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Problem_12_18

# Problem_12_18 - 2 E ⋅ I ⋅ k L ⋅ 2 = π 2 E ⋅ 1 2 L...

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Problem 12.18 Given: A steel (E=200 GPa) column with a solid circular cross section has clamped ends, is 2.50 m long and must support a load of 40.0 kN. Find: a) the minimum required diameter using a factor of safety of SF = 2.00 b) What is the minimum value for the proportional limit for the column to buckle elastically? Solution: a) E 200 GPa := L 2.5 m := SF 2.0 := P 40 kN := I π d 4 64 = P cr π 2 E I k L
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Unformatted text preview: 2 E ⋅ I ⋅ k L ⋅ ( ) 2 = π 2 E ⋅ 1 2 L ⋅ ⎛ ⎝ ⎞ ⎠ 2 π d 4 ⋅ 64 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ = P cr SF P ⋅ 80 kN = := d P cr 1 2 L ⋅ ⎛ ⎝ ⎞ ⎠ 2 ⋅ 64 ⋅ π 3 E ⋅ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 1 4 33.702 mm = := b) The minimum value for the proportional limit is: σ PL P cr π d 2 ⎛ ⎝ ⎞ ⎠ 2 ⋅ 89.68 MPa = :=...
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