工程力学 (3)

1、12 轴向拉压弯 曲 扭 转3q4内力内力 右截面正向右截面正向 左截面正向左截面正向 微段变形（正）微段变形（正）yxx顺时针错动顺时针错动FS向上凹向上凹M56解：解：1）求约束反力：）求约束反力： 受力如图。受力如图。2) 截面法求内力截面法求内力( 取坐标如图取坐标如图) 0 xa:aaaA B 3F 45 yx0F 2MMFSxFFN0FN=0; FS=- -F; M=- -Fx7a x2a: 2a x3a: 2) 截面法求内力截面法求内力0 xa: FN=0; FS=- -F; M=- -FxaaaA B 3F 45 yx0F MMFSxFFN0MMFSxFFN03FFMMFSxF

2、FN0 3F F3FFN=- -F；FS=3F- -F=2FM=3F(x-a)- -Fx=F(2x-3a)FN=- -F；FS=3F- -F- -3F=- -F M=-Fx+3F(x-a) -3F(x-2a) =F(3a-x)8a x2a: FN=- -F； FS=2F M=F(2x-3a)0 xa: FN=0; FS=- -F; M=- -F x 2a x = =maxLEIbLFabLEIbLFbzzA6)(6)(22+ += =- -= =q qLEIaLFabLEIbLLbLFbzzB6)(6)(33222+ += =- - - -= =q q61LEIbLFabzA6)( +-=qL

3、EIaLFabzB6)( +=q是否在是否在0 x a)3(62221LbxLEIFbz-+=q)()(33622222LbbaxLxLEIFbz-+-=qax0Lxa 3/ )(221bLx- -= =62)(62221LbxLEIFbxyz-+=-+=bxLaxLbxLEIFbxyz32222)()(6ax0Lxa )3(36222222maxLbbLbLLEIFbyz-+-=)(23182222LbbLLEIFbz-=zLEIbLFb39)(322-=63LEIbLFabzA6)( + +- -= =q qLEIaLFabzB6)(+ += =q q3/ )(221bLx- -= =zL

4、EIbLFby39)(322max- -= =zABEIFL162= = =q qq q64CBAqaaaD651( )( )dzzEIEI y xM xxCq=+12( )d dzEI yM xx xC xC=+( )SEIyF x =( )EIyq x =( )( )zEI y xM x=66l/4xyBAFlCFAl/2l/4FD(a)xyBAFCFA1FB1l/2(b)l/4xyBAFA2FB2FD(c)67l/4xyBAFlCFAl/2l/4FDLEIbLFabzA6)( + +- -= =q qLEIaLFabzB6)(+ += =q q)(6222LbxLEIFbxyz- -+

5、+= =ax 0)4/3(6222/LbLEIFbL/2 yzLx- -= = =68LEIbLFabzA6)( + +- -= =q qLEIaLFabzB6)(+ += =q qxyBAFCl/2l- - -4/3(6222/LbLEIFbL/2 yzLx- -= = =)69LEIbLFabzA6)( + +- -= =q qLEIaLFabzB6)(+ += =q qxyBAFD3l/4l- - -4/3(6222/LbLEIFbL/2 yzLx- -= = =)70 xyBAFCl/2lxyBAFD3l/4ll/4xyBAFCl/2l/4FDl- - - -71；maxyymaxqq

6、72LAxBymm5 .82m1025. 8101202104532363=-b326max322s sMbbhWbhz = = = =73zBEIqLy84-=25084maxLyEIqLyyzB=EqLbbbbhIz82503212)2(123433=mm2 .89m1092. 8210200825031010324933=-bLAxBy74maxmaxss=zWM),(maxmaxyLfEIMyz=75 = =AzdAyI2maxsmaxs76变形变形体静体静力学力学问题问题研究对象研究对象受力图受力图平衡方程平衡方程求反力？求反力？静不定静不定物理物理方程方程几何几何方程方程静定静定求求内力内力应力应力求求变变形形物物理理求求位位移移几几何何联立求解联立求解反力、内反力、内力、应力、力、应力、变形、位变形、位移等移等弹塑性问题弹塑性问题 物理方程不同物理方程不同77Bxy78Bxy7980BxylAxBy81xyBAL3L/