Question

Q2. A hollow mild steel column has 25mm thickness and is 20cm in external diameter with 6m length. The modulus value for mild steel is 1.2 x 106 N/mm² and this column is hinged at both ends find the Euler's crushing load and compare the load with the crushing load given by the Rankine's formula Determine for what length these two formulae will give the same crushing load? Finally Calculate the critical stresses for the series of strut having slenderness ratio of 40, 80, 120 and 200 for the same above conditions.

Answer 1

Given:

Part A

D= external dia = 200 mm,

t= thickness = 25 mm

d= internal dia = D-2t=200-2*25=150 mm

Area = A= *D2 - 12) = (2002 – 1502) = 13744.46mm?

- 1504) = 53.68 * 106mm

k = radius of gyration = VĀ= 13744.46 53.68 * 106 = 62.49mm

Both ends hinged.hence, leff = 1 = 6000mm

$\lambda =Slenderness \: ratio=\frac{le}{k}=\frac{6000}{62.49}=96.01$

1. fc * A PR = 1 + a2

$P_{R}=\frac{320*13744.46}{1+\frac{1}{7500}(96.01 ^2)}$

The following table shows the value of fe and a for different materials. Material f N/mm² a= Wrought iron 250 9000 Cast iron 550 1600 Mild steel 320 7500 Timber 50

Part B

$P_{E}=\frac{\pi ^{2}EI}{l_{eff}^{2}}$

Both ends hinged.hence, leff = 1 = 6000mm

$P_{E}=\frac{\pi ^{2}1.2*10^6*53.68*10^6}{6000^{2}}$

Its an assumption that the buckling capacity of a column obtained by Euler’s theory and that obtained by Rankine-Gordon’s formula are independent of each other.

Rankine-Gordon’s formula is for maximum buckling load of a column is

Where Pe = Euler crippling load Pc = Crushing load or Yield point load in Compression PR = Actual load to cause failure or Rankine load Where

Part C

Rankine’s formula intuitively makes sense because, for a short column compressive strength governs and for a long column, buckling governs. We can see the idea that both these formula are independent is incorrect.

Rankine’s load and Euler’s load can be virtually equal when the length of the column is very long, in which case 1/Pe is very large and hence 1/Pc can be neglected. i.e., Pr=Pe

Part D

The Rankine loads for following struts having slenderness ratios given are:

1. $\lambda = 40$

$P_{R}=\frac{320*13744.46}{1+\frac{1}{7500}(40 ^2)}$

The critical stress is given by,

$\sigma _{CR}=\frac{P_{R}}{A}=\frac{3624.91*10^3}{13744.46}=263.73\: N/mm^2$

2. $\lambda = 80$

$P_{R}=\frac{320*13744.46}{1+\frac{1}{7500}(80 ^2)}$

The critical stress is given by,

$\sigma _{CR}=\frac{P_{R}}{A}=\frac{2373.14*10^3}{13744.46}=172.66\: N/mm^2$

3. $\lambda = 120$

$P_{R}=\frac{320*13744.46}{1+\frac{1}{7500}(120 ^2)}$

The critical stress is given by,

$\sigma _{CR}=\frac{P_{R}}{A}=\frac{1506.24*10^3}{13744.46}=109.58\: N/mm^2$​​​​​​​

4. $\lambda = 200$

$P_{R}=\frac{320*13744.46}{1+\frac{1}{7500}(200 ^2)}$

The critical stress is given by,

$\sigma _{CR}=\frac{P_{R}}{A}=\frac{694.45*10^3}{13744.46}=50.52\: N/mm^2$