由图1受力分析求支座反力FAy、FAz、FCy、FCz:
?MCz(F)?FAy?l1?l2??FEya?MDz?FHyl3?0
?M?M?MAz(F)?FCy?l1?l2??FEy?l1?l2?a??MDz?FHy?l1?l2?l3??0(F)?FAz?l1?l2??FEza?MDy?FHzl3?0Cy
Ay(F)?FCz?l1?l2??FEz?l1?l2?a??MDy?FHz?l1?l2?l3??0解上面的方程，则有:
FAy=2076.31N，FAz=-1327.13N，FCy=-6971.58N，FCz=2868.42N
根据已知分别作出Y、Z方向的剪力图与弯矩图，如下图所示：
６
由剪力图及弯矩图可知C点为危险点且：
Mc?15752?8402=1785 N?m
Me?124.14N?m
A. 根据第三强度理论校核（忽略剪力）：
MeMc????2WW
?r3?(W?Mc2?Me2????W
?32D3(1??4))
?2代入数据解得：D1?5.29?10m
B. 由刚度对轴进行校核：
?iMci利用图乘法???对各点进行刚度校核：
EIi?1n1) 根据D点刚度计算轴径，在D点分别沿y、z轴加一单位力,有弯矩图如下:
７
1?1??0.58?1204.26?349.242
2Mc1??0.135?0.093?2?1204.26?0.15?180.64Mc21???0.135?0.17??0.153 21?0.15??1575.00?1204.26??27.81 2
?3?Mc32???0.17?0.135??0.135=0.1583?4?810?0.17?137.7
Mc4?1?0.17=0.085 21?5??0.17??1575.00?810.00??65.025
22Mc5??0.17=0.113
3
８
fDy
?iMci182.48????1Mc1??2Mc2??3Mc3??4Mc4??5Mc5?EIEIi?1EI5??
1?1??0.58?769.74?223.22
22Mc1??0.135=0.09
3?2?769.74?0.15?115.46
Mc2?31???0.135?0.17?=0.153 21??0.15??840.00?769.74??5.2722??0.17?0.135??0.135=0.158 3Mc3??4?432.00?0.17?73.44
Mc4?12?0.17=0.085
９
?5?1?0.17??840.00?432.00??34.68 2Mc52??0.17=0.113
3?iMci148.72???1Mc1??2Mc2??3Mc3??4Mc4??5Mc5??EIEIEIi?15fDz?????fD?f2Dy?f2Dz95.80???fD??3.3?10?4m EIE?210?109Pa
?4I?D(1??4)?0.04D4
6495.80?D?210?109?0.04?3.3?10?44
?D2?7.67?10?2m
2) 根据E点刚度计算轴径，在E点分别沿y、z轴加一单位力,有弯矩图如下:
?1?1?0.58?1204.26?349.24 2 １０