find the deflection and rotation for fixed-fixed beam on a point away from the force
3 ビュー (過去 30 日間)
古いコメントを表示
I'm trying to get the deflection and rotation of a point located away from the line of force's action, the problem is to fined the deflection and the rotation at x=1m and x=0.5m of a fixed-fixed beam has a force in the middle of it (at x=1), knowing that its length =2m, and the force 240N, I got the answers at x=1 but I'm struggeling for x=0.5, I think I need to do interpolation but I'm not sure.
I need help
here is the code I made and the question figure
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1490502/image.png)
syms L;
EI=1;
x1=1;
x2=0.5;
F=240;
%%
for c = 1:2
if c==1
L=1;
%element 1
k1= [ 12, 6*L, -12, 6*L;
6*L, 4*L^2, -6*L, 2*L^2;
-12, -6*L, 12, -6*L;
6*L, 2*L^2, -6*L, 4*L^2]
EA_L_k1=(EI / L^3) *k1
%element 2
k2= [ 12, 6*L, -12, 6*L;
6*L, 4*L^2, -6*L, 2*L^2;
-12, -6*L, 12, -6*L;
6*L, 2*L^2, -6*L, 4*L^2]
EA_L_k2=(EI / L^3) *k2
%Calculate stiffness matrix
K = sym(zeros(6,6));
K(1:4, 1:4) = k1;
K(3:6, 3:6) = K(3:6,3:6) + k2
% boundary condition inducates that u1=v1=u3=v3=0 , p2x=0 p2y=10 kN
F=[0; 0; 240; 0; 0; 0]
k_new=K(3:4,3:4)
u=inv(k_new)*F(3:4,1)
else
%I'm struggeling here I tried this but the answers are wrong
L=0.5 %at x=0.5
%element 1
k1= [ 12, 6*L, -12, 6*L;
6*L, 4*L^2, -6*L, 2*L^2;
-12, -6*L, 12, -6*L;
6*L, 2*L^2, -6*L, 4*L^2]
EA_L_k1=(EI / L^3) *k1
%Calculate stiffness matrix
K = k2;
p=[0;0;240;0;0;0]
f=p(3:4,1)
k_new=K(3:4,3:4)
u=inv(k_new)*f
end
end
0 件のコメント
回答 (1 件)
Umang Pandey
2023 年 10 月 4 日
Hi Mohammad,
I understand that you want to find the deflection and rotation at the middle of the fixed-fixed beam. You can refer to the Page - 18/23 of following document, which demonstrates the same problem:
Best,
Umang
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Calculus についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!