Problem 2: The material used for an eccentrically loaded column (L=5\": P - 20,0

Question

Problem 2: The material used for an eccentrically loaded column (L=5": P - 20,000 lbs) and rectangular cross section 3-x 6" has a yield point 'p 30,000 psi. The deflection curve is r(x)=e[tan(kL /2)sin(kr) +cos(kx)-where k-/P7E7, and moment in the column. (a) Show that the maximum moment in M-re+ dol is the the column is v(x) M = Pe sec[(L/2.)./P/EA] , where ,,-JF/A (b) Show that the maximum compressive | | +te sec-/(P / EU) | | , (c) Determine the maximum value um shear stress theory. (I-bh3/12); Safety factor N ec stress is given by of the eccentricity e using the maxim 1.25, E = 30 × 106 psi.

Explanation / Answer

a) the moment in column is given by P{e+e[tan(kL/2)sin(kx)+cos(kx)-1]}

Maximum moment in column will be at midspan,i.e., x=L/2

Therefore, P{e+e[tan(kL/2)sin(kL/2)+cos(kL/2)-1]}=Pe{1+tan(kL/2)sin(kL/2)+cos(kL/2)-1}=Pe[tan(kL/2)sin(kL/2)+cos(kL/2)] = Pe[sin2(kL/2)/cos(kL/2) + cos(kL/2)]=Pe[(sin2(kL/2)+cos2(kL/2)]/cos(kL/2)]=Pe*sec(kL/2)=Pe*sec[sqrt(P/EI)*L/2]=Pe*sec[sqrt(P/EAr2)*L/2]

=Pe*sec[L/2r*sqrt(P/AE)]

b)maximum compressive stress =stress due to axial load+stress due to bending moment

Maximum compressive stress = P/A + M/S=P/A + Pec*sec[L/2r*sqrt(P/AE)]/Ar2

=(P/A)[1+ec*sec[L/2r*sqrt(P/AE)]/r2]=(P/A){1+(ec/r2)*sec[L/2r*sqrt(P/AE)]}

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