彈性力學(xué)吳家龍課后答案khda

1、答案網(wǎng)(http:/www. daanwang. com)O莒案網(wǎng)（原廉后轄案網(wǎng)）最生業(yè)的課匠刃題苞案知享社區(qū)數(shù)材課后答案丨練習(xí)冊(cè)答案I期末考總答采I實(shí)驗(yàn)報(bào)告答案6.了2 =+廠*. +口、嚴(yán) rJ + J阿1120x10 Pab = b / + b tn2 + a n2 + 2r mn + 2r In + 2r Im = 265 x IO5 Pav 1yKx;rv=A/；2-(T； =1090xl05 Pa10.Z, =bj , /IT = fv. =r/iZ = Jf： + 宀 /；： = Jg(b： + b； + CT：)12.a. =a./2 +b、”d +a./r 二一(b. + r

2、% + a邊界條件底面rn =0滿足-q AB = -q答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)斜面B = -p、g4=0n = Zv =設(shè)沿坐標(biāo)軸的方向余弦為a,加，“)I = cos(90 + /?) = sin p. m = cos(l 80 -fl) /i =0Z. =/ + /=0 y=-tan P x將b,a.rn代入邊界條件的表達(dá)式，可解得:心 qtan 0 _ 02-13邊界條件鉛垂面6扁=-Qgy 召 |xO=0斜面A=o4=設(shè)v的方向余弦為(人n )1 = cos pm = -sin p zi =

3、0 = tan pyZ，,=TtZ+rMf、=Tj + b嚴(yán)由上面兩式可解得c = Ctgp (pg -D = pgclg卩-pg，方向(A fn9打)的伸長(zhǎng)率公式:3-9 (1)設(shè)主應(yīng)變?yōu)榉?心將坐林軸選為應(yīng)變主方向。任意方向v的伸長(zhǎng)率為& =+ e2nr + r3/r3 =冷3(廠 +/ +/I2) 6只需將坐標(biāo)軸乙方向定為斫方向.與&垂直的平面內(nèi)取y方向 則坐標(biāo)軸的方向?yàn)閼?yīng)變主方向。與上題同理町證兩“ *當(dāng)V的方向?yàn)閖方向時(shí)取最小值幣，v的方向在.0,平面內(nèi)吋，取最大值E當(dāng) l-2 Sy 時(shí),同理 e2 y 若主應(yīng)變弓=、=則任意方向?yàn)橹鲬?yīng)變方向任意方向 =&答案網(wǎng)(http:/www.

4、 daanwang. com)答案網(wǎng)(http:/www. daanwang. com)az oydz ox=5 6avdudy答案網(wǎng)(http:/www. daanwang. com)3-11= 2Ay + 2Ax,ddy3-12故該應(yīng)變分量不滿足應(yīng)變協(xié)調(diào)方程故該應(yīng)變分量不可能發(fā)生。耳+空av2 d2y2答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)duSv y=dy ” = Tx + /；(y E:.v = -vy + /2Uz)3ir:=dz. w =年于+ /；(“, y) 2E答案網(wǎng)(http:/www. daan

5、wang. com)答案網(wǎng)(http:/www. daanwang. com)&色匹壘 dx dy dx答案網(wǎng)(http:/www. daanwang. com)Z ciy + gl(z)9f2=c2xg2z9 q =-)g嚴(yán) QZ + , D = 一D，述骨-詈+警知。+G+ &2 Sg2 = DZ + E2 D2 = -C3u = 十 x + c+ qz + 厶 E等+警+哆小該應(yīng)變狀態(tài)為平面應(yīng)變狀態(tài)，應(yīng)滿足應(yīng)變協(xié)調(diào)方程dx2內(nèi) 2a 12/ +12坊 +12於 +2 珂+3/C, +C02.C, = 4 A, + d = 2C2；=rx-v(rx+rJ = O& = +,+ S： =7(

6、S + b J卜l-2v q 2(1-v)答案網(wǎng)(http:/www. daanwang. com)v. =+ ayy + a;e. + t亠 + TJy + 匚必)答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)4- 3將應(yīng)力主方向設(shè)為坐標(biāo)軸方向答案網(wǎng)(http:/www. daanwang. com)1-V柱坐標(biāo):b =久& + 2Gg ppb* =久& + 2G 竹a.=人。+ 2Ge. J 二 G.Ev(1+vXl 2)E2(1+ v)由方程(4-12)知人= .= 0 故坐標(biāo)軸方向?yàn)閼?yīng)變主方向I卜各向同性體不具仔這種

7、性質(zhì).極坐標(biāo)廠心_性).=丄(b, - vq,)(平面應(yīng)力)2(1+ v) * =匚.4-4各向同性體.廣義胡克定律的形式不變對(duì)平面應(yīng)變問題，需將&換為總答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)0 = +.+ 二答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)球坐標(biāo):(rr = Z0 + 2Ger= A0 + 2G% b* =朋 + T: = % 咲=GfTr0 = 0丫詢A=竺_(l + vXl-2v)E2(1+ v)答案網(wǎng)(http:/www. d

8、aanwang. com)答案網(wǎng)(http:/www. daanwang. com)53根據(jù)普加原理.山圖示受力情況對(duì)假設(shè)該組應(yīng)力顯然滿足平衡方程 邊界條件：卜 |“=06 |= ?2ryr | v=11f = 可見邊界條件滿足。B （6 + 6）= 0滿足應(yīng)變協(xié)調(diào)條件（上下邊界）（左右邊界）故該組應(yīng)力適合做本問題的解+E答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)dvQvq, + qE V = _ : W y + u)E色+色* =S = 0dy dx - G &(y) = Cy + D /,(x)= -Ca + EE5

9、4由該半無(wú)限體的受力特征，可知物體在水平而內(nèi)應(yīng)力均勻分布，可設(shè)a =/（：）,水平 面的應(yīng)變分星:為o, = cy =y =0水平面在變形后仍為平面，不發(fā)牛翹曲.故yri=rrv=o5 = -v（av + 礦司Ik云丿= vV2V26/ + 2VV2V-V2V答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)V2V2t/=-lv2v對(duì)平面應(yīng)力問題，還應(yīng)滿足協(xié)調(diào)方程答案網(wǎng)(http:/www. daanwang. com)a, i + vd2u dx2dy2d2 d1du &u、 dy? +活 +V2V=-v:(V:/+2V)AV

10、2V2t/=（v-l）V2v6- 4根據(jù)材料力學(xué)的解答MvF沖= T- 2hli（i）3F現(xiàn)在來(lái)校核該解答足否滿足彈性力學(xué)基木方程碎+龍?bào)谩?? 2h h1m+b/0滿足平衡微分方程+b, ) = 0 滿足 L6vy 方程邊界條件=0上面一.二兩式顯然滿足3F2A 3h2I)卜3 b d crv /礦+訂-2云亍冷匕+6)答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)（2）根據(jù)材料力學(xué)的解答3qx仁 4v2 n12答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. co

11、m)故不滿足平衡微分方答案網(wǎng)(http:/www. daanwang. com)為了滿足平衡方程，需給6加上一項(xiàng)g(y)。S =O + g(y)2/3/r答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)再看該應(yīng)力解是否滿足應(yīng)力協(xié)調(diào)方程6爐&2卅)hy丿24 嚴(yán)。故不滿足協(xié)調(diào)條件 試在右端加一項(xiàng)/(y)護(hù)，y + f(y)為使應(yīng)力滿足協(xié)調(diào)條件需有+6) = 0邊界條件-q可得rh =0 滿足4 + y2 + Ey * = Eh =0/. E = 04q y5 _ y/? 536q滬C+滬、評(píng)4/)2h麗-八6- 5應(yīng)力由重力產(chǎn)生

12、，與朋成止比，應(yīng)力分是應(yīng)具有下面的形式pgN(a兒 y)n 43丄工丄C 2. D 3u =x +x +xy + y6 2 2 丿 6Wt/=O, (/滿足雙調(diào)和函數(shù)答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)d2U=Cx + Dyy2u=Ax + By答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)J = _ Pgx = _Bx _ Cy _ pgxdxdy邊界條件：/(x),取應(yīng)力函數(shù)為U(x9 y) =- = yf (x).字=-f(x)y2 + g

13、(x) U = l/(x)y2 + g(x)y + 傾y)Oydy 26&(p6 at4駕v +空+ 2空gav4 a? at2 /. f(x) = CjX3 + C2x2 + Cyx + C4(p(x) = D,x + D2xgM-邑+三疋+魚尸+ E3x10 6 6 2一三十 +b +x2+EvJy + D/+D2x2 I 10662 丿答案網(wǎng)(http:/www. daanwang. com).-.crt=(c/+cy+C + C4)y 唁骼 rx =Fxy - Fyxo = (6C| a* + 2C )y (2C. 2C.t + Etx + )y + 6Z)rv + 2D、答案網(wǎng)(ht

14、tp:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)q. =(3C|X PE 仏尺疋| p&h +2C2x + C3)y2 +232 丿答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)6 h*、gh4V將q代入以上邊界條件可解得:竺-丄3 口 g23加、F-帀T2h26 =r = 6 Ax v-l 24w + 6Dxvx dy2&uav =r = 6 Aw、+ 6Cxy + 6.vQu dx_ (9Ar2 v2 + 3Ay4 + 3Cx2 +3Dv2 + F) dxdv 9答

15、案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)邊界條件:4=A=wzC=d4仿q(、LT答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)4h 80/6-9取應(yīng)力函數(shù)為U+Bxy+Cxy3V2V26/ = 0滿足協(xié)調(diào)條件 d2U必人“t, = -rr = 6 Ay + oCxy篇 530)應(yīng)力邊界條件“dy = F. j；rJIwOy2dV于(Fx一 Sy 2昭2d2刊a答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. da

16、anwang. com)dzud2u= =ad2U匸匸(a)、(b)即為(6-13)式。故(6-13), (6-14)nJ*得出柿同的應(yīng)力解答。只需在應(yīng)力曲數(shù)相應(yīng)地増減(審+曲)項(xiàng)_L+2-!1q1 v p 1 V答案網(wǎng)(http:/www. daanwang. com)x _ ab(q、-qj廠 _ qxa2 - q2b2、；c =,/r 一(T2廣 一 tr)a(l-v4 2 *)ua =Eb2 +a2 vb( - v2)2alA/ = uh 一叫應(yīng)力邊界條件位移邊界條件Aav+2Cp */r答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanw

17、ang. com)邊界條件:心0令J走可得:E9答案網(wǎng)(http:/www. daanwang. com)(1 - 5(A-k)a +(R(l-2v) + l)b1 (l-k)gd2 (1-R)/+(R(l_2v) + l)Fl + (l_2v)Rp_(l_R)l21 + (1_2 嘆/ (I ；:)L2l+(l-2v)-?-(l-)(T =ab2Hl + (l-2vUp-(l-Ar)7- 6假設(shè)在離圓孔中心距離為b的地方，應(yīng)力分布已經(jīng)和沒有圓孔的情況完全一樣，建立 極坐標(biāo)系將應(yīng)力進(jìn)行分解答案網(wǎng)(http:/www. daanwang. com)= -rsina =rsina % = rcos

18、 20可設(shè)應(yīng)力函數(shù)U = /(Q)sin2&,j空+丄藝一篤L。dP Pg P )P pdp p ),C/. f (p) = Ap + Bp + r + DP邊界條件解得:I dll 1 d2U Frp dp p d(p:d2U“2從竽+ siWP P)=(2A + 12B” +作j1 du papsin 2 卩)=(2 A + 6Bp2-馬 sin 20Ja 6C 4D 一宀+h韻=02A6Ba2- - = O 一 2答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)2(b: -a2)(b2a2)2= (1 +詈jrsin2

19、卩 j2_Z+7八cClf (b4 4-a4 + aF )W 一小令一=0 貝 1得人=2. B = 0t C = t D = ar b22當(dāng) q = “ 0 = 或 0 =兀 土 h寸.有bgx=4r44 w /nvM7-7設(shè)尖劈內(nèi)任一點(diǎn)的應(yīng)力正比于分布枚荷與p, a., p冇關(guān)，應(yīng)力具冇qMpa.（p）的 形式.N具有L的0次量綱。應(yīng)力歯數(shù)U應(yīng)為L(zhǎng)的2次星綱，可設(shè)U “fW、d2U 1 dU 1 d2U=0解得f(p) = C| + C.p + Cy cos2? + C4 sin-., 、一 2Cj + 2C(p - 2C. cos 2卩 一 2C4 sin 2(pp dp p dtp%

20、=竺=2C, + 2C,0 + 2C3 cos 2(p + 2C4 sin 2psin a二 q2sina答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)a = (/cot a +cos 2(p*sin a=7+2C7- 9假設(shè)在離圓孔中心距離為b的地應(yīng)力分布已經(jīng)和沒有圓孔的情況完全一樣。 吹令2Cp 0邊界條件：解得：A= !b*若令0則A = -a2 h5 二qa02h2(17-10分析該板的受力恬況，可假定惻環(huán)在變形后仍保持為同心圜.同心圜之間不發(fā)生擠壓,故b。在同心圜環(huán)上應(yīng)力均勻分布竺 =0。d（p:.U =CP +

21、Cw普。dp”二川必+從卩)Z(p) = ci fi (p) = cii du 1 d2 1+ :rP dp 如-)=0 S = o 0應(yīng)力邊界條件tp-q C-lrqb1dp2dui /、存邊界條件：“丄產(chǎn)。/. /()=0up = 0件二 g（Q）答案網(wǎng)(http:/www. daanwang. com)r2(1+ v) b2=(IPTFrE P丄_/3 _ _、T. = aG 寸=My(x-a) dv a%45* “(、rM-x2 +Tk a dy 2a2一 ax + y37=-j4y2(x-a)2 +(-x2 +|ar + /)2當(dāng) y = OHj. r取最大值 仏=15“M/(2/)

22、叩 些二丄蟲丄丄理 p dp ZG dp1 卩 ZG p dp F3i(pdv _ 1 d1dzdp G dpdz.dz ZGdz2 卩 ZGdp1a = k0+ f，b = Ev/(/v)(l 2v)p ep答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)z dpdz.竺L +竺L +冬二=邂+ 2空、dp dz p 、+ ke + pd000 t 1 叭ap茁 dz2 P P)p6p1 一 V=01 - 2v理+吆+2(業(yè)p dpydpdzdp dz+鳳丄業(yè)+空 dz p dpdz dp dz=0應(yīng)力分屋滿足軸對(duì)稱問題的平

23、衡微分力程(10-*)a 210-2 體力為零時(shí) (10-13)可化為一占+二二f 一 = 0dr1 r dr rdir 2 du, 2ur _ 1d02d+ 一二看苕節(jié)怎2d*2 d0drdr故方程(io 13滿足10-3 叫込Ev_ 1 cl1/( 2d$ZG drdr2 r dr=0dur2u,門=0 = cdrrcD+3r3C iEC 2D)1 +V=0(1 + v)(l-2v)CDur = r + 3r2答案網(wǎng)(http:/www. daanwang. com)邊界條件可解得c=卅一加r_ E 一 (1 +心1-2-)2 Ev2(1+ v)答案網(wǎng)(http:/www. daanwan

24、g. com)答案網(wǎng)(http:/www. daanwang. com)卜討卅“）TWqb -q(么 -宀滬2(R-戻)嚴(yán)Aio-5取拉梅應(yīng)變勢(shì)中=+ y1ZG= - + 2-B,a,= 丫 / l-2v十22邊界條件：（空）嚴(yán)=0可解得A=薄、B =4b小 1+vr+ 2a3 l-2v4b，I 2 1 + a3 * l-2vU r = ZGqb1+v2戻卜P+l-2vi qyy 2b1 +口丿(J =(J2b Hva3 1 -2v1 +vql-2v2b3 Hvay l-2vqlr11 + v2b1 + v 才 1 - 2va3 1 2v2P & h = (S )ay 1 -2v10-6應(yīng)力分

25、布軸對(duì)稱，0 = 0（Q）觀察（7-13）的解.設(shè)拉梅應(yīng)變勢(shì)e = Al“p+Bp，丄理p ig dp12G2Bp、w =PI2Hdz=0答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)I26+ 2B10-7/?(=/?2= R、由(10-5)S+4B(10-6)十瓦v A_，4B-7 + 2設(shè)在（占斗2）#總圭仏代入邊界條件,bp(p.a q、bpb = 一佈解得A二航cr - If kybq 一” M-F) j =叫- 4(竹 _ q J A p cr 一 b cr 一pR； = R；=O-,A + B=,A-B=-0 t

26、RR不能采用書上算法l!-2 (a)6啦(741叭卜=011 -6平面應(yīng)力問題平衡方程理+吒=0dv dy答案網(wǎng)(http:/www. daanwang. com)答案網(wǎng)(http:/www. daanwang. com)本構(gòu)方程EEaT Eg、 + 1 - v 1 - vEaT E、1 v 1 - v22(1 + %答案網(wǎng)(http:/www. daanwang. com)得位移表示的平衡方程E da dT Ev d2u ；+ ； -1 - v* 1 - v dx 1一芮 dxdy 2(1 +v)r=0答案網(wǎng)(http:/www. daanwang. com)(d2u d2A Ev d2u

27、Ea dT E d2u 門+麗丿+L碩廠口礦 L喬“假設(shè)存在勢(shì)函數(shù)0（x,v）,使M二學(xué)+二學(xué)，代入上式OXOV滿足式恥但不一定滿足邊界條件）易證中滿足應(yīng)變協(xié)調(diào)方程等+等二孟i + vdx2=0ddv=0i ayi+vav2若黑+礬八式即恒等w(b)e ay Edr Ev1 - v2 dx2 1 - v 1 - v2 dy 2峯竿+各票1 -v dv* 1 -v 1 一芮 ” e d2燈(1 + v) Oxdy由 11-3 可知b： =ri =0.空二土刃一丄空 dx vv dr代入式害T將。Wr/V卜e2呎-REv為了滿足邊界條件，再構(gòu)造一組齊次方稈解v*6 = X + a： = -EaT() + EaT込=a； + b； = 0,心=(V. a1 =3 a/5,. a = ?同理 v = i時(shí)，a = ? c, =? c, =?c3=?141 F角 d” + Jids = JbQ&jdvb“&”dv = b”+ S)dv = Jb“也 jdu=J(6內(nèi))妙- j,hJ&i,dv(竹+6丿也加+ (/ -b,.)牝d$=0由于血，