The Euler bucking load for a long slender column with fixed end condition is 10000 N. If the end condition are changed to fixed free, Euler bucking load (in N) would be

5000

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2500

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1250

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625

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Solution

The correct option is D 625
$P_{c r} = \frac{π^{E} I}{L_{c}^{2}}$

$L_{c} = 2 L f o r f i x e d - f r e e e n d c o n d i t i o n s$

$= \frac{L}{2} f o r f i x e d - f i x e d e n d c o n d i t i o n s$

$P_{1} L_{1}^{2} = P_{2} L_{2}^{2}$

$P_{2} = P_{1} (\frac{L_{1}}{L_{2}})^{2} = P_{1} (\frac{L}{2 \times 2 L})^{2} = \frac{P_{1}}{16} = \frac{10000}{16} = 625 N$

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The Euler bucking load for a long slender column with fixed end condition is 10000 N. If the end condition are changed to fixed free, Euler bucking load (in N) would be

The correct option is D 625Pcr=πEIL2c Lc=2L forfixed−free end conditions =L2forfixed−fixed end conditions P1L21=P2L22 P2=P1 (L1L2)2=P1 (L2×2L)2=P116=1000016=625N