Chemical Engineering - Mechanics of Solids and Structures II

Full exam

Test on 14/02/2019 Student (Name, Surname)_________________________________________ Id. Number (Matr.):___________ Please note that: correct solution of exercise 1 is mandatory to pass the exam; you are required to write in English. Please explain all what you want to do with your computations or you will be penalized. Exercise 1 We need to install a tall vertical distillation column with outer diameter D 0 = 4.2 m (including insulation of 0.3 m in thickness) and total height H = 42 m. The available specifications are: External diameter of the vessel d = 3.6 m. Thickness of vessel shell t = 5 mm. Weight density of vessel shell s=76.00 kN/m 3. Weight density of insulation i = 5.0 kN/m 3. Load per unit of vertical length of the external access ladder: w l = 0.30 kN/m. Load per unit of vertical length of a 280 mm outer diameter pipe: w p =0.60 kN/m. Load per unit of vertical length equivalent to tray weight, including liquid, (computed assuming a spacing 0.6 m of the trays and a weight of the trays of 1.1 kN/m 2): w t = 18.65 kN/m. Permissible material stress of vessel shell adm = 160 MPa. Young Modulus of vessel shell material E = 210 GPa. Operating temperature and pressure – 160°C and 0.4 MPa. Wind force acting below the lower 20 m q w1 = 3.0 kN/m, wind force acting above the lower 20 m q w2 = 3.0 kN/m (see picture). 1) After having schematized the distillation column as vertical cantilever, and considering only the bending stiffness of a circular hollow cross-section of external diameter d = 3.6 m and thickness t = 5 mm. (i.e. considering as providing stiffness only the section of the vessel), evaluate through an application of the Virtual Work Principle (VWP) the HORIZONTAL DISPLACEMENT at the top of the distillation column. 2) Select 4 points on the base circumference of the shell (d = 3.6 m) and check there the material of the vessel using the Von Mises criteria for the normal and tangential stressed due to: internal PRESSURE, self WEIGHT of the distillation column and of appendages (ladder, insulation, external pipe) and WIND forces. Note that: only the shell is providing resistance to the structure. Remember that the second area modulus for the ring having external diameter d e and internal diameter d i is: I 0 = (/64) (d e4 - d i4). Figure 1. Exercise 2 Answer the following theoretical question: 1. Demonstrate the equation for the value of the Euler critical load of a beam restrained by a hinge and a roller. Solution Exercise 1 The structure is statically determinate, so we just need to use the Muller-Breslau equation, and we can disregard axial and shear deformability since EA = inf. And GA = inf, by the data given. Having denoted with f the rotation at the top, with q 1 the lateral load over h 1, with q 2 that over h 2, with M 1 the moment due to a unit moment applied at the top, with M 0 the moment due to the lateral load q 1 and q 2, the equation of the PVW becomes: 1*f = integrate(M 1(x)*M 0(x)/EJ, x, 0, h1+h2) H 42 m D0 4.2 m d 3.6 m t 5 mm gamma_s 76 kN/m^3 gamma_i 5 kN/m^3 w_l 0.3 kN/m ladder w_p 0.6 kN/m pipe w_t 18.65 kN/m trays+liquid p 0.4 N/mm^2 pressure q_w 3 kN/m q_zi 18.37831702 kN/m q_zs 4.291729724 kN/m q_zl 0.3 kN/m q_zp 0.6 kN/m q_zt 18.65 kN/m q_zTot 42.22004675 kN/m N 1773.241963 kN M 2646 kNm V 126 kN A_s 0.056470128 m^2 I0_s 91227844639 mm^4 r_i 1.795 m^2 sigma_nN -31.40141572 N/mm^2 sigma_np 71.8 N/mm^2 sigma_nM 52.06272294 N/mm^2 sigma_rp 143.6 N/mm^2 tau_nrV 4.468751605 N/mm^2 sigma_n1 -11.66413866 sigma_r1 143.6 tau_nr1 0 sigma_VM 149.7731032 sgima_n2 92.46130722 sigma_r2 143.6 tau_nr2 0 126.0658939 sigma_n3 40.39858428 sigma_r3 143.6 tau_nr3 4.468751605 128.4977748