You want to know the vertical deflection of the rod, which will pull the ends of the
rod inward toward the red arrow.
This is a center point loaded simple beam.
The shear diagram plots as two horizontal rectangles ( up 0.5 on the left, down 1.0
in the middle, up 0.5 on the right.
The bending moment diagram plots like the low pitched roof of a house ( 2 back to back
The next integration is a quadratic equation ( K Y^2 )
The next integration is what you want, a cubic equation. ( K/3 Y^3 )
The distance that was originally horizontal now lies along the
curved surface of the cubic equation.
You can model this using two aluminum yard sticks ( or meter sticks )
in parallel over two fulcrums ( like engineers scales ). One is horizontal
and unloaded, the other is center point loaded, and deflects more and more
with increasing load.
Because the rod is a rectangle, the top half gets shorter ( inside the curve ),
while the outside ( bottom ) gets longer, but the average length does not change,
it just lies along the curved line of the cubic equation. ( the neutral axis
of the rectangular rod bends along the "Minus" K/3 Y^3 line ). ( deflects downward )
Note if you change to a uniformly distributed load, the curve would be a fourth order
equation ( minus K/12 Y^4 equation ).
Coins, or steel washers equally spaced along the aluminum yard ( meter )
stick would make a good distributed load.
Remember the center point must move down vertically, so the distance the
rod moves inward from both ends is symmetrical.
Using a one kilogram mass at a distance of 0.5 meters from both ends should work
if it does not bend the meter stick in half.
The deflection would also be dependent upon the height and width of the rectangle,
and the strength of the material selected. You would need to look up the
Modulus of elasticity of aluminum, or of steel, or wood depending upon
what material is used in the rod.
Golden, Colorado, USA