北工大結(jié)構(gòu)力學(xué)本科教學(xué)課件張延慶版

1、11.1 概 述11.2 用靜力法確定臨界荷載11.3 用能量法確定臨界荷載11.5彈性支撐壓桿的方程11.7組合壓桿的結(jié)構(gòu)的計(jì)算一、第一類問(wèn)題(分支點(diǎn)失穩(wěn))p 2 EIP=Pcr-臨界荷載l 2P < P平衡隨遇平衡lEIcrP = PcrP > P不平衡crPPq-第一類問(wèn)題兩種平衡狀態(tài):軸心受壓和彎曲、壓縮。完善體系不平衡狀態(tài)在任意微小外界擾動(dòng)下失去穩(wěn)定性稱為失穩(wěn)(屈曲).11.1概 述-兩類問(wèn)題PP二、第二類問(wèn)題(極值點(diǎn)失穩(wěn))第二類問(wèn)題三、分析大撓度理論。小撓度理論。偏心受壓有初曲率靜力法能量法四、度計(jì)算中,一個(gè)體系產(chǎn)生彈性變形時(shí),確定其變形狀態(tài)所需的在幾何參數(shù)的數(shù)目,稱為

2、P度。PPEI = ?EI = ?j無(wú)限度1個(gè)度2個(gè)度非完善體系一、一個(gè)度體系?M AP= 0kj ?j - Pl sin j = 0小撓度、小位移情況下：EI = ?lsin j=&jjAkj(k- Pl )j = 0jj ? 0kjjkj - Pl = 0-kj1方程（特征方程）= kj抗轉(zhuǎn)彈簧Pcr/ l-臨界荷載11.2用靜力法確定臨界荷載Pk二、N（以2?M B?M A度體系ky1y1EI = ?度體系為例）ky1 ? l + P( y2lkBky- y1 ) = 0= 0= 02lyky2 ? l + ky1 ? 2l - Py1 = 02A(kl - P) y1 + Py

3、2 = 0 (2lk - P) y1 + kly2 = 0P1kl - P2kl - PPkl= 0-方程1.618kl(kl - P) - P(2lk - P) = 0= 0.382klPcry2- 3klP + k 2l 2 = 0-臨界荷載P2P = 3 ?5 kl = ?2.618kl= -1.618?-失穩(wěn)形式2?0.382kly1x三、無(wú)限度體系PPQ彈性曲線的微分方程為QEIy?(x) = -M ( x)M = py - Q(l - x)EIy?(x) = -Py + Q(l - x)EIlyxMy得PQA + Q l = 0PQy? (x) +y =(l - x)或P EIEI

4、Bn -= 0P=n2令EIA cos nl + B sin nl = 0Qy? ( x) + n2 y = n2(l - x)P通解為y(x) = A cos nx + B sin nx + Q (l - x)10cos nl0nsin nll-1 = 00方程P- nl cos nl + sin nl = 0tan nl = nl由邊界條件y(0) = 0, y?(0) = 0, y(l ) = 0xy(nl) = nly(nl) = tan nlyPPQQEIlyxMnlp23p5py得A + Q l = 022PQBn -= 0PA cos nl + B sin nl = 010cos

5、 nl0nsin nllnl = 4.493tan nl = 4.485經(jīng)試算-1 = 00方程P= n2EIcr- nl cos nl + sin nl = 0tan nl = nl= ( 4.493)2 EI = 20.19EI / l 2lPP= 3EIkjEIlkjkjEIl 1k = 3EI練習(xí):簡(jiǎn)化成具有彈簧支座的壓桿Pl 3kPPPEA = ?= 6EIEIkljEIEIEIlkjEIEIll11.3具有彈性支座壓桿的撓曲線的微分方程為EIy?(x) = -M ( x)M = Py - Q(l - x)EIy?(x) = -Py + Q(l - x)PPQQIlxMkjy?Mn2

6、Ql = k j= 0PjAk?jj=令方程EIk jj10cos nl0nsin nlkj/ Py? ( x) + n2 y =通解為(l - x)EI ? l- (kj / Pl +1) = 0k jj0y(x) = A cos nx + B sin nx +(l - x)y(0) = 0, y?(0) = jPlnl1 + EI (nl)2, y(l ) = 0邊界條件tan nl =A + kj j = 0P kjkj l解方程可得nl的最小正根+1)j = 0Bn - (P= n2EIPlcrA cos nl + B sin nl = 0EyA若 kj = 0tan nl = 0si

7、n nl = 0nl = pPPPQQIllEIMxp 2 EIky=Pcrjl 2kj ?jPkj = ?tan nl = nl= 20.19EI / l方程若10cos nl0nsin nlkj/ PEIl- (k/ Pl +1) = 02Pcrj0nl1 + EI (nl)2tan nl =kj l解方程可得nl的最小正根P= n2EIcrEyA若 kj = 0tan nl = 0sin nl = 0nl = pPPnl tan nl = kj lEIllEIEIp 2 EI=Pcrkjl 2P若 kj = ?tan nl = nlPEIl= 20.19EI / l2PcrEIEI (nl)3tan nl = nl -kl 3例:求圖示剛的臨界荷載.PPPPPPll正對(duì)稱失穩(wěn)稱失穩(wěn)正對(duì)稱失穩(wěn)時(shí)Pkjnl1 + EI (nl)2kjtan nl =1kj lnl= 2EI= 4EI / lk=jPl / 21 + (nl )2 / 4P= n2 EI = 14.67 EI / l 2nl = 3.83crI1 = 2III例:求圖示剛的臨界荷載.PPPPPPll正對(duì)稱失穩(wěn)稱失穩(wěn)稱失穩(wěn)時(shí)Pkjkjk ljnl tan nl = 12EI1=