4-13 求例3-5中的单位长度扭转角。已知G=80Gpa。 解：
已知：h=100mm，b=45mm，T=2kN·m，G=80GPa；
??MxG?hb3其中，?和h/b有关h/b?2.2，插值h/b?2.0，??0.229，h/b?2.5，??0.2492.2?2.0??0.229?（0.249?0.229）?0.2372.5?2.02?103?20???1.157?10rad/m?0.66/m9380?10?0.237?0.1?0.045
4-14 用积分法求图4-22所示各梁的挠曲线方程和转角方程，并求最大挠度和转角。各梁EI均为常数。 （a）解：
由挠曲线方程：
MABxM(x)??MEIZ???M(x)dx?C???Mdx?C??Mx?C当x?0，?A?0，故C?0故：???MxEIZM-?max??B??MlEIZ11EIw??Mx2?Cx?D??Mx2?D22当x?0时，wA?0，所以D?0Mx2?w??2EI
（b）解：
qwmaxMl2?wB??2EIM(x)?q(xl?x2)2q/212ql8q/2qEIZ???M(x)dx?C??(xl?x2)dx?C2q111?x2l?x3q?C?qlx2?qx3?C464611EIZw?qlx3?qx4?Cx?D1224当x?0时，w?0,?D?0lql3当x?时，??0，从而，C??2245ql4ql3则,wmax?wl???max???A??B?384EI24EI2Z
4-15 用叠加法求图4-23所示梁的?C及wB。设EI均为已知常数。 （a）解：
求：?C,wB.（一）求?CM?ql2qc?C??C1??C2??C3??C4??C1(B')??C2(B')??C3(B')??C4(C')ql30.5ql2lql3ql3????(?)?2EIEI6EIEIql3??6EIBM?ql2(1)(B')qql（二）求wBwB?wB1?wB2?wB3?wB4(ql)l3?(0.5ql2)l2??ql2l2?1ql4?????)???????(?3EI2EI??2EI?8EI?45ql4?1111?ql?????????24EI?3428?EI
4-15(b)解
(2)(B')+12ql2(3)(B')qw'(4)（一）?CFlBFC?C??C1??C2FlFl2l3Fl????2EIEI2EI22(1)Fl（二）求 wBFl3Ml2wB?wB1?wB2???3EI2EIFl3Fl3Fl3????3EI2EI6EI
C(2)
4-16 用叠加法求图4-24所示梁的最大挠度和最大转角。 4-16(a) 解
（一）最大转角2EIACFCCEIFBFlFBFl?B??max??1FB??1B??2BFl2Fl?lFl2111Fl25Fl2?????(???)??2(2EI)2EI2EI422EI4EI（二）最大挠度FlFFlwB?w1FC?w1C??1C?l??1C?l?w2BFl3Fl?l2Fl2(Fl)lFl3?????l??l?3(2EI)2(2EI)2(2EI)2EI3EI11111Fl33Fl3?(?????)??64423EI2EI
4-16（b）解
qADlCl/2qwB1θθB2B1(1)?BBwB2θB3lq()3ql32?B1????6EI48EIql3?B2?24EI12ql?lql38?B3????3EI24EIwB3?B??B1??B2??B3(2)wBql3??48EIll??B3?22wB?wB1?wB2?wB3?wB1??B1?lq()4ql42wB1????8EI128lql3lql4wB2??B2????224EI248EIlql3lql4wB3??B3??????224EI248EIql4故:wB??128(3)经过验算，wD?wB?A,?Ｃ??B?max??B,wmax?wB
4-16（c）解
qAB(1)求?Ad?A??(qdx)x(l?x)(l?l?x)q??x(l?x)(2l?x)dxdxx(3/8)ql-(1/8)ql+
6EIl6EIlll???q26EIl?x(l?x)(2l?x)dx??q?6EIl??l2x?lx3?14x4?2A0??0q93??43ql6EIl364l??128EI(2)求?Bd?(qdx)x(l?x)(l?x)B?6EIl?q6EIlx(l?x)(l?x)dxl?q23q7347qlB??206EIl(lx?x)dx?6EIl64l?384EI(3)wmax根据Q,M图，wmax?wC?3qx2qlxlM???x??82x?[0,2]?1??8qlxx?[l2,l]M[0，lmax位于2]??11312xEI?M(x)dx?C1?EI?(8qlx?2qx)dx?C1?1EI(316qlx2?16qx3)?Cl1x?[0,2]x?0时，?3ql33ql3当0??128EI因此C1??128EI所以?1EI(316qlx?16qx3)?3ql32x?128EI1313wEI48qlx343qlx?(?24ql)?128EIx则w??5ql4l2768EI当?x?0时，wx?wmax即：x?0.46l代入：w?5.04ql4max?768
4-16(d)
?max???A??c qql ql3ql35ql3A?A??A(q)??A(ql)?????C 24EI16EI48EIB wmax?wBl/2l/2 5ql4ql413ql4wB?wB(q)?wB(ql)????? 题4-16(d)图
38448EI384
4-16(e)
q ql13?max??B?? 6EIACBl2 l1ql14ql13wmax?wC?wB??B?l2???l2
8EI6EI 题4-16(e)图
ql13??(3l1?4l2)
24EI
4-16(f)
34 qlqlq?1??w1?? 6EI8EID A3q3ql33ql4CB3?2?(2l)?w2? 128EI16EI16EI2ll 12qql?2l ql3ql42?3????w3???w 13EI3EI3EIq
5ql3w2??max??D??1??2??3??
163EI
(1/2)ql213ql4 wmax?wD?w1?w2?w3???3w348EI
题4-16(f)图
4-17 工字形截面Ⅰ的20b简支梁受载如图4-25所示，E?200GPa，求最大挠度。 解：
10kNIx?2500cm4?2500?104mm44kN/m
E?200GPa AB435qlFl Cwc??? 38448EI3m3m 5?(4?103/103)?(6?103)4?? 384?200?103?2500?104 题4-17图 10?103?(6?103)3?
48?200?103?2500?104
12?22.5mm