Question

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. Question 9.7 machenical properties of solid

Answer 1

total mass support by cylindrical column(m) = 50000 kg
so, total weight support by cylindrical column= mg = 50000×9.8 N
                    = 490000 N
loading for each cylindrical support = 490000/4 = 122500 N
inner radius of each column(r1) = 30cm = 0.3 m
outer radius of each column (r2) = 60 cm = 0.6m
so, CSA of each column = π(r₂² -r₁²)
                       = 3.14 (0.6²-0.3²)
                       = 3.14×0.27 m²

young's modulus (Y) = 2×10¹¹ N/m²

we know,
young's modulus = stress/strain
so,
so, the compressional strain of each column = F/AY
                = 122500/(3.14×0.27)2×10¹¹
                =7.22 ×⁻⁷ 

Answer 2

the answer is =7.22×10^-7

Explanation:

________now explanation.....

total mass support by cylindrical columns (m)=50000kg

so,total weight support by cylindrical columns=50000×9.8N........

loading for each cylindrical support=490000/4=122500N......

inner radius of each column=0.3m...

outer radius of each column=0.6m...

So,curve surface area of each column=π【R^2-r^2】

=3.14×(0.6^2-0.3^2)

=3.14×0.27m^2.......

Now, Young's modulus (y)=2×10^11N/m^2......

soo,,

we know..........

the Young's modulus=stress/stain.....

_________so,the congressional strain of each column=

F/Ay.....

• _________----after calculation......

122500/(3.14×0.27)×2×10^11.....

_____--------7.22×10^-7......

___________________hope this helps to all of u friends!!