Question

Derive the relation of the critical load of a long column pinned at both ends as shown

where E, I, and L are Young’s modulus, moment of Inertia and the length of the

column respectively.

Answer 1

Derivation for the relation of the critical load of a long column pinned at both ends In long column there will be buckling hence bending stress is high com pored to direct stress. Euler derived an equation for the long column based on bending stress for the buckling load. Assumptions in the Euler's theory • Initially column should be straight- • cross-section should be uniform throughout. Material should be homogeneous I Isotropic self weight is neglected · Axial compression is negligible Failure only due to buckling. Sign conversions for bending moments: 1: -ve tve Thi Positive negabive

y deflection of the column at X. p. Moment at x=) M=-py Esde -py Solving differential equation | X che 02- Assume Er = 4² y"+k+y =0 (D²+k²)y=0 General formula for the differential equation y = complementang function+ Particular (CF) Integral (PI) CF: Auxiliary equation is m²+k²=0 m-o iki General formula for m=m, Im :-) CF=em. Acosmx +B sinmx) CP = A coska + Bsinka PI=0.

We know that X=o, Y=0 =) A=0 when xal, then Y=0 = 0 = Bsin(kl). y = Acos(kx) + Bsin (kx) If sin(kl) = 0 =) klant Kl= 0, 1, 21, 30.... the least significant value Kl= K²2²=172 = 1² = 17² p= DEI Take the ez & I e Young's Modulus Moment of Inertia length of the column.