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/EX12.1/example12_1.sce | 48 +++++++++ 2087/CH12/EX12.10/example12_10.sce | 20 ++++ 2087/CH12/EX12.11/example12_11.sce | 161 ++++++++++++++++++++++++++++++ 2087/CH12/EX12.12/example12_12.sce | 51 ++++++++++ 2087/CH12/EX12.2/example12_2.sce | 105 ++++++++++++++++++++ 2087/CH12/EX12.3/example12_3.sce | 161 ++++++++++++++++++++++++++++++ 2087/CH12/EX12.4/example12_4.sce | 198 +++++++++++++++++++++++++++++++++++++ 2087/CH12/EX12.5/example12_5.sce | 31 ++++++ 2087/CH12/EX12.6/example12_6.sce | 45 +++++++++ 2087/CH12/EX12.7/example12_7.sce | 15 +++ 2087/CH12/EX12.8/example12_8.sce | 40 ++++++++ 2087/CH12/EX12.9/example12_9.sce | 28 ++++++ 12 files changed, 903 insertions(+) create mode 100755 2087/CH12/EX12.1/example12_1.sce create mode 100755 2087/CH12/EX12.10/example12_10.sce create mode 100755 2087/CH12/EX12.11/example12_11.sce create mode 100755 2087/CH12/EX12.12/example12_12.sce create mode 100755 2087/CH12/EX12.2/example12_2.sce create mode 100755 2087/CH12/EX12.3/example12_3.sce create mode 100755 2087/CH12/EX12.4/example12_4.sce create mode 100755 2087/CH12/EX12.5/example12_5.sce create mode 100755 2087/CH12/EX12.6/example12_6.sce create mode 100755 2087/CH12/EX12.7/example12_7.sce create mode 100755 2087/CH12/EX12.8/example12_8.sce create mode 100755 2087/CH12/EX12.9/example12_9.sce (limited to '2087/CH12') 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