Problem 5 - Solve this problem using Matlab
Figure below shows a beam of length L, cross-sectional area A, and moment of inertia I. Young's modulus for the material is designated as E. The left end of the beam is hinged and the other end rests against a smooth wall (friction is neglected), at a height H above the horizontal. As load P is increased from zero, the right end slides downward a distance X because of deformation of the beam to buckle as an Euler column. As the beam buckles, it assumes a bow shape as shown in the figure below. Because the beam supports axial loads only, the load P = 0 is at the undeformed position X = 0 and also at the horizontal position X = H.
Un-deformed Position
Deformed Position
Using simple engineering mechanics, we can estimate the axial force in the beam prior to buckling as a function of the displacement X to be X - 2HX.
The axial force under which the beam buckles as an Euler column is Iax = H^2/12.
The actual force F in the beam under a given deflection X, where 0 ≤ X ≤ H, is the smaller force calculated from the equations above; F = min(F1, F).
Knowing the force F in the beam, we can determine the load P from statics to be (H - X) * sqrt(r^2 + X^2) - 2HX.
Consider now a specific case:
H = 0.25m, L = 2.5m, A = 25 * 10^-4 m^2, I = 75 * 10^-10 m^4, E = 70 * 10^9 N/m^2.
Using the quadratic interpolation method, find the maximum load Pmax and the corresponding X. (NOTE: Starting from the initial design variable Xo = 0, calculate the upper bound, b, for the optimal solution in the quadratic interpolation code.)