{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 29 Prob 9_141-1

# HW 29 Prob 9_141-1 - PROBLEM 9.141 The machine element...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM 9.141 The machine element shown is fabricated from steel. Determine the mass moment of inertia of the assembly with respect to (a) the x axis, (27) the y axis, (0) the z axis. (The density of steel is 7850 kg/rn3 .) SOLUTION First compute the mass of each component. We have m = Per Then m1 = (7850 kg/m3)(7z(0.08 m)2(0.04 m)] = 6.31334 kg m2 = (7850 kg/m3)[7z(0.02 m)2 (0.06 m)] = 0.59188 kg m3 = (7850 kg/m3)[7z(0.02 m)2 (0.04 m)] = 0.39458 kg Using Figure 9.28 and the parallel-axis theorem, we have (a) Ix :(Ix)1+(1x)2 _(Ix)3 = {\$(6-31334 kg)[3(0.08)2 + (0.04021 m2 + (6.31334 kg)(0.02 my} +{1i2(0-59188 kg)[3(0.02)2 + (0.06)2] m2 + (0.59188 kg)(0.03 my} .—{1i2(0.39458 kg)[3(0.02)2 + (004)21 m? + (0.39458 kg)(0.02 my} = [(10.94312 + 2.52534) + (0.23675 + 0.53269) — (0.09207 + 0.15783)]><10'3 kg - m2 = (13.46846 + 0.76944 — 0.24990) ><10‘3 kg - m2 =13.988OO><10‘3 kg-m2 or 1x=13.99><10‘3 kg-m2 4 PROBLEM 9.141 (Continued) (b) 1y:(1y)1+(1y)2—(Iy)3 = 56-31334 kg)(0.08 my] + E(0.59188 kg)(0.02 m)2 + (0.59188 kg)(0.04 Inf] — E(0.39458 kg)(0.02 m2) + (0.39458 kg)(0.04 my] =[(20.20269) +(0.11838 + 0.94701) — (0.07892 + O.63133)]><10‘3 kg - m2 = (20.20269 +1.06539 — 0.71025)><10‘3 kg - m2 = 20.55783x10‘3 kg - m2 or [y =20.6><10'3kg-m2 4 (C) Iz =(Iz)1+(Iz)2 _(Iz)3 _{%(0.39458 kg)[3(0.02)2 +(0.04)2] m2 5.0103958 kg)[(0.04)2 + (0.032] m2} = [(10.94312 + 2.52534) + (0.23675 +1.47970) — (0.09207 + 0.78916)]><10‘3 kg - m2 = (l3.46846+1.71645—0.88123)><10‘3 kg-m2 =14.30368><10—3 kg-m2 or [Z =14.30><10'3kg-m2 4 To the Instructor: The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the solutions of Problems 9.142 through 9.145. From Figure 9.28: . 1 Cylznder (I x ) cyl = 2 mcyla2 1 (1y)cy1 = (Iz)cy1 =1—2n’lcyl(3a2 +L2) PROBLEM 9.141 (Continued) Symmetry and the deﬁnition of the mass moment of inertia (I = lrzdm) imply 1 (I)semicylinder = E (I) cylinder 1 1 (Ix )sc = ‘2‘ [’2— mcylazj d 1 — I — 1 1 2 2 an (y)SC—( z)sc‘§ Emcylc’a +1“) 1 "1 However, msc = 9:le 1 2 Thus, (Ix )Sc = Emsca % i l 1 2 2 ’2. 442' x and (1y)SC =(IZ)sc =1—2msc(3a +L ) 3‘7 X Also, using the parallel axis theorem ﬁnd where x' and z’ are centroidal axes. ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online