Let the sides of the triangle be $a$, $b$, and $c.$
Then by the triangle inequality,
$$a+b>c$$ $$b+c>a$$ $$c+a>b$$
We also have that $a+b+c=15.$
The longest side of this triangle must be shorter than $15/2=7.5$. However, we have to round this up to get an integer, so the longest side must be shorter than $8.$ This is because the remaning sides must sum to less than $8,$ or $7,$ and we won't have a triangle!
Now, to focus on how short the longest side might be, we know that if $a=b=c$ then all of the sides must be the longest side. This is possible if we have side lengths of $5, 5, 5.$ However, if the longest side is $4,$ the remaining sides must sum to $15-4=11.$ We must have more than a 4 for the longest side, as $4+4+4 = 12.$ Let the longest side be l. So, $5\le l \le 7.$
Listing out the possibilities with 5 as the longest side, $$(5,5,5).$$ With 6, $$(6,3,6)~(6,4,5)$$ And with 7, $$(7,1,7) ~ (7,2,6) ~ (7,3,5) ~ (7,4,4)$$ we have a grand total of $$\boxed{7}$$ possibilities for non-congruent triangles.
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