MECH 314 Dynamics of Mechanisms February 7, 2011 Mid-Term Test-I, , 16H05-17H25

Transcription

1 MECH 314 Dynamics of Mechanisms February 7, 2011 Mid-Term Test-I, , 16H05-17H25 1 General Description of a Design Problem Examine Fig. 1 carefully. It shows the design of a crank rocker 4-bar mechanism and an offset slider crank. The equilateral triangle BCC is inscribed in the unit circle. Note that in this case φ = δ was chosen. Furthermore the crank centre A(x A, y A ) was placed so that y A is half of the radius, i.e., the distance from origin O to the rocker joint at B. In addition to the limiting singular positions of the rocker arm tip and slider wrist pin C and C, two other crank angle positions are shown. In one of these the rocker arm tip is halfway along its arc, in the other, the slider wrist pin is halfway along its straight line trajectory. a. What is the time ratio Q? (It is identical for both mechanisms.) b. If the actual mechanisms to be constructed have a crank throw radius AD = s = 10dm what is the length of the coupler (or connecting rod) CD = q? In the case of the crank rocker, what is the length of the rocker arm BC = p? c. If the crank AD = s turns at a constant angular velocity of 10rdn/s CCW what is the angular velocity ω p of the rocker arm BC 1 when the arm is at its halfway position? What is the linear velocity vector v C2 of the slider wrist pin C 2 when it is at its halfway position? C 4 C 1 C C 2 C 3 C O D 13 D 24 s=10 D A MT1TR12e D B Figure 1: A Crank Rocker and Offset Slider Crank Design

2 2 Bonus Question Show clearly on the diagram where one should place a spring so that the slider will begin its quick-return stroke without the mechanism locking up in a singularity. This can be indicated by neatly drawing a small arrow where and in the direction that the spring force is to be applied. 2 Statics Find the torque Γ A that must be supplied at the crank pin A if it must overcome a force of 50N, to the right, applied to the slider wrist pin C 3 in the position shown between the limit, labeled C and the mid-stroke position halfway along the slider path. Point C 3 is on the slider path somewhat to the left of C. 3 Instant Centre Using a Cartesian frame with origin on O and axes x- to the right and y- upward, find the instant centre of the coupler D 24 C 4 = q of the crank rocker when the tip C 4 is on the arc somewhat to the right of the leftmost limit C.

3 Solutions to Midterm-I, MECH 314, The following Maple worksheet should be consulted together with the graphical solution on the last page. > restart: Parameters by inspection:- s=r/2=10, q=3r/2=3s=30, p=rsqrt(3)=2sqrt(3)s, A(-sqrt(3)s,-s), C1(0,p-R)=(0,2(sqrt(3)-1)s), C2(0,s) > s:=10:ka:=evalf((xd+sqrt(3)*s)^2+(yd+s)^2-s^2); ka := (xd ) 2 + (yd + 10.) ka is circle centred on A of radius s. > kc1:=evalf(xd^2+(yd-2*(sqrt(3)-1)*s)^2-(3*s)^2); kc1 := xd 2 + (yd ) kc1 is circle centred on C1 of radius q. > kc2:=xd^2+(yd-s)^2-(3*s)^2; kc2 := xd 2 + (yd 10) kc2 is circle centred on C2 of radius q. > g1:=collect(simplify(ka-kc1),yd); g1 := xd yd > r1:=collect(resultant(g1,kc1,xd),yd): > yd1:=solve(r1,yd):yd13:=yd1[1]; yd13 := > xd13:=solve(subs(yd=yd13,g1)); xd13 := > g2:=collect(simplify(ka-kc2),yd); g2 := xd yd > r2:=collect(resultant(g2,kc2,xd),yd): > yd2:=solve(r2,yd):yd24:=yd2[1]; yd24 := > xd24:=solve(subs(yd=yd24,g2)); xd24 := Circles ka and kc1 are intersected to yield D13. Circles ka and kc2 are intersected to yield D24. > kd1:=(xc-xd13)^2+(yc-yd13)^2-(3*s)^2; kd1 := (xc ) 2 + (yc ) kd1 is circle centred on D13 of radius q. It is intersected by line y-s=0 to yield C3. 1

4 > yc3:=s;xc3:=solve(subs(yc=yc3,kd1))[1]; yc3 := 10 xc3 := > kd2:=(xc-xd24)^2+(yc-yd24)^2-(3*s)^2;kb:=xc^2+(yc+2*s)^2-(2*sqrt(3)*s )^2; kd2 := (xc ) 2 + (yc ) kb := xc 2 + (yc + 20) kd2 is circle centred on D24 of radius q. It is intersected by circle kb, centred at B of radius p, to yield C4. > g3:=collect(simplify(kb-kd2),yd); g3 := yc xc > r3:=collect(resultant(g3,kb,xc),yc):yc4:=solve(r3,yc)[1]; yc4 := > xc4:=solve(subs(yc=yc4,g3)); xc4 := > e1:=-rad1y*os-rd1c1y*oq+rbc1y*omp;e2:=rad1x*os+rd1c1x*oq-rbc1x*omp; E1:=subs(os=10,rAD1y=yD13+s,rD1C1y=2*sqrt(3)*s-2*s-yD13,rBC1y=2*sqrt(3 )*s,e1); E2:=subs(os=10,rAD1x=xD13+sqrt(3)*s,rD1C1x=-xD13,rBC1x=0,e2); e1 := rad1y os rd1c1y oq + rbc1y omp e2 := rad1x os + rd1c1x oq rbc1x omp E1 := ( ) oq omp E2 := oq > omp:=solve(resultant(e1,e2,oq)); omega p is omp. omp := As regards the bonus, note that the slider-crank is singular when wrist-pin is at C. A spring, pushing it to the right, may help matters by preventing a lock-up at the beginning of the short froward return stroke with CCW crank rotation. This problem does not affect the crabk-rocker. > e3:=-rad2y*os-rd2c2y*oq-vc2x;e4:=rad2x*os+rd2c2x*oq; E3:=subs(os=10,rAD2y=yD24+s,rD2C2y=s-yD24,e3);E4:=evalf(subs(os=10,rAD 2x=xD24+sqrt(3)*s,rD2C2x=-xD24,e4)); e3 := rad2y os rd2c2y oq vc2x e4 := rad2x os + rd2c2x oq E3 := oq vc2x E4 := oq > vc2x:=solve(resultant(e3,e4,oq)); vc2x :=

5 Mid way at C2 slider velocity is 100dm/s to the left. > Gamma:=rAD13y*fC3x-rAD13x*fC3y:fC3x:=kD*rD3C3x:fC3y:=kD*rD3C3y:kD:=50 /rd3c3x: rad13x:=xd13+sqrt(3)*s:rad13y:=yd13+s:rd3c3x:=xc3-xd13:rd3c3y:=yc3-yd1 3: > evalf(gamma); A CCW torque of Nm must be applied at A. > with(linalg):mgad:=matrix(3,3,[w,x,y,1,-sqrt(3)*s,-s,1,xd24,yd24]);mg BC:=matrix(3,3,[w,x,y,1,0,-2*s,1,xC4,yC4]); Warning, the protected names norm and trace have been redefined and unprotected w x y MgAD := MgBC := w 1 x 0 y Forming lines on AD24 and BC4... > MIC:=matrix(3,3,[W,X,Y,det(minor(MgAD,1,1)),-det(minor(MgAD,1,2)),det (minor(mgad,1,3)),det(minor(mgbc,1,1)),-det(minor(mgbc,1,2)),det(minor (MgBC,1,3))]); MIC := W X Y and then intersecting these locates the instant centre of the coupler C4D24 at (xic,yic) on B. > IC0:=det(minor(MIC,1,1));IC1:=-det(minor(MIC,1,2));IC2:=det(minor(MIC,1,3)); IC0 := IC1 := IC2 := > xic:=evalf(ic1/ic0);yic:=evalf(ic2/ic0); xic := yic :=

6 p =3.514rdn/s(CCW) v =12.173m/s C1 C v C/D 0 * v D1 =10m/s s=10 v D1 A v D2 v C/D v C2 =10m/s 0 D v D2 =10m/s D C 4 C 1 f C3 =50N Time ratio Q= N C 2 C 3 C q=30 O p= dm Nm(CCW) supplied at A MT1TR12e B Instant centre of coupler at C is on B! 4 Bonus: apply spring force * Figure 1: Graphical Solutions 4