MATLAB有限元分析與應(yīng)用.ppt

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1、2020/9/7,1,,第三章 MATLAB有限元分析與應(yīng)用,,,,,,3-1 彈簧元,結(jié)構(gòu)分析編程及軟件應(yīng)用,3-2 線性桿元,3-3 二次桿元,3-4 平面桁架元,3-5 空間桁架元,3-6 梁元,2020/9/7,2,,3-1 彈簧元,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,1、有限元方法的步驟：,離散化域,形成單剛矩陣,集成整體剛度矩陣,引入邊界條件,求解方程,后處理,2020/9/7,3,,,,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、基本方程,3-1 彈簧元,彈簧元是總體和局部坐標(biāo)一致的一維有限單元,每個彈簧元有兩個節(jié)點(diǎn)（node）,,,,,單剛矩陣為：,,總剛矩陣：,,結(jié)構(gòu)方程：,單元節(jié)點(diǎn)力：,

2、,2020/9/7,4,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、MATLAB函數(shù)編寫,3-1 彈簧元,%SpringElementStiffness This function returns the element stiffness %matrix for a spring with stiffness k. %The size of the element stiffness matrix is 2 x 2.,3.1 單元剛度矩陣的形成,y = k -k ; -k k;,function y = SpringElementStiffness(k),2020/9/7,5,,,,結(jié)構(gòu)分析編程

3、及軟件應(yīng)用,3、MATLAB函數(shù)編寫,3-1 彈簧元,%SpringAssemble This function assembles the element stiffness % matrix k of the spring with nodes i and j into the % global stiffness matrix K. % This function returns the global stiffness matrix K % after the element stiffness matrix k is assembled.,,,,3.

4、2 整體剛度矩陣的形成,K(i,i) = K(i,i) + k(1,1); K(i,j) = K(i,j) + k(1,2); K(j,i) = K(j,i) + k(2,1); K(j,j) = K(j,j) + k(2,2); y = K;,function y = SpringAssemble(K,k,i,j),2020/9/7,6,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、MATLAB函數(shù)編寫,3-1 彈簧元,%SpringElementForces This function returns the element nodal force % vector given the e

5、lement stiffness matrix k % and the element nodal displacement vector u.,,,,3.3 節(jié)點(diǎn)載荷計(jì)算,y = k * u;,function y = SpringElementForces(k,u),2020/9/7,7,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,4、實(shí)例計(jì)算分析應(yīng)用,3-1 彈簧元,,,,如圖所示二彈簧元結(jié)構(gòu)，假定k1=100kN/m，k2=200kN/m，P=15kN。 求：系統(tǒng)的整體剛度矩陣； 節(jié)點(diǎn)2、3的位移； 節(jié)點(diǎn)1的支反力； 每個彈簧的內(nèi)力,解：,步驟1：離散化域,2020/9/7,

6、8,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,4、實(shí)例計(jì)算分析應(yīng)用,3-1 彈簧元,,,,步驟2：形成單元剛度矩陣,k1=SpringElementStiffness(100);,k1 = 100 -100 -100 100,k2=SpringElementStiffness(200)；,k2 = 200 -200 -200 200,調(diào)用 function y = SpringElementStiffness(k)函數(shù),2020/9/7,9,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,4、實(shí)例計(jì)算分析應(yīng)用,3-1 彈簧元,,,,步驟3：集成整體剛度矩陣,調(diào)用 function y = SpringAssembl

7、e(K,k,i,j)函數(shù),n=3; K = zeros(n,n);,K = SpringAssemble(K,k1,1,2),K = 0 0 0 0 0 0 0 0 0,K = SpringAssemble(K,k2,2,3),K = 100 -100 0 -100 100 0 0 0 0,K = 100 -100 0 -100 300 -200 0 -200 200,2020/9/7,10,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,4、實(shí)例計(jì)算分析應(yīng)用,3-1 彈簧元,,,,步驟4：引入邊界條件,,已知邊界條件：,,2020/9/7,11,,

8、,,結(jié)構(gòu)分析編程及軟件應(yīng)用,5、實(shí)例計(jì)算分析應(yīng)用,3-1 彈簧元,,,,步驟5：解方程,,U=zeros(2,1); F=0;15; K = K(2:3,2:3); U=KF,U=inv(K)*F,K(1,:)=; K(:,1)=;,,U = 0.1500 0.2250,2020/9/7,12,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,5、實(shí)例計(jì)算分析應(yīng)用,2-1 彈簧元,,,,步驟6：后處理,,,U=0;U,U = 0 0.1500 0.2250,F=K*U,F = -15.0000 0.0000 15.0000,u1=U(1:2); f1=SpringElementForces(k1

9、,u1);,f1 = -15.0000 15.0000,u2=U(2:3); f2=SpringElementForces(k2,u2);,f2 = -15.0000 15.0000,2020/9/7,13,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,5、實(shí)例計(jì)算分析應(yīng)用,3-1 彈簧元,,,,,,k1=SpringElementStiffness(100); k2=SpringElementStiffness(200); n=3; K=zeros(n,n); K=SpringAssemble(K,k1,1,2); K=SpringAssemble(K,k2,2,3); U=zeros(2,1); F=0

10、;15; K = K(2:3,2:3); KK=K; U=KF U=0;U; F=K*U; u1=U(1:2); f1=SpringElementForces(k1,u1) u2=U(2:3); f2=SpringElementForces(k2,u2),2020/9/7,14,,,,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,1、基本方程,3-2 線性桿元,線性桿元也是總體和局部坐標(biāo)一致的一維有限單元，用線性函數(shù)描述,每個線性桿元有兩個節(jié)點(diǎn)（node）,,,,,單剛矩陣為：,,總剛矩陣：,結(jié)構(gòu)方程：,單元節(jié)點(diǎn)力：,2020/9/7,15,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%Line

11、arBarElementStiffness This function returns the element % stiffness matrix for a linear bar with % modulus of elasticity E, cross-sectional % area A, and length L. The size of the % element stiffness matrix is 2 x 2.,,,2.1 單元剛度矩陣的形成,y = E*A/L -E*A/L ; -E*A

12、/L E*A/L;,function y = LinearBarElementStiffness(E,A,L),3-2 線性桿元,2020/9/7,16,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%LinearBarAssemble This function assembles the element stiffness % matrix k of the linear bar with nodes i and j % into the global stiffness matrix K. % This function ret

13、urns the global stiffness % matrix K after the element stiffness matrix % k is assembled.,,,,2.2 整體剛度矩陣的形成,K(i,i) = K(i,i) + k(1,1); K(i,j) = K(i,j) + k(1,2); K(j,i) = K(j,i) + k(2,1); K(j,j) = K(j,j) + k(2,2); y = K;,function y =LinearBarAssemble(K,k,i,j),3-2 線性桿元,2020/9/7,17,,,,結(jié)構(gòu)

14、分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%LinearBarElementForces This function returns the element nodal % force vector given the element stiffness % matrix k and the element nodal % displacement vector u.,,,,2.3 節(jié)點(diǎn)載荷計(jì)算,y = k * u;,function y = LinearBarElementForces(k,u),3-2 線性桿元,2020/9

15、/7,18,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%LinearBarElementStresses This function returns the element nodal % stress vector given the element stiffness % matrix k, the element nodal displacement % vector u, and the cross-sectional area A.,,,,2.4 節(jié)點(diǎn)應(yīng)力計(jì)算,y = k * u/A;,function y

16、 = LinearBarElementStresses(k, u, A),3-2 線性桿元,2020/9/7,19,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,如圖所示二線性桿元結(jié)構(gòu)，假定E=210MPa，A=0.003m2，P=10kN， 節(jié)點(diǎn)3的右位移為0.002m。 求：系統(tǒng)的整體剛度矩陣； 節(jié)點(diǎn)2的位移； 節(jié)點(diǎn)1、3的支反力； 每個桿件的應(yīng)力,解：,步驟1：離散化域,3-2 線性桿元,2020/9/7,20,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,步驟2：形成單元剛度矩陣,k1=LinearBarElementStiffness(E,A

17、,L1),k2=LinearBarElementStiffness(E,A,L2),調(diào)用 function y = LinearBarElementStiffness(E,A,L)函數(shù),3-2 線性桿元,2020/9/7,21,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,步驟3：集成整體剛度矩陣,調(diào)用 function y = LinearBarAssemble(K,k,i,j)函數(shù),n=3; K = zeros(n,n),K = LinearBarAssemble (K,k1,1,2),K = 0 0 0 0 0 0 0 0 0,K = LinearBarA

18、ssemble (K,k2,2,3),3-2 線性桿元,2020/9/7,22,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,步驟4：引入邊界條件,,已知邊界條件：,3-2 線性桿元,2020/9/7,23,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,步驟5：解方程,,U=zeros(1,1); U3=0.002 F=-10; K = K(2,2) 105000 K0 = K(2,3); -630000 U=K(F-K0*U3),U =0.0012,3-2 線性桿元,2020/9/7,24,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,步驟6：后處

19、理,,U=0;U;0.002,U = 0 0.0012 0.0002,F=K*U,F = -500.0000 -10.0000 510.0000,u1=U(1:2); f1= LinearBarElementForces(k1,u1) sigma1=LinearBarElementStresses(k1, u1, A),u2=U(2:3); f2= LinearBarElementForces(k2,u2) sigma2=LinearBarElementStresses(k2, u2, A),3-2 線性桿元,2020/9/7,25,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用

20、,,,,,,E=210E6; A=0.003; L1=1.5; L2=1; k1= LinearBarElementStiffness(E,A,L1); k2= LinearBarElementStiffness(E,A,L2); n=3; K = zeros(n,n); K = LinearBarAssemble (K,k1,1,2); K = LinearBarAssemble (K,k2,2,3); U=zeros(1,1); U3=0.002; F=-10;,3-2 線性桿元,KK=K; K=K(2,2); K0=K(2,3); U=K(F-K0*U3); U=0;U;U3; F=K

21、K*U u1=U(1:2); f1= LinearBarElementForces(k1,u1) sigma1=LinearBarElementStresses(k1, u1, A) u2=U(2:3); f2= LinearBarElementForces(k2,u2) sigma2=LinearBarElementStresses(k2, u2, A),,2020/9/7,26,,,,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,1、基本方程,3-3 二次桿元,二次桿元也是總體和局部坐標(biāo)一致的一維有限單元，用二次方程描述,每個線性桿元有三個節(jié)點(diǎn)（node）,,,,單剛矩陣為：,總剛矩陣：,結(jié)構(gòu)方程：,單

22、元節(jié)點(diǎn)力：,2020/9/7,27,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%QuadraticBarElementStiffness This function returns the element % stiffness matrix for a quadratic bar % with modulus of elasticity E, % cross-sectional area A, and length L. % The size of the element stiff

23、ness % matrix is 3 x 3.,,,2.1 單元剛度矩陣的形成,y = E*A/(3*L)*7 1 -8 ; 1 7 -8 ; -8 -8 16;,function y = QuadraticBarElementStiffness(E,A,L),3-3 二次桿元,2020/9/7,28,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%QuadraticBarAssemble This function assembles the element stiffness % matrix k of the quadratic b

24、ar with nodes i, j % and m into the global stiffness matrix K. % This function returns the global stiffness % matrix K after the element stiffness matrix % k is assembled.,,,,2.2 整體剛度矩陣的形成,K(i,i) = K(i,i) + k(1,1); K(i,j) = K(i,j) + k(1,2); K(i,m) = K(i,m) + k(1,

25、3); K(j,i) = K(j,i) + k(2,1); K(j,j) = K(j,j) + k(2,2);,function y =QuadraticBarAssemble(K,k,i,j,m),3-3 二次桿元,K(j,m) = K(j,m) + k(2,3); K(m,i) = K(m,i) + k(3,1); K(m,j) = K(m,j) + k(3,2); K(m,m) = K(m,m) + k(3,3); y = K;,2020/9/7,29,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%QuadraticBarElementForces This functio

26、n returns the element nodal % force vector given the element stiffness % matrix k and the element nodal % displacement vector u.,,,2.3 節(jié)點(diǎn)載荷計(jì)算,y = k * u;,function y = QuadraticBarElementForces(k,u),3-3 二次桿元,2020/9/7,30,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%QuadraticBarElem

27、entStresses This function returns the element % nodal stress vector given the element % stiffness matrix k, the element nodal % displacement vector u, and the % cross-sectional area A.,,,,2.4 節(jié)點(diǎn)應(yīng)力計(jì)算,y = k * u/A;,function y = QuadraticBarElementStresses

28、(k, u, A),3-3 二次桿元,2020/9/7,31,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,如圖所示雙二次桿元結(jié)構(gòu)，假定E=210MPa，A=0.003m2 求：系統(tǒng)的整體剛度矩陣； 節(jié)點(diǎn)2、3、4、5的位移； 節(jié)點(diǎn)1的支反力； 每個桿件的應(yīng)力,解：,3-3 二次桿元,2020/9/7,32,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,,,E=210E6; A=0.003; L=2; k1= QuadraticBarElementStiffness(E,A,L); k2= QuadraticBarElementStiffness(E,A,L

29、); n=5; K = zeros(n,n); K =QuadraticBarAssemble(K,k1,1,3,2); K =QuadraticBarAssemble(K,k2,3,5,4); U=zeros(4,1); F=5;-10;-7;10;,KK=K; K=K(2:n,2:n); U=KF; U=0;U; F=KK*U; u1=U(1);U(3);U(2); f1= QuadraticBarElementForces(k1,u1); sigma1=QuadraticBarElementStresses(k1, u1, A); u2=U(3);U(5);U(4); f2=Quadra

30、ticBarElementForces(k2,u2); sigma2=QuadraticBarElementStresses(k2, u2, A);,,3-3 二次桿元,2020/9/7,33,,,,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,1、基本方程,3-4 平面桁架元,平面桁架元是既有局部坐標(biāo)又有總體坐標(biāo)二維有限元，用線性函數(shù)描述,每個平面桁架元有二個節(jié)點(diǎn)（node）,,,,單剛矩陣為：,總剛矩陣：,結(jié)構(gòu)方程：,單元節(jié)點(diǎn)力：,2020/9/7,34,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%PlaneTrussElementLength This function returns

31、 the length of the % plane truss element whose first node has % coordinates (x1,y1) and second node has % coordinates (x2,y2).,,2.1 計(jì)算單元長度,y = sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1));,function y = PlaneTrussElementLength(x1,y1,x2,y2),3-4 平面桁架元,2020/9/7,35,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)

32、編寫,%PlaneTrussElementStiffness This function returns the element % stiffness matrix for a plane truss % element with modulus of elasticity E, % cross-sectional area A, length L, and % angle theta (in degrees). % The size of the element stiffness % matrix is

33、4 x 4.,,,2.2 單元剛度矩陣的形成,x = theta*pi/180; C = cos(x); S = sin(x); y = E*A/L*C*C C*S -C*C -C*S ; C*S S*S -C*S -S*S ; -C*C -C*S C*C C*S ; -C*S -S*S C*S S*S;,function y = PlaneTrussElementStiffness(E,A,L, theta),3-4 平面桁架元,2020/9/7,36,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%PlaneTrussAssemble This function assembl

34、es the element stiffness % matrix k of the plane truss element with nodes % i and j into the global stiffness matrix K. % This function returns the global stiffness % matrix K after the element stiffness matrix k is assembled.,,,,2.3 整體剛度矩陣的形成,K(2*i-1,2*i-1) = K(2*i-1,2*i-1) + k(

35、1,1); K(2*i-1,2*i) = K(2*i-1,2*i) + k(1,2); K(2*i-1,2*j-1) = K(2*i-1,2*j-1) + k(1,3); K(2*i-1,2*j) = K(2*i-1,2*j) + k(1,4); K(2*i,2*i-1) = K(2*i,2*i-1) + k(2,1); K(2*i,2*i) = K(2*i,2*i) + k(2,2); K(2*i,2*j-1) = K(2*i,2*j-1) + k(2,3); K(2*i,2*j) = K(2*i,2*j) + k(2,4);,function y =PlaneTrussAssemble(K

36、,k,i,j),K(2*j-1,2*i-1) = K(2*j-1,2*i-1) + k(3,1); K(2*j-1,2*i) = K(2*j-1,2*i) + k(3,2); K(2*j-1,2*j-1) = K(2*j-1,2*j-1) + k(3,3); K(2*j-1,2*j) = K(2*j-1,2*j) + k(3,4); K(2*j,2*i-1) = K(2*j,2*i-1) + k(4,1); K(2*j,2*i) = K(2*j,2*i) + k(4,2); K(2*j,2*j-1) = K(2*j,2*j-1) + k(4,3); K(2*j,2*j) = K(2*j,2*j

37、) + k(4,4); y = K;,3-4 平面桁架元,,2020/9/7,37,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%PlaneTrussElementForce This function returns the element force % given the modulus of elasticity E, the % cross-sectional area A, the length L, % the angle theta (in degrees), and the % element nodal disp

38、lacement vector u.,,,2.4 節(jié)點(diǎn)載荷計(jì)算,x = theta * pi/180; C = cos(x); S = sin(x); y = E*A/L*-C -S C S* u;,function y = PlaneTrussElementForce(E,A,L,theta,u),3-4 平面桁架元,2020/9/7,38,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%PlaneTrussElementStress This function returns the element stress % given the modulus of ela

39、sticity E, the % the length L, the angle theta (in % degrees), and the element nodal % displacement vector u.,,,,2.5 節(jié)點(diǎn)應(yīng)力計(jì)算,x = theta * pi/180; C = cos(x); S = sin(x); y = E/L*-C -S C S* u;,function y = PlaneTrussElementStress(E,L,theta,u),3-4 平面桁架元,2020/9/7,39,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例

40、計(jì)算分析應(yīng)用,,,,如圖所示平面桁架結(jié)構(gòu)，假定E=210MPa，A=0.0004m2 求：系統(tǒng)的整體剛度矩陣； 節(jié)點(diǎn)2的水平位移； 節(jié)點(diǎn)3的水平豎向位移； 節(jié)點(diǎn)1、2的支反力； 每跟桿件的應(yīng)力,3-4 平面桁架元,2020/9/7,40,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,1、基本方程,3-5 空間桁架元,空間桁架元是既有局部坐標(biāo)又有總體坐標(biāo)三維有限元，用線性函數(shù)描 述。各單元之間通過鉸接系統(tǒng)連接，只能傳遞力，而不能傳遞彎矩,每個桁架元有二個節(jié)點(diǎn)（node）,,,,,2020/9/7,41,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,1、基本方程,,3-5 空間桁架元,,,,總剛矩陣：,結(jié)

41、構(gòu)方程：,單元節(jié)點(diǎn)力：,單剛矩陣為：,2020/9/7,42,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%SpaceTrussElementLength This function returns the length of the % space truss element whose first node has % coordinates (x1,y1,z1) and second node has % coordinates (x2,y2,z2).,,2.1 計(jì)算單元長度,y = sqrt((x2-x1)*(x2-x1) + (y2-y

42、1)*(y2-y1) + (z2-z1)*(z2-z1));,function y = SpaceTrussElementLength(x1,y1,z1,x2,y2,z2),3-5 空間桁架元,2020/9/7,43,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%SpaceTrussElementStiffness This function returns the element % stiffness matrix for a space truss % element with modulus of elasticity E,

43、 % cross-sectional area A, length L, and % angles thetax, thetay, thetaz % (in degrees). The size of the element % stiffness matrix is 6 x 6.,,,2.2 單元剛度矩陣的形成,x = thetax*pi/180; u = thetay*pi/180; v = thetaz*pi/180; Cx = cos(x); Cy = cos(u); Cz = cos(v)

44、; w = Cx*Cx Cx*Cy Cx*Cz ; Cy*Cx Cy*Cy Cy*Cz ; Cz*Cx Cz*Cy Cz*Cz; y = E*A/L*w -w ; -w w;,function y = SpaceTrussElementStiffness(E,A,L,thetax,thetay,thetaz),3-5 空間桁架元,2020/9/7,44,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%SpaceTrussAssemble This function assembles the element stiffness % matrix k of the

45、space truss element with nodes % i and j into the global stiffness matrix K. % This function returns the global stiffness % matrix K after the element stiffness matrix % k is assembled.,,,,2.3 整體剛度矩陣的形成,K(3*i-2,3*i-2) = K(3*i-2,3*i-2) + k(1,1); K(3*i-2,3*i-1) = K(3*i-2,3*i-1) + k

46、(1,2); K(3*i-2,3*i) = K(3*i-2,3*i) + k(1,3); K(3*i-2,3*j-2) = K(3*i-2,3*j-2) + k(1,4); K(3*i-2,3*j-1) = K(3*i-2,3*j-1) + k(1,5); K(3*i-2,3*j) = K(3*i-2,3*j) + k(1,6); K(3*i-1,3*i-2) = K(3*i-1,3*i-2) + k(2,1); K(3*i-1,3*i-1) = K(3*i-1,3*i-1) + k(2,2); K(3*i-1,3*i) = K(3*i-1,3*i) + k(2,3); K(3*i-1,3*j

47、-2) = K(3*i-1,3*j-2) + k(2,4); K(3*i-1,3*j-1) = K(3*i-1,3*j-1) + k(2,5); K(3*i-1,3*j) = K(3*i-1,3*j) + k(2,6);,function y =SpaceTrussAssemble(K,k,i,j),3-5 空間桁架元,2020/9/7,45,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,,,,2.3 整體剛度矩陣的形成,3-5 空間桁架元,K(3*j-1,3*i-2) = K(3*j-1,3*i-2) + k(5,1); K(3*j-1,3*i-1) = K(3*j-1,3*i

48、-1) + k(5,2); K(3*j-1,3*i) = K(3*j-1,3*i) + k(5,3); K(3*j-1,3*j-2) = K(3*j-1,3*j-2) + k(5,4); K(3*j-1,3*j-1) = K(3*j-1,3*j-1) + k(5,5); K(3*j-1,3*j) = K(3*j-1,3*j) + k(5,6); K(3*j,3*i-2) = K(3*j,3*i-2) + k(6,1); K(3*j,3*i-1) = K(3*j,3*i-1) + k(6,2); K(3*j,3*i) = K(3*j,3*i) + k(6,3); K(3*j,3*j-2) = K

49、(3*j,3*j-2) + k(6,4); K(3*j,3*j-1) = K(3*j,3*j-1) + k(6,5); K(3*j,3*j) = K(3*j,3*j) + k(6,6); y = K;,K(3*i,3*i-2) = K(3*i,3*i-2) + k(3,1); K(3*i,3*i-1) = K(3*i,3*i-1) + k(3,2); K(3*i,3*i) = K(3*i,3*i) + k(3,3); K(3*i,3*j-2) = K(3*i,3*j-2) + k(3,4); K(3*i,3*j-1) = K(3*i,3*j-1) + k(3,5); K(3*i,3*j) =

50、K(3*i,3*j) + k(3,6); K(3*j-2,3*i-2) = K(3*j-2,3*i-2) + k(4,1); K(3*j-2,3*i-1) = K(3*j-2,3*i-1) + k(4,2); K(3*j-2,3*i) = K(3*j-2,3*i) + k(4,3); K(3*j-2,3*j-2) = K(3*j-2,3*j-2) + k(4,4); K(3*j-2,3*j-1) = K(3*j-2,3*j-1) + k(4,5); K(3*j-2,3*j) = K(3*j-2,3*j) + k(4,6);,,2020/9/7,46,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLA

51、B函數(shù)編寫,%SpaceTrussElementForce This function returns the element force % given the modulus of elasticity E, the % cross-sectional area A, the length L, % the angles thetax, thetay, thetaz % (in degrees), and the element nodal % displacement vector u.,,2.4 節(jié)點(diǎn)載荷計(jì)算,x = thet

52、ax * pi/180; w = thetay * pi/180; v = thetaz * pi/180; Cx = cos(x); Cy = cos(w); Cz = cos(v); y = E*A/L*-Cx -Cy -Cz Cx Cy Cz*u;,function y = SpaceTrussElementForce(E,A,L,thetax,thetay,thetaz,u),3-5 空間桁架元,2020/9/7,47,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%SpaceTrussElementStress This function returns the elem

53、ent stress % given the modulus of elasticity E, the % length L, the angles thetax, thetay, % thetaz (in degrees), and the element % nodal displacement vector u.,,,,2.5 節(jié)點(diǎn)應(yīng)力計(jì)算,x = thetax * pi/180; w = thetay * pi/180; v = thetaz * pi/180; Cx = cos(x); Cy = cos(w); Cz = cos(v);

54、 y = E/L*-Cx -Cy -Cz Cx Cy Cz*u;,function y = SpaceTrussElementStress(E,L,thetax,thetay,thetaz,u),3-5 空間桁架元,2020/9/7,48,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,如圖所示空間桁架結(jié)構(gòu)，假定E=210MPa，A14=0.001m2 A24=0.002m2，A34=0.001m2，P=12kN 求：系統(tǒng)的整體剛度矩陣； 節(jié)點(diǎn)4的水平位移； 節(jié)點(diǎn)3的水平豎向位移； 節(jié)點(diǎn)1、2、3的支反力； 每跟桿件的應(yīng)力,3-5 空間桁架元,2020/9/7,4

55、9,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,1、基本方程,3-6 梁元,梁元是總體坐標(biāo)與局部坐標(biāo)一致的二維有限元，用線性函數(shù)描 述。各單元之間通過鉸接系統(tǒng)連接，只能傳遞力，而不能傳遞彎矩,每個梁元有二個節(jié)點(diǎn)（node）,,,,,單剛矩陣為：,,,,總剛矩陣：,結(jié)構(gòu)方程：,單元節(jié)點(diǎn)力：,2020/9/7,50,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%BeamElementStiffness This function returns the element % stiffness matrix for a beam % element with

56、 modulus of elasticity E, % moment of inertia I, and length L. % The size of the element stiffness % matrix is 4 x 4.,,,2.1單元剛度矩陣的形成,y = E*I/(L*L*L)*12 6*L -12 6*L ; 6*L 4*L*L -6*L 2*L*L ; -12 -6*L 12 -6*L ; 6*L 2*L*L -6*L 4*L*L;,function y = BeamElementStiffness(E,I,L),3-

57、6 梁元,2020/9/7,51,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%BeamAssemble This function assembles the element stiffness % matrix k of the beam element with nodes % i and j into the global stiffness matrix K. % This function returns the global stiffness % matrix K after the element stiffness matrix %

58、 k is assembled.,,,,2.2 整體剛度矩陣的形成,K(2*i-1,2*i-1) = K(2*i-1,2*i-1) + k(1,1); K(2*i-1,2*i) = K(2*i-1,2*i) + k(1,2); K(2*i-1,2*j-1) = K(2*i-1,2*j-1) + k(1,3); K(2*i-1,2*j) = K(2*i-1,2*j) + k(1,4); K(2*i,2*i-1) = K(2*i,2*i-1) + k(2,1); K(2*i,2*i) = K(2*i,2*i) + k(2,2); K(2*i,2*j-1) = K(2*i,2*j-1) + k(2

59、,3); K(2*i,2*j) = K(2*i,2*j) + k(2,4);,function y =BeamAssemble(K,k,i,j),3-6 梁元,K(2*j-1,2*i-1) = K(2*j-1,2*i-1) + k(3,1); K(2*j-1,2*i) = K(2*j-1,2*i) + k(3,2); K(2*j-1,2*j-1) = K(2*j-1,2*j-1) + k(3,3); K(2*j-1,2*j) = K(2*j-1,2*j) + k(3,4); K(2*j,2*i-1) = K(2*j,2*i-1) + k(4,1); K(2*j,2*i) = K(2*j,2*

60、i) + k(4,2); K(2*j,2*j-1) = K(2*j,2*j-1) + k(4,3); K(2*j,2*j) = K(2*j,2*j) + k(4,4); y = K;,2020/9/7,52,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%BeamElementForces This function returns the element nodal force % vector given the element stiffness matrix k % and the element nodal displacement vector u.,2.

61、4 節(jié)點(diǎn)載荷計(jì)算,y = k * u;,function y = BeamElementForces(k,u),3-6 梁元,2020/9/7,53,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%BeamElementShearDiagram This function plots the shear force % diagram for the beam element with nodal % force vector f and length L.,2.4 繪制剪力圖,x = 0 ; L; z = f(1) ; -f(3); hold on; ti

62、tle(Shear Force Diagram); plot(x,z); y1 = 0 ; 0; plot(x,y1,k),function y = BeamElementShearDiagram(f, L),3-6 梁元,2020/9/7,54,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,2、MATLAB函數(shù)編寫,%BeamElementMomentDiagram This function plots the bending moment % diagram for the beam element with nodal % force vector f and length L.,2.4 繪制彎矩圖,x = 0 ; L; z = -f(2) ; f(4); hold on; title(Bending Moment Diagram); plot(x,z); y1 = 0 ; 0; plot(x,y1,k),function y = BeamElementMomentDiagram(f, L),3-6 梁元,2020/9/7,55,,,,結(jié)構(gòu)分析編程及軟件應(yīng)用,3、實(shí)例計(jì)算分析應(yīng)用,,,,3-6 梁元,