initial commit / add all books

12 files changed, 903 insertions, 0 deletions

diff --git a/2087/CH12/EX12.1/example12_1.sce b/2087/CH12/EX12.1/example12_1.sce
new file mode 100755
index 000000000..0a38fde25
--- /dev/null
+++ b/2087/CH12/EX12.1/example12_1.sce
@@ -0,0 +1,48 @@
+

+

+//example 12.1

+//calculate average hydraulic gradient

+//uplift presuures and thickness of floor at 6m, 12m and 18m from u/s

+clc;funcprot(0);

+//given

+rho=2.24;             //relative density of material

+gamma_w=9.81;         //unit weigth of water

+L=22;                 //total length

+lc=(2*6)+L+(2*8);     //length of creep

+hg=4/lc;              //hydraulic gradient

+mprintf("avearge hydraulic gradient=%f.",hg);

+//at 6 m from u/s

+x=6;

+lg=(6*2)+x;

+h1=4*(1-lg/50);       //unbalanced head

+up=gamma_w*h1;

+t=4*h1/(3*(rho-1));

+up=round(up*100)/100;

+t=round(t*100)/100;

+mprintf("\n\nuplift at 6 m from u/s=%f kN/square metre.",up);

+mprintf("\nthickness at 6 m from u/s=%f m.",t);

+

+//at 12 m from u/s

+x=12;

+lg=(6*2)+x;

+h1=4*(1-lg/50);       //unbalanced head

+up=gamma_w*h1;

+t=4*h1/(3*(rho-1));

+up=round(up*100)/100;

+t=round(t*100)/100;

+mprintf("\n\nuplift at 12 m from u/s=%f kN/square metre.",up);

+mprintf("\nthickness at 12 m from u/s=%f m.",t);

+

+//at 18m from u/s

+x=18;

+lg=(6*2)+x;

+h1=4*(1-lg/50);       //unbalanced head

+up=gamma_w*h1;

+t=4*h1/(3*(rho-1));

+up=round(up*10)/10;

+t=round(t*100)/100;

+mprintf("\n\nuplift at 18 m from u/s=%f kN/square metre.",up);

+mprintf("\nthickness at 18 m from u/s=%f m.",t);

+

+

+

diff --git a/2087/CH12/EX12.10/example12_10.sce b/2087/CH12/EX12.10/example12_10.sce
new file mode 100755
index 000000000..6c01786f2
--- /dev/null
+++ b/2087/CH12/EX12.10/example12_10.sce
@@ -0,0 +1,20 @@
+

+

+//example 12.10

+//calculate critical exit gradient and factor of safety of system

+clc;funcprot(0);

+//given

+b=60;       //length of floor

+H=6;        //static head of weir

+d=6;        //downstream depth of pile

+n=0.3;      //porousity of soil particles

+G=2.7;      //relative density of soil particles

+

+alpha=b/d;

+lambda=(1+(1+alpha^2)^0.5)/2;

+Ge=H/(d*%pi*(lambda)^0.5);

+e=n/(1-n);

+chg=(G-1)/(1+e);

+f=chg/Ge;

+f=round(f*100)/100;

+mprintf("critical exit gradient=%f.\nfactor of safety of system=%f.",chg,f);

diff --git a/2087/CH12/EX12.11/example12_11.sce b/2087/CH12/EX12.11/example12_11.sce
new file mode 100755
index 000000000..f5c75adff
--- /dev/null
+++ b/2087/CH12/EX12.11/example12_11.sce
@@ -0,0 +1,161 @@
+

+

+//example 12.11

+//design a vertical drop weir on Bligh's theory

+//test floor by Khosla's theory

+clc;funcprot(0);

+//given

+Q=2800;                    //maximum flood discharge

+hfl=285;                   //H.F.L before construction

+hw=278;                    //minimum water level

+fsl=284;                   //F.S.L of canal

+c=12;                      //coefficient of creep

+flux=1;                    //allowable afflux

+Ge=1/6;                    //permissible exit gradient

+rho=2.24;                  //specific gravity of concrete

+

+//Hydraulic calculation

+L=4.75*Q^0.5;

+q=Q/L;

+q=round(q*10)/10;

+mprintf("Hydraulic calculation:");

+mprintf("\ndischarge per unit width of river=%f cumecs.",q);

+f=1;

+R=1.35*(q^2/f)^(1/3);

+R=round(R*100)/100;

+mprintf("\nregime scour depth=%f m.",R);

+V=q/R;                   //regime velocity

+vh=V^2/(2*9.81);        //velocity head

+l_down=hfl+vh;

+l_up=l_down+flux;

+hfl_up=l_up-vh;

+hfl_down=hfl-0.5;

+hfl_down=round(hfl_down*100)/100;

+mprintf("\nactual d/s H.F.L allowing 0.5 m for retrogation=%f m.",hfl_down);

+K=(q/1.7)^(2/3);

+cl=l_up-K;               //crest level

+cl=round(cl*100)/100;

+mprintf("\ncrest level=%f m.",cl);

+pl=fsl+0.5;              //pond level

+s=hfl_down-cl;                //heigth of shutter

+mprintf("\nheigth of shutter=%f m.",s);

+rl_up_pile=hfl_up-1.5*R;    //R.L of bottom u/s pile

+d_up_cut=hw-276;           //depth of upstream cut-off

+mprintf("\ndepth of upstream cut-off=%f m.",d_up_cut);

+mprintf("\n provide concrete cut off 2 m depth.");

+rl_bot_ds=hfl_down-2*R;

+Hs=hfl_down-hw;            //seepage head

+Hc=cl-hw;                  //heigth of crest

+mprintf("\nR.L of gates crest=%f m.",Hs);

+mprintf("\nHeigth of crest=%f m.",Hc);

+

+//design of weir wall

+d=hfl_up-cl;

+a=d/(rho)^0.5;

+a=3*d/(2*rho);            //from sliding consideration

+a=s+1;                    //from practical consieration

+a=a+1;

+mprintf("\n\ndesign of weir wall:")

+mprintf("\nprovide top width of %i m.",a);

+Mo=9.81*Hs^3/6;                //overtirning moment

+//equating the moment of resistance to overturning moment and putting the values we get

+y=poly([-1.084,0.020,0.039],'x','c');

+b=roots(y);

+//we get b= - 5.5347261 and 5.0219056

+//taking

+b=5;

+//when weir is submerged

+C=0.58;

+d=(q^2/((2*C/3)^2*2*9.81))^(1/3);

+Mo=9.81*d*Hc^2/2;

+//from equation of moment of resistence we get

+y=poly([-77.55,3,1],'x','c');

+b=roots(y);

+//we get b= - 10.433085 and 7.4330846

+//taking

+b=8;

+mprintf("\nbottom width=%i m.",b);

+

+//design of impervious and pervious aprons

+C=12;

+L=C*Hs;

+mprintf("\n\ndesign of impervious and pervious aprons:");

+mprintf("\ntotal creep length=%i m.",L);

+l1=2.21*C*(Hs/13)^0.5;

+l1_=l1+1;

+mprintf("\nlength of downstream impervious apron=%i m.",l1_);

+d1=hw-276;

+d2=hw-271;

+l2=L-l1-(b+2*d1+2*d2);

+mprintf("\nlength of upstream impervious apron=%i m.",l2);

+l3=18*C*(Hs*q/975)^0.5;

+mprintf("\ntotal length of d/s apron=%i m.",l3);                //calculation is wrong in book

+l=l3-l1;

+le=l/2;

+le=round(le*100)/100;

+mprintf('\nprovide filter of length %f m. and launching apron of length %f m.',le,le);

+t=d2*10^0.5/le;

+mprintf("\nthickness of launching apron in horizontal position=%f m.",t);

+mprintf("\nprovide launching apron of thickness 1.5 m.");

+T=2*d1;

+V=d1*10^0.5;

+ta=V/T;

+ta=round(ta*10)/10;

+mprintf("\nthickness of apron in horizontal position=%f m.",ta);

+Hr=Hs-Hs*(4+33+8)/L;

+t=4*Hr/(3*(rho-1));

+t=round(t*10)/10;

+mprintf('\nprovide thickness of %f m from d/s of weir wall to point 6 m from it.',t);

+Hr=Hs-Hs*(4+33+8+6)/L;

+t=4*Hr/(3*(rho-1));

+t=round(t*10)/10;

+mprintf("\nprovide thickness of %f m from 6 m to 12 m from d/s end of weir wall.",t);

+Hr=Hs-Hs*(4+33+8+12)/L;

+t=4*Hr/(3*(rho-1));

+t=round(t*10)/10;

+mprintf("\nprovide thickness of %f m for rest of length of weir floor.",t);

+

+//check by khosla's theory

+b=33+8+19;            //total horizontal length of impervious floor

+d=7;                  //depth of downstream pile

+alpha=b/d;

+n=0.14;                //n=1/%pi*(lambda)^0.5;

+Ge=Hs*n/d;

+mprintf("\n\ncheck by Khosla theory:");

+mprintf("\nexit gradient=%f. < 1/6\n hence safe",Ge);

+alpha_=d/b;

+fic1=0.83;fid1=0.88;

+corec_c1=(fid1-fic1)*100/2;

+bdash=b;

+d=2;D=7;

+C1=19*(D/bdash)^0.5*(d+D)/b;

+fic1=fic1*100+corec_c1+C1;

+Pc=Hs*fic1/100;                        //pressure head at C

+alpha_=d/b;

+fie2=0.31;fid2=0.21;

+corec_e1=(fie2-fid2)*1.7*100/7;

+bdash=b;

+d=7;D=2;

+C1=19*(D/bdash)^0.5*(d+D)/b;

+fie2=fie2*100-corec_e1-C1;             //in book 3.53 value is wrong

+Pe=Hs*fie2/100;                         //pressue head at E

+//assuming linear variation of pressure for intermediate points

+Pa=Pc-(Pc-Pe)*(33+8)/b;

+t=Pa/1.24;

+Pa=round(Pa*100)/100;

+t=round(t*100)/100;

+mprintf("\npressure at d/s of weir wall=%f m.",Pa);

+mprintf("\nthickness at d/s of weir wall=%f m. < thickness by Bligh theory;\nhence safe.",t);

+Pb=Pc-(Pc-Pe)*(33+8+6)/b;

+t=Pb/1.24;

+Pa=round(Pa*100)/100;

+t=round(t*100)/100;

+mprintf("\npressure at 6 m from d/s of weir wall=%f m.",Pb);

+mprintf("\nthickness at 6m from d/s of weir wall=%f m. < thickness by Bligh theory;\nhence safe.",t);

+Pc=Pc-(Pc-Pe)*(33+8+12)/b;

+t=Pc/1.24;

+Pa=round(Pa*100)/100;

+t=round(t*100)/100;

+mprintf("\npressure at 12 m from d/s of weir wall=%f m.",Pc);

+mprintf("\nthickness at 12m from d/s of weir wall=%f m. > thickness by Bligh theory;\nhence unsafe.",t);

+mprintf("\nhence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.");

diff --git a/2087/CH12/EX12.12/example12_12.sce b/2087/CH12/EX12.12/example12_12.sce
new file mode 100755
index 000000000..f42021a1f
--- /dev/null
+++ b/2087/CH12/EX12.12/example12_12.sce
@@ -0,0 +1,51 @@
+

+

+//example 12.12

+//calculate

+//number of gates required for the barrage

+//head regulator if each gate has 10 m clear span(neglect end contractions and approach velocity)

+//length and R.L of basin floor if silting basin is provided downstream of barrage

+clc;funcprot(0);

+//given

+Lmax=212;           //maximum reservior level

+Lp=211;             //pond level

+hfl=210;            //downstream high flood level in the river

+Qmax=3500;          //maximum design flood discharge

+Lcrest=207;         //crest level of the barrage

+Lcrest_r=208;       //crest level of head regulator

+Cd=2.1;             //coefficient of discharge for barrage

+Cd_r=1.5;           //coefficient of discharge for head regulator

+rbl=205;            //river bed level

+Q=500;              //design discharge of main canal

+

+//design of water way for barrage during flood

+H=Lmax-Lcrest;

+L=Qmax/(Cd*H^1.5);

+//which gives L=149.07.

+//provide 15 bays of 10m clear span

+mprintf("nunmber of gates for the barrage=15.");

+

+//design of waterway for canal head regulator

+H=Lp-Lcrest_r;

+L1=Q/(Cd_r*H^1.5);

+//which gives L=64.2

+//hence provide 7 bays of 10 m each

+mprintf("\n\nnunmber of gates for the head regulator=7.");

+

+//design of stilling basin

+Hl=Lmax-hfl;

+q=Qmax/L;

+yc=(q^2/9.81)^(1/3);

+Z=Hl/yc;

+//since Z<1

+Y=1+0.93556*Z^0.368;

+y2=Y*yc;

+Lc=5*y2;

+Lc=round(Lc*10)/10;

+mprintf("\n\nLength of cistern=%f m.",Lc);

+Ef2=yc*(Y+1/(2*Y^2));

+j=hfl-Ef2;

+j=round(j*10)/10;

+mprintf("\nR.L of cistern=%f m.",j);

+

+

diff --git a/2087/CH12/EX12.2/example12_2.sce b/2087/CH12/EX12.2/example12_2.sce
new file mode 100755
index 000000000..b2956c85d
--- /dev/null
+++ b/2087/CH12/EX12.2/example12_2.sce
@@ -0,0 +1,105 @@
+

+

+//example 12.2

+//calculate uplift pressure and exit gradient

+//check whether section is safe against overturning and piping

+clc;funcprot(0);

+//given

+b=54;            //width of section

+D1D2=16;         //distance between points D1 and D2

+D2D3=37;         //distance between points D2 and D3

+

+//first pipe line

+//taking data from figure

+d=105-97;

+b1=0.5;

+alpha=b/d;

+//from the curves we get

+fic1=0.665;

+fid1=0.76;

+fie1=1;

+t=105-104;                      //floor thickness

+corec=(fid1-fic1)*100*t/d;          //correction for floor thickness

+//for pile no. 2

+D=104-97;

+d=104-97;

+bdash=16;

+C=19*(D/bdash)^0.5*(d+D)/b;    //correction for pile no. 2

+fic1=fic1*100+corec+C;         //corrected pressures

+

+//intermedite pipe line

+d=105-97;

+b1=16.5;

+alpha=b/d;

+r=b1/b;               //ratio b1/b

+//from the curves we get

+fic2=0.52;

+fie2=0.725;

+fid2=0.615;

+corec_c1=(fid2-fic2)*100*t/d;

+corec_e1=(fie2-fid2)*100/d;

+

+//for pile no. 1

+C1=C;

+d=104-97;

+bdash=37;

+D=104-95;

+C2=19*(D/bdash)^0.5*(d+D)/b;

+//correction due to slope

+corec_e2=3.3;                //from table 12.4

+//correction is negative due to upwrd slope

+l=4;               //horizontal length of slope

+corec_c2=corec_e2*l/bdash;

+

+fie2=fie2*100-corec_e1-corec_e2;

+fic2=fic2*100+corec_c1+C2-corec_c2;

+

+//pile no. 3 at d/s end

+d=103.5-95;

+alpha_=d/b;

+//for curves

+fie3=0.35;fid3=0.242;

+corec_t=(fie3-fid3)*100*(103.5-102)/d;

+

+//correction for interference at pile no. 2

+d=102-95;

+D=102-97;

+C3=19*(D/bdash)^0.5*(d+D)/b;

+fie3=fie3*100-corec_t-C3;

+

+point=['C1' 'C2' 'E2' 'E3'];            //Point

+P=[fic1 fic2 fie2 fie3];                //pressure percent

+P_=[3.55 2.78 3.39 1.58];               //pressure head

+mprintf("Points         Pressure percent        Pressure head");

+for i=1:4

+    P(i)=round(P(i)*10)/10;

+    mprintf("\n%s                %f               %f",point(i),P(i),P_(i));

+end

+

+//check for floor thickness

+Pa=P_(2)-((P_(2)-P_(4))*6.5/37);

+Pb=P_(2)-((P_(2)-P_(4))*24/37);

+Pc=P_(2)-((P_(2)-P_(4))*30/37);

+rho=2.24;                              //specific gravity of concrete

+ta=Pa/(rho-1);

+tb=Pb/(rho-1);

+tc=Pc/(rho-1);

+ta=round(ta*100)/100;

+tb=round(tb*100)/100;

+tc=round(tc*100)/100;

+mprintf("\n\nThickness required at A=%f m.",ta);

+mprintf("\nThickness required at B=%f m.",tb);

+mprintf("\nThickness required at C=%f m.",tc);

+t=103.5-102;

+mprintf("\nThickness provided=%f m.",t);

+mprintf("\nFloor thickness at B and C are adequate");

+

+//exit gradient

+H=108.5-103.5;               //seepage head

+d=103.5-95;                 //depth cut-off

+//from exit gradient curve

+alpha=6.35;

+lambda=(1+(1+alpha^2)^0.5)/2;

+Ge=H/(d*%pi*lambda^0.5);

+mprintf("\n\nexit gradient=%f.",Ge);

+mprintf("\n it is less than permissible exit gradient  < 1/6\nHence safe..");

diff --git a/2087/CH12/EX12.3/example12_3.sce b/2087/CH12/EX12.3/example12_3.sce
new file mode 100755
index 000000000..4140d389f
--- /dev/null
+++ b/2087/CH12/EX12.3/example12_3.sce
@@ -0,0 +1,161 @@
+

+

+//example 12.3

+//design a vertical drop weir on Bligh's theory

+//test floor by Khosla's theory

+clc;funcprot(0);

+//given

+Q=2800;                    //maximum flood discharge

+hfl=285;                   //H.F.L before construction

+hw=278;                    //minimum water level

+fsl=284;                   //F.S.L of canal

+c=12;                      //coefficient of creep

+flux=1;                    //allowable afflux

+Ge=1/6;                    //permissible exit gradient

+rho=2.24;                  //specific gravity of concrete

+

+//Hydraulic calculation

+L=4.75*Q^0.5;

+q=Q/L;

+q=round(q*10)/10;

+mprintf("Hydraulic calculation:");

+mprintf("\ndischarge per unit width of river=%f cumecs.",q);

+f=1;

+R=1.35*(q^2/f)^(1/3);

+R=round(R*100)/100;

+mprintf("\nregime scour depth=%f m.",R);

+V=q/R;                   //regime velocity

+vh=V^2/(2*9.81);        //velocity head

+l_down=hfl+vh;

+l_up=l_down+flux;

+hfl_up=l_up-vh;

+hfl_down=hfl-0.5;

+hfl_down=round(hfl_down*100)/100;

+mprintf("\nactual d/s H.F.L allowing 0.5 m for retrogation=%f m.",hfl_down);

+K=(q/1.7)^(2/3);

+cl=l_up-K;               //crest level

+cl=round(cl*100)/100;

+mprintf("\ncrest level=%f m.",cl);

+pl=fsl+0.5;              //pond level

+s=hfl_down-cl;                //heigth of shutter

+mprintf("\nheigth of shutter=%f m.",s);

+rl_up_pile=hfl_up-1.5*R;    //R.L of bottom u/s pile

+d_up_cut=hw-276;           //depth of upstream cut-off

+mprintf("\ndepth of upstream cut-off=%f m.",d_up_cut);

+mprintf("\n provide concrete cut off 2 m depth.");

+rl_bot_ds=hfl_down-2*R;

+Hs=hfl_down-hw;            //seepage head

+Hc=cl-hw;                  //heigth of crest

+mprintf("\nR.L of gates crest=%f m.",Hs);

+mprintf("\nHeigth of crest=%f m.",Hc);

+

+//design of weir wall

+d=hfl_up-cl;

+a=d/(rho)^0.5;

+a=3*d/(2*rho);            //from sliding consideration

+a=s+1;                    //from practical consieration

+a=a+1;

+mprintf("\n\ndesign of weir wall:")

+mprintf("\nprovide top width of %i m.",a);

+Mo=9.81*Hs^3/6;                //overtirning moment

+//equating the moment of resistance to overturning moment and putting the values we get

+y=poly([-1.084,0.020,0.039],'x','c');

+b=roots(y);

+//we get b= - 5.5347261 and 5.0219056

+//taking

+b=5;

+//when weir is submerged

+C=0.58;

+d=(q^2/((2*C/3)^2*2*9.81))^(1/3);

+Mo=9.81*d*Hc^2/2;

+//from equation of moment of resistence we get

+y=poly([-77.55,3,1],'x','c');

+b=roots(y);

+//we get b= - 10.433085 and 7.4330846

+//taking

+b=8;

+mprintf("\nbottom width=%i m.",b);

+

+//design of impervious and pervious aprons

+C=12;

+L=C*Hs;

+mprintf("\n\ndesign of impervious and pervious aprons:");

+mprintf("\ntotal creep length=%i m.",L);

+l1=2.21*C*(Hs/13)^0.5;

+l1_=l1+1;

+mprintf("\nlength of downstream impervious apron=%i m.",l1_);

+d1=hw-276;

+d2=hw-271;

+l2=L-l1-(b+2*d1+2*d2);

+mprintf("\nlength of upstream impervious apron=%i m.",l2);

+l3=18*C*(Hs*q/975)^0.5;

+mprintf("\ntotal length of d/s apron=%i m.",l3);                //calculation is wrong in book

+l=l3-l1;

+le=l/2;

+le=round(le*100)/100;

+mprintf('\nprovide filter of length %f m. and launching apron of length %f m.',le,le);

+t=d2*10^0.5/le;

+mprintf("\nthickness of launching apron in horizontal position=%f m.",t);

+mprintf("\nprovide launching apron of thickness 1.5 m.");

+T=2*d1;

+V=d1*10^0.5;

+ta=V/T;

+ta=round(ta*10)/10;

+mprintf("\nthickness of apron in horizontal position=%f m.",ta);

+Hr=Hs-Hs*(4+33+8)/L;

+t=4*Hr/(3*(rho-1));

+t=round(t*10)/10;

+mprintf('\nprovide thickness of %f m from d/s of weir wall to point 6 m from it.',t);

+Hr=Hs-Hs*(4+33+8+6)/L;

+t=4*Hr/(3*(rho-1));

+t=round(t*10)/10;

+mprintf("\nprovide thickness of %f m from 6 m to 12 m from d/s end of weir wall.",t);

+Hr=Hs-Hs*(4+33+8+12)/L;

+t=4*Hr/(3*(rho-1));

+t=round(t*10)/10;

+mprintf("\nprovide thickness of %f m for rest of length of weir floor.",t);

+

+//check by khosla's theory

+b=33+8+19;            //total horizontal length of impervious floor

+d=7;                  //depth of downstream pile

+alpha=b/d;

+n=0.14;                //n=1/%pi*(lambda)^0.5;

+Ge=Hs*n/d;

+mprintf("\n\ncheck by Khosla theory:");

+mprintf("\nexit gradient=%f. < 1/6\n hence safe",Ge);

+alpha_=d/b;

+fic1=0.83;fid1=0.88;

+corec_c1=(fid1-fic1)*100/2;

+bdash=b;

+d=2;D=7;

+C1=19*(D/bdash)^0.5*(d+D)/b;

+fic1=fic1*100+corec_c1+C1;

+Pc=Hs*fic1/100;                        //pressure head at C

+alpha_=d/b;

+fie2=0.31;fid2=0.21;

+corec_e1=(fie2-fid2)*1.7*100/7;

+bdash=b;

+d=7;D=2;

+C1=19*(D/bdash)^0.5*(d+D)/b;

+fie2=fie2*100-corec_e1-C1;             //in book 3.53 value is wrong

+Pe=Hs*fie2/100;                         //pressue head at E

+//assuming linear variation of pressure for intermediate points

+Pa=Pc-(Pc-Pe)*(33+8)/b;

+t=Pa/1.24;

+Pa=round(Pa*100)/100;

+t=round(t*100)/100;

+mprintf("\npressure at d/s of weir wall=%f m.",Pa);

+mprintf("\nthickness at d/s of weir wall=%f m. < thickness by Bligh theory;\nhence safe.",t);

+Pb=Pc-(Pc-Pe)*(33+8+6)/b;

+t=Pb/1.24;

+Pa=round(Pa*100)/100;

+t=round(t*100)/100;

+mprintf("\npressure at 6 m from d/s of weir wall=%f m.",Pb);

+mprintf("\nthickness at 6m from d/s of weir wall=%f m. < thickness by Bligh theory;\nhence safe.",t);

+Pc=Pc-(Pc-Pe)*(33+8+12)/b;

+t=Pc/1.24;

+Pa=round(Pa*100)/100;

+t=round(t*100)/100;

+mprintf("\npressure at 12 m from d/s of weir wall=%f m.",Pc);

+mprintf("\nthickness at 12m from d/s of weir wall=%f m. > thickness by Bligh theory;\nhence unsafe.",t);

+mprintf("\nhence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.");

diff --git a/2087/CH12/EX12.4/example12_4.sce b/2087/CH12/EX12.4/example12_4.sce
new file mode 100755
index 000000000..f688c2648
--- /dev/null
+++ b/2087/CH12/EX12.4/example12_4.sce
@@ -0,0 +1,198 @@
+

+

+//example 12.4

+//design a slopeing glacis

+clc;funcprot(0);

+//given

+q=10;                     //maximum discharge intensity on weir crest

+hfl=255;                  //H.F.L before construction of weir

+rb=249.5;                 //R.L of river bed

+pl=254;                   //pond level

+s=1;                      //heigth of crest shutter

+dhw=251.5;                //anticipated downstream water level in river when water is dischrging with pond level upstream

+br=0.5;                   //bed retrogression

+f=0.9;                    //Laecey silt factor

+Ge=1/7;                   //permissible exit gradient

+flux=1;                   //permissible afflux

+

+cl=pl-s;                   //crest level

+mprintf("crest level=%f m.",cl);

+K=(q/1.7)^(2/3);

+tel_up=cl+K;

+tel_up=round(tel_up*100)/100;

+mprintf("\nelevation of u/s T.E.L=%f m.",tel_up);

+R=1.35*(q^2/f)^(1/3);

+R=round(R*10)/10;

+mprintf("\nregime scour depth=%f m.",R);

+V=q/R;                     //regime velocity

+vh=V^2/(2*9.81);           //velocity head

+hfl_up=tel_up-vh;

+tel_down=hfl+vh;

+flux=hfl_up-hfl;

+flux=round(flux*100)/100;

+mprintf("\nafflux=%f. which is near to permissible",flux);

+hfl_down=hfl-br;             //downstream H.F.L after retrogression

+tel_down=tel_down-br;        //downstream T.F.L after retrogression

+Hl=tel_up-tel_down;          //loss of head in flood

+Hl=round(Hl*100)/100;

+mprintf("\nloss of head in at high flood=%f m.",Hl);

+K=pl-cl;              //head over crest

+q_=1.7*(K)^1.5;

+Hl_=pl-dhw;             //loss of head

+mprintf("\nloss of head=%f m.",Hl_);

+Ef2=4.3;

+Ef2_=1.7;              //from Blench curve

+jump=tel_down-Ef2;

+jump_=251.5-Ef2_;      //level at which jump will form

+Ef1=Ef2+Hl;

+Ef1_=Ef2_+Hl_;

+D1=1.03;

+D1_=0.15;                //calculated from Ef1 and Ef1_ respectively

+D2=3.96;D2_=1.68;       //calculated from Ef2 and Ef2_ respectively

+hj=D2-D1;

+hj_=D2_-D1_;            //heigth of jump

+concrete=5*hj;

+concrete_=5*hj_;         //length of concrete floor

+mprintf("\n\nHydraulic jump calculation:");

+mprintf("\nheigth of jump for high flood condition=%f m.",hj);

+mprintf("\nlength of concrete floor for high flood condition=%f m.",concrete);

+mprintf("\nheigth of jump for pond level condition=%f m.",hj_);

+mprintf("\nlength of concrete floor for high pond level condition=%f m.",concrete_);

+

+cw=2;                    //crets width

+us=2;                    //upstream slope

+ds=3;                    //downstream slope

+l=15;

+mprintf("\n\n upstream slope of glacis=%i:1.",us);

+mprintf("\ndownstream slope of glacis=%i:1.",ds);

+mprintf("\nhorizontal length of floor beyond the toe=%i m..",l);

+

+R=6.5;

+sh_up=hfl_up-1.5*R;

+sh_down=hfl_down-2*R;

+sh_up=round(sh_up*100)/100;

+mprintf("\nR.L of bottom of upstream sheet pile=%f m.",sh_up);

+mprintf("\nR.L of downstream sheet pile=%f m.",sh_down);

+mprintf("\nprovide intermediate sheet pile at d/s toe of glacis.");

+Hs=pl-249.6;                       //maximum percolation head

+d=249.6-sh_down;                   //depth of d/s cut-off

+n=Ge*d/Hs;                          //n=1/(%pi*lambda^0.5);

+//from khosla exit gradient curve

+alpha=1.5;

+b=alpha*d;

+mprintf("\n\nlength of impervious floor=%f m.",b);

+fl=(2*(253-249.5))+2+(3*(253-249.6))+15;

+us=36-fl;

+mprintf("\nlength of floor already provide=%f m.",fl);

+mprintf("\nwhich is more than required from permissible exit gradient.\nno upstream floor is required.");

+mprintf("\nprovide %f m upstream floor so that total length becomes 36 m.",us);

+alpha_1=0.089; 

+alpha_2=0.225;            //alpha_=1/alpha

+b1=21;

+alpha=4.44;

+mprintf("\n\nPressure percent at points:");

+point=['C1' 'D1' 'C2' 'E2' 'D2' 'D3' 'E3'];

+bc=[72 82 31.5 45.5 58.5 29 44];

+crt=[3.1 0 3.5 0 -3.2 0 0 -3.6];

+crs=[0 0 0 0 2.3 0 0 0];

+cri=[3.7 0 6.4 0 -2.4 0 -6.4];

+mprintf("\nPoints       Before correction            After correction");

+for i=1:7

+    after(i)=bc(i)+crt(i)+crs(i)+cri(i);

+    mprintf("\n%s                   %i                       %f",point(i),bc(i),after(i));

+end

+Hs=254-249.6;               //no flow condition

+Hs_=256.13-254.5;           //high flood condition

+Hs__=254-251.5;             //flow at pond level

+mprintf("\n\nelevation of subsoil H.G above datum:");

+mprintf("\nno flow condition:");

+fie1=1*Hs;

+fid1=0.82*Hs;

+fic1=0.788*Hs;

+fie2=0.552*Hs;

+fid2=0.455*Hs;

+fic2=0.414*Hs;

+fie3=0.34*Hs;

+fid3=0.29*Hs;

+fic3=0;

+fie1=round(fie1*100)/100;fid1=round(fid1*100)/100;fic1=round(fic1*100)/100;

+fie2=round(fie2*100)/100;fid2=round(fid2*100)/100;fic2=round(fic2*100)/100;

+fie3=round(fie3*100)/100;fid3=round(fid3*100)/100;fic3=round(fic3*100)/100;

+mprintf("\nfie1=%f.;fid1=%f.;fic1=%f.\nfie2=%f.;fid2=%f.;fic2=%f.\nfie3=%f.;fid3=%f.;fic3=%f.",fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);

+mprintf("\nhigh flood condition:");

+fie1=1*Hs_;

+fid1=0.82*Hs_;

+fic1=0.788*Hs_;

+fie2=0.552*Hs_;

+fid2=0.455*Hs_;

+fic2=0.414*Hs_;

+fie3=0.34*Hs_;

+fid3=0.29*Hs_;

+fic3=0;

+fie1=round(fie1*100)/100;fid1=round(fid1*100)/100;fic1=round(fic1*100)/100;

+fie2=round(fie2*100)/100;fid2=round(fid2*100)/100;fic2=round(fic2*100)/100;

+fie3=round(fie3*100)/100;fid3=round(fid3*100)/100;fic3=round(fic3*100)/100;

+mprintf("\nfie1=%f.;fid1=%f.;fic1=%f.\nfie2=%f.;fid2=%f.;fic2=%f.\nfie3=%f.;fid3=%f.;fic3=%f.",fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);

+mprintf("\nflow at pond level:");

+fie1=1*Hs__;

+fid1=0.82*Hs__;

+fic1=0.788*Hs__;

+fie2=0.552*Hs__;

+fid2=0.455*Hs__;

+fic2=0.414*Hs__;

+fie3=0.34*Hs__;

+fid3=0.29*Hs__;

+fic3=0;

+fie1=round(fie1*100)/100;fid1=round(fid1*100)/100;fic1=round(fic1*100)/100;

+fie2=round(fie2*100)/100;fid2=round(fid2*100)/100;fic2=round(fic2*100)/100;

+fie3=round(fie3*100)/100;fid3=round(fid3*100)/100;fic3=round(fic3*100)/100;

+mprintf("\nfie1=%f.;fid1=%f.;fic1=%f.\nfie2=%f.;fid2=%f.;fic2=%f.\nfie3=%f.;fid3=%f.;fic3=%f.",fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);

+

+mprintf("\n\nPrejump profile:");

+mprintf("\nhigh flood condition:");

+dist=[3 6 8.4];                 //distance

+glacis=[252 251 250.32];        //R.L of glacis

+D1=[1.3 1.15 1.03];

+mprintf("\nEf1              D1");

+for i=1:3

+    Ef1(i)=256.25-glacis(i);

+    mprintf("\n%f         %f",Ef1(i),D1(i));

+end

+mprintf("\npond level flow:");

+dist=[3 6 9 9.6];             //distance

+glacis=[252 251 250 249.9];       //R.Lof glacis

+D1=[0.31 0.23 0.16 0.15];

+mprintf("\nEf1              D1");

+for i=1:4

+    Ef1(i)=254-glacis(i);

+    mprintf("\n%f         %f",Ef1(i),D1(i));

+end

+

+

+rho=2.24;

+Uf=4;                           //unbalanced head for high flood condtion

+Us=2.56;                        //unbalanced static head

+Hf=2*Uf/3;

+t=Hf/(rho-1);

+t=round(t*10)/10;

+mprintf("\n\nfloor thickness at the point of formation of hydraulic jump=%f m.",t);

+Uf=2.9;                           //unbalanced head for high flood condtion

+Us=2.2;                        //unbalanced static head

+Hf=2*Uf/3;

+t=Us/(rho-1);

+t=round(t*10)/10;

+mprintf("\nfloor thickness at the point of formation of hydraulic jump at the pond level condition=%f m.",t);

+P=1.5;                        //pressure head at d/s end of floor

+t=P/(rho-1);

+t=round(t*10)/10;

+mprintf("\n\nfloor thickness at downstream side of sloping glacis=%f m.",t);

+D=rb-sh_up;                 //depth of u/s scour hole above bed level

+a=1.5*D;

+a=round(a*10)/10;

+mprintf("\n\nminimum length of upstream launching apron=%f m.",a);

+mprintf("\nprovide 1.5 m thick apron for length of 5 m.");

+D=249.6-241.5;

+a=1.5*D;

+mprintf("\n\nminimum length of downstream launching apron=%f m.",a);

+mprintf("\nprovide 1.5 m thick apron for length of 12 m.");

+

diff --git a/2087/CH12/EX12.5/example12_5.sce b/2087/CH12/EX12.5/example12_5.sce
new file mode 100755
index 000000000..2cc29b121
--- /dev/null
+++ b/2087/CH12/EX12.5/example12_5.sce
@@ -0,0 +1,31 @@
+

+

+//example 12.5

+//calculate uplift pressure at the junction of inner faces of pile with weir floor using Khosla theory

+clc;funcprot(0);

+//given

+b=16;       //total length of floor

+d=5;        //depth of downstream pile

+D=4;        //depth of upstream pile

+H=2.5;      //head created by weir

+

+//pressure at E

+alpha=b/d;

+lambda=(1+(1+alpha^2)^0.5)/2;

+fie=acos((lambda-2)/lambda)/%pi;

+C=19*(D/b)^0.5*((d+D)/b);

+fie=fie*100-C;

+P=H*fie/100;

+P=round(P*1000)/1000;

+mprintf("Pressure at E=%f m.",P);

+

+//pressure at C1

+alpha=b/D;

+lambda=(1+(1+alpha^2)^0.5)/2;

+fie=acos((lambda-2)/lambda)/%pi;

+fic=1-fie;            //by principle reversibility of flow

+C=19*(d/b)^0.5*((d+D)/b);

+fic=fic*100+C;

+P=fic*H/100;

+P=round(P*1000)/1000;

+mprintf("\n Pressure at C=%f m.",P);

diff --git a/2087/CH12/EX12.6/example12_6.sce b/2087/CH12/EX12.6/example12_6.sce
new file mode 100755
index 000000000..e2fcf7869
--- /dev/null
+++ b/2087/CH12/EX12.6/example12_6.sce
@@ -0,0 +1,45 @@
+

+

+//example 12.6

+//calculate floor thickness at mid length and at junction with u/s and d/s cut-off walls

+clc;funcprot(0);

+//given

+b=13;         //length of floor

+d=2;          //depth of downstream wall

+D=1.5;       //depth of upstream cut-off

+rho=2.24;     //relative density

+H=1.5;

+

+//at junction of d/s cut-off with floor

+alpha=b/d;

+lambda=(1+(1+alpha^2)^0.5)/2;

+fie=acos((lambda-2)/lambda)/%pi;

+C=19*(D/b)^0.5*((d+D)/b);

+fie=fie*100-C;

+P=H*fie/100;

+t=P/(rho-1);

+t=round(t*10)/10;

+mprintf("floor thickness at junction of d/s cut-off with floor=%f m.",t);

+

+//at junction of u/s cut-off with floor

+alpha=b/D;

+lambda1=(1+(1+alpha^2)^0.5)/2;

+fie=acos((lambda1-2)/lambda1)/%pi;

+fic=1-fie;            //by principle reversibility of flow

+C=19*(D/b)^0.5*((d+D)/b);

+fiec=fic*100+C;

+P=fiec*H/100;

+t=0.3;              //this the uplift will be counter balanced by downward weigth of impounded water

+mprintf("\nfloor thickness at junction of u/s cut-off with floor=%f m.",t);

+

+//at mid-length

+P=(1.08+0.489)/2;            //assuming linear variation

+t=P/(rho-1);

+t=round(t*100)/100;

+mprintf("\nfloor thickness at mid-length=%f m.",t);

+

+//exit gradient

+G=H/(d*%pi*(lambda)^0.5);

+G=round(G*1000)/1000;

+//since G<0.18

+mprintf("\n G=%f. <0.18./nfloor is safe against failure by piping.",G);

diff --git a/2087/CH12/EX12.7/example12_7.sce b/2087/CH12/EX12.7/example12_7.sce
new file mode 100755
index 000000000..d6b20376d
--- /dev/null
+++ b/2087/CH12/EX12.7/example12_7.sce
@@ -0,0 +1,15 @@
+

+

+//example12.7

+//calculate heigth of weir to be built

+clc;funcprot(0);

+//given

+B=30;            //stream width

+D=3;             //stream depth

+V=1.25;          //mean velocity

+Cd=0.95;         //discharge coefficient

+Q=B*D*V;

+C=2*Cd*(2*9.81)^0.5/3;

+x=4-(Q/(C*B))^(2/3);

+x=round(x*1000)/1000;

+mprintf("heigth of weir to be built=%f m.",x);

diff --git a/2087/CH12/EX12.8/example12_8.sce b/2087/CH12/EX12.8/example12_8.sce
new file mode 100755
index 000000000..4dd5a64f1
--- /dev/null
+++ b/2087/CH12/EX12.8/example12_8.sce
@@ -0,0 +1,40 @@
+

+

+//example 12.8

+//calculate uplift pressure at two cut-off

+clc;funcprot(0);

+//given

+b=50;       //length of floor

+d=8;        //depth of downstream pile

+D=8;        //depth of upstream pile

+H=5;        //effective head 

+tu=1;        //floor thickness at upstream

+td=2;        //floor thickness at downstream

+

+//downstream cut-off

+alpha=b/d;

+lambda=(1+(1+alpha^2)^0.5)/2;

+fie=acos((lambda-2)/lambda)/%pi;

+fid=acos((lambda-1)/lambda)/%pi;

+Ct=(fie-fid)*td/d;

+C=19*(D/b)^0.5*((d+D)/b);

+fie=fie*100-C-Ct*100;

+P=H*fie/100;

+P=round(P*100)/100;

+mprintf("Pressure at downstream cut-off=%f m.",P);

+

+//upstream cut-off

+fie=acos((lambda-2)/lambda)/%pi;

+fid=acos((lambda-1)/lambda)/%pi;

+fic1=1-fie;

+fid1=1-fid;

+Ct=(fic1-fid1)*td/d;

+C=-19*(D/b)^0.5*((d+D)/b);

+fic1=fic1*100-C-Ct*100;

+P=H*fic1/100;

+P=round(P*100)/100;

+mprintf("\nPressure at upstream cut-off=%f m.",P);

+G=H/(d*%pi*(lambda)^0.5);

+mprintf("\nExit Gradient=%f.",G);

+

+

diff --git a/2087/CH12/EX12.9/example12_9.sce b/2087/CH12/EX12.9/example12_9.sce
new file mode 100755
index 000000000..b838621b4
--- /dev/null
+++ b/2087/CH12/EX12.9/example12_9.sce
@@ -0,0 +1,28 @@
+

+

+//example 12.9

+//calculate depth of downstream cut-off

+clc;funcprot(0);

+//given

+Q=1000;        //discharge of river

+L=256;         //crest length of diversion

+f=1.1;         //silt factor

+seg=1/6;       //safe exit gradient

+hfl=103;       //high flood level

+cf=100;        //reduced level of downstream concrete floor

+H=2.4;         //maximum static head of weir

+b=40;          //length of concrete floor

+

+q=Q/L;

+R=1.35*(q^2/f)^(1/3);

+rld=hfl-1.5*R;

+d=cf-rld;

+d=round(d*100)/100;

+mprintf("depth of downstream cut-off=%f m.",d);

+

+alpha=b/d;

+lambda=(1+(1+alpha^2)^0.5)/2;

+G=H/(d*%pi*(lambda)^0.5);

+//since G<seg

+mprintf("\n G=%f. <1/6./nfloor is safe against failure by piping.",G);

+