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Q1:Problem 1, 19pts An initially straight beam is bent into a circle with radius R as shown in the figure. Material fibers that are perpendicular to the axis of the undeformed beam are assumed to remain perpendicular to the axis after deformation, and the beam's thickness and the length of its axis are assumed to be unchanged. Under these conditions the deformation can be described 2₁ = (R-X₂) sin(¹), 2₂-R- (R-X2₂) cos( (a) Calculate the deformation gradient field in the beam, expressing your answer as a function of X₁, X₂,, and as components in the basis e₁,e₂, e shown. (b) Calculate the Green Lagrange E* strain field in the beam. (c) Calculate the infinitesimal strain field E in the beam. (d) Compare the values of Lagrange strain and infinitesimal strain for two points that lie at (X₁ = 0, X₂=h) and (X₁ = L, X₂-0). Explain briefly the physical origin of the difference between the two strain measures at each point. Recommend maximum allowable values of h/R and L/R for use of the infinitesimal strain measure in modeling beam deflections. (e) Calculate the deformed length of an infinitesimal material fiber that has length lo and orientation e, in the undeformed beam. Express your answer as a function of X₂.See Answer
Q2:Problem 5, 21pts Given the following motion in rectangular Cartesian coordinates: X(X, t) = 3X3e₁-X₁е₂ - 2X₂03 (a) Determine the deformation gradient F, right Cauchy-Green tensor C, the right stretch tensor U, and the stretch invariants I₁, I2, (b) Determine the Green Lagrange strain tensor E* = (C-1). (c) Determine the ratio of deformed volume to initial volume J./nProblem 6; 20 pts A circular cylindrical bar of length / hangs vertically under gravity force from the ceiling. Let the axis coincide with the axis of the bar and point downward, and let the point (x1, x2, 3) = (0,0,0) be fixed at the ceiling. (a) Verify that the following stress field satisfies the equation of equilibrium in the presence of the gravity force: T₁1=pg(1-₁), all other T₁ = 0. 3 (b) Verify that the boundary conditions of zero surface traction on the lateral surface and the lower end face are satisfied. (c) Obtain the resultant force of the surface traction at the upper end face.See Answer
Q3:According to the Cayley-Hamilton theorem, a second-order tensor should satisfy its own characteristic equation, thusSee Answer
Q4:Using the results of Problem 1 with A = F (the deformation gradient), we find thatSee Answer
Q5:Let A be a given general second-order tensor and consider that we need to satisfy V.A.v>0See Answer
Q6:The behavior of an isotropic non-linear material is described by Green elastic expression and where a, b, c are constants. Please ( a ) derive the stress-strain relations for this material; ( b ) show that the constitutive equations are reduced to those of the isotropic linear elastic material for c = 0; ( c ) find the elastic moduli E and for this case. See Answer
Q7:The equilibrium equation for the linear elastic one-dimensional rod (as shown in Fig. 1), with body force neglected, is d/dx EA du/dx= 0 Please solve u1 at x=1 by Galerkin’s approach. See Answer
Q8:Assume is a symmetric second order tensor, please prove that satisfy the equilibrium equation when , Fi=0See Answer
Q9:The stress tensor at a point under the working condition is given by See Answer
Q10:Please calculate or explain that (a) EijkErjk = ? (b)EijkUjUk=?See Answer
Q11: Assuming that =find Kp and Ka such that both of the closed loop poles (roots of the characteristic equation) are at -= m = 1 = 1 and that g = 10 (instead of 9.81, to make the numbers nice),mp1.See Answer
Q12: d. A slightly more complicated controller has the form D(s) = Kp + Kas. Derive the closed-looptransfer function when this controller is in use.See Answer
Q13: c. Is there a value of K, that can make the system stable? Why or why not? (While we have not yetdiscussed controller design or stability much, you should be able to answer this based on thematerials covered already.)See Answer
Q14: Example 2.7 in the 7th edition of the FPE textbook describes a model for an inverted cart-and-pendulum system that is similar to the Segway scooter. Let's take a more detailed look at what is neededto balance an inverted pendulum or Segway. a. First, start with the equations of motion for this system that are given in the book (equation 2.30): \begin{array}{l} \left(I+m_{p} l^{2}\right) \ddot{\theta}^{\prime}-m_{p} g l \theta^{\prime}=m_{p} l \ddot{x} \\ \left(m_{t}+m_{p}\right) \ddot{x}+b \dot{x}-m_{p} l \ddot{\theta}^{\prime}=u \end{array} Derive the transfer function from the cart input U(s)to the angle O'(s).See Answer
Q15: f. Using Matlab, plot the unit step response of your closed loop system with these gains to a stepchange in reference input. Use zero initial conditions. (Submit a printout of your Matlab code andyour plot.) What happened? Did you get the response you want?See Answer
Q16: b. Now assume you have a feedback loop closed around the system with a controller D(s) as in thefollowing block diagram: Derive the closed loop transfer function from reference 0,(s) to pendulum angle O'(s), assumingthat the controller is a simple gain, D(s) = Kp. You should also assume that b 0 (makes thingscleaner for now).%3DSee Answer

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