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x=0.43m
x=0.7m
3) 程序3:
#include
#include
#define MAX 100 main() {
double F[MAX],a[MAX],b[MAX],l,Mx[MAX],vx[MAX],x,E,I,MX,VX; int n,i; MX=0; VX=0;
printf(\
scanf(\
for (i=1;i<=n;i++) {
printf(\ scanf(\ }
if (x>l||x<0)
printf(\ else {
for (i=1;i<=n;i++) {
if (0<=x&&x<=a[i]) {
Mx[i]=b[i]*F[i]*x/l;
vx[i]=-F[i]*b[i]*x*(l*l-x*x-b[i]*b[i])/(6*E*I*l); }
if (a[i] Mx[i]=a[i]*F[i]*(l-x)/l; vx[i]=-F[i]*b[i]*x*(l*(x-a[i])*(x-a[i])*(x-a[i])/ b[i]+(l*l-b[i]*b[i])*x-x*x*x)/(6*E*I*l); } MX=MX+Mx[i]; VX=VX+vx[i]; } printf(\ } getch(); } 例题:已知:n=3, F1=1kN,a1=0.8m,b1=0.2m; l=1m,F2=2kN,a2=0.5,mb2=0.5m; F3=1.5kN,a3=0.35m,b3=0.65m. 求:x=0.8m时的弯矩与挠度。 n ?Fi i?1ai l bi 4-1)程序: #include #define MAX 100 main() { double M[MAX],a[MAX],b[MAX],l,Mx[MAX],vx[MAX],x,E,I,MX,VX; int h,k; MX=0; VX=0; printf(\ scanf(\ for (k=1;k<=h;k++) { printf(\
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