答辩人：任飞第一部分：选题及要求第二部分：理论求解（2）自重单独作用附材料力学求解位移采用的MATLAB代码如下：clcclearsymsFLEA1A2xpgF=30000;L=2.1;E=2.1*10^11;A1=pi*0.15^2;A2=pi*0.05^2;g=9.8;p=7800;l1=F*L/(E*(A1-A2))*(log(A1)-log(A2))lx=p*g/(3*E)*x*(A2*L/(A2*L+(A1-A2)*x)+(A2*L/(A2*L+(A1-A2)*x))^0.5+1);l2x=@(x)eval(lx);l2=quad(l2x,0,2.1,0.00001)l=l1+l2%answerl=1.0918e-005（2）弹性力学的模型建立和分析本题中锥角为5.45°误差为-0.16%,则根据材料力学结论可推算出弹性力学近似的位移L3=1.0918e-005*（1+0.0016）=1.0935e-005第三部分：MATALB有限元编程求解（二）单元的划分和节点的编号轴对称问题的弹性矩阵和几何矩阵如图注意：轴对称单元是三维实体，因此这里的积分为三重积分。但由于被积函数与角度无关。于是单元被近似地当成了常应变单元。sk.m代码如下（单刚形成及放入总纲）functionK=sk(r1,z1,r2,z2,r3,z3,i,j,k,n)E=210*10^9;u=0.3;D=E*(1-u)/((1+u)*(1-2*u))*[1u/(1-u)0.5*(1-2*u)/(1-u)0u/(1-u)1u/(1-u)00.5*(1-2*u)/(1-u)u/(1-u)100000.5*(1-2*u)/(1-u)];squ=[1r1z11r2z21r3z3];A=det(squ);a1=r2*z3-r3*z2;a2=r3*z1-r1*z3;a3=r1*z2-r2*z1;b1=z2-z3;b2=z3-z1;b3=z1-z2;c1=r3-r2;c2=r1-r3;c3=r2-r1;r0=(r1+r2+r3)/3;z0=(z1+z2+z3)/3;B=1/A*[b10b20b300c10c20c3(a1+b1*r0+c1*z0)/r00(a2+b2*r0+c2*z0)/r00(a3+b3*r0+c3*z0)/r00c1b1c2b2c3b3];KK=pi*r0*A*B'*D*B;k1=zeros(n);l1=2*i-1;l2=2*i;l3=2*j-1;l4=2*j;l5=2*k-1;l6=2*k;k1([l1l2l3l4l5l6],[l1l2l3l4l5l6])=KK;K=k1;（四）体力及表面力的形成surf.m（下端面表面力等效节点力）functionF=surf(r1,z1,r2,z2,r3,z3,i,j,k,n)N=30000;S=pi*0.05^2;q=N/S;l=((z3-z2)^2+(r3-r2)^2)^0.5;squ=[1r1z11r2z21r3z3];A=det(squ);f1=0;f2=-pi*l*q*(2*r2+r3)/3;f3=-pi*l*q*(r2+2*r3)/3;FF=[];form=1:nFF=[FF0];endl1=2*i;l2=2*j;l3=2*k;FF(1,l1)=f1;FF(1,l2)=f2;FF(1,l3)=f3;F=FF;（五）主程序求出杆下端的位移ef3=equal(0,-0.7325,0,-1.4505,0.1151,-0.7325,3,5,4,16);ef4=equal(0.1151,-0.7325,0.0000,-1.4505,0.0809,-1.4505,4,5,6,16);ef5=equal(0.0000,-1.4505,0.0000,-2.1000,0.0809,-1.4505,5,7,6,16);ef6=equal(0.0809,-1.4505,0.0000,-2.1000,0.0500,-2.1000,6,7,8,16);F1=ef1+ef2+ef3+ef4+ef5+ef6;F2=surf(0.0809,-1.4505,0.0000,-2.1000,0.0500,-2.1000,6,7,8,16);F=F1+F2%划行划列从支撑情况可知位移编号为12345913的位移均为0，划去得kk=K([78101112141516],[78101112141516]);f=F([78101112141516]);u=inv(k)*f'wy=u(6)%运算后得出杆下端位移为wy=1.1103e-005第四部分：结论分析和心得体会（二）心得体会Thankyou!