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 2 is ZERO, so α2 = 0. I asked you to find the accelertion of point B and the angular acceleration of links 3 & 4 (they are the same).
193.62n m
m
n
11.17◦ (a) Analytical Acceleration. Since there is sliding relative to a rotating body (link 3 relative to body 4), you must use the “1 point moving on a body” equation. Since you know the path of A relative to body 4, that equation is written as aA = aA4 + 4 aA + 2ω 4 × 4 vA
(1)
The acceleration of point A (left side of (1)) can be found from point O2 using aA = aO2 + α2 ×rAO2 + ω 2 × (ω 2 × rAO2 ) |{z} |{z} =0
(2)
=0
= 233.8 i − 135 j m/s
2
(3)
The three terms on the right of (1) will now be found in turn. 1. Term aA4 Point 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 +α4 × rAO4 + ω 4 × (ω 4 × rAO4 ) |{z}
(4)
=0
= (−0.0375α4 i − 0.19α4 j) + (91.94 i − 18.15 j) m/s
2
(5)
2. Term 4 aA . This relative acceleration is along line AB, so just like (6) in the velocity solution, we can express this as an unknown magnitude and a known direction: 4
aA =
4
aA (cos 11.17◦ i − sin 11.17◦ j) =
4
aA (0.9811 i − 0.1937 j)
(6)
3. Term 2ω 4 × 4 vA . This Coriolis term can be computed from the velocity results; since ω4 = 22 rad/s and 4 vA = − 1453 mm/s, −1453 4 2ω 4 × vA = 2(22 k) × (cos 11.17◦ i − sin 11.17◦ j) (7) 1000 = −12.38 i − 62.7 j m/s
2
(8)
OK, gathering up all the terms and putting (1) in matrix form, I got 0.0375 0.9811 α4 154.24 = 4 −0.19 −0.1937 aA −54.15
(9)
Solving (9) with MATLAB, I got 2
α4 = 120rad/s
2
=⇒ α3 = α4 = 120 k rad/s
1
(10)
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 on a body” equation along with the (now known) α3 and ω 3 : aB = aA + α3 × rBA + ω 3 × (ω 3 × rBA ) = 53.2 i − 50.4 j m/s
2
(11) (12)
(b) ADAMS Analysis. I used exactly the same ADAMS model as for the velocity analysis, and just plotted the angular 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 my ADAMS 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 2,000 rpm CW. I asked you to just find the angular acceleration of link 3, and the acceleration of point B. Angular acceleration α2 = 0. The following 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 using only “two points on a body.” First find the acceleration of A: aA = aO2 + α2 ×rAO2 + ω 2 × (ω 2 × rAO2 ) |{z} |{z}
(13)
=0
=0
= −7311 i ft/s
2
(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 + ω 3 × (ω 3 × rBA ),
(15)
where α3 × rBA = −0.4867α3 i + 0.1143α3 j ft/s ω 3 × (ω 3 × rBA ) = −152 i − 647 j ft/s
2
2
(16) (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 0.866
0.4867 −0.1143
aB −7463 = α3 −647
(21)
Solving with MATLAB, I got 2
aB = −2440ft/s , α3 = −12, 827 rad/s
2
(22)
and the corresponding vectors are aB = −1220 i − 2113 j ft/s α3 = −12, 827 k rad/s
3
2
2
(23) (24)
ME 314
Acceleration HW
September 28, 2011
ADAMS Acceleration. Again, I used the same ADAMS model; just plotted the angular acceleration of link 3, and the acceleration of point B.
-12,825
The configuration in the problem statement corresponds to 0◦ crank angle, which is a value of 2
(α3 )ADAMS = −12, 825 k rad/s
(25)
This is in almost exact agreement with the analytical value of −12, 827 k.
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/s
2
Again, this is in almost exact agreement with the analytical value of −1220 i − 2113 j.
4
(26)