不定跨连续梁各种荷载与不同端约束下弯矩计算程序(The calculation p条rocedure of bending moment under different load and different end of continuous beam).doc

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1、不定跨连续梁各种荷载与不同端约束下弯矩计算程序(The calculation procedure of bending moment under different load and different end of continuous beam)#包含iostream 使用名称空间;双MINVERSE(双 ,int n);双mmult(双 , 双B,int n);国际main()int,i,j;常量双EI = 20000;cout “请输入连续梁跨数:“ endl;CIN;双L n ;cout “请从左向右依次输入每跨跨长,用空格隔开:(1代表4米,2代表6米,3代表8米)“ endl;

2、对于(i = 1;i = n;i + +)CIN;开关(j)案例1:l = 4;中断;案例2:l = 6;中断;案例3:l = 8;中断;默认值:cout “对不起,输入错误” endl;打破;cout “第”我+ 1 “跨的长度为”我我 “米” endl; /输入跨长双EI n ;cout “请从左向右依次输入每跨抗弯刚度,用空格隔开:(EI = 2.0x10000千牛平方米。1代表EI,2代表1.5ei,3代表2ei)“ endl;为(i = 0;i ;+(+)CIN;开关(j)案例1:EI I = EI;中断;案例2:EI I = 1.5 * EI;中断;案例3:EI I = 2 * E

3、I;中断;默认值:cout “对不起,输入错误” endl;打破;cout “第”我+ 1 “跨的抗弯刚度为” EI 我 “千牛平方米” endl; /输入抗弯刚度双k n + 1 n + 1 ;对于(i = 0;i = n;i + +)对于(j0;jn;j + +)如果(J = = i-1)K 我 J = 2EI JL J.;否则如果(j = i = j = J = 0)k i 4 = * i i ;如果(J = =我& 1 = j = n-1)K 我 J = 4EI L + 4 * EI 我L 我;如果(J = =我和J = = N)K 我 J = 4EI L ;否则,如果(j = i =

4、 1)k i 2 = EI i i ;否则k i = 0; /计算整体刚度矩阵cout “整体刚度矩阵为:“ endl;对于(i = 0;i + 1;i + +)对于(j0;j n + 1;j + +)cout SEtW商店(7) K 我 J ;cout endl; /输出整体刚度矩阵双FP的 2 ,FQ的 2 ;/定义各杆固端弯矩双p;cout “请从左向右依次输入每跨跨中荷载,用空格隔开:(0代表无,1代表40kn,2代表50kN,3代表80kn)“ endl;为(i = 0;i ;+(+)CIN;开关(j)案例0:pi0;断裂;案例1:pi40;断裂;案例2:pi50;断裂;案例3:pi

5、80;断裂;默认值:cout “对不起,输入错误” endl;打破;cout “第”我+ 1 “跨的跨中荷载为” P 我 “千牛” endl;FP 0 = 8 i ;FP 我的 1 = P 我我我 / 8 / /计算各杆固端弯矩; /输入跨中荷载双q n ;cout “请从左向右依次输入每跨均布荷载,用空格隔开:(0代表无,1代表15knm,2代表30knm,3代表20kn / M)“ endl;为(i = 0;i ;+(+)CIN;开关(j)案例0:q I0;断裂;案例1:q I15;断裂;案例2:q I30;断裂;案例3:q I20;断裂;默认值:cout “对不起,输入错误” endl;

6、打破;Cout No. i+1 the cross load is qi kN/m endl;Fqi1=-qi*li*li/12;Fqi2=qi*li*li/12; / / calculate rod fixed end moment/ input loadDouble Fn+1; / / define summing point load vectorF0=-fP00-fq00;Fn=-fPn-11-fqn-11;For (i=1; in; i+)Fi=- (fPi-11+fPi0+fqi-11+fqi0);/ calculate the total nodal load vectorCou

7、t the total nodal load vector is: endl;For (i=0; in+1; i+)CoutFiendl;/ output summary point load vectorInt, a, b;Double Jn+1; / / define summing point angular displacement vectorDouble jjn2; / / define each link node displacementCout enter a constraint (1 represents an articulation, and 2 represents

8、 a solid connection): ab;If (a=1&b=1)J=mmult (minverse (K, n+1), F, n+1);For (i=0; in; i+)Jji0=Ji;Jji1=Ji+1;/ right left hinge hingeElse if (a=2&b=1)Double Kbnn;For (i=0; in; i+)For (j=0; jn; j+)Kbij=Ki+1j+1;Double Fbn;For (i=0; in; i+)Fbi=Fi+1;Double Jbn;Jb=mmult (minverse (Kb, n), Fb, n);For (i=1;

9、 in; i+)Jji0=Jbi-1;Jji1=Jbi;Jj01=0;Jj01=Jb0;/ right left fixed hingeElse if (a=1&b=2)Double Kann;For (i=0; in; i+)For (j=0; jn; j+)Kaij=Kij;Double Fan;For (i=0; in; i+)Fai=Fi;Double Jan;Ja=mmult (minverse (Ka, n), Fa, n);For (i=0; in-1; i+)Jji0=Jai;Jji1=Jai+1;Jjn-10=Jan-1;Jjn-11=0;/ left hinge fixed

10、 rightElse if (a=2&b=2)Double Kabn-1n-1;For (i=0; in-1; i+)For (j=0; jn-1; j+)Kabij=Ki+1j+1;Double Fabn-1;For (i=0; in-1; i+)Fabi=Fi+1;Double Jabn-1;Jab=mmult (minverse (Kab, n-1), Fab, n-1);For (i=1; in-1; i+)Jji0=Jabi-1;Jji1=Jabi;Jj00=0;Jj01=Jab0;Jjn-10=Jabn-2;Jjn-11=0;/ left right solid solidElse

11、 cout excuse me, enter an error. endl.;Double Mn2; / / define the rod end momentDouble k22; / / define the rod stiffness matrixFor (i=0; in; i+)K00=4*EIi/li; k01=2*EIi/li;K11=4*EIi/li; k10=2*EIi/li;Mi0=k00*jji0+k01*jji1+fqi0+fPi0;Mi1=k10*jji0+k11*jji1+fqi1+fPi1;Cout No. i+1 rod end bending moment is

12、 endlMi0endlMi1endl;Return 0;Double minverse (double M, int n) / / matrix inversionDouble, Mnn, Nnn, Ln2*n, a;Int, I, J, k=0;For (i=0; in; i+)For (j=0;j2;如果(j )如果(i)!= j)n i = 0;否则n i = 1; i ;另外我我 J = n 我【N】;对于(k = 0;k;n;k +)如果(l k = k = 0)对于(i = k;i;+)如果(l k != 0)对于(j = k;j2;n;j +)a = l j ;l i ; i ;打破;继续;对于(i = k + 1;i = 0;我-)a = l k ;对于(j0;j2;n;j +) i ;为(i = 0;i ;+(+)a = i i ;对于(j = i;j2;n;j +)l i ;为(i = 0;i ;+(+)对于(j0;j;n;j +)n i ;cout SEtW商店(10) n 我 J ;cout endl;返回(n);双mmult(双 , 双B,int n)/矩阵与向量相乘运算双A,n,n,n,n;int,j,k;为(i = 0;i ;+(+)对于(k = 0;k;n;k +)c i ;返回(c);

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