Question

Find the diameter of a short cylindrical cast iron post to
support a compressive load of 200 kN. Assume the F.S.
as 10, and ultimate compressive strength of cast iron as
550 N/mm²

WRONG ANSWER WILL BE REPORTED​

Answer 1

For a Cylindrical Cast Iron Post -

Given :-

$\sf{Compressive \: load \: (P) = 200 \: kN = 200\times 10^{3} \: N}$

$\sf{Compressive \: strength \: (F) = 550 \: N/mm^{2}}$

To Find :-

$\sf{Diameter \: of \: this \: cylindrical \: iron \: post = ?}$

Using Formula :-

$\sf{Compressive \: strength=\dfrac{Compressive \: load}{Cross \: section \: Area}}$

Solution :-

$\sf{F =\dfrac{P}{A}}$

$\sf{A =\dfrac{P}{F}}$

$\sf{A =\dfrac{200\times 10^{3}}{550}}$

$\sf{A =\dfrac{200\times 10^{2}}{55}}$

$\sf{A =\dfrac{40\times 10^{2}}{11}}$

$\sf{A = 3.64\times 10^{2} \: mm^{2}}$

We know that :-

$\sf{Cross \: Section \: Area \: (A) =\dfrac{\pi d^{2}}{4}}$

So ,

$\sf{\pi d^{2} = 4(3.64\times 10^{2})}$

$\sf{\pi d^{2} = 4\times 364)}$

$\sf{\dfrac{22}{7}\times d^{2} = 4\times 364}$

$\sf{d^{2} =4\times 364\times \dfrac{7}{22}}$

$\sf{d^{2} =\dfrac{4\times 364}{3.14}}$

$\sf{d^{2} =\dfrac{400\times 364}{314}}$

$\sf{d^{2} = 400\times 1.16}$

$\sf{d^{2} =4\times 116}$

$\sf{d =\sqrt{464}}$

$\sf{d =4\sqrt{29}}$

$\sf{d =4\times 5.39}$

$\sf{d = 21.56 \: mm}$

$\sf{d = 21.6 \: mm}$

Result :-

$\sf{Diameter \: of \: this \: cylindrical \: iron \: post \: is \: 21.6 \: mm.}$