A long steel column of uniform cross section is subjected to a crippling load of $10\text{ kN}$ as per the Euler designed condition. If the length of the column is reduced to half, then what would be the maximum Euler's crippling load?
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Because the column length term is squared in the denominator of Euler's formula, any change in length impacts the buckling load inversely by the square of that change:
- Double the length $\rightarrow$ Critical load drops to $\frac{1}{4}$th.
- Halve the length $\rightarrow$ Critical load increases by $4$ times.
Step 1: Write Euler's load as a proportion.
Euler's crippling load depends on column length only through \( P \propto \dfrac{1}{L^2} \); the material, cross-section and end conditions stay unchanged here.
Step 2: Halve the length.
For the new length \( L_2 = L_1/2 \), \( \dfrac{1}{L_2^2} = \dfrac{1}{(L_1/2)^2} = \dfrac{4}{L_1^2} \), which is 4 times the old value of \( 1/L^2 \).
Step 3: Scale the load by the same factor.
\[ P_2 = 4 \times 10 = 40\text{ kN} \]
\[ \boxed{P_2 = 40\text{ kN}} \]