梁橫向彎曲振動(dòng)的振型正交性及振型疊加法

1、梁橫向彎曲振動(dòng)的振型正交性 TOC o 1-5 h z d2U (尤,t)d2d2U (X, t)_./ 、m+ EI ( x)=p ( x, t)dt 23x8x2/齊次方程為:Su ( x, t) 828 2u ( x, t)m+ EI ( x) = 08t 28x 28x 28 2U ( X, t8 2U ( X, t)828t 28x 2EI (X)8 2U ( X, t)8x 2根據(jù)分離變量法，設(shè)：U(X, t) =4 (x)0(t)，可得:q(t) + w 2q(t) = 0d2d 2(x)iEI ( x) = mw 29 ( x)dx2dx2上式即為分析頻率和振型的特征方程。設(shè)對(duì)

2、于第，、j兩階頻率，有：d 2d 2(x)id EI (x) d-i = mw 2, (x)d 2d 2, ( x).d EI (x) j 弋=mw 29. (x)上面第一式兩邊乘以9 .(x)，并沿桿長積分得：j9. (x) EI ( x)，=mw.29, (x)9 . ( x)J, 9 (X匹EI (X)d 29 (X) dx = Jz mw 29 (x)9 (x)dx = w2 Jz m9 (x)9 (x)dx 0 jdx 2dx 20 i i ji 0 i jW (x)企EI (x) dW (x)企EI (x) d2(x)dxo jdx 2dx 2=JW (x)d(dEI(x)d)o

3、j dxdx 2=* (x)d EI (x)絆也1 _Jg EI (x)絆坦網(wǎng)(x)j dxdx 2o dxdx 2jo入，、d、d2 (x) 1 dd2 (x) d (x)=8 (x)EI(x)i -J EI(x)i一jdxj dxdx 2o dxdx 2dx_ . z xx d28 (x) 1Ld8 (x)力 d28 (x)= 8 (x)_;-EI(x)了 J jdEI(x) 了j dx dx 2 o dxdx 20、d 5、d28 (x)】1d8 (x)d28 (x)】1=8 3)丁 EI (x) -i - EI (x) -ij dx dx 2 dxdx 2+ J 舊(x) 4 dd d

4、x 2dx=8 (x)d EI (x) ) 1j dx dx 20d8 .、d28 (x) 1 Jd28 (x) d28 .j EI (x) i + J EI (x) ij dxdxdx 2odx 2dx 20對(duì)于基本邊界條件，有:8 (x)d EI (x)寫回j dxdx 21=o0迎U EI (x)地)dxdx2i=o則有:J18d 2 Ed 28 (x)dx /d 28 (x) d 28,(x) dx m 2J1 88dx8 (x)EI (x)idx EI (x)ijux m 2 J m8 (x)8 (x)dx0 jdx 2dx 20dx 2dx 2i 0 i jd d 28 d 28

5、j5(x)87 (x)dxJ18 (x)企EI(x) d瑋jdx = J1 EI(x) i、j idx 2dx 20 dx 2dx 2兩式相減得到：( 2 一 2)Jlm8 (x)8,(x)dx = 0當(dāng)或2時(shí)，有:J 1m8 (x)8 (x)dx = 0i j令： J 1m8 (x)8 (x)dx = M為振型i對(duì)應(yīng)的廣義質(zhì)量。0 i ii上面為振型關(guān)于質(zhì)量的正交性代入有：ld 2d 2巾(x)/ d 2d 2小(x)ld 2巾(x)d 2巾(x)9 (x)EI (x)idx = J 9 (x)EI (x)jdx = J EI (x)ijdx = 0o j dx 2dx 20 i dx 2d

6、x 20dx 2dx 2d 2d 26 (x)d 26 (x) d 26 (x),令：Jl9 (x)廠EI(x)T)dx = J，EI(x)-y_)_)dx = K 為振型i 對(duì)應(yīng)的廣義 0 idx 2dx 20dx 2dx 2i剛度。上面為振型關(guān)于剛度的正交性梁橫向彎曲振動(dòng)的振型疊加法將u(x, t)=工9 (x)q (t)代入微分方程得： i ii=1工m9 (x)q (t) + 工牛EI(x) d氣q (t) = p(x, t) TOC o 1-5 h z i idx 2dx 2i=1i=1上式兩邊同乘以9 (x)，并沿桿長積分得： jl9 (x)工m9 (x)q (t)dx + jl9

7、 (x)E-EI(x)必巾i(x) q (t)dx = l9 (x) p (x, t )dx0 ji i0 jdx2dx2i0 j、i=ii=1工 jl9 (x)m9 (x)q (t)dx + 工 jl9 (x)-EI(x)(x)q (t)dx = l9 (x)p(x,t)dx0 j i i0 jdx 2dx 2i 0 ji=1i=1根據(jù)振型的正交性有：M q (t) + K q (t) = P (t) = j l9 (x)p(x, t)dx， j = 1,2,.j j j j j 0 j得到無窮多個(gè)單自由度系統(tǒng)的運(yùn)動(dòng)方程。對(duì)于初始位移條件：u (x,0)=工 9 (x)q (0)i = 1上式兩邊同乘以m9 (x)，并沿桿長積分得：jjlm甲(x)u(x,0) dx = jlm