MGM S JAWAHARLAL NEHRU ENGINEERING COLLEGE N-6, CIDCO, AURANGABAD LAB MANUALS SUB: COMPUTER LAB-II CLASS: S.E.CIVIL

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1 MGM S JAWAHARLAL NEHRU ENGINEERING COLLEGE N-6, CIDCO, AURANGABAD LAB MANUALS (Procedure for conduction of Practical/Term Work) SUB: COMPUTER LAB-II CLASS: S.E.CIVIL Prepared by Ms.V.S.Pradhan Lab In charge

2 Exercise 1: Program for setting out of curves #include"stdio.h" #include"conio.h" #include"math.h" void main() { int r,angle,n,i; float y,x,t,ct,ch,cl,ct2,l,o,od,a; clrscr(); printf("\n Enter radious->"); scanf("%d",&r); printf("\nenter angle->"); scnaf("%d",&angle); printf("\nangle tan (angle)->"); x = (3.1417/180) * (angle/2); y = tan(x); printf("%d %f",angle,y); t1 = r * y; printf("\n\n tangent length = %f",t1); ct1=ch-t; printf("\n\nchainage of T1 %f",ct); c1 = ( *r*angle/180); printf("in curve length = %f",cl); ct2 = ct+cl; printf("\n\n chainage of T2 = %f",ct2); l=2*r*sin(x); printf("\n\nlength of l.c. = %f",l); o=r-sqrt((r*r)-(l/2)*(l/2));

3 printf("\n\n mid ordinate l.c. = %f",o); printf("\n\nenter no:-"); for(i=0;i<n;i++) printf("enter offset distance->"); scanf("%f",&od); g = sqrt((r*r)-(od-od))-(r-o); printf("\n\n A = %f",a); getch(); }

4 Exercise 2: Program for setting out of curves #include"stdio.h" #include"conio.h" #include"math.h" void main() { int rs, rl,as,al,pi,pi,t1,n; float xl,xs,yl,ys,ts,tl,bd,be,de,ts,tl,wl,ws,zl,zs,bl,ml,ms,cll,cls,cl,cs,ds,dl,ct1,ct2,ct3,isc,no,ifc,cc,fsc; clrscr(); printf("\nenter small radius->"); scanf("%d",&rs); printf("\nenter long radius->"); scanf("%d",&rl); printf("\nenter small angle->"); scanf("%d",&as); printf("\nenter long angle->"); scanf("%d",&al); xl=(3.1417/180)*(al/2); xs=(3.1417/180)*(as/2); yl=tan(xl); ys=tan(xs); ml=sin(xl); ms=sin(xs); printf("%d %f",al,yl,ml); printf("%d %f",as,ys,ms); t1=rl*yl; ts=rs*ys; printf("\n\nlong tangent length = %f",tl);

5 printf("\n\nsmall tangent length = %f",ts); de = (tl+ts); printf("\n\nlength of t2 = %f",de); wl = (3.1417/180)*al; ws = (3.1417/180)*as; zl = sin(wl); zs = sin(ws); printf("%d %f",al,wl); printf("%d %f",as,ws); bl=sin(wl+ws); printf("%d %f %f %f",al,wl,as,ws); be = (de*(zl/bl)); printf("\n\nlength of be = %f",be); bd = de*(zs/bl); printf("\n\nlength of db = %f",bd); ts = (bd+tl); printf("\n\ntotal langent length of small curve Ts = %f",ts); Tl=(be+ts); printf("\n\ntotal tangent length of long curve Tl = %f",tl); cls = (ws * rs); printf("\n\n shourt curve length = %f",cls); cll = (wl*rl); printf("\n\nshort curve length =%f",cll); cs = 2*rs*ms; cl = 2*rl*ml; printf("\n\n chort of shourt curve = %f",cs); printf("\n\nchour of long curve = %f",cl); ds = (1718*cs)/rs; dl = (1718/cl)/rl;

6 printf("\n\ndeflection angle of short curve = %f",ds); printf("\n\ndeflection angle of long curve = %f",dl); printf("\nenter point of intersection->"); scanf("%d",&pi); ct1=pi-tl; ct2=ct1+cls; printf("\n\nchainage of T1 = %f",ct1; ct2= ct1+cll; printf("\n\nchainage of T2 = %f",ct2); ct3 = ct2+cll; printf("\n\nchainage of T3 = %f",ct3); printf("\nenter round up value of ct1->"); scanf("%d",&t1); isc = t1-ct1; printf("\n\nlength of initial sub chord = %f",isc); printf("\nenter peg interval pi-"); scanf("%d",&pi); no = ((ct2-t1)/pi); printf("\n\nmultiplying factor for full chord = %f",no); printf("\nenter round up value of n ->"); scanf("%d",&n"); ifc = n*pi' printf("\n\n length of full chord = %f",lfc); cc=ct1+lfc; printf("\n\nchainage covered = %f",cc); fsc=ct2-cc; printf("\n\nlength of final sub chord = %f",fsc);

7 Exercise3: Program for estimation of head loss #include"stdio.h" #include"conio.h" #include"math.h" void main() { float l1,d1,q1,z1,l2,d2,z2,z2,l3,d3,f,hf1,hf2,ph,q3,q2,v1,v2,v3; clrscr(); printf("enter the value of q1,d1,f,l1,z1,l2,d2,d3,l3"); scanf("%f %f %f %f %f %f %f %f %f",&q1,&d1,&f,&l1,&z1,&l2,&d2,&d3,&l3); printf("\nvelocity = %f",v1); hf1=(4*f*l1*v1*v1)/(z*d1*9.81); printf("\nhead loss due to friction = %f",hf1); ph = z1-hf1; printf("\bpizometric head = %f",ph); hf2=z2-ph; printf("\nhead loss due to friction 2 = %f",hf2); v2=sqrt((hf2*d2*2*9.8)/(4*f*12)); printf("\n\nvelocity = %f",v2); q2=v2*(3.1417/4)*d2*d2); q3=q1+q2; printf("\n\ndischarge = %f",q3); v3=q3/((3.1417/4)*(d3*d3)); printf("\nvelocity v3 = %f",v3); hf3=(4*f*l3*v3*v3)/(d3*2*9.81); printf("\nhead loss due to friction 3 = %f",hf3); z3=ph-hf3; printf("\nht. of water in reservoir 3 = %f", z3);

8 getch(); }

9 Exercise 4: Program for estimation of Power Transmitted #include"stdio.h" #include"conio.h" #include"math.h" void main() { int h,n,q,no,w; float P,NS; clrscr(); printf("\nenter head H->"); scanf("%d",&h); printf("\nenter speed N->"); scanf("%d",&n"); printf("\nenter discharge q ->"); scanf("%d",&q); printf("\nenter overall efficiency no ->"); scanf("%d",&no); P=(no*9810*h*q); printf("\n\n power transmitted P = %f",p); ns = (n*sqrt(p))/(sqrt(sqrt(2*2*2*2*2))); printf("\n\n specific speed NS = %f",ns); getch(); }

10 Exercise 5: Program to compute a function at a given ---value using the Langrange Interpolation Method Algorithm: 1) Read n, the number of data points. 2) Read a[i] and corresponding function values b[i] for i=1 to n. 3) Read intermediate data value at which the polynomial should be evaluated. 4) Let sum=0. Sum will hold the result finally. 5) For i=1 to n, do the following steps. a) Initialize prod=1 b) For j=1 to n, if I does not equal j then c) Prod=prod*aa-a[i]/a[i]-a[j] d) Sum=sum+prod*b[i] and go to 5(a). 6) Print out the polynomial as sum. 7) Stop

11 Exercise 6: Solution of Simultaneous Equations using the Gauss Elimination Technique Algorithm: 1) Read n, the no of simultaneous equations. 2) Read row wise matrix a[ n,n+1], the matrix and right hand side constants. 3) Start the elimination process. A) For k=1 to (n-1) do the following. For i= (k+1) to n do steps (a) and (b) a) S=a[i,k]/a[k,k]. b) For j=(k+1) to (n+1) do a[i,j]= a[i,j]-s* a[k,j] B) For i= (k+1) to n, a[i,k]/=0 C) Go to start of step 3. 4) Start the back substitution process as follows: i) d[n] = a [n,n+1]/a[n,n]. ii) For m=1 to (n-10 do steps (a), (b), (c) and (d). a) Sum=0 b) I=n-m c) For j=(i+1) to n, Sum=Sum+ a[i,j]*d[i,j]. d) d[i] =(a[i,n+1]-sum)/ a[i,j] 5) For i=1 to n, print all x[i] 6) Stop

12 Exercise 7: Program for Fourth order Runge Kuttas Method to solve ordinary differential equations Algorithm: 1) Read n, the number of data intervals (n+1 are the data points) 2) Read a{1] and a[n+1], the first and last data points of the entire interval. 3) Compute h, the step size as h= (a[n+1]-a[1])/n 4) Generate the intermediate points for i=2 to n as a[i] = a[i-1]+h 5) Read b[1] the initial value of the function at a[1]. 6) Hence for i=2 to (n+1), generate the corresponding y values as a) K1 = f(a[i-j],b[i-1] b) K2= f(a[i-j] + 0.5*h, b[i-1]+0.5*h*k1) c) K3= f(a[i-j] + 0.5*h, b[i-1]+0.5*h*k2) d) K4= f(a[i-j] +h, b[i-1]+h*k3) e) b[i]= b[i-1]+((1/6)*k1+2*k2+2*k3+k4)*h 7) Print out x and y values for i=1 to (n+1) 8) Stop

13 Exercise 8: Program for Newton Raphsons Method Algorithm: 1) Declare x,y,a,b,i 2) X=(a+b)/2 3) For ( i= 1: i<=10; i++) 4) Y=(x-(f(x)/d(x)) 5) X=y 6) Stop