ME 314 Acceleration HW September 28, 2011

Acceleration HW Solution

Problem 3.15: The figure is shown below, with my numerical lengths and angles. The angular acceleration of link 2is ZERO, so α2 = 0. I asked you to find the accelertion of point B and the angular acceleration of links 3 & 4 (theyare the same).

n193.62 mm

11.17!

n

(a) Analytical Acceleration. Since there is sliding relative to a rotating body (link 3 relative to body 4), you mustuse the “1 point moving on a body” equation. Since you know the path of A relative to body 4, that equation iswritten as

aA = aA4 + 4aA + 2ω4 × 4vA (1)

The acceleration of point A (left side of (1)) can be found from point O2 using

aA = aO2︸︷︷︸=0

+ α2︸︷︷︸=0

×rAO2 + ω2 × (ω2 × rAO2) (2)

= 233.8 i− 135 j m/s2

(3)

The three terms on the right of (1) will now be found in turn.

1. Term aA4Point A4 is fixed to body 4, so can be related to point O4 (which is fixed). The “2 points on a body”

acceleration equation for these two points is

aA4= aO4︸︷︷︸

=0

+α4 × rAO4+ ω4 × (ω4 × rAO4

) (4)

= (−0.0375α4 i− 0.19α4 j) + (91.94 i− 18.15 j) m/s2

(5)

2. Term 4aA. This relative acceleration is along line AB, so just like (6) in the velocity solution, we can express thisas an unknown magnitude and a known direction:

4aA = 4aA(cos 11.17◦ i− sin 11.17◦ j) = 4aA(0.9811 i− 0.1937 j) (6)

3. Term 2ω4 × 4vA. This Coriolis term can be computed from the velocity results; since ω4 = 22 rad/s and 4vA = − 1453mm/s,

2ω4 × 4vA = 2(22k)×(−1453

1000

)(cos 11.17◦ i− sin 11.17◦ j) (7)

= −12.38 i− 62.7 j m/s2

(8)

OK, gathering up all the terms and putting (1) in matrix form, I got[0.0375 0.9811−0.19 −0.1937

] [α44aA

]=

[154.24−54.15

](9)

Solving (9) with MATLAB, I got

α4 = 120rad/s2

=⇒ α3 = α4 = 120k rad/s2

(10)

1

ME 314 Acceleration HW September 28, 2011

But what I asked for was the acceleration of B, which can be related to the acceleration of A using the “2 points ona body” equation along with the (now known) α3 and ω3:

aB = aA +α3 × rBA + ω3 × (ω3 × rBA)

= 53.2 i− 50.4 j m/s2

(11)

(12)

(b) ADAMS Analysis. I used exactly the same ADAMS model as for the velocity analysis, and just plotted theangular acceleration of link 3 and the acceleration of point B. These plots are shown below.

Since it’s difficult to see the acceleration values at 150◦, I took the following ADAMS screenshots to PROVE myADAMS results agrees with the analytical (angular acceleration α3 first, then X and Y accelerations of point B):

They agree!

2

ME 314 Acceleration HW September 28, 2011

Problem 3.23: The figure is shown at right; crank 2 rotates at 2,000rpm CW. I asked you to just find the angular acceleration of link 3,and the acceleration of point B. Angular acceleration α2 = 0. Thefollowing dimensions are given:

RAO2= RCA = 2 in

RBA = RCB = 6 in

RDC = 5 in

In addition, the following angles are needed:

6 AB from horizontal = 76.78◦

6 AC from horizontal = 22.82◦

6 CD from horizontal = 54◦

(a) Analytical Acceleration. Since the only sliding is relative to a fixed body, this problem can be solved usingonly “two points on a body.”

First find the acceleration of A:

aA = aO2︸︷︷︸=0

+ α2︸︷︷︸=0

×rAO2+ ω2 × (ω2 × rAO2

) (13)

= −7311 i ft/s2

(14)

Points A and B are both on link 3; the direction of aB is known (it’s along line BO2), and the direction of α3 is known(k direction). So we can relate

aB = aA +α3 × rBA + ω3 × (ω3 × rBA), (15)

where

α3 × rBA = −0.4867α3 i + 0.1143α3 j ft/s2

(16)

ω3 × (ω3 × rBA) = −152 i− 647 j ft/s2

(17)

Also, since we know the direction of aB (refer to (19) in the velocity solution), we can write

aB = 0.5aB i + 0.866aB j (18)

Equate (15) and (18), with (16) and (17) substituted in, you get the following two equations:

i : 0.5aB = −7311− 0.4867α3 − 152 (19)

j : 0.866aB = 0.1143α3 − 647 (20)

Put (19) and (20) in matrix form, [0.5 0.4867

0.866 −0.1143

] [aBα3

]=

[−7463−647

](21)

Solving with MATLAB, I got

aB = −2440ft/s2, α3 = −12, 827 rad/s

2(22)

and the corresponding vectors are

aB = −1220 i− 2113 j ft/s2

α3 = −12, 827k rad/s2

(23)

(24)

3

ME 314 Acceleration HW September 28, 2011

ADAMS Acceleration. Again, I used the same ADAMS model; just plotted the angular acceleration of link 3, andthe acceleration of point B.

-12,825

The configuration in the problem statement corresponds to 0◦ crank angle, which is a value of

(α3)ADAMS = −12, 825k rad/s2

(25)

This is in almost exact agreement with the analytical value of −12, 827k.

The ADAMS plot of the acceleration of point B is shown below:

-1220-2114

At the crank angle of 0◦ the value for the acceleration is:

(aB)ADAMS = −1220 i− 2114 j ft/s2

(26)

Again, this is in almost exact agreement with the analytical value of −1220 i− 2113 j.

4