ME 3313 FALL 2014 FINAL EXAM

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1.

(a)(5%) The device on the left is Peaucelliers exact straight-line mechanism.It is designed to move a point P along a straight line. It is therefore a (circle one)

Function Generator.

Path Generator.

Motion Generator.

(b)(5%) The device on the right is an analog computer which, for each value of theinput angle , delivers the measured output x = R cos. It is therefore a (circle one)

Function Generator.

Path Generator.

Motion Generator.

2.(10%) A GCRR 4-bar mechanism has link-lengths {R1,R2,R3,R4} = {4, 1, 3.43, 2.12},and is assembled in the open configuration. Find its minimum transmission angle.

3. Consider the six-link mechanism shown :

(a)(5%) In the left-hand picture, the joints are labeled in the standard way. Use thisinformation to compute the mobility of the mechanism.

(b)(5%) In the right-hand picture, loops and loop-vectors for a Position Analysis ofthe mechanism are declared. The variables of the analysis are 2, 3, 4, 5, 7 -the orientations of the loop-vectors so numbered - and d5, the magnitude of R5. Thecorresponding loop equations, given below, are proposed as a Position Analysis forthe mechanism. Circle whichever of the three statements below is true.

d2 cos 2 + d3 cos 3 d4 cos 4 d1 = 0,d2 sin 2 + d3 sin 3 d4 sin 4 = 0,

d7 cos 7 + d5 cos 5 d6 = 0,d7 sin 7 + d5 sin 5 = 0.

The Position Analysis is complete, and contains no unnecessary equations.

The Position Analysis is complete, but contains unnecessary equations.

The Position Analysis is incomplete.

4.(10%) For the six-bar mechanism shown, find the Instant Center I4,6. Clearly label allInstant Centers used in the construction.

5.(10%) The heavy-duty cutter shown is a four-bar mechanism with input link AC, couplerBE, and output link DE. In the given x y coordinates, the Instant Centers I1,4 = D,I1,2 = A, and I2,4 are located, respectively, at points (0,0), (64,37), and (80.06,46.28). Ahorizontal input force Fin is applied at C. The output force Fout = Fout

.2419 +.9703 .The velocity of its point of application is Vout = !4

48 5 = !45 + 48 . Thus,

the power of the output force is Pout = FoutVout = 45.3649!4Fout. Develop the PowerIdentity for this system. Use the above information on Instant Centers to relate !2 and!4. Find the Mechanical Advantage of the cutter.

6.(10%) The FBD for link 3 of a mechanism is shown above. The link has massM3 = 2.5 kg,and moment of inertia IG3 = 13kgm2. At the instant of concern, R23 = {3.4 3.8 }m,R43 = {4.6 + 2.2 }m, RP = {.3 + .4 }m, AG3 = {6 + 4 }m/s2, 3 = 4 rad/s2x.The known vertical force FP is given to be FP = 10 N. Using the xy horizontal-verticalcoordinate system, and versions F32, F43, of the joint forces, write down the three scalarkinetic equations for the link in numerically explicit form.

7.(10%) The above plot shows the normalized torque TN = T12 TAVE required to drivea certain mechanism at a constant input-link angular velocity !2 = 80 rad/s. The averagetorque is TAVE = 63.518Nm, and the areas of the lobes between the T12 plot and theTAVE line are A1 = 1.1175106Nm, A2 = 1.2180106Nm, A3 = 1.1161106Nm,A4 = 1.6067106Nm, A5 = 5.9119105Nm. It is desired to drive this system by meansof a motor operating near constant torque TAVE at approximately constant angular velocity!avg = 80 rad/s. Given a value k = 0.018 for the coecient of fluctuation, estimate themoment of inertia of a flywheel that will accomplish this.

8.(10%) The wheel and tire assembly shown has been run at 10 rad/s on a dynamic balanc-ing machine. The force measured in the left bearing has a peak of 0.6N at a phase angleof 75 with respect to the zero reference line on the tire. The force measured in the rightbearing has a peak of 0.4N at a phase angle of 60 with respect to the zero reference lineon the tire. The center distance between the two bearings on the machine is 25 cm. Theleft edge of the wheel rim is 10 cm from the centerline of the closest bearing. The wheel is17.5 cm wide at the rim. The wheel-rim diameter is 37.5 cm. The tire is to be balancedby attaching masses mA,mB to the rim in correction planes A, B at angles A, B to thezero reference line on the tire. Draw the FBD for the balanced assembly, and find mA andA. (Hint: Use the vector equation developed in class for the inertia force FA.)

9.

(a)(4%) Consider the cam shown. In a consistent system of units, for a particularvalue of , s() = 0.35, () = 1.3. If RP = 1.8, " = 0.1, find the corresponding valueof the pressure-angle .

(b)(2%) Is that value of in the acceptable range?(circle one)

YES NO

(c)(2%) A force-closed cam with translating follower is designed for a roller head ofradius Rf = .08m. The minimum radius of curvature of the pitch curve turns out tobe min = .065m. This situation gives rise to the problem of (circle one)

Crossover Shock.

Undercutting.

Overturning Moment.

none of the above.

(d)(2%) The translating follower of a force-closed cam is fitted with a mushroomfollower. This means that: (circle one)

The follower displacement is based on a pitch curve.

Friction is negligible.

Overturning Moment is of concern.

none of the above.

10. For the gear shown:(a)(3%) A gearset has = 14.5, rg = 3.75 in, ag = .125 in, rp = 1.5 in, ap = .125 in,and pb = .3689 in. Find the length of action Z, and the contact ratio mp.

(b)(3%) A hypothetical gear-set has Np = 12, Ng = 30, k = 0.8, = 20. Check forinterference.

(c)(2%) The primary advantage of using an involute gear-tooth profile is: (circle one)

Backlash is minimized.

Enhanced tooth strength.

Velocity ratio is exactly constant.

none of the above.

(d)(2%) The Base Circle separates the face of a gear tooth from its flank. Thisstatement is (circle one)

TRUE FALSE

FINAL GUIDELINES

On Problem #1: The level at which you need to think about which kind of synthesis isthat of the design task. What did the designer set out to accomplish? The only possiblescores on this problem are 0, 5, and 10. See that yours is 10!

On Problem#2: Remember the definition of Transmission Angle (Norton (5th Ed.), p.204).You need to check both extrema of 12.

On Problem #3(b): Youre not asked to justify your conclusion, just to circle one. Youcan figure it out by appeal to a counting principle, or by spotting what, if anything, ismissing or is unneeded.

On Problem #4: The mechanism has 6 links, so there are 4 Kennedy lines passing througheach IC. For instance, I2,5 lies on the lines joining I1,2 to I1,5, I2,3 to I3,5, I2,4 to I4,5, I2,6to I5,6. To locate I2,5, you need to draw two of these lines. Clearly indicate what ICs andKennedy lines you are using. If one of your lines is correct, we can give you partial creditfor that.

On Problem #5: Youre given all the bits, and the tricky calculation - finding Pout - isdone for you. All you have to do is write down the power identity, and do the trig to figureout what

!2!4

is.

HINT: |RPQ| =q(xP xQ)2 + (yP yQ)2

On Problem #6: Three scalar equations - the x and y components of the force equation,and the z component of the moment equation about CG3. Apart from the x and ycomponents of the unknown joint forces F32 and F43, everything in these equations shouldbe an explicit number.

HINT: T = R F = Rx +Ry Fx + Fy = RyFx +RxFykOn Problem #8: Youre only asked for mA and A, so take moments about correction-plane B, or, equivalently, (and more eciently) use the derived expression for the inertiaforce FA.

On Problem #9: Calculate using equation (8.31d) on page 456 of Norton (5th Ed.).

On Problem #10: Calculate Z and mp using equations (9.2) on page 486 of Norton (5thEd.), and (9.6a) on page 494 of Norton (5th Ed.) respectively. The interference formulawas given in Lecture. If the denominator of a certain quotient expression in Np is negative,or if it is positive, and the expression exceeds Ng, there is no interference.