Week #8 : Applications with Laplace Goals: Solving application ODEs using Laplace transforms Forced Spring/mass system Deformation of beams under load Transport and diffusion of contaminants in groundwater 1
Spring System with Piecewise Forcing - Example 1-1 Spring System with Piecewise Forcing - Example 1 Problem. Describe the scenario for a spring/mass system defined by the differential equation { my + cy 10t 0 t < 3 + ky = 0 t 3 Can this system be solved in a straightforward way using our earlier solution techniques?
Spring System with Piecewise Forcing - Example 1-2 Predict the motion for the spring/mass using the values given, if the mass starts at equilibrium at t = 0. { y + 4y 10t 0 t < 3 + 20y = 0 t 3
Spring System with Piecewise Forcing - Example 1-3 y + 4y + 20y = { 10t 0 t < 3 0 t 3
Spring System with Piecewise Forcing - Example 1-4 y + 4y + 20y = { 10t 0 t < 3 0 t 3
Spring System - Example 1 - Animation Spring System - Example 1 - Animation - 1
Spring System - Example 2-1 Spring System - Example 2 Problem. Consider a mass-spring system with mass 1 kg, spring constant k = 13 N/m, and damping constant c = 4 N m s 1. Predict the motion of the mass if it is: displaced 1 m from equilibrium and released at t = 0, then at time t = π, the forcing function F ext = 40 sin(3t) is applied. 3
Spring System - Example 2-2 y + 4y + 13y = 40 3 sin(3t)u π
Spring System - Example 2-3 y + 4y + 13y = 40 3 sin(3t)u π
Spring System - Example 2-4 y + 4y + 13y = 40 3 sin(3t)u π
Spring System - Example 2 Animation Spring System - Example 2 Animation - 1
Beam Deflection - Example 1-1 Beam Deflection - Example 1 Recall that the deflection of a horizontal beam under load satisfies EIy (4) = p(x) where y(x) is the deflection (distance away from a straight line), p(x) is the loading in N/m at point x along the beam, E is the elasticity of the material of the beam, and I is the moment of inertia of the beam, derived from the crosssectional size and shape.
Beam Deflection - Example 1-2 EIy (4) = p(x) Problem. Predict the shape under load in the cantilevered beam shown below. Use the boundary conditions y(0) = 0, y (0) = 0, y (4) = 0 and y (4) = 0. 4m 3m 600 N/m We will build this beam out of a pine 2x10, I = 4.1 10 5 m 4, and E = 9 10 9 N/m.
Beam Deflection - Example 1-3 4m 3m
Beam Deflection - Example 1-4 4m 3m
Beam Deflection - Example 1-5 4m 3m
Beam Deflection - Example 2-1 Problem. Predict the shape under load of the beam below. This beam is supported and clamped to be horizontal at both ends. 1m 1m 2m 9600 N/m For this beam, the boundary conditions are: y(0) = 0, y (0) = 0, y(4) = 0 and y (4) = 0. The beam is a pine 2x10, so I = 4.1 10 5 m 4, and E = 9 10 9 N/m.
Beam Deflection - Example 2-2 1m 1m 2m
Beam Deflection - Example 2-3 1m 1m 2m
Beam Deflection - Example 2-4 1m 1m 2m
Beam Deflections - Animation Beam Deflections - Animation - 1 4m 3m 600 N/m y(x) = { 1 ( EI 2250x 2 300x 3) 0 x < 1 ( 2250x 2 300x 3 + 25(x 1) 4) 1 x < 4 1 EI
Beam Deflections - Animation - 2 Example 2 1m 1m 2m 9600 N/m y(x) = 1 EI 1 EI 1 EI ( 2725x 2 1087.5x 3) 0 x < 1 ( 2725x 2 1087.5x 3 + 400(x 1) 4) 1 x < 2 ( 2725x 2 1087.5x 3 + 400(x 1) 4 400(x 2) 4) 2 x 4
Pollutant Transport and Diffusion - Introduction - 1 Pollutant Transport and Diffusion - Introduction
Pollutant Transport and Diffusion - Introduction - 2 One-Dimensional Model High Conc. Possible Water Flow v or Diffusion Direction Low Conc. Non-Porous Clay Porous Rock or Sand Non-Porous Clay Steady-State Concentration DA c (x) + Av c(x) = f(x)
Pollutant Transport and Diffusion - Introduction - 3 DA c (x) + Av c(x) = f(x) Comments: the velocities of water through rock tend to be very slow by our usual standards. The c(x) solution we obtain is a steady-state concentration along the layer, based on known constant concentrations (usually measured at the ends of some interval). For simplicity, we will use A = 1m 2 for all our examples.
Pollutant - Example 1 D c (x) + v c(x) = f(x) Pollutant - Example 1-1 Problem. Predict the steady-state concentration gradient along a 1000 m layer of sand, if C(0) = 50 ppm and C(1000) = 0 ppm. f(x) = 0 (no outside effects) Leave v and D as constants for now. c(x) 50 40 30 20 10 0 x 0 200 400 600 800 1000
Pollutant - Example 1-2 c (x) v D c(x) = 0
Pollutant - Example 1-3 c (x) v D c(x) = 0
Pollutant - Example 1-4 Sketch graphs of c(x) for various v/d ratios. c(x) 50 40 30 20 10 0 x 0 200 400 600 800 1000 c(x) 50 40 30 20 10 0 x 0 200 400 600 800 1000
Pollutant - Example 1-5 c(x) 50 40 30 20 10 0 x 0 200 400 600 800 1000
Pollutant - Example 2 - Part 1-1 Pollutant - Example 2 DAc (x) + Av c(x) = f(x) f(x) represents some change to the natural diffusion process. Problem. Study the units of this equation to find the units of f(x). Recall: v is a velocity, A is an area, and D is a diffusion coefficient, with units m 2 /s.
Pollutant - Example 2 - Part 1-2 DAc (x) + Av c(x) = f(x) In this form of the DE, the units of f(x) are (ppm/s) m 2. If f(x) = 0, this represents If f(x) > 0 at some interior point,
Pollutant - Example 2 - Part 2-1 Dc (x) + v c(x) = f(x) (using A = 1) Problem. Predict the steady concentration c(x) over the interval x = 0... 1000 m if 0 0 t < 200 f(x) = 1 10 ( 11) 300 t < 500 0 500 t 1000 Use D = 2 10 8 m 2 /s (diffusion in sand), v = 0 (no groundwater flow), and c(0) = 0 and c(1000) = 0 boundary conditions. c(x) Pollutant Source x 0 200 400 600 800 1000
Pollutant - Example 2 - Part 2-2 c(x) Pollutant Source x 0 200 400 600 800 1000
Pollutant - Example 2 - Part 2-3 c(x) Pollutant Source x 0 200 400 600 800 1000
Pollutant - Example 2 - Part 2-4 c(x) Pollutant Source x 0 200 400 600 800 1000
c(x) = Pollutant - Example 2 - Part 2-5 6 100 x 0 x 300 6 100 x 4000 1 (x 300)2 300 x 500 6 100 x 4000 1 (x 300)2 + 4000 1 (x 500)2 500 x 1000 Pollutant Source c(x) (ppm) 20 18 16 14 12 10 8 6 4 2 0 x (m) 0 200 400 600 800 1000