1. ## Plastic modulus

Hello I was wondeing if anybody could help me work this out. I am an Structural Engineering student btw so feel free to have a laugh at my mathematical incompetence. I have an exam tomorrow where I will most likely have to find the plastic modulus of a triangle or trapezuim, this is the axis that cuts a shape into two equal areas. If you had a triangle what height below the apex would you draw this line to cut the shape in two? I have tried using similar triangles and some algebra, but I am just going round in circles (or triangles ) Thanks

2. Originally Posted by gjwo1
Hello I was wondeing if anybody could help me work this out. I am an Structural Engineering student btw so feel free to have a laugh at my mathematical incompetence. I have an exam tomorrow where I will most likely have to find the plastic modulus of a triangle or trapezuim, this is the axis that cuts a shape into two equal areas. If you had a triangle what height below the apex would you draw this line to cut the shape in two? I have tried using similar triangles and some algebra, but I am just going round in circles (or triangles ) Thanks
Hello,

I've attached a sketch.

The area of the larger triangle is calculated by:

$A=\frac12 \cdot B \cdot H$

The area of the smaller triangle is calcutlated by:

$a=\frac12 \cdot b \cdot h$

You know that 2a = A. That means:

$\frac12 \cdot B \cdot H= 2\left(\frac12 \cdot b \cdot h \right)$ . After a few steps of transformation you'll get:

$\frac{b}{B} \cdot \frac{h}{H}=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}$

That means: Draw a parallel to B with $h = \frac{1}{\sqrt{2}}\cdot H$

3. Thanks ear-b, thats saved me lots of time I think I will take that formulae into the exam in my head. Hopefully they will just give us some sort of symetrical shape