Distributed Forces: Centroids and Centers of Gravity

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1 Distriuted Forces: Centroids nd Centers of Grvit

Introduction Center of Grvit of D Bod Centroids nd First Moments of Ares nd Lines Centroids of Common Shpes of Ares Centroids of Common Shpes of Lines Composite Pltes nd Ares Smple Prolem 5.

2 Introduction Center of Grvit of D Bod Centroids nd First Moments of Ares nd Lines Centroids of Common Shpes of Ares Centroids of Common Shpes of Lines Composite Pltes nd Ares Smple Prolem 5.1 Determintion of Centroids Integrtion Theorems of Pppus-Guldinus Smple Prolem 5.7 Distriuted Lods on Bems Smple Prolem 5.9 Center of Grvit of D Bod: Centroid of Volume Centroids of Common D Shpes Composite D Bodies Smple Prolem 5.1 Smple Prolem

3 The erth eerts grvittionl force on ech of the prticles forming od. These forces cn e replce single equivlent force equl to the weight of the od nd pplied t the center of grvit for the od. The centroid of n re is nlogous to the center of grvit of od. The concept of the first moment of n re is used to locte the centroid. Determintion of the re of surfce of revolution nd the volume of od of revolution re ccomplished with the Theorems of Pppus-Guldinus. 5 -

4 Center of grvit of plte Center of grvit of wire M W W M W dw W dw 5-4

5 Centroid of n re Centroid of line W dw At t A da da Q first moment wit h respect to A da Q first moment wit h respect to W dw L L dl L dl dl 5-5

An re is smmetric with respect to n is BB if for ever point P there eists point P such

The first moment of n re with respect to line of smmetr is zero.

lines of smmetr, its centroid lies t their intersection.

6 An re is smmetric with respect to n is BB if for ever point P there eists point P such tht PP is perpendiculr to BB nd is divided into two equl prts BB. The first moment of n re with respect to line of smmetr is zero. If n re possesses line of smmetr, its centroid lies on tht is If n re possesses two lines of smmetr, its centroid lies t their intersection. An re is smmetric with respect to center O if for ever element da t (,) there eists n re da of equl re t (-,-). The centroid of the re coincides with the center of smmetr. 5-6

7 5-7

8 5-8

9 Composite pltes X Y W W W W Composite re X Y A A A A 5-9

10 SOLUTION: Divide the re into tringle, rectngle, nd semicircle with circulr cutout. Clculte the first moments of ech re with respect to the es. For the plne re shown, determine the first moments with respect to the nd es nd the loction of the centroid. Find the totl re nd first moments of the tringle, rectngle, nd semicircle. Sutrct the re nd first moment of the circulr cutout. Compute the coordintes of the re centroid dividing the first moments the totl re. 5-10

11 Find the totl re nd first moments of the tringle, rectngle, nd semicircle. Sutrct the re nd first moment of the circulr cutout. Q Q mm mm 5-11

12 Compute the coordintes of the re centroid dividing the first moments the totl re. X A A mm mm X 54.8 mm Y A A mm mm Y 6.6 mm 5-1

13 A A da da dd dd el el da da Doule integrtion to find the first moment m e voided defining da s thin rectngle or strip. A A el el da d da d A A da el el da d d A A el el da r 1 cos da r 1 sin r r d d 5-1

14 SOLUTION: Determine the constnt k. Evlute the totl re. Determine direct integrtion the loction of the centroid of prolic spndrel. Using either verticl or horizontl strips, perform single integrtion to find the first moments. Evlute the centroid coordintes. 5-14

15 5-15 SOLUTION: Determine the constnt k. 1 1 or k k k Evlute the totl re. 0 0 d d da A

16 5-16 Using verticl strips, perform single integrtion to find the first moments d d da Q d d da Q el el

17 5-17 Or, using horizontl strips, perform single integrtion to find the first moments d d d da Q d d d da Q el el

18 Evlute the centroid coordintes. A Q 4 4 A Q

19 Surfce of revolution is generted rotting plne curve out fied is. Are of surfce of revolution is equl to the length of the generting curve times the distnce trveled the centroid through the rottion. A L 5-19

20 Bod of revolution is generted rotting plne re out fied is. Volume of od of revolution is equl to the generting re times the distnce trveled the centroid through the rottion. V A 5-0

SOLUTION: Appl the theorem of Pppus-Guldinus to evlute the volumes or revolution for the rectngulr rim section nd the inner cutout section. The outside dimeter of pulle is 0.

21 SOLUTION: Appl the theorem of Pppus-Guldinus to evlute the volumes or revolution for the rectngulr rim section nd the inner cutout section. The outside dimeter of pulle is 0.8 m, nd the cross section of its rim is s shown. Knowing tht the pulle is mde of steel nd tht the densit of steel is kg m determine the mss nd weight of the rim. Multipl densit nd ccelertion to get the mss nd ccelertion. 5-1

22 SOLUTION: Appl the theorem of Pppus-Guldinus to evlute the volumes or revolution for the rectngulr rim section nd the inner cutout section. Multipl densit nd ccelertion to get the mss nd ccelertion kg m mm 10 m mm 60.0 kg 9.81m s m V m 60.0 kg W mg W 589 N 5 -

23 W L 0 wd da A A distriuted lod is represented plotting the lod per unit length, w (N/m). The totl lod is equl to the re under the lod curve. OP W L dw OPA da A 0 A distriuted lod cn e replce concentrted lod with mgnitude equl to the re under the lod curve nd line of ction pssing through the re centroid. 5 -

24 SOLUTION: A em supports distriuted lod s shown. Determine the equivlent concentrted lod nd the rections t the supports. The mgnitude of the concentrted lod is equl to the totl lod or the re under the curve. The line of ction of the concentrted lod psses through the centroid of the re under the curve. Determine the support rections summing moments out the em ends. 5-4

25 SOLUTION: The mgnitude of the concentrted lod is equl to the totl lod or the re under the curve. F 18.0 kn The line of ction of the concentrted lod psses through the centroid of the re under the curve. X 6 kn m 18 kn X.5 m 5-5

26 Determine the support rections summing moments out the em ends. 6 m 18 kn.5 m 0 M A 0 : B B 10.5 kn 6 m 18 kn6 m.5 m 0 M B 0 : A A 7.5 kn 5-6

27 Center of grvit G W j W j r r G G W W W j r W j j rw j dw rgw rdw Results re independent of od orienttion, W V dw W For homogeneous odies, W V nd dw dv dv V dw zw dv zv zdv zdw 5-7

28 5-8

29 Moment of the totl weight concentrted t the center of grvit G is equl to the sum of the moments of the weights of the component prts. X W W Y W W Z W zw For homogeneous odies, X V V Y V V Z V zv 5-9

30 SOLUTION: Form the mchine element from rectngulr prllelepiped nd qurter clinder nd then sutrcting two 1-in. dimeter clinders. Locte the center of grvit of the steel mchine element. The dimeter of ech hole is 1 in. 5-0

31 5-1

32 X V V 4.08 in 5.86 in X in. Y V V in 5.86 in Y in. Z zv V in 5.86 in Z in. 5 -

STATICS VECTOR MECHANICS FOR ENGINEERS: and Centers of Gravity. Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

STATICS VECTOR MECHANICS FOR ENGINEERS: and Centers of Gravity. Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr. 007 The McGrw-Hill Compnies, Inc. All rights reserved. Eighth E CHAPTER 5 Distriuted VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinnd P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Wlt Oler Tes Tech