工程力学期末复习题-答案

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

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