In Prob. 24.16, the basic differential equation of the elastic curve for a uniformly loaded beam was formulated as E I d2 y dx2 = wLx 2 wx2 2 Note that the right-hand side represents the moment as a function of x. An equivalent approach can be formulated in terms of the fourth derivative of deflection as E I d4 y dx4 = w For this formulation, four boundary conditions are required. For the supports shown in Fig. P24.16, the conditions are that the end displacements are zero, y(0) = y(L) = 0, and that the end moments are zero, y(0) = y(L) = 0. Solve for the deflection of the beam using the finite-difference approach (x = 0.6 m). The following parameter values apply: E = 200 GPa, I = 30,000 cm4 , w = 15 kN/m, and L = 3 m. Compare your numerical results with the analytical solution given in Prob. 24.16.

PY 205 Daniel Dougherty Week 2 Notes Chapter 3 - Kinematics in two or three dimensions Vectors and scalars – velocity is how fast and in what direction the particle is moving o Magnitude – vector quantity o Scalar quantities are specified by numbers and units Addition of vectors – graphical methods o above D = displacement vectors...