同济大学 材料力学 习题解答4(练习册P75-P82)

1、材料力学,四,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,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,a =

2、l,斜直线,p75 41-1-2,qa,qa2,d,fs 图,形状,水平线,斜直线,fsb左 = - qa,fb = qa,fc = qa,mb = 0,fy = 0,fsb右 = qa,fsc = - qa,m 图,形状,抛物线,ma = 0,md = qa2,mb = - qa2,mc = 0,mb = - qa2,突变,特殊点,抛物线,e,斜直线,斜直线,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 = q

3、a2,mb = qa2,md = 0,mb = qa2,m 图,形状,mc = qa2,me = qa2,突变,按比例,p80 43-1,c,d,fd = 20 kn,fc = 10 kn,md = 0,fy = 0,m1,fc1.5,15 knm,8103 cm4,ya = - 5 cm,yb = 10 cm,- 9.375 mpa,18.75 mpa,反力,指定截面,形心主轴,p80 43-2,e,g,a,b,c,d,i 32 a,fe = 10 kn,fg = 50 kn,me = 0,fy = 0,m1,10 knm,iz = 11080 cm4,h = 32 cm,t = 1.5 c

4、m,查表,ya = - 16 cm,- yd,yb = - 14.5 cm,- yc,- 14.44 mpa,sd = 14.44 mpa,- 13.09 mpa,sc = 13.09 mpa,指定截面,导学篇,附录b-3 p380中,i 32a,p81 43-3,a,b,i 20 a,c,d,m 图,ma = 0,fy = 0,mmax,wz = 236.9 cm3,查表,s,f 56.9 kn,则,导学篇,附录b-3 p380中,i 20a,f = 57 kn,p81 43-4,a,b,m 图,mmax,80 knm,s,41.6 cm,取,28 cm,判定,h = 42 cm,p82 4

5、4-1,c,d,a,b,fd = 20 kn,fc = 10 kn,md = 0,fy = 0,fs1 = 10 kn,8103 cm4,b = 12 cm,sza = 450 cm3,szb = 0,0.469 mpa,tb = 0,剪力方向,t 分布,指定截面,mpa,p82 44-2,21.15 mpa,14.39 mpa,85 mm,负面积法,c,3752 cm4,组合法,sz max,264.5 cm3,tw = 4 cm,21.15 mpa,分布,腹板,抛物线分布,20 mm,115 mm,d,14.39 mpa,td,t 分布规律,tmax ) 位置,p84 45-1,a,b,d

6、,c,m 图,ma,20 knm,20 knm,20 knm,mb,- 20 knm,mmax,20 knm,s,210-3 m3,1,2,3,28.8 cm,h,取 h = 288 mm,b = 144 mm,22.9 cm,a,取 a = 229 mm,27.3 cm,d,取 d = 273 mm,三种截面形状的材料用量比,a1 : a2 : a3,1 : 1.26 : 1.41,危险截面,p84 45-2,a,b,c,d,m 图,12 knm,8 knm,sc max,sc max,st max,st max,b 截面,c 截面,fb = 36 kn,fa = 12 kn,b 截面,c 截面,mc,12 knm,mb,- 8 knm,几何性质,147 mm,yc,i