内容发布更新时间 : 2024/12/23 8:43:56星期一 下面是文章的全部内容请认真阅读。
FPOx+ 1.5FPMz15 答:
12A 2qlEIB
ll22+
qA EIB ll22=
12ql2qA EIB ll22资料
FPFP22Ox+ 0.5FP?3?0.5a?Mz1?B??2ql2?12lql3EI??4EI12ql2?l?w2???2??l3ql4B??2EI??B?2??16EI ql3?B?6EIwql4B?8EI
ql3ql3ql3?B??4EI?6EI??12EI3ql4ql4ql4wB??16EI?8EI??16EI
16 答:
A EIB Pl2?B??2EIllP22+ qA EIB
ll22= qA EIB llP2217 答:
(1) 静定基
M0A EIB
ll22+
资料
Pl3wB??3EI
3q??l????2?ql3B?6EI?48EI4q??l??w?2?l7ql4B?8EI??B2?384EI
ql3?Pl2B?48EI?2EI7ql4Pl3wB?384EI?3EI
?l?2M?w?2??Ml00l3M20lB?2EI?2EI2?8EI
A EIB RBl2=
l2RBl3wB??3EI
M0A EIB RBl2(2) 变形协调条件
l23M0l2RBl3l2?9M0?8RBl?wB???8EI3EI24EI
l2?9M0?8RBl??0 wB?24EIRB?9M0(↑) 8l(3) 平衡分析
MARAxRAyRB
?Fx?0RAx?0
9M0(↓) 8lM0(顺时针) 8?Fy?0RAy?RB?0RAy??RB???MA?0MA?M0?RBl?0MA?M0?RBl??18 答:
资料
(1)
??a??FPcos??lFPsin??lbh2?b2h66 ?6lFPh2b2?bcos??hsin??(2)
??a??6lFPh2b2?bcos??hsin???0 ??tan?1bh
19 答:
如图所示,危险点在煤气罐外表面。
?pDm?4??0 ?pDt?2??0 所以
?1??t?pD2? ?pD2??m?4? ?3?0
根据第三强度理论
?1??3??s
pD2???2??0.28s p?sD?2?32?170?2.975MPa
资料
?m?t