1、 由此得,還可知,T=0可見(jiàn)，情況與題(a)及(b)相同，故結(jié)論亦同。A-11試計(jì)算圖a所示矩形截面對(duì)AA軸的慣性矩。題AJ1圖解：選坐標(biāo)系如圖b所示，得沁=8/-3/亠沁Li曲3Iy:=0+(2a-3a)(-a)=-9a4S1X10=f-,V13cos&=13從而有13根據(jù)轉(zhuǎn)軸公式,人+.I、22二cos20+sin2&將相關(guān)表達(dá)式代入上式，得A-12圖示矩形截面，試確定A點(diǎn)的主軸方位及截面對(duì)該主軸的慣性矩。題A-12圖解：坐標(biāo)取如圖A-12,并設(shè)邊長(zhǎng)Z?=20mm,h=30nun,于是有/=+/?/?(-)2=(丄x0030 x0020+0020 x0030 x00107)m4=gOOxl

2、OYn122742/.=+/?/(-)2=(x0.020 x0.0303+0.020 x0.030 x0.0152)m4=1.800 xl0-7m12212/v.=(Zj/1)(|)(_)=-2L=_lx(0.020)2x(0.030)2m4=-9.00 xl0_sm4圖A-12依據(jù)主軸方位與主慣性矩公式，得2a=arctan一=60.945,a=-30.471.800 x10&00 xl(psr8.00 xWs+1.800 x108.00 xWs-1.800 x107,”。川。、22(900 xl0Y)sin(60945)m4=270 xl0Ym22(900 xl0Y)sin(60945)m

3、4=270 xl0Ym4=270 xl04mm48.00 xl0-s+1.800 xl0-78.00 xl0-s-1.800 xl0-cos(_60.94y)+22(-9.00 xl0s)sm(-60.945)m4=233xl07m4=233xl0imi4A-13試求圖示各截面的主形心軸位置及主形心慣性矩。題AJ3圖(a)解：坐標(biāo)示如圖A3a,C為截面形心。80 x10 x404-50 x10 x5_3m_2654xl0.3m80 x10+50 x1070 x10 x5+60 x10 x30 x心1A-3乙=()x10m=16.MxlOmc70 x10+60 x10/.-10 x83+80 x

4、10 x(4026.54)2+50 x134-50 x10 x(26.545)，jxl0-i2m4=g078x07m410 x603.70 xl03人一+60 x10 x(3016.54)-+70 xl0 x(16.54-5)2xl0-12m4=3.878xl0-7m4/、.產(chǎn)80 x10 x(26.5440)(516.54)+50 x10 x(26.545)x(10+516.54)xl0-12m4-3.23lxlO-7m4由2x3.231xW7(“ctan2a=1.539(&078-3.878)x10-7得2&=56.98,a=2&49心28.5最后得到(3.878+8.078)xl0-7亠

5、(3.878-&078)xl0-7古鈣og01,2_2E*+3.231xl0_7sin56.98m4Z-=2.13xl0_7m4=2.13xl05nim4,/3=9.83xl0_7m4=9.83xl05nim解：坐標(biāo)示如圖A3b,有圖A-13bj0.030 x0.0404“介a-74L=m=1.600 x10m12z,0.040 x0.0103,0.040 x0.020,1AA/1AnAOAI、.=+1+x0.040 x0.020 x12362(0.005+0020)2x2m4-1.300 x10-7m4x0.040 x0.020 x(0.0200040)(0.005+0020)x2m4=-6.22xl0_sm4由(込=2x(-622x10-)1.600 x10-7-1.300 x107得2=-76.44,a=-38.22-38.2從而得61_1.300 x107+1.600 x10-71.300 x10-7-1.600 x10-7飛羽。)+