MAE2323 Dynamics
(218pts)
Homework #7
Due October 24, 2024
MAE 2323 DYNAMICS Homework #7 (218pts) with solved solutions University of Texas, Arlington Grade P4.42 and P4.51. S4.49 (33pts) Find the inertia dyadic about the mass center for the thin plate in Figure 4.56. A square void is centered at the center of the plate. The thin plate has a density of ρ = 50kg/m2. ^ Y ^ X ^ Y 45o ^ X 300mm
Figure 1: Thin Plate. The inertia dyadics for the large and small plates are: mc L 2 ^ ^ ^AY ^A ICC = XAYA + Y 12 (4.33,A.11,A.12,A.13)
mcL2 ^ ^ ZAZA 6 BZB mBw2 Z ^ ^ + 6 +
mBw2 ^ ^ ^BY^B XBXB + Y 12 (4.33,A.11,A.12,A.13) We need to convert frame B to frame A: ◦ ◦ ◦ ◦ ^ B = cos(45 ) X ^ A − sin(45 ) Y ^A Y ^ B = cos(45 ) Y ^ A + sin(45 ) X ^A X IBB
=
2
IBB
=
mBw 12
◦ ^ ◦ ^ cos(45 )X A − sin(45 )YA ◦
◦
^ A + sin(45 )X^A cos(45 )Y IBB
=
(9pts) (9pts)
^ B = Z^ A (10pts) Z
◦ ^ ◦ ^ cos(45 )X A − sin(45 )YA ◦
◦
^ A + sin(45 )X^A cos(45 )Y
+ mBw2 ^ ^ + ZAZA 6
(4pts)
mBw2 ◦ ^ ^ ◦ ^ ^ 2 2 ◦ ◦ ^ AY ^A + Y ^AX^A + cos (45 )X X AXA + sin (45 )YAYA − cos(45 ) sin(45 ) 12 ◦ ◦ ◦ ◦ ^ AY ^A + Y ^AX^A + cos2 (45 )Y X ^ AY ^A + sin 2(45 )X^AX^ A + cos(45 ) sin(45 ) 2 mBw ^ ^ ZAZA 6 2 mBw2 ^ ^ ^AY ^A + mBw Z ^ AZ^A IBB = XAXA + Y (2pts) 6 12
Finally IAA = IBB − ICC
(4pts)
(4.96)
mc L 2 − m B w 2 12 Now we evaluate the terms IAA =
^ XAY^A + Y^AY^ A
+
mcL2 − mBw2 ^ ^ ZAZA 6
(2pts)
mC = 50kg/m2 · (1.4m)2 = 98kg mB = 50 98kg(1.4m) − 4.5kg(0.3m) 12 2
IAA =
2
kg/m2
· (0.3m)
2
= 4.5kg
^ XAX^A + Y^AY^ A
^ A + Y^AY^A IAA = 15.97kgm2 ^XAX
(2pts) (2pts)
98kg(1.4m) − 4.5kg(0.3m) 6 2
+
+ 31.95kgm2 Z^AZ^A
2
^ AZ ^A Z (2pts)
S4.52 (34pts) In Section 4.3.1 we found the moment of inertia of the block in Figure 4.57 about the ^ZA axis 1