00:02
Question number 141.
00:04
We can divide the element into three slender, so the mass moment of inertia of the old element can be written as ix equal ix1, ix2, minus ix3.
00:31
And iy equal iy1 plus iy2 minus ix3.
00:44
Iy3 and i z equal i z1 plus i z2 minus i z z okay note that all three strangers are above are below xx plan so we will drive expressions for moments of inertia of this particular case from figure nine twenty eight we know that moments of inertia about central axis r i x equal i z equal m over 12 multiply 3a squared plus l squared and i y equal m over 2 a squared now let's take just one half of the slender because of the symmetry moments of inertia that half or i x dash equal i z dash equal m over 24 multiply 3 a square plus l squared i y equal i y dash equal m over 4 a squared note that mass of the newslender is m a dash equal half m so why is it is it's high it will equal half nf so we can rise the exhibitions above as ix dash equal i z dash equal m dash over 12 multiply 3 a squared plus l squared and iy equal m -dash over 2 a squared the mass of the new slender oh okay uh the density of the steel is row equal 7 .850 multiply 10 over minus 10 power minus 6 kilogram clear millimeter cube the mass of the slender 1 is m1 equal v1 row equal 80 squared multiplied by multiply 40 multiply 7 multiply 7 .5 0 multiplied 10 power negative 6 equal 6 .6 .6 .3 .3 kigrate.
05:12
Using the integrations, we wrote in step 2, the mass moments of inertia of part 1 are ix1 equal iz1 equal 6 .313 over 12, multiply 3 .8 squared plus 4 .4 squared equal 30 and 468 milligram millimeter squared.
06:02
Iy1 equal 6 .313 over 2, multiply 8 squared, equal 8 squared, equal 3133 over 2, multiply 8 squared, equal 20 and 202 km gram millimeter and so on the mass of slender 2 is m2 equal b2 row equal 20 squared 5 multiply by multiply 60 multiply 7 .850 multiply 7 .850 multiply 10 power negative 6 equal 0 .592 kilogram...