From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 2087/CH12/EX12.3/example12_3.sce | 161 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 161 insertions(+)
 create mode 100755 2087/CH12/EX12.3/example12_3.sce

(limited to '2087/CH12/EX12.3')

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.");
-- 
cgit