哈工大 材料力学 MATLAB 梁 上机 大作业.doc

本程序只支持静定结构的梁（左端悬臂梁、简支梁）函数输入格式：beamsolver(L,EI,supports,loads,maxdx);参量的输入格式：L=10.0EI=2e8supports=f,0,v,10.0-左端悬臂梁supports=p,2.0,r,8.0 -简支梁loads=f,2.0,1000,m,4.0,500,d,7.0,9.0,3.0,20,100maxdx=0.01输出为V,M,vy，x的一维行向量和三张坐标图-图1:剪力V图,图2:弯矩M图,图3:挠度vy图.范例：悬臂梁：纯受集中力：beamsolver(10.0,2e8,f,0,v,10.0,f,2.0,1000,0.01)纯受集中矩：beamsolver(10.0,2e8,f,0,v,10.0,m,4.0,500,0.01)纯受分布力（格式一）：beamsolver(10.0,2e8,f,0,v,10.0,d,7.0,9.0,3.0,20,100,0.01)纯受分布力（格式二）：beamsolver(10.0,2e8,f,0,v,10.0,d,(7.0:0.01:9.0);(0:0.01:2.0).2,0.01)受混合力：beamsolver(10.0,2e8,f,0,v,10.0,f,2.0,1000,m,4.0,500,d,7.0,9.0,3.0,20,100,0.01)简支梁：纯受集中力：beamsolver(10.0,2e8,p,2.0,r,8.0,f,5.0,1000,0.01)纯受集中矩：beamsolver(10.0,2e8,p,2.0,r,8.0,m,4.0,500,0.01)纯受分布力（格式一）：beamsolver(10.0,2e8,p,2.0,r,8.0,d,7.0,9.0,3.0,20,100,0.01)纯受分布力（格式二）：beamsolver(10.0,2e8,p,2.0,r,8.0,d,(7.0:0.01:9.0);(0:0.01:2.0).2,0.01)受混合力：beamsolver(10.0,2e8,p,2.0,r,8.0,f,5.0,1000,m,4.0,500,d,7.0,9.0,3.0,20,100,0.01)以上范例输出的剪力V图、弯矩M图都经过笔算检验完全正确。悬臂梁的挠度经检验为正确的，简支梁的挠度有误（整个程序的唯一错误）。function x,V,M,vy=beamsolver(L,EI,supports,loads,maxdx)%材料力学上机课作业（满分）%本DOC文件由M文件转换而成，可能有格式错误%班级;1036*(专业班级1036*)%支持两种梁的模型:%1.左端悬臂梁%2.一端pin一端roller的简支梁%输入格式：beamsolver(L,EI,supports,loads,maxdx);%例如：L=10.0,EI=2e8,supports=f,0,v,10.0,loads=f,2.0,1000,m,4.0,500,d,7.0,9.0,3.0,20,100,maxdx=0.01%输出为三张坐标图,图1:剪力V图,图2:弯矩M图,图3:挠度vy图.%fixed-freeif (supports11=f & supports21=v) n=1; M=linspace(0,0,L/maxdx); V=linspace(0,0,L/maxdx); while n=length(loads)%P-fixed-free- if loadsn1=f x=linspace(maxdx,L,L/maxdx); %将x分为小量 n1=loadsn2(1)/maxdx; %确定x=L1处对应的下标 V1=linspace(loadsn2(2),loadsn2(2),n1); %第一段剪力赋值 V2=zeros(1,L/maxdx-n1); %第二段剪力赋值 M1=-loadsn2(2)*(loadsn2(1)-x(1:n1); %第一段弯矩赋值 M2=zeros(1,L/maxdx-n1); %第二段弯矩赋值 M=M+M1,M2; %全梁的弯矩 V=V+V1,V2; %全梁的剪力 end%M-fixed-free- if loadsn1=m x=linspace(maxdx,L,L/maxdx); %将x分为小量 n1=loadsn2(1)/maxdx; %确定x=L1处对应的下标 M1=linspace(loadsn2(2),loadsn2(2),n1); %第一段弯矩赋值 M2=zeros(1,L/maxdx-n1); %第二段弯矩赋值 M=M+M1,M2; %全梁的弯矩 end%q-fixed-free- if loadsn1=d x=linspace(maxdx,L,L/maxdx); %将x分为小量 n1=loadsn2(1)/maxdx; %确定x=L1处对应的下标 n2=loadsn2(2)/maxdx; %确定x=L2处对应的下标 k=length(loadsn2)-4; A=size(loadsn2); if A(1)=1 %如果是第一种q的输入情况 ks=1; q2=linspace(loadsn2(4),loadsn2(4),n2-n1+1); while ks=k k0=loadsn2(4+ks); q2=q2+k0*(x(n1:n2)-loadsn2(3); ks=ks+1; end q1=zeros(1,n1-1); q3=zeros(1,L/maxdx-n2); q0=q1,q2,q3; q=fliplr(q0); %求出q Vq0=cumsum(q)*maxdx; Vq=fliplr(Vq0); %求出剪力V Mq0=-cumsum(Vq0)*maxdx; Mq=fliplr(Mq0); %求出弯矩M end if A(1)=2 %如果是q的第二种输入情况 q2=loadsn2(2,:); %表示出q2 n1=loadsn2(1,1)/maxdx; %定出q开始的坐标 B=length(loadsn2); %求q的长度 n2=loadsn2(1,B)/maxdx; %定出q结束的坐标 q1=zeros(1,n1-1); q3=zeros(1,L/maxdx-n2); q0=q1,q2,q3; %求出q0 q=fliplr(q0); Vq0=cumsum(q)*maxdx; Vq=fliplr(Vq0); %求出剪力V Mq0=-cumsum(Vq0)*maxdx; Mq=fliplr(Mq0); %求出弯矩M end M=M+Mq; %全梁的弯矩 V=V+Vq; %全梁的剪力 end n=n+1; end V=V M=M A=cumsum(M)*maxdx/EI; %对弯矩积分求转角 vy=cumsum(A)*maxdx %对转角积分求挠度 subplot(3,1,1),plot(x,V),grid; %绘图 subplot(3,1,2),plot(x,M),grid; subplot(3,1,3),plot(x,vy),grid; end%分割线-%pin-roller-if (supports11=p | supports11=r)&(supports21=p | supports21=r) n=1; np=supports12/maxdx; nr=supports22/maxdx; M=linspace(0,0,L/maxdx); V=linspace(0,0,L/maxdx); while n=length(loads)%P-pin-roller- if loadsn1=f x=linspace(maxdx,L,L/maxdx); %将x分为小量 n1=loadsn2(1)/maxdx; %确定x=L1处对应的下标 Nr=loadsn2(2)*(loadsn2(1)-supports12)/(supports22-supports12); if n1np & n1nr %力作用在支架右边 V1=zeros(1,np-1); V2=linspace(loadsn2(2)-Nr,loadsn2(2)-Nr,nr-np); V3=linspace(loadsn2(2),loadsn2(2),n1-nr); V4=zeros(1,L/maxdx-n1+1); V=V+V1,V2,V3,V4; M1=zeros(1,np-1); M2=(loadsn2(2)-Nr)*(x(np:nr-1)-supports12); M3=-loadsn2(2)*(loadsn2(1)-x(nr:n1-1); M4=zeros(1,L/maxdx-n1+1); M=M+M1,M2,M3,M4; end end%M-pin-roller- if loadsn1=m x=linspace(maxdx,L,L/maxdx); %将x分为小量 n1=loadsn2(1)/maxdx; %确定x=L1处对应的下标 if n1np & n1nr %弯矩作用在支架右边 Np=loadsn2(2)/(supports22-supports12); Nr=Np V1=zeros(1,np-1); V2=linspace(Np,Np,nr-np); V3=zeros(1,L/maxdx-nr+1); V=V+V1,V2,V3; M1=zeros(1,np-1); M2=Np*(x(np:nr-1)-supports12); M3=linspace(loadsn2(2),loadsn2(2),n1-nr); M4=zeros(1,L/maxdx-n1+1); M=M+M1,M2,M3,M4; %全梁的弯矩 end end%q-pin-roller- if loadsn1=d x=linspace(maxdx,L,L/maxdx); %将x分为小量 n1=loadsn2(1)/maxdx; %确定x=L1处对应的下标 n2=loadsn2(2)/maxdx; %确定x=L2处对应的下标 k=length(loadsn2)-4; A=size(loadsn2); if A(1)=1 %如果是第一种q的输入情况 ks=1; q2=linspace(loadsn2(4),loadsn2(4),n2-n1+1); while ks=k k0=loadsn2(4+ks); q2=q2+k0*(x(n1:n2)-loadsn2(3); ks=ks+1; end q1=zeros(1,n1-1); q3=zeros(1,L/maxdx-n2); q0=q1,q2,q3; F=q0*maxdx; %等效为很多个力作用在杆上 nF=length(F); %找出F的最后一点值得坐标 X=1; while X=nF if F(X)=0 Nr=F(X)*(X-np)/(nr-np); if Xnp & Xnr %力作用在支架右边 V1=zeros(1,np-1); V2=linspace(F(X)-Nr,F(X)-Nr,nr-np); V3=linspace(F(X),F(X),X-nr); V4=zeros(1,L/maxdx-X+1); V=V+V1,V2,V3,V4; M1=zeros(1,np-1); M2=(F(X)-Nr)*(x(np:nr-1)-supports12); M3=-F(X)*(X*maxdx-x(nr:X-1); M4=zeros(1,L/maxdx-X+1); M=M+M1,M2,M3,M4; end end X=X+1; end end if A(1)=2 %如果是q的第二种输入情况 q2=loadsn2(2,:); %表示出q2 n1=loadsn2(1,1)/maxdx; %定出q开始的坐标 B=length(loadsn2); %求q的长度 n2=loadsn2(1,B)/maxdx; %定出q结束的坐标 q1=zeros(1,n1-1); q3=zeros(1,L/maxdx-n2); q0=q1,q2,q3; %求出q F=q0*maxdx; %等效为很多个力作用在杆上 nF=length(F); %找出F的最后一点值得坐标 X=1; while X=nF if F(X)=0 Nr=F(X)*(X-np)/(nr-np); if Xnp & Xnr %力作用在支架右边 V1=zeros(1,np-1); V2=linspace(F(X)-Nr,F(X)-Nr,nr-np); V3=linspace(F(X),F(X),X-nr); V4=zeros(1,L/maxdx-X+1); V=V+V1,V2,V3,V4; M1=zeros(1,np-1); M2=(F(X)-Nr)*(x(np:nr-1)-supports12);