initial commit / add all books

17 files changed, 982 insertions, 0 deletions

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")

+