Notifications
Clear all
0
08/02/2022 3:26 pm
Topic starter
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 × 103 kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column.
[Use Y = 2.0 × 1011 Pa, g = 9.8 m/s2]
(1) 3.60 × 10–8
(2) 2.60 × 10–7
(3) 1.87 × 10–3
(4) 7.07 × 10–4
1 Answer
0
08/02/2022 3:30 pm
(2) 2.60 × 10-7
Explanation:
Force on each column = \(\frac{mg}{4}\)
Strain = \(\frac{mg}{4AY}\)
= \(\frac{50 \times 10^3 \times 9.8}{4 \times \pi(1 - 0.25) \times 2 \times 10^{11}}\)
= 2.60 × 10-7
This post was modified 3 years ago 2 times by admin