Answer to Question #154891 in Physics for aRIF HASAN JOY

Question #154891

A 50 x 100 mm timber is used as a column with fixed ends. Determine the minimum length required at which Euler’s formula can be used if E= 10 GPa and critical stress up to proportional limit is 30 MPa. What safe load can be carried with a FS of 2 if the length is 2.5 m


1
Expert's answer
2021-01-11T11:33:33-0500

The Euler buckling load is


Pcr=π2EI(KL)2σcrA=π2EI(KL)2P_{cr}=\frac{\pi^2EI}{(KL)^2}\to \sigma_{cr}A=\frac{\pi^2EI}{(KL)^2}


L=π2EIK2σcrAL=\sqrt{\frac{\pi^2EI}{K^2\sigma_{cr}A}}


I=10050312=1.04167106(mm4)=1.04167106(m4)I=\frac{100\cdot50^3}{12}=1.04167\cdot10^6(mm^4)=1.04167\cdot10^{-6}(m^4)


A=10050=5000(mm2)=5103(m2)A=100\cdot50=5000(mm^2)=5\cdot10^{-3}(m^2)


L=π2EIK2σcrA=3.142101091.041671060.52301065103=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) . Answer


σcr=π2EIA(KL)2=3.142101091.041671065103(0.52.5)2=13.1(MPa)\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)


Pcr=σcrA=65500(N)=65.5(kN)P_{cr}=\sigma_{cr}A=65500(N)=65.5(kN)


safe load 65.5/2=32.75(kN)65.5/2=32.75(kN) . Answer







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