Question

r of solid circular cross section is 50 mm in diameter. The bar is pinned at each end and subject to axial compression. If the proportional limit of the material is 210 MPa and E 200 GPa, determine the minimum length for which Euler's formula is valid. Also, determine the value of the Euler buckling load if the column has this minimum length. Ans. 1.21 m, 412 kN 10.20. The column shown in Fig. 10-10 is pinned at both ends and is free to expand into the opening at the upper end. The bar is steel, is 25 mm in diameter, and occupies the position shown at 16°C. Determine the temperature to o6pc ond F- 200 GPa Neglect the

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Answer 1

"Steel bar diameter = 50 mm limit of proportionality = 210 Mpa (E) youngs modulus - 200 GPa = 200 times 10^3 Mpa Euler's buckling load P = pi^2 epsilon I/lc^2 Moment of inertia in terms of radius of gyration I = A(r_min)^2 vert minimum Radius of gyration gamma_min = squareroot I/A (r_min)^2 = I/A I = A (r_min)^2 vert Substitute (I) form in euler's formula P = pi^2 epsilon A (gamma_min)^2/lc^2 P = pi^2 epsilon A/lc^2/r_min^2 P = Pi^2 epsilon A/(lc/r_min)^2 lc/r_min = effective length of column/minimum radius of gyration lc/r_min = slenderness ration rightwardsdoublearrow ""lambda"" Then P = pi^2 epsilon A/lambda^2 P/A = pi^2 epsilon/lambda^23 P/A = critical stress nothing but limit of proportionality sigma_cr = pi^2 epsilon/lambda^2 \r\n sigma_cr = pi^2 epsilon/lambda^2 substitute the given values in above formula 210 = pi^2 times 200 times 10^3/lambda^2 lambda = squareroot pi^2 times 200 times 10^3/210 lambda = 96.95 lambda = lc/r_min rightwardsdoublearrow r_min = squareroot I/A I = pi/64 (D)^4 = pi/64 (50)^4 = 306796.15 mm^4"