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1、材料力学,习题解答(三),P71 39-1,Mn图,Mnmax = 4.5 kNm,= 227.3 cm3,= 19.80 MPa, t ,= 1193 cm4,= 0.27 /m, q ,满足强度与刚度条件,j14 = j34 + j23 + j12,= - 0.162 ,= - 2.8310-3 rad,刚度校核,扭转角,S,P71 39-2,Mn = T0,= 10.505 Nm, t ,= 9.63 mm, j = 1,= 19.8 mm,取,或, q ,Nk = 0.33 kW,d = 20 mm,P72 39-3,圆轴:,= 196.3 cm3,= 70 MPa,Mn = WPtm
2、ax,T0 =,= 13.74 kNm,螺栓:,FS =, t ,= 7.3,令,取,n = 8,P72 39-4,Mn图,Mn1 = TA,Mn2 = TA - T0,jAB = jAC + jCB,m = 0,TB = T0 - TA,反力：受力图,P110 58-1(b),负面积法,A1 = 800 cm2,A1,A2,A3,A2 = - 25p cm2,A3 = - 56.25p cm2,y1 = 20 cm,y2 = 30 cm,y3 = 10 cm,= 21.8 cm,对称性,zC = 10 cm,= 76396 cm4,组合法,另求：,(106667+2592)cm4,-(491
3、+5281)cm4,-(2485+24606)cm4,错的多!,对于过形心水平轴的惯性矩,P111 58-3,O,= 3.30R4,P282：,Iz = IzC+Ab2,平行移轴公式,附录A 第四项,P111 58-4,Iz1 = Iy1 = 179.51 cm4,A1 = 19.26 cm2,z0 = 2.84 cm,Iz = 2Iz1,= 359 cm4,Iy = 2(Iy1+A1a12),= 927 cm4,查表：,(型钢),不会?,等边角钢,组合法,导学篇 附录B-1 P374中,L 10010,P73 40-1(a),FS1,= q2 - F,= 10 kN,M1,= F4 - 2q
4、1,= 20 kNm,FS2,= - F,= - 10 kN,M2,= F2,= 20 kNm,FS3,= FS2,= - 10 kN,M3,= M2,= 20 kNm,FS1 + F - q2 = 0,Fy = 0,F4 - M1 - 2q1 = 0,mA = 0,求指定截面的内力,截面法!,相邻截面,(正负号规定),P73 40-1(b),FA = 40 kN,FC = 80 kN,mA = 0,Fy = 0,FS1,= FA,= 40 kN,M1,= 0,FS2,= FA,= 40 kN,M2,= FA2,= 80 kNm,FS3,= FA - 3q2,= - 20 kN,M3,= FA
5、4 - 3q21,= 100 kNm,特殊位置截面,P73 40-1(c),a = l,FA = 5 kN,FB = 15 kN,mA = 0,Fy = 0,FS1,= - FA,= - 5 kN,M1,= - FA2,= - 10 kNm,FS2,= F,= 10 kN,M2,= - F1,= - 10 kNm,FS3,= F,= 10 kN,M3,= 0,相邻截面,P73 40-1(d),a = l,FD = 10 kN,Fy = 0,FS1,= - F,= - 10 kN,M1,= - F1,= - 10 kNm,FS2,= - FD,= - 10 kN,M2,= FD2,= 20 kN
6、m,FS3,= - FD,= - 10 kN,M3,= FD1,= 10 kNm,相邻截面,P73 40-1(e),FC = 5 kN,FE = 5 kN,mC = 0,Fy = 0,FB = 7.5 kN,FA = 2.5 kN,mB = 0,Fy = 0,FS1,= - FA,= - 2.5 kN,M1,= - FA1,= - 2.5 kNm,FS2,= FB - FA,= 5 kN,M2,= FB1 - FA3,= 0,FS3,= F - FE,= 5 kN,M3,= FE3 - F1.5,= 0,特殊位置截面,相邻截面,多跨梁,(有主次之分结构),P74 40-2-1,F,2F,2Fa
7、,3Fa,AB段：,BC段：,= - F,M(x) = 2F(2a - x) - F(a - x),= 3Fa - Fx,FS(x) = F - 2F,M(x) = 2F(2a - x),= 4Fa - 2Fx,FS(x) = - 2F,M 图,FS 图,按比例,建立坐标,M图 正确性？,P74 40-2-2,qa,AB段：,BC段：,M(x) = - qa(2a- a-x),= qax - qa2,FS(x) = qa,M(x) = q(2a-x) (2a-x),= - qx2 + 2qax - 2qa2,FS(x) = q (2a-x),= 2qa - qx,= - q(2a-x)2,M
8、图,FS 图,光滑,P74 40-2-3,FA = F,FB = F,mA = 0,Fy = 0,AC段：,DB段：,CD段：,= F,M(x) = FAx,= Fx,FS(x) = FA,M(x) = FAx - F(x-a),= Fa - Fx,FS(x) = FA-F,= - F,= F,FS(x) = FB,M(x) = -FB(3a-x),= Fx - Fa,M 图,FS 图,P74 40-2-4,F,Fa,F,Fa,FA = F,FC = 2F,mA = 0,Fy = 0,AB段：,CD段：,BC段：,= -F,M(x) = -FAx,= -Fx,FS(x) = -FA,M(x)
9、= -FAx + M0,= Fa - Fx,FS(x) = -FA,= -F,FS(x) = F,M(x) = -F(3a-x),= Fx - 3Fa,M 图,FS 图,P75 41-1-1,qa,形状：,FS 图,M 图,斜直线,斜直线,FSA = 0,FSB = qa,FSC = 0,FSB = qa,MA = 0,MD = qa2,MB左 = qa2,MB右 = - qa2,ME = - qa2,MC = 0,D,E,形状：,抛物线,抛物线,简便方法,微分关系,(突变),(特殊点),(形状.方向),P75 41-1-2,qa,qa2,D,FS 图,形状：,水平线,斜直线,FSB左 = -
10、 qa,FB = qa,FC = qa,mB = 0,Fy = 0,FSB右 = qa,FSC = - qa,M 图,形状：,斜直线,抛物线,MA = 0,MD = qa2,MB = - qa2,MC = 0,MB = - qa2,(突变),(特殊点),P75 41-1-3,FD = qa,FA = qa,mD = 0,Fy = 0,FS 图,形状：,水平线,水平线,斜直线,FSB左 = qa,FSB右 = - qa,FSC = - qa,FSD = - qa,MA = 0,MC = qa2,MB = qa2,MD = 0,MB = qa2,M 图,形状：,斜直线,斜直线,抛物线,MC = qa2,ME = qa2,E,(突变),(按比例),P78 42-2-1,M 图(M0),M 图,M 图(F),FS 图(M0),FS 图,FS 图(F),叠加法,(过程),(单个荷载),P78 42-2-2,M 图(M0),M 图,M 图(q),FS 图(M0),FS 图,FS 图(q),(抛物线,(抛物线 - 中点 或 极值点),
