MATLAB

Question

我想要实现的是一个经典的弯矩分布图，可能看起来像这样：

我尝试使用面积、xy 和条形图，最后一个最接近我需要的 - 但它仍然不是我可以接受的。我可以使用任意形式的数据。

Answer 1

我会手动解决它生成你想要的行。

%some example plot
x1 = -3;
x2 = 2;
upfun = @(x) -1/10*(x-x1).*(x-x2);
downfun = @(x) 1/5*(x-x1).*(x-x2);
%set slope you want. Inf for vertical lines
slope=inf;

x_dense = linspace(x1,x2,100);
x_sparse = linspace(x1,x2,20);

%plotting it without the stripes. nan is used not to have unintended lines connecting first and second function
plot([x_dense ,nan,x_dense ],[upfun(x_dense),nan,downfun(x_dense)])

x_stripes=nan(size(x_sparse).*[3,1]);
y_stripes=nan(size(x_sparse).*[3,1]);

if slope==inf
    %vertical lines, no math needed to know the x-value.
    x_stripes(1,:)=x_sparse;
    x_stripes(2,:)=x_sparse;
else
    %intersect both functions with the sloped stripes to know where they
    %end
    for stripe=1:numel(x_sparse)
        x_ax=x_sparse(stripe);
        x_stripes(1,stripe)=fzero(@(x)(upfun(x)-slope*(x-x_ax)),x_ax);
        x_stripes(2,stripe)=fzero(@(x)(downfun(x)-slope*(x-x_ax)),x_ax);
    end
end
y_stripes(1,:)=upfun(x_stripes(1,:));
y_stripes(2,:)=downfun(x_stripes(2,:));
x_stripes=reshape(x_stripes,1,[]);
y_stripes=reshape(y_stripes,1,[]);
plot([x_dense ,nan,x_dense,nan,x_stripes],[upfun(x_dense),nan,downfun(x_dense),nan,y_stripes])

坡度示例=1

斜率示例=inf

Answer 2

虽然更通用，可用于倾斜条纹，但这里有一个更简单的解决方案，使用 stem，没有标记和基线：

x1 = -3;
x2 = 2;
upfun = @(x) -1/10*(x-x1).*(x-x2);
downfun = @(x) 1/5*(x-x1).*(x-x2);

x_dense = linspace(x1,x2,100);
x_sparse = linspace(x1,x2,20);

%// plot outline
plot(x_dense,upfun(x_dense),'b-',x_dense,downfun(x_dense),'b-');
hold on;
%// plot stripes
stem(x_sparse,upfun(x_sparse),'b','marker','none','showbaseline','off');
stem(x_sparse,downfun(x_sparse),'b','marker','none','showbaseline','off');

结果：

MATLAB - xy曲线下的条纹区域（弯矩分布）

MATLAB - striped area under the xy curve (bending moment distribution)

plot

matlab-figure