# ch12-2 - SECTION 12.6 Polar Moments of Inertia 15 Polar...

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Problem 12.6-2 Determine the polar moment of inertia ( I P ) C with respect to the centroid C for a circular sector (see Case 13, Appendix D). Solution 12.6-2 Polar moment of inertia SECTION 12.6 Polar Moments of Inertia 15 Polar Moments of Inertia Problem 12.6-1 Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D) Solution 12.6-1 Polar moment of inertia P OINT C ( CENTROID ) FROM C ASE 5: ( I P ) c 5 bh 144 (4 h 2 1 3 b 2 ) P OINT A( APEX ): I P 5 bh 48 ( b 2 1 12 h 2 ) 5 bh 144 h 2 1 3 b 2 ) 1 bh 2 ¢ 2 h 3 2 I P 5 ( I P ) c 1 A ¢ 2 h 3 2 h C b A y 2/3 h P OINT O ( ORIGIN ) FROM C ASE 13: ( a 5 radians) ( I P ) o 5 a r 4 2 A 5 a r 2 P OINT C ( CENTROID ): 5 r 4 18 a (9 a 2 2 8 sin 2 a ) ( I P ) C 5 ( I P ) O 2 Ay 2 5 a r 4 2 2 a r 2 ¢ 2 r sin a 3 a 2 y 5 2 r sin a 3 a y C r O x aa y Problem 12.6-3 Determine the polar moment of inertia I P for a W 8 3 21 wide-flange section with respect to one of its outermost corners. Solution 12.6-3 Polar moment of inertia W 8 3 21 I 1 5 75.3 in. 4 I 2 5 9.77 in. 4 A 5 6.16 in. 2 Depth d 5 8.28 in. Width b 5 5.27 in. I x 5 I 1 1 A ( d / 2) 2 5 75.3 1 6.16(4.14) 2 5 180.9 in. 4 I y 5 I 2 1 A ( b / 2) 2 5 9.77 1 6.16(2.635) 2 5 52.5 in. 4 I P 5 I x 1 I y 5 233 in. 4 x C 11 2 2 y O

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16 CHAPTER 12 Review of Centroids and Moments of Inertia Problem 12.6-4 Obtain a formula for the polar moment of inertia I P with respect to the midpoint of the hypotenuse for a right triangle of base b and height h (see Case 6, Appendix D). Solution 12.6-4 Polar moment of inertia P OINT C FROM C ASE 6: ( I P ) c 5 bh 36 ( h 2 1 b 2 ) P OINT P: 5 bh 24 ( b 2 1 h 2 ) I P 5 bh 36 ( h 2 1 b 2 ) 1 bh 2 ¢ b 2 1 h 2 36 5 b 2 36 1 h 2 36 5 b 2 1 h 2 36 d 2 5 ¢ b 2 2 b 3 2 1 ¢ h 2 2 h 3 2 A 5 bh 2 I P 5 ( I P ) c 1 Ad 2 h h /2 C b b /3 b /2 h /3 P d = CP Problem 12.6-5 Determine the polar moment of inertia ( I P ) C with respect to the centroid C for a quarter-circular spandrel (see Case 12, Appendix D). Solution 12.6-5 Polar moment of inertia P OINT O FROM C ASE 12: A 5 ¢ 1 2 p 4 r 2 y 5 (10 2 3 p ) r 3(4 2 p ) I x 5 ¢ 1 2 5 p 16 r 4 P OINT C ( CENTROID ): C OLLECT TERMS AND SIMPLIFY : (by symmetry) ( I P ) c 5 2 I x C 5 r 4 72 ¢ 176 2 84 p 1 9 p 2 4 2 p I y C 5 I x C I x C 5 r 4 144 ¢ 176 2 84 p 1 9 p 2 4 2 p 2 ¢ 1 2 p 4 ( r 2 ) B 2 3 p ) r 2 p ) R 2 Ix c 5 I x 2 Ay 2 5 ¢ 1 2 5 p 16 r 4 x yy C y r O C x C x
SECTION 12.7 Products of Inertia 17 Products of Inertia Problem 12.7-1 Using integration, determine the product of inertia I xy for the parabolic semisegment shown in Fig. 12-5 (see also Case 17 in Appendix D). Solution 12.7-1 Product of inertia Product of inertia of element dA with respect to axes through its own centroid equals zero. dI xy 5 product of inertia of element dA with respect to xy axes d 1 5 xd 2 5 y / 2 Parallel-axis theorem applied to element dA : xy 5 0 1 ( dA )( d 1 d 2 ) 5 ( ydx )( x )( y / 2) I xy 5 # dI xy 5 h 2 2 # b 0 x ¢ 1 2 x 2 b 2 2 dx 5 b 2 h 2 12 5 h 2 x 2 ¢ 1 2 x 2 b 2 2 dx dA 5 y dx 5 h ¢ 1 2 x 2 b 2 dx y h O x b dx x y /2 dA y 5 h ( 1 2 ) x 2 b 2 Problem 12.7-2 Using integration, determine the product of inertia I xy for the quarter-circular spandrel shown in Case 12, Appendix D.

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ch12-2 - SECTION 12.6 Polar Moments of Inertia 15 Polar...

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