The shortest distance (d) to the edge of the rectangular-plate is equal to 66.65 mm.
Given the followin data:
Thickness (length) of plate = 10 mm to m = 0.001 m.
Width of plate = 200 mm to m = 0.2 m.
c = [tex]\frac{0.2}{2}[/tex] = 0.1 m.
How to calculate the shortest distance.
First of all, we would determine the area of the rectangular-plate and its moment of inertia.
For area:
[tex]A=LW\\\\A=0.001 \times 0.2\\\\A= 0.002 \;m^2[/tex]
For moment of inertia:
Mathematically, the moment of inertia of a rectangular-plate is given by this formula:
[tex]I=\frac{b^3d}{12} \\\\I=\frac{0.2^3 \times 0.01}{12} \\\\I=\frac{0.008 \times 0.01}{12}\\\\I=\frac{0.0008 }{12}\\\\I=6.67 \times 10^{-6}\;m^4[/tex]
The compressive stress of a rectangular-plate with respect to axial and bending stress is given by this formula:
[tex]\sigma = \frac{P}{A} -\frac{Mc}{I} \\\\\sigma = \frac{P}{0.002} -\frac{P(0.1-d)\times 0.1}{6.67 \times 10^{-6}} \\\\\sigma = \frac{P}{0.002} -\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}} \\\\\frac{P}{0.002}=\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}}\\\\500P=\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}}\\\\3.335 \times 10^{-3}P=0.01P-0.1Pd\\\\[/tex]
[tex]0=0.01P-0.1Pd-3.335 \times 10^{-3}P\\\\0=(0.01-0.1d-3.335 \times 10^{-3})P\\\\0.01-0.1d-3.335 \times 10^{-3}=0\\\\0.1d=0.01-3.335 \times 10^{-3}\\\\d=\frac{6.665\times 10^{-3}}{0.1} \\\\d=6.665\times 10^{-2}\;m\\\\[/tex]
d = 66.65 mm.
Read more on moment of inertia here: https://brainly.com/question/3406242
