Question

A column is to support a statue concentrically that weighs 200 kN. The cross section of the column is a regular hexagon with a hole in its center. The diameter of the hole is 75mm. If the stress in the column is limited to 250 MPa, what is the measure of one side of the hexagon? Answer should be in mm.

Answer 1

Solution:- the values given in the question are as follows:

column support concentrically weight(P)=200 kN , or 200*10^3 N

diameter of hole(d)=75 mm

stress in column=250 MPa , or 250 N/mm^2

Calculating side of hexagon:-

let side of hexagon column is a, shown in figure.

area of hexagon(A)={3(3)^0.5*a^2}/2

where, a=side of hexagon

area of hole=(Pi/4)*d^2

area of hole=(3.14/4)*75^2=4415.625 mm^2

area of hole=4415.625 mm^2

area of column=area of hexagon-area of hole

area of column={(3(3)^0.5)/2}*a^2-4415.625

stress of column=load on column/area of column

stress of column=(200*10^3)/[{(3(3)^0.5)/2}*a^2-4415.625] , (Eq-1)

stress in column=250 N/mm^2 pu in equation-(1) and calculate the value of a

250=(200*10^3)/[{(3(3)^0.5)/2}*a^2-4415.625]

250*[{(3(3)^0.5)/2}*a^2-4415.625]=(200*10^3)

649.519*a^2-1103906.25=200*10^3

649.519*a^2=1303906.25

a^2=1303906.25/649.519

a^2=2007.495

a=44.805 mm

side of hexagon(a)=44.805 mm

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