同濟大學(xué)材料力學(xué)習(xí)題解答(練習(xí)冊P9P)

1、習(xí)題解答習(xí)題解答( (七七) ) 50- -2t txT0 a at txWPMn=Mn = T0= 2.5 kNmWP = 16p pD3= 42.41 cm3= 58.95 MPas s30 = - -t t sin( (60) )s s- -60 = - -t t sin( (- -120) ) = - - 51.05 MPa= 51.05 MPa1Ee e = ( (s s- -nsns) ) = 319106正應(yīng)變正應(yīng)變 （線應(yīng)變）（線應(yīng)變）單元體單元體無量綱無量綱 50- -3s sxa aF60s s = 0= t ts sAF=2s sxs s- -30 = + + cos(

2、(- -60) )2s sx= s sx432s sxs s60 = + + cos( (120) )2s sx= s sx41s sz = 01Ee e = ( (s s- -nsns) ) = s sx4E3- -n ns sx = e e- -304E3- -n n= 160 MPaF = s sxA = 50 kN則則而而 50- -4 z yABCDF0.1ll0.1lqEFt ts st ts sM 圖圖( (kNm) )4541.841.8210FS 圖圖( (kN) )21020882088Iz = 5278 cm4 h = 250 mm t = 13 mm Wz = 422.

3、72 cm3 b = 118 mm tw = 10 mmSz max* * = 246.3 cm3 * * Mmax = 45 kNms s = = 106.5 MPaFS = 208 kN= 98.0 MPa= FS maxSz maxIztw* * M = 41.8 kNmFS max = 210 kN= WzMmax= 72.35 MPat t = = 88.70 MPas s1 = = 129.2 MPas s3 = = - - 40.5 MPa= 210 kN =)()()(212132322214ssssssssr313ssssr= 169.7MPa= 153.5 MPa 全面校核

4、全面校核內(nèi)力圖內(nèi)力圖：導(dǎo)學(xué)篇導(dǎo)學(xué)篇 51- -1Fl bA l h CBFa aj jb bFy = Fcosj jFz = Fsinj j= 9.66 kN= 2.59 kN14Mz = Fy2l = 7.24 kNm14My = Fz2l = 1.94 kNm = 1104 cm4 = 5625 cm4s s = y z IzMzIyMy( (1) )s s = 0 令令tana a =z0y0=Iy MzIz My= tanj jbh2( )= 0.476a a = 25.5 = 1103 cm3 = 750 cm3= + +WzMzWyMys st max = s sc max= 9.

5、83 MPa( (2) )Fy( (2l) )348EIzfy =Fz( (2l) )348EIyfz = 5.43 mm= 2.59 mmf = = fy2 + + fz 2= 6.02 mmtanb b =fyfz= tana a b b = a a = 25.5f 中性軸中性軸拉壓區(qū)拉壓區(qū) 51- -2F lA lCBF a a：b = 20 cmz0 = 5.69 cmIz = 4554.55 cm4 Iy = 1180.04 cm4 危險截面危險截面：跨中跨中Fy = Fz22= F = 17.68 kNMy = Mz = = 17.68 kNms s = y z IzMzIyMys

6、 s = 0 令令tana a =z0y0= 3.86a a = 75.5zB = - - z02yB = 0= - - 80.47 mms sB = 120.6 MPas sA = 146.2 MPa= 60.95 mmzA = b- - z0222= 141.4 mmyA = b22 51- -3zyx2FlFN = 2F = 20 kN23h2Mz = Fl - - 2F 12My = Fl + + 2F b2= 7.46 kNm= 5.6 kNm= 72 cm2s s = + + + + y + + z IzMzIyMyAFNs sA = 2.78 + + 51.81 + + 77.7

7、8s sB = 2.78 + + 51.81 - - 77.78s sC = 2.78 - - 51.81 - - 77.78s sD = 2.78 - - 51.81 + + 77.78= 132.4 MPa= 23.2 MPa= 126.8 MPa= 28.8 MPa = 144 cm3 = 72 cm3 斜彎曲斜彎曲偏心受壓偏心受壓Fs s = WzMzWyMyAFN 51- -4FFeeFN = FA = bhM = FeWz = 6bh2s s = WzMzAFNmax+- -min= ( (1 ) ) h6ebhF+- -e e 左左 = Ee e 右右 = E= ( (1- -

8、) ) h6eEbhF= ( (1+ + ) ) h6eEbhFF = ( ( e e左左 + + e e右右 ) ) Ebh2e = h6e e右右 + + e e左左e e右右 - - e e左左= 625 kN= 25 mm討論：討論：( (1) )s smax 0s smin 0 0( ( e 0 整個截面：整個截面：( ( e ) )h6- - 虎克定律虎克定律偏心受拉偏心受拉 52- -2hbD風(fēng)壓力的合力：風(fēng)壓力的合力：底截面：底截面：14F = p p pD2 M = F h Mn = F b = 314 Nm= 235.6 Nms sr3 = = M2 + + Mn2W1=

9、= h2 + + b2d3( (1- -a a4) )8pD21- -a a4a a 0.3086- - d d = = ( (1- -a a ) )d 2.64 mm- - 取取 彎扭組合彎扭組合( (圓桿圓桿) ) - - s s 危險截面危險截面d d = 2.7 mm 52- -3lllABCDF3F4F1F2Mn 圖圖My 圖圖Mz 圖圖反力反力 內(nèi)力圖內(nèi)力圖危險截面危險截面B：Mz= 1.092 kNmMy= 3 kNms sr4 = = M2 + + 0.75Mn2W1彎扭組合彎扭組合- - s s M2 + + 0.75Mn2 s s - - M2 + + 0.75Mn2 s

10、s 323p p= 69.6 mm取取F4F1內(nèi)力圖內(nèi)力圖 彎扭組合彎扭組合( (圓軸圓軸) )：d = 7.0 cm 52- -4T T M M t ts ss s = WMW = 32p pd3t t = WPTWP = 16p pd3e e = Es ss s = Ee e 0 ：M = Ws s = 105 MPa= 278 Nm2s ss sa a = + + cos2a a ( t t ) sin2a a2s s2s ss s = + + t t2s ss s - - = - - t t1Ee e = ( (s s- - nsns- - ) ) 1 - -n n2= e e+ t

11、t 1 + +n nE1 - -n n2t t = ( (e e- - e e ) )1 + +n nE= 40.4 MPaT = WPt t = 214 NmP101 53- -3 5m 7m 9m2m 4m 4m ( (5) )2p p2EI10.70.52150.770.592214( (4.9) )2p p2EI( (4) )2p p2EI( (4.5) )2p p2EIP101 53- -4 Fcr圓圓Fcr方方= I圓圓I方方= 64p pd4121a43p p= 0.955 P100 53- -2 Es sPl lP = p p IminAimin =：Imin = 225.9

12、cm4 A = 42.1 cm2 iminm mll lmax = l lP= 2.32 cm= 216= 99.35 = 42.3 MPa = 178 kNFcr = s scrA= 178 kNP102 54- -1hb( (1) )( (2) ) 24 cm2 = 32 cm4 = 72 cm4 IzAiz = IyAiy =23=3= 1.155 cm = 1.732 cm m my = 0.5 m mz = 1 l lz l lyl lz l lPiym myll ly =izm mzll lz = 132.8 = 99.6 Es sPl lP = p p= 99.35 = 269

13、kN l l相同相同s scr相同相同iym myl=izm mzl=b12h12121= 2hb P102 54- -2FFABCDaaaaFCFN1FN1Bs scr = 304 - - 1.12l l直線公式直線公式 Es sPl lP = p p= 99.35 ba- -s sSl lS = 57.1l lim m al l = l lPl lS l lP = 100= 79.6 MPaFcr = As scr= 400 kN nw FcrF AB段先失穩(wěn)段先失穩(wěn) F = 160 kNP104 55- -1 lBCA2bbFS SMC = 0 im m ll l = 1 cm IAi = d4= 80 Es sPl lP = p p= 99.35 ba- -s sSl lS = 57.1l l l lPl lS s scr = 304 - - 1.12l l直線公式直線公式：Fcr = As scr= 214.4 MPa= 269.4 kN