Question

Could i get help writing the code for this question in matlab?

An internal combustion engine slider-crank mechanism is shown in the figure. Crank AB rotates in selected clockwise positive direction as shown. Piston position is Y=AD. o(t) is angular position of the crank, 0(t) is angular velocity of the crank, ő is angular acceleration of the crank.

Answer 1

MatLab Script

1 2 - clear all close all clc 3 - 4 5 - 1 = 150; r = 50; Omega = 5; 6 - 7 - 8 9 Length of the Connecting rod BD, mm Length of the Crank AB, mm Angular velocity of the crank AB, rad/s - - 10 n = 1/r; time = 0:0.01: (4*pi) /omega; t = time* omega; time vector Theta vector 11 - 12 13 - d = sqrt (n^2 sin(t).^2); 14 15 16 y = r* (cos(t) + d); Displacemment of the piston, mm 17 18 - v = -r*omega* ( sin(t) + sin(2*t)./(2*d) ); Velocity, mm/s 19 20 21 a = -r* omega^2 *( cos(t) + cos (2*t)./d + (sin(2*t).^2)./(4*0.43)); } acceleration of piston, mm/s^2 22 23 24 25 26 - 27 28 29 Plot piston diplacement vs time figure plot (time, y); xlabel('Time (s)'); vlabel ('Displacement (mm)'); title('Displacement vs Time'); grid on & label x-axis label y-axis $ title Turn the grids on 30 31 32 -

33 34 35 36 37 38 39 40 $ plot piston velocity vs time figure plot (time, v); xlabel('Time (s)'); ylabel ('Velocity (mm/s)'); title('Velocity vs Time'); grid on - 41 42 43 44 45 46 - 47 $ plot piston acceleration vs time figure plot (time, a); xlabel ('Time (s)'); vlabel('Acceleration (mm/s^2)'); title('Acceleration vs Time'); grid on 48 49 50

MatLab Output

Displacement vs Time 200 190 180 170 160 Displacement (mm) 150 140 130 120 110 100 0 0.5 1 2 2.5 3 1.5 Time(s)

Velocity vs Time 300 200 100 Velocity (mm/s) -100 -200 -300 0 0.5 2 2.5 3 1.5 Time(s)

Acceleration vs Time 1000 500 0 Acceleration (mm/s2) -500 -1000 -1500 -2000 0 0.5 1 2 2.5 3 1.5 Time(s)

Equations:

D 8 B A Referring to the figure shown, displacement of the piston as the crank rotates through an angle e from the inner dead centre, can be written as y = AD = (l cos + r cos ) Or y = r =r{cos $ + cose) = r(n cos $ + cos e) Where we have substituted, = n From the trigonometric relation coso = V1 - sina o From the figure I sin = rsin or r sin sin e sin = n Substituting this

sin , 2 cos 1-sines n? - sin n Substituting this in the equation for displacement, y y=r(n; y=rin-vna - sin2 @ + cos Or =rſcose + Vn2 – sin? 0] Velocity of the piston can be obtained by differentiating above equation with respect to time. v= y = dy dy de dt de dt. dy (0) dᎾ Substituting for y ý = (r (cos e + Vn? – sin? e)] (6) Or j = rė (-sin sin + -2 sin cos e 2 n2 - sin2 @ or sin 20 v = y = -ré sin + 2vn? - sinº e -rö (sin

Differentiating the above equation with respect to time, we get the acceleration of the piston dý de a = y = dt. dy (0) de Substituting the expression for y sin 20 -résine + *2Vn? – sinta)](0) Or į = -ri2 2 [cose + 2 cos 20 sin 20 (-sin 20)] 2Vn2 - sin2 4(n? - sin? A) 3/2 a = y = -rô-{cos e + cos 20 sina 20 Vn? - sin? '4(n? - sin? 0)3/2 + if w is the angular velocity of the crank (i.e. 8), then at any time t, 6 = wt. Substituting this the above equation can be expressed as function of time as follows y(t) =r (cos(wt) + Vn2 – sin?(wt)] sin(2wt) v(t) = y(t) = -rw sin(wt) + 2/n? - sin?(wt) a(t) = y = -rw-cos(wt) + »{{cos(et) + cos(2wt) sin?(wt) n? - sin?(wt) '4(n? - sin?(wt))3/2 +

(Note: Variable 'd' is used in the code just for the ease of writting the equations in code.)