ME 302 THEORY of MACHINES II SPRING 2013 HOMEWORK 2 SOLUTION Due date: xx

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1 ME 302 THEORY of MACHINES II SPRING 2013 HOMEWORK 2 SOLUTION Due date: xx NOTE: You are asked to solve only the first and second problems. Put your homework report in the box provided in front of room C-204 in xx until 15:40. PROBLEM 1 F C y B 3 2 T k A 4 θ 3 O θ 2 x s 1 Figure 1 Consider the motion of the mechanism shown in figure 1 corresponding to the full cycle rotation of the input crank(link 2) i.e. 0 o θ o. a) Perform a position analysis of the mechanism for input joint variable 2 and plot the variations of 3 and s 1 with respect to 2. Solution: Loop Closure equation b 2 e iθ 2 = s 1 + b 3 e iθ 3 (1) Real part b 2 cosθ 2 = s 1 + b 3 cosθ 3 (2) Imaginary part b 2 sinθ 2 = b 3 sinθ 3 (3) Rearrange the real and imaginary parts as follows: b 2 cosθ 2 + s 1 = b 3 cosθ 3 (4) b 2 sinθ 2 = b 3 sinθ 3 (5) Take the square root of the equations (4) and (5) and then sum them up: s s 1 b 2 cosθ 2 + b 2 2 b 3 2 = 0 6

2 For convenience; Equation (6) becomes as The solution of s 1 is B = 2b 2 cosθ 2 C = b 2 2 b 3 2 s Bs 1 + C = 0 s 1 = B+σ 1 B 2 4C 2 (7) Here, σ 1 = ±1 is the "closure sign" of the second loop. Note that s 1 should be positive and this is possible, if σ 1 = +1. As for θ 3 ; Hence, we get, cosθ 3 = (b 2 cosθ 2 + s 1 )/b 3 = x 3 /b 3 sinθ 3 = b 2 sinθ 2 /b 3 = y 3 /b 3 θ 3 = atan2 y 3, x 3 (8)

3 b) Perform velocity and acceleration analyses of the mechanism and plot the velocities and accelerations of link 3 and link 4 with respect to 2 ( 3 vs. 2 and 3 vs. 2, s 1 vs. 2 and s 1 vs. 2 ). Solution To perform velocity analysis, take the derivative of the equations (2) and (3) respectively, Real part: b 2 sinθ 2 θ 2 = s 1 b 3 sinθ 3 θ 3 Imaginary part : b 2 cosθ 2 θ 2 = b 3 cosθ 3 θ 3 Real and imaginary parts can be written in matrix form as follows: 1 b 3 sinθ 3 s 1 = b 2sinθ 2 θ 0 b 3 cosθ 3 θ 3 b 2 cosθ 2 2 Hence, Then, s 1 θ 3 = 1 b 3 cosθ 3 b 3 sinθ 3 b 3 cosθ b 2 sinθ 2 b 2 cosθ 2 θ 2 s 1 = 1 cosθ 3 cosθ 3 b 2 sinθ 2 + sinθ 3 b 2 cosθ 2 θ 2 = b 2 sin θ 2 θ 3 cos θ 3 θ 2 (9) θ 3 = 1 b 3 cosθ 3 b 2 cosθ 2 θ 2 (10) g 32 = 1 b 3 cosθ 3 b 2 cosθ 2 and g 12 = b 2 sin θ 2 θ 3 cos θ 3 that To perform accelaration analysis, take derivative of the s 1 and θ 3respectively, such s 1 = b 2 θ 2 θ 3 cos θ 2 θ 3 cosθ 3 + sin θ 2 θ 3 sinθ 3 θ 3 cos 2 θ 3 θ 2 + b 2 sin θ 2 θ 3 cosθ 3 θ 2 θ 3 = b 2 b 3 sinθ 2 cosθ 3 θ 2 + cosθ 2 sinθ 3 θ 3 cos 2 θ 3 θ 2 + cosθ 2 cosθ 3 θ 2

5 c) Plot the x and y components of the linear velocity and acceleration of the mass center of link 3 with respect to 2. Solution r G3 = s 1 + c 3 e iθ 3 r x,g3 = s 1 + c 3 cosθ 3 r y,g3 = c 3 sinθ 3 To perform velocity analysis, take the first derivative as follows; rx,g3 = s 1 c 3 sinθ 3 θ 3

6 ry,g3 = c 3 cosθ 3 θ 3 To perform acceleration analysis, take derivative of the velocities as follows; rx,g3 = s 1 c 3 cosθ 3 θ 3 2 c 3 sinθ 3 θ 3 ry,g3 = c 3 sinθ 3 θ c 3 cosθ 3 θ 3

7 d) Derive the required torque relation (T), which is necessary to rotate the crank at a constant speed of 2, by using principle of virtual work method. Plot the variations of the torque T with respect to 2 (Neglect frictional and gravitational effects on the system). F C y r y,c B I 3 θ 3 I 2 θ 2 r y,g3 F s A 4 3 m 4 s 1 θ 3 m 3 rx,g3 m 3 ry,g3 r x,g3 2 T O θ 2 x s 1 Virtual work expression can be written as:

8 δu = Tδθ 2 I 2 θ 2δθ 2 Fδr y,c m 3 ry,g3δr y,g3 m 3 rx,g3δr x,g3 I 3 θ 3δθ 3 F s + m 4 s 1 δs 1 = 0 Where θ 2 = 0 r y,c = b 2 + d 2 sinθ 2 δr y,c = (b 2 + d 2 )cosθ 2 δθ 2 δθ 3 = 1 b 3 cosθ 3 b 2 cosθ 2 δθ 2 δs 1 = b 2 sin θ 2 θ 3 cosθ 3 δθ 2 δr y,g3 = c 3 cosθ 3 δθ 3 δr x,g3 = δs 1 c 3 sinθ 3 δθ 3 F s = k s 1 θ 2 s 1 θ 2 = π 2 N from s1 vs teta2 grap, s 1 θ 2 = π 2 = m

9 NOTE : You have to submit hardcopy of your codes with your graphs. Numerical Data: I) Dimensions of the mechanism: b 2 = OB = 15 cm b 3 = AB = 100 cm d 2 = BC = 10 cm All mass centers are in the middle of each corresponding link i.e. c 2 = OG 2 = 12.5 cm c 3 = AG 3 = 50 cm II) III) Masses and inertias: m 2 = 6 kg m 3 = 5.4 kg m 4 = 9 kg I G2 = kg.m 2 I G3 = 0.04 kg. m 2 All other masses and inertias are assumed to be zero. Loading force is applied to point C vertically and always downwards. Magnitude of the force changes with the angle of crank ( i.e θ 2 ) as follows:

10 Magnitude of Force [N] θ 2 [deg] IV) Spring is unstreched when θ 2 = 90 o and k = 100 N/m. V) Rotational speed of link 2 2 = 100 rpm counterclockwise (constant) PROBLEM 2 k G θ T(t) F(t) x The radius, mass and inertia of the disk are R, m and I G respectively. No slip conditions prevail. Spring is unstreched when x = 0. By using virtual work method, derive equation of motion. Solution kx mx Iθ G θ T(t) F(t) x

11 Virtual work expression can be written as Where x = θ + θ o R expression, we get; Finally, F t δx kxδx mx δx + T t δθ Iθ δθ = 0 δx R = δθ x R = θ F t δx kxδx mx δx + substituting into the virtual work T t δx R m + I T t x + kx = F t + R2 R Ix δx R 2 = 0 MATLAB code for Problem 1: clc clear all close all syms x; b2=0.15; b3=1; d2=0.10; c2=0.125; c3=0.50; m2=6; m3=5.4; m4=9; I2=0.317; I3=0.04; k=100; w2=100*pi/30; teta2=0:0.001:2*pi; B=2*b2*cos(teta2); C=b2^2-b3^2; %% s1 and teta3 s1=(-b+sqrt(b.^2-4*c))./2; x3=b2*cos(teta2)+s1; y3=b2*sin(teta2)/b3; teta3=atan2(y3,x3); %% s1dot and teta3dot g12=b2.*sin(teta2-teta3)./cos(teta3); g32=b2/b3*cos(teta2)./cos(teta3); s1dot=g12.*w2; teta3dot=g32.*w2; %% s1doubledot and teta3doubledot s1doubledot=b2./(cos(teta3).^2).*((w2-teta3dot).*cos(teta2- teta3).*cos(teta3)+sin(teta2-teta3).*sin(teta3).*teta3dot).*w2; teta3doubledot=((b2./b3)./(cos(teta3).^2)).*(- sin(teta2).*cos(teta3).*w2+cos(teta2).*sin(teta3).*teta3dot).*w2; %% rxg3dot, rxg3doubledot, ryg3dot and ryg3doubledot rxg3dot=-s1dot-c3*sin(teta3).*teta3dot; ryg3dot=c3*cos(teta3).*teta3dot;

12 rxg3doubledot=-s1doubledot-c3*cos(teta3).*teta3dot.^2- c3*sin(teta3).*teta3doubledot; ryg3doubledot=-c3*sin(teta3).*teta3dot.^2+c3*cos(teta3).*teta3doubledot; %% virtual work del_teta2=1; del_ryc=(b2+d2).*cos(teta2).*del_teta2; del_teta3=1./(b3.*cos(teta3)).*b2.*cos(teta2).*del_teta2; del_s1=b2.*sin(teta2-teta3)./cos(teta3).*del_teta2; del_ryg3=c3*cos(teta3).*del_teta3; del_rxg3=-del_s1-c3*sin(teta3).*del_teta3; Fs=k*(s ); F=abs(220*(heaviside(teta2-pi/2)-heaviside(3*pi/2-teta2))); %% Torque T=F.*del_ryc+m3.*ryG3doubledot.*del_ryG3+m3.*rxG3doubledot.*del_rxG3+I3.*te ta3doubledot.*del_teta3+(fs+m4.*s1doubledot).*del_s1; %% PLOTs teta2=teta2*180/pi; figure(1) plot(teta2,s1) ylabel('s1 [m]') title('s1 vs. teta2') figure(2) plot(teta2,teta3) ylabel('teta3 [rad]') title('teta3 vs. teta2') figure(3) plot(teta2,teta3dot) ylabel('teta3dot [rad/s]') title('teta3dot vs. teta2') figure(4) plot(teta2,s1dot) ylabel('s1dot [m/s]') title('s1dot vs. teta2') figure(5) plot(teta2,teta3doubledot) ylabel('teta3doubledot [rad/s^2]') title('teta3doubledot vs. teta2') figure(6)

13 plot(teta2,s1doubledot) ylabel('s1doubledot [m/s^2]') title('s1doubledot vs. teta2') figure(7) plot(teta2,rxg3dot) ylabel('horizantol velocity of the mass center of the link 3 [m/s]') title('rxg3dot vs. teta2') figure(8) plot(teta2,ryg3dot) ylabel('vertical velocity of the mass center of the link 3 [m/s]') title('ryg3dot vs. teta2') figure(9) plot(teta2,rxg3doubledot) ylabel('horizantol accelaration of the mass center of the link 3 [m/s^2]') title('rxg3doubledot vs. teta2') figure(10) plot(teta2,ryg3doubledot) ylabel('vertical accelaration of the mass center of the link 3 [m/s^2]') title('ryg3doubledot vs. teta2') figure(11) plot(teta2,t) ylabel('torque [Nm]') title('torque vs. teta2') figure(12) plot(teta2,fs) ylabel('fs [N]') title('spring force vs. teta2')

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