?iMci128.07fEz?-??-?1Mc1??2Mc2??3Mc3??4Mc4??5Mc5?-EIEIi?1EI5??fE?fEy?fEz?2240.60EI?[fE]?3.5?10?4m
E?210?109Pa
I??644D4(1??4)?0.04D4
?D40.60?210?109?0.04?3.5?10?4
D3?6.08?10?2m
3) 根据C点刚度计算轴径，在C点处加一单位力偶,有弯矩图如下:
1?1???0.18?122.53??11.032 ?2??1122.53??122.53??2.5222978.63
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?31?122.53?????0.55-0.15-?1068.92?191.80??2?2978.63?
?4?1068.92?0.15?160.34
?51??0.15??1575.00?1068.92??37.96
2Mc1?Mc2?Mc3?Mc4?Mc5?1
?cy?iMci1384.23????1Mc1??2Mc2??3Mc3??4Mc4??5Mc5?EIEIEI i?15??
?1??2?388.05??0.55-0.15??155.22?31?0.18?388.05?34.922
1???0.55-0.15???810.42-388.05??84.472
?4?810.42?0.15?121.56
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1?5??0.15??840.00?810.42??2.22
2Mc1?Mc2?Mc3?Mc4?Mc5?1
?cz????iMci1398.40???1Mc1??2Mc2??3Mc3??4Mc4??5Mc5??EIEIEIi?15???c????2cy2cz553.49??[?c]?0.0028radEI
E?210?109Pa
I??64D4(1??4)?0.04D4
?D4?553.49210?109?0.04?0.0028
?D4?6.95?10?2m12综上所述：D?max?D，D，D，D??7.49?1034
?2m
C. 疲劳强度校核：
若不计键槽对抗弯截面系数的影响，则危险截面处抗弯截面系数：
???D332(1??4)?3.39?10?5m?3
由弯矩M不变可知该循环为对称循环，则有：
?max???minM??W1575.002?840.002Pa?52.65MPa?53.39?10
Wp?2Wz?2W?6.78?10?5?maxMX124.14??Pa?1.83MPa?5WP6.78?10
查表确定铣加工的键槽危险截面处疲劳强度的影响系数：
K??1.60K??1.88???0.75???0.73??1.8
２８
n?则：???1K??max????420MPa1.60?52.65MPa0.75?1.8?6.73
n????1K??max????2?240MPa?91.67 1.88?1.83MPa0.73?1.82n??n?n?n??n??6.71?3
故满足强度条件。
五.循环计算程序
２９
#include
#define pi 3.141592654 #define ip 0.017453292
float L1,L2,L3,a,b,A0,n,P,i,R,Fhy,Fhz, Fay,Faz,Fcy,Fdz,Fdy,Fcz,Fey,Fez, Me,Mby,Mbz,Mdy,Mdz, Mcy,Mcz,Mey,Mez,Md,Mc, D1,D2,D3,D4,D,Xs,w,sjwyb, Fby=0,Fbz=0, E=210000000000,yl=150,fzby,fzbz, nde=0.00035,ndd=0.00033,zjc=0.0028,wyb=0.00005; int pd=0; void zaihe() {
float Ft,Fr,An=20.0,Bn=10.0; Me=9549*P/n; Ft=Me/R;
Fr=Ft*tan(An*ip)/(cos(Bn*ip)); Fey=Ft*sin(A0*ip)-Fr*cos(A0*ip); Fez=Ft*cos(A0*ip)+Fr*sin(A0*ip); Mdy=Fhz*b; Mdz=Fhy*b; }
void waili()
{ Fay=(Fhy*L3+Mdz-Fey*a-Fby*L2)/(L1+L2);
Fcy=(-Fhy*(L1+L2+L3)-Mdz-Fey*(L1+L2-a)-Fby*L1)/(L1+L2); Faz=(-Fhz*L3-Mdy-Fez*a-Fbz*L2)/(L1+L2);
Fcz=(Fhz*(L1+L2+L3)+Mdy-Fez*(L1+L2-a)-Fbz*L1)/(L1+L2); Mby=Fay*L1; Mbz=Faz*L1;
Mey=Fay*(L1+L2-a)+Fby*(L2-a); Mez=Faz*(L1+L2-a)+Fbz*(L2-a);
Mcy=Fay*(L1+L2)+Fby*L2+Fey*a; Mcz=Faz*(L1+L2)+Fbz*L2+Fez*a; /*对于静定情况B点受力为0*/ }
void qiangdu() {
float wb,wc,we,temp; wb=sqrt(Mby*Mby+Mbz*Mbz); wc=sqrt(Mcy*Mcy+Mcz*Mcz); we=sqrt(Mey*Mey+Mez*Mez); if (wb>wc&&wb>we)
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