Solution:
Solution
the maximum stress will be at the bottom of the pillar$$stress = \dfrac{F}{A}$$
weight of the pillar is $$volume \times density = length \times Area \times density = 20 \times A \times 2.5 \times 10^3$$
the weight at the bottom of the pillar is $$5A \times 10^4 \times g+ 5 \times 10^5$$
compressive stress given is $$16 \times 10^6$$
therefore
$$16 \times 10^5 =\dfrac{ 5A \times 10^5 + 5 \times 10^5}{A}$$
$$A = 0.45m^2$$