(b)
(c)
基本结构为: A
q l
C EI=常数
B D l q
A 4m 60kN
3m C EI=常数
B 5m
4m D 9 Fp49 Fp49 Fp49 Fp49 Fp2 M图 整体结构M图
解:根据对称性,考虑1/4结构:
q
q X1 1
q2l 8 1 M1 Mp
?11?1?l?l?1??1?2??EI?2?EI
1?1lql2lql2?ql2??1p????1???1????EI?32828?12EI
?11X1??1pql2?0?X1??12
M?M1X1?Mp
ql2ql2ql2 242424
ql 122ql2 12ql2 12 (d)
解:取1/4结构: q
基本结构为: q
X2 l
A B q l
D E q
ql2ql2 M 2424F
EI=常数
C l X1 1 q
1 q2l2l 1
2l2
q2
M1 1 M2 Mp
?1?l22?l311?EI???2?l?3????3EI ??1??1l2212?EI?2?l?1?????2EI
?22?1?EI?l?2?1?1?l?1?1????3l2EI ?1?1ql23?ql41p??EI???3?l?2?l?4?????8EI ?1?1ql2?ql32p?EI???3?l?2?1????6EI ??l3l2?X?ql4?5?3EI12EIX2?8EI?0??X1?ql?23??12 ???l2EIX1?3l2EIX2?ql6EI?0?1??X2?36ql2ql2 ql2 9 ql2936 ql236 ql2 ql 2 36 9 ql236 ql2 362(e) ql 50kN 9
E 2I F I I mql26 9 C 2I D I I m6 A B 9m
M
(f)
4FP G H I 2a D E F 2a
A a 取1/2结构:
B a a a C ( BEH杆弯曲刚度为2EI,其余各杆为EI )
2FpFp FFp
Fp FFp F = +
2FpFp Fp Fp FFp F ① ② ②中弯矩为0。
考虑①:反对称荷载作用下,取半结构如下:
Fp
Fp2 2Fp2 Fp2 = + Fp
Fp 2 ④ 2 ④中无弯矩。 F ③Fp2 FpFFp FpFp2 考虑③:
2 Fp2aFp2 Fp2
Fp2Fp ? FpFp2Fp2Fp22aa 弯矩图如下:
aa Fp2Fp2Fp2a 2a pF 2 a Fpa a pFpa 2aFFp2a
pFaFp2 2Fp2Fpa aFp2Fp2aaa
aFp2Fp2a Fp2aa
相关推荐: