
+kd1=((-b1+sqrt((b1*b1)-(4*a*c)))/(2*a))
+disp("mm",kd1,"kd=")
+disp("As kd>Df, the assumption kd</Df is incorrect")
+disp("Now solving Eq.4.40 for kd>/Df")
+a1=bw/2
+b12=CSA+mAst+Df*(bf-bw)
+c1=(-CSA*dd)-(mAst*d)-(Df*Df*(bf-bw)/2)
+D1=((b12*b12)-(4*a1*c1))
+kd12=(-b12+sqrt(D1))/(2*a1)
+disp("mm",kd12,"kd=")
+k=kd12/d
+disp("mm",k,"Therefore, k =")
+kb=(280)/(280+(3*sigmast))
+disp(kb,"For a balanced WSM section with sigmast=130MPa, kb=")
+disp("As k=0.3496<kb, the section is under reinforced whereby fst=sigmast=130MPa")
+disp("Considering the linear stress distribution fc=")
+fc= (sigmast/m)*(kd12/(d-kd12))
+disp(fc)
+z1=bf*kd12*(d-(kd12/3))
+z2=(bf-bw)*(((kd12-Df)/kd12)^2)*(d-Df-((kd12-Df)/3))
+z3=CSA*fc*((kd12-dd)/kd12)*(d-dd)
+Mall=((0.5*fc*(z1-z2))+z3)/10^6
+disp(Mall)
+
+
diff --git a/3761/CH4/EX4.9/Ex4_9.sce b/3761/CH4/EX4.9/Ex4_9.sce
new file mode 100644
index 000000000..413f806e7
--- /dev/null
+++ b/3761/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,80 @@
+disp("Example 4.9")
+disp("Grade of Steel,fy = Fe415","Grade of Concrete,fck = M20","D=600mm","d=550mm","b=300mm","Bars used = 4 - 25 dia")
+b=300
+d=550
+D=600
+fck=20
+Ast=%pi*4*25*25/4
+disp("mm^2",Ast,"Ast=")
+disp("For Fe415 Steel,")
+Es=2*10^5
+fy=415
+Est=0.87*fy/Es
+xumaxd=(0.0035/(0.0055+Est))
+disp(xumaxd,"xumax/d")
+xumax=xumaxd*d
+disp("mm",xumax,"xu,max=")
+disp("Assuming, xu</xu,max and applying the force equilibrium condition Cu=Tu")
+xu= (0.87*fy*Ast)/(0.362*fck*b)
+disp("mm",xu,"xu")
+disp("xu>xu,max, 326.3mm>263.5mm")
+disp("As xu>xu,max steel would not have yielded accordingly the strain compatibility method is adopted to obtain the correct value of xu")
+disp("FIRST CYCLE")
+disp("1. Assume xu = (xu+xu,max)/2 ")
+xu1=(xu+xumax)/2
+disp("mm",xu1,"xu,1=")
+disp("2. Strain Compatibility = Est = 0.0035*(d/xu1-1)")
+//Est=strainst, ephselon st
+Est =0.0035*(d/xu1-1)
+disp(Est,"Est=")
+disp(Est,"Interpoating for value of fst, corresponding to strain ,Fe415 and Est = ")
+disp("For strain, 0.00276 fst = 351.8 and for strain >/0.00380 fst=360.9  From table 3.2")
+fst1=351.8
+fst2=360.9
+disp("fst= ")
+fst=fst1+((fst2-fst1)*((Est*10^5-276)/(380-276)))
+disp("MPa",fst,"fst=")
+disp("Cu=Tu")
+xu2=fst*(Ast/(0.362*fck*b))
+disp("mm",xu2,"xu,2=")
+
+disp("SECOND CYCLE")
+disp("Assume xu= ")
+xu3=(xu2+xu1)/2
+disp("mm",xu3,"xu,3=")
+Est1=0.0035*(d/xu3-1)
+disp(Est1,"Est=")
+disp(Est,"Interpoating for value of fst, corresponding to strain ,Fe415 and Est = ")
+disp("For strain, 0.00276 fst = 351.8 and for strain 0.00241 fst=342.8  From table 3.2")
+fst4=351.8
+fst3=342.8
+fst11=(fst3+(fst4-fst3)*((Est1*10^5-241)/(276-241)))
+disp("MPa",fst11,"fst1=")
+
+disp("Cu=Tu")
+fact=Ast/(0.362*fck*b)
+//disp(fact)
+xu4=fst11*(fact)
+disp("mm",xu4,"xu,4=")
+
+
+disp("THIRD CYCLE")
+disp("1.Assume xu=")
+xu5=(xu4+xu3)/2
+disp("mm",xu5,"xu,5=")
+Est2=0.0035*(d/xu5-1)
+disp(Est2, "Est=")
+disp(Est,"Interpoating for value of fst, corresponding to strain ,Fe415 and Est = ")
+disp("For strain, 0.00276 fst = 351.8 and for strain 0.00241 fst=342.8  From table 3.2")
+fst4=351.8
+fst3=342.8
+fst12=(fst3+(fst4-fst3)*((Est2*10^5-241)/(276-241)))
+disp("MPa",fst12,"fst2=")
+
+disp("Cu=Tu")
+fact=Ast/(0.362*fck*b)
+//disp(fact)
+xu6=fst12*(fact)
+disp("mm",xu6,"xu,6=")
+disp("Therefore, the final value of xu may be takaen as xu=315mm")
+
-- 
cgit