VIEWS: 42 PAGES: 12 POSTED ON: 12/15/2009 Public Domain
Theory of Machines (ME 220) Solution to Tutorial sheet 2 Solution :1a) a p o2 , o4 b Figure 1 The angular velocity of the link O2A is given ω2 = 100 rad/sec The velocity of link O2A vao2 = 100 × 0.075 = 7.5m / s The configuration diagram has shown in figure 3 to a convenient scale. Writing the vector equation, Or Or vbo2 = vba + vao2 vbo4 = vao2 + vba Take the vector vao2 to a convenient scale in the proper direction and sense vba is perpendicular to BA, draw a line perpendicular to BA through a; vbo4 is perpendicular to BO4, draw a line perpendicular to BO4 through O4; The intersection of the two lines locates the point b. 1) The angular velocity of the coupler AB is v 4.34 ωba = ba = = 57.86 rad/sec BA 0.075 2) vap is perpendicular to AP, draw a line perpendicular to AP through a; vbp is perpendicular to BP, draw a line perpendicular to BP through b; The intersection locates the point p v p = O4p = 9.18 m/s Solution: 1b) p b o2 , o4 a Figure 2 The angular velocity of the link O2A is given ω2 = 100 rad/sec The velocity of link O2A vao2 = 100 × 0.10 = 10m / s The configuration diagram has shown in figure 2 to a convenient scale. Writing the vector equation, Or Or vbo2 = vba + vao2 vbo4 = vao2 + vba Take the vector vao2 to a convenient scale in the proper direction and sense vba is perpendicular to BA, draw a line perpendicular to BA through a; vbo4 is perpendicular to BO4, draw a line perpendicular to BO4 through O4; The intersection of the two lines locates the point b. 1) The angular velocity of the coupler AB is v 6.45 ωba = ba = = 129 rad/sec BA 0.05 2) vap is perpendicular to AP, draw a line perpendicular to AP through a; vbp is perpendicular to BP, draw a line perpendicular to BP through b; The intersection locates the point p v p = O4p = 7.69 m/s Solution 2 Figure 3 Firstly find out all instantaneous centre of rotation using Aronhold-Kennedy theorem. These are shown in Figure 3 along with the construction lines (shown in green). The velocity of crank OA is va = ω × OA = 6 × 0.15 = 0.9m / s Referring to Figure 4 below, the instant centre P12, P14, and P24 must be in a straight line according to the Kennedy-Aronhold theorem. The absolute velocity of link 2 and 4 is common about P24. First consider instant center P24 as a point of link 2. The velocity Va can be found from w2 using velocity difference equation about P12, and the velocity of P24 can be found from while using graphical construction shown in figure . Now consider P24 as a point of link 4 rotating about P14, knowing V P24, we can find the velocity of any other point of link 4. Similarly we can find out the velocity of slider B. 1) The velocity of slider B is 0.79 m/s 2) The angular velocity of link BD is ⎛ 25 − 12 ⎞ w5 = w2 ⎜ ⎟ ⎝ 25 − 15 ⎠ ⎛ 10.9 ⎞ w5 = 6 ⎜ ⎟ ⎝ 82.25 ⎠ w5 = 0.795rad / s Please note that if we were only interested in finding the velocity of the slider, the method of instantaneous centres allows us to determine that immediately by locating I26 and using that to relate the velocity of body 2 with that of body 5 as shown in Figure 5. Figure 4 Solution I using ICs Figure 4 Solution II using ICs Figure 6 Solution: 3 D C 185 185 20o 35 O 180 240 E A 90 G 210 Q 20o B 45 0.34m/s d oV 0.98m/s b c f (b) Figure 7 a e 0.79m/s The velocity of point A 2π N × 0.09 = 0.943m / s 60 OE 0.125 VE = VA = 0.943 × = 1.309m / s OA 0.09 VA = VB = VF QB 0.045 = 1.309 × = 0.982m / s QF 0.06 Obtain the VB graphically as follows: Take vector VA to a convenient scale (fig 7) produce oa to e such that oe/oa = OE/OA. Rotate oe to of so that of is perpendicular to QF. Mark point b on qf such that qb/qf = QB/QF. Now, VCA is perpendicular to AC, draw a line perpendicular to AC through a; VCB is perpendicular to BC, draw a line perpendicular to BC through b; These intersect at the point c and VC can be computed. Further VDC is perpendicular to DC, draw a line perpendicular to DC through c; VD is horizontal, draw a horizontal line OV; The point of intersection of these locates d. The magnitude of VD can be estimated graphically to be 0.34 m/s, ωbc = vbc 0.12 = = 0.649 rad/sec (clockwise) BC 0.185 vdc 1.0 = = 4.17 rad/s (counter clockwise) DC 0.24 ωcd = Solution-4 The angular velocity of the link O2A is given ω2 = 20 rad/sec The velocity of link O2A VA2 = VA3 = 20 x 0.04 = 0.8 m/s The configuration diagram has shown in Figure 9 to a convenient scale. Writing the vector equation, VB4 = VD + VB4D VB3 = VA3 + VB3A3 VB3 = VB4 + VB3/4 and also Note that both VB3A3 and VB3/4 are vertical direction (also perpendicular to A3B3), therefore the point b4 is the intersection of horizontal line from O2 and vertical line from a2. Having determined point , one can draw the velocity image of the wheel. Link 3 rotates at the same rate as link 4 (due to the slot constraint) has the same angular velocity and therefore its velocity image is scaled with the same factor as the wheel. We can sketch the image of link 3 as we know the absolute velocity of point A3. This completes the velocity diagram. B 25 y 40 37.5o x 25 D 8 A 25 C O2 VB4=0.5 m/s b4 0v, d VB3/4=0.5 m/s b3, c4 VA=0.8 m/s VB3/A3=0.14 m/s a2,a3 Figure 8 The absolute velocity of point B on link AB is 0.707 m/s and its velocity with respect to rolling wheel is 0.5 m/s. The motion of link 3 can be better visualized by locating the instantaneous centre of rotation of link3 (with respect to ground), i.e., I13. The instantaneous centre can be found out using Aronhold-Kennedy theorem as shown below in Figure 9. Figure 9 (End)