The Euler buckling load is
P c r = π 2 E I ( K L ) 2 → σ c r A = π 2 E I ( K L ) 2 P_{cr}=\frac{\pi^2EI}{(KL)^2}\to \sigma_{cr}A=\frac{\pi^2EI}{(KL)^2} P cr = ( K L ) 2 π 2 E I → σ cr A = ( K L ) 2 π 2 E I
L = π 2 E I K 2 σ c r A L=\sqrt{\frac{\pi^2EI}{K^2\sigma_{cr}A}} L = K 2 σ cr A π 2 E I
I = 100 ⋅ 5 0 3 12 = 1.04167 ⋅ 1 0 6 ( m m 4 ) = 1.04167 ⋅ 1 0 − 6 ( m 4 ) I=\frac{100\cdot50^3}{12}=1.04167\cdot10^6(mm^4)=1.04167\cdot10^{-6}(m^4) I = 12 100 ⋅ 5 0 3 = 1.04167 ⋅ 1 0 6 ( m m 4 ) = 1.04167 ⋅ 1 0 − 6 ( m 4 )
A = 100 ⋅ 50 = 5000 ( m m 2 ) = 5 ⋅ 1 0 − 3 ( m 2 ) A=100\cdot50=5000(mm^2)=5\cdot10^{-3}(m^2) A = 100 ⋅ 50 = 5000 ( m m 2 ) = 5 ⋅ 1 0 − 3 ( m 2 )
L = π 2 E I K 2 σ c r A = 3.1 4 2 ⋅ 10 ⋅ 1 0 9 ⋅ 1.04167 ⋅ 1 0 − 6 0. 5 2 ⋅ 30 ⋅ 1 0 6 ⋅ 5 ⋅ 1 0 − 3 = 1.65 ( m ) L=\sqrt{\frac{\pi^2EI}{K^2\sigma_{cr}A}}=\sqrt{\frac{3.14^2\cdot 10\cdot 10^9\cdot 1.04167\cdot10^{-6}}{0.5^2\cdot 30\cdot10^6\cdot 5\cdot10^{-3}}}=1.65(m) L = K 2 σ cr A π 2 E I = 0. 5 2 ⋅ 30 ⋅ 1 0 6 ⋅ 5 ⋅ 1 0 − 3 3.1 4 2 ⋅ 10 ⋅ 1 0 9 ⋅ 1.04167 ⋅ 1 0 − 6 = 1.65 ( m ) . Answer
σ c r = π 2 E I A ( K L ) 2 = 3.1 4 2 ⋅ 10 ⋅ 1 0 9 ⋅ 1.04167 ⋅ 1 0 − 6 5 ⋅ 1 0 − 3 ( 0.5 ⋅ 2.5 ) 2 = 13.1 ( M P a ) \sigma_{cr}=\frac{\pi^2EI}{A(KL)^2}=\frac{3.14^2\cdot 10\cdot 10^9\cdot 1.04167\cdot10^{-6}}{5\cdot10^{-3}(0.5\cdot2.5)^2}=13.1(MPa) σ cr = A ( K L ) 2 π 2 E I = 5 ⋅ 1 0 − 3 ( 0.5 ⋅ 2.5 ) 2 3.1 4 2 ⋅ 10 ⋅ 1 0 9 ⋅ 1.04167 ⋅ 1 0 − 6 = 13.1 ( MP a )
P c r = σ c r A = 65500 ( N ) = 65.5 ( k N ) P_{cr}=\sigma_{cr}A=65500(N)=65.5(kN) P cr = σ cr A = 65500 ( N ) = 65.5 ( k N )
safe load 65.5 / 2 = 32.75 ( k N ) 65.5/2=32.75(kN) 65.5/2 = 32.75 ( k N ) . Answer
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