Statik Centroids And Centers of Gravity VECTOR MECHANICS FOR ENGINEERS: STATICS Eighth Edition Ferdinand P. Beer Ferdinand P. Beer Ferdinand P. Beer E. Russell Johnston, Jr. E. Russell Johnston, Jr. E. Russell Johnston, Jr. & R.C. HIBBELER & R.C. HIBBELER & R.C. HIBBELER Lecture Notes: Lecture Notes: Lecture Notes: Dr. N. MEYDANLIK Dr. N. MEYDANLIK Dr. N. MEYDANLIK T T Trakya rakya rakya University University University CHAPTER © All rights reserved. 5 Yay Yay ı ı l l ı ı kuvvetler kuvvetler : : A A ğı ğı rl rl ı ı k Merkezleri k Merkezleri © All rights reserved to bana. Eighth Edition 5 - 2 Contents Introduction Center of Gravity of a 2D Body Centroids and First Moments of Areas and Lines Centroids of Common Shapes of Areas Centroids of Common Shapes of Lines Composite Plates and Areas Sample Problem 5.1 Determination of Centroids by Integration Sample Problem 5.4 Theorems of Pappus-Guldinus Sample Problem 5.7 Distributed Loads on Beams Sample Problem 5.9 Center of Gravity of a 3D Body: Centroid of a Volume Centroids of Common 3D Shapes Composite 3D Bodies Sample Problem 5.12© All rights reserved to bana. Eighth Edition 5 - 3 • Mass is a property of a physical object that quantifies the amount of matter it contains. • Weight is a force that results from the action of gravity on matter. • Center of gravity (CG) - Point which locates the resultant weight of a physical object or system of particles. • Center of mass (CM) - Point where the mass of a physical object or system of particles can be assumed to be concentrated. • Centroid - Point which defines the geometric center of a physical object or system of particles. Definitions: © All rights reserved to bana. Eighth Edition 5 - 4 Note: The center of gravity and the center of mass coincide when the gravitational field the object is subjected to has the same magnitude and direction everywhere. On Earth, in general engineering applications, the assumption of a uniform gravity field is appropriate to use. The centroid will coincide with the CM and CG only if the material composing the physical object is homogeneous.© All rights reserved to bana. Eighth Edition 5 - 5 e zaman Ağırlık Merkezi/Kü tle Merkezini hesaplama ihtiyacı duyarız? Statikte genellikle cismin toplam kütlesi ile ilgileniriz ve cismin toplam ağırlığını tekil bir kuvvet olarak göstermek isteriz.??? Su deposu taşıyıcı ayakların tasarımı Aracın keskin virajlarda devrilmeye karşı daha emniyetli olması için kütle merkezi nerde olmalı? © All rights reserved to bana. Eighth Edition 5 - 6 Center of Center of Mass Mass (K (Kü ütle Merkezi) tle Merkezi) -Cismi sonsuz sayıda maddesel noktadan meydana gelmiş bir sistem olarak gözönüne alırız. x Use the concept of equivalent loads to locate the centre of mass: The total mass located at the centre of mass must create a moment about any point equal to the moment created by all the individual particle masses about that same point. For example, to find coordinate of centre of mass, consider moment about y-axis: W=Mg dW=?g dV ? ? = = V V dV g x ~ Mg x dW x ~ W x ?© All rights reserved to bana. Eighth Edition 5 - 7 z , y , x -Kütle merkezinin koordinatları, ; M dV z ~ z , M dV y ~ y , M dV x ~ x V V V ? ? ? = = = ? ? ? (Yerçekimi kuvvetini heryerde aynı kabul ederek) Center of Center of Mass Mass (K (Kü ütle Merkezi) tle Merkezi) The centre of gravity is a point which locates the resultant weight of a system of particles or body, and is generally the same as the centre of mass (as long as the gravity field is assumed to have the same magnitude and direction everywhere –appropriate for most engineering applications) © All rights reserved to bana. Eighth Edition 5 - 8 Cent Centroid roid (Geometri Merkezi) (Geometri Merkezi) The centroid C is a point which defines the geometric centre of an object. The centroid coincides with centre of mass or centre of gravity only if: the material of the body is homogenous (density or specific weight is constant throughout the body). V dV z ~ z , V dV y ~ y , V dV x ~ x V V V ? ? ? = = = z , y , x -Geometri merkezinin koordinatları, ;© All rights reserved to bana. Eighth Edition 5 - 9 If an object has an axis of symmetry, then the centroid of the object lies on that axis. Cent Centroid roid (Geometri Merkezi) (Geometri Merkezi) x © All rights reserved to bana. Eighth Edition 5 - 10 Understanding the difference between Center of Area (centroid) and Center of Mass: by sticking a coin on this surface, the CM will be changed!© All rights reserved to bana. Eighth Edition 5 - 11 Consider an extruded homogeneous volume as shown -this has a constant cross section along an axis, shown as Area A X c and Y c for the volume will be located at the centroid(X c , Y c ) of the Area A Cent Centroid roid of an of an Area Area (Alan (Alanı ın Geometri Merkezi) n Geometri Merkezi) To find X C , for example, consider the moment about the y-axis : ?? ?? = = A T A T ) dy dx ( x ~ ) A ( x veya ) dy dx t g ( x ~ ) A t g ( x ? ? Alan A T Alan A T © All rights reserved to bana. Eighth Edition 5 - 12 Cent Centroid roid of an of an Area Area (Alan (Alanı ın Geometri Merkezi) n Geometri Merkezi) Bir alanın geometri merkezinin koordinatları, önceki ifade tekrar düzenlenerek , T A A A T A A A A dA y ~ dA dy dx y ~ Y A dA x ~ dA dy dx x ~ X ? ? ?? ? ? ?? = = = = benzer şekilde ; Q x : x eksenine göre 1. alan momenti (statik moment) T x T y A Q Y A Q X = = Q y : y eksenine göre 1. alan momenti (statik moment) Q x : x eksenine göre 1. alan momenti (statik moment) SONU SONUÇ Ç : Sabit kalınlıklı, homojen bir düzlemsel bölgenin geometri merkezinin koordinatları , şeklinde elde edilir.© All rights reserved to bana. Eighth Edition 5 - 13 ? ? ? ? ? ? = = = L L L L L L dL dL z z dL dL y y dL dL x x ~ ~ ~ Cent Centroid roid of of Line Line (Bir Uzunlu (Bir Uzunluğ ğun Geometri Merkezi) un Geometri Merkezi) © All rights reserved to bana. Eighth Edition 5 - 14 Bir alan Bir alanı ın geometri merkezinin n geometri merkezinin integrasyonla integrasyonla belirlenmesi i belirlenmesi iç çin ad in adı ımlar mlar 1. Önce genel bir noktada (x,y) uygun diferensiyel eleman (dA) seçilir. Bunun için eğer y, x cinsinden kolaylıkla ifade ediliyorsa ,e.g., y=x 2 +1, düşey eleman tercih edilir, aksi durumda yatay eleman. 2. dA diferensiyel elemanı dx (veya dy) diferensiyelleri cinsinden ifade edilir © All rights reserved to bana. Eighth Edition 5 - 15 İki boyutlu cismin ağırlık merkezi • Center of gravity of a plate ? ? ? ? ? ? = ? = = ? = dW y W y W y M dW x W x W x M y y • Center of gravity of a wire • Alanın ağırlık merkezi de cismin ağırlık merkezine benzer. Ağırlık merkezini bulmak için alanın birinci momenti (statik moment) kullanılır. Plağı n tane küçük elemana böldüğümüzü kabul edelim © All rights reserved to bana. Eighth Edition 5 - 16 Centroids and First Moments of Areas and Lines ( ) ( ) x to respect with moment first Q dA y A y y to respect with moment first Q dA x A x dA t x At x dW x W x x y = = = = = = = = ? ? ? ? ? ? • Centroid of an area ( ) ( ) ? ? ? ? = = = = dL y L y dL x L x dL a x La x dW x W x ? ? • Centroid of a line© All rights reserved to bana. Eighth Edition 5 - 17 First Moments of Areas and Lines • Eğer bir alan BB’ eksenine göre simetrik ise her P noktasına karşılık bir P’ noktası vardır, PP’ doğrusu BB’ eksenine diktir ve BB’ doğrusu tarafından ikiye bölünmüştür. • Simetri eksenine göre alanın 1. momenti sıfırdır. • Eğer bir alanın simetri ekseni varsa ağırlık merkezi bu eksen üzerindedir • Dki simetri ekseni varsa, centroid onların kesişimindedir. • An area is symmetric with respect to a center O if for every element dA at (x,y) there exists an area dA’ of equal area at (-x,-y). • The centroid of the area coincides with the center of symmetry. © All rights reserved to bana. Eighth Edition 5 - 18 Centroids can be found using three methods: 1. Composites – If an object can be divided up into relatively simple shapes with known centroids, then the centroid of the entire object can be found using the weighted average of the centroids of the composites. 2. Integration – If the area, volume, or line of an object can be described by a mathematical equations, then the centroid can be determined through integration. 3. Solid modeling software – Software such as AutoCAD can be used to construct 3D models of objects. The software can also determine the centroid of the objects (as well as volumes, moments of inertia and other mass properties). This is not a required element of this course, but an example will be provided. Determining Determining centroids centroids of objects of objects© All rights reserved to bana. Eighth Edition 5 - 19 Centroids Centroids of common shapes of common shapes r x' y, y' C x ) 2 ( 3 4 0 2 2 1 ? ? ? + = = = = r P r A r y x r x' y' C x y ) 2 ( 3 4 1 2 4 1 ? ? ? + = = = = r P r A r y x h x' y' C b x y c h b A h y c b x 2 1 3 1 3 1 ) ( = = + = © All rights reserved to bana. Eighth Edition 5 - 20 Centroids of Common Shapes of Areas© All rights reserved to bana. Eighth Edition 5 - 21 Centroids of Common Shapes of Areas © All rights reserved to bana. Eighth Edition 5 - 22 Centroids of Common Shapes of Lines© All rights reserved to bana. Eighth Edition 5 - 23 Composite Plates and Areas • Composite plates ? ? ? ? = = W y W Y W x W X • Composite area ? ? ? ? = = A y A Y A x A X © All rights reserved to bana. Eighth Edition 5 - 24 Sample Problem 5.1 Yukardaki taralı alanın, x and y eksenlerine göre 1. momentlerini ve ağırlık merkezinin yerini bulunuz. SOLUTION: • Divide the area into a triangle, rectangle, and semicircle with a circular cutout. • Compute the coordinates of the area centroid by dividing the first moments by the total area. • Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout. • Calculate the first moments of each area with respect to the axes.© All rights reserved to bana. Eighth Edition 5 - 25 Sample Problem 5.1 3 3 3 3 mm 10 7 . 757 mm 10 2 . 506 × + = × + = y x Q Q • Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout. © All rights reserved to bana. Eighth Edition 5 - 26 Sample Problem 5.1 2 3 3 3 mm 10 13.828 mm 10 7 . 757 × × + = = ? ? A A x X mm 8 . 54 = X 2 3 3 3 mm 10 13.828 mm 10 2 . 506 × × + = = ? ? A A y Y mm 6 . 36 = Y • Compute the coordinates of the area centroid by dividing the first moments by the total area.© All rights reserved to bana. Eighth Edition 5 - 27 CONCEPT QUIZ 1. Based on the typical centroid information available in handbooks, what are the minimum number of segments you will have to consider for determing the centroid of the given area? 1, 2, 3, or 4 3cm 1 cm 1 cm 3cm © All rights reserved to bana. Eighth Edition 5 - 28 ATTENTION QUIZ 1. A rectangular area has semicircular and triangular cuts as shown. For determining the centroid, what is the minimum number of pieces that you can use? A) Two B) Three C) Four D) Five 2. For determining the centroid of the area, two square segments are considered; square ABCD and square DEFG. What are the coordinates (x, y ) of the centroid of square DEFG? A) (1, 1) m B) (1.25, 1.25) m C) (0.5, 0.5 ) m D) (1.5, 1.5) m ~ ~ 2cm 2cm 2cm 4cm x y A 1m 1m y E F G C B x 1m 1m D 1 2 3 A 2 -A 1 -A 3 Örnek SWF1, 2© All rights reserved to bana. Eighth Edition 5 - 29 Composite Plates and Areas Example: Find the centroid of the area shown below. N i i 1 1 2 2 3 3 i 1 N 1 2 3 i i 1 N i i 1 1 2 2 3 3 i 1 N 1 2 3 i i 1 x A x A + x A + x A x = A + A + A A y A y A + y A + y A y = A + A + A A = = = = · · · · = · · · · = ? ? ? ? A 3 A 2 A 1 (x 3 , y 3 ) (x 2 , y 2 ) (x 1 , y 1 ) 1) Ağırlık merkezini bildiğin parçalara böl. 2) Her bir parçanın ağırlık merkezini bul. 3) Verilen şeklin ağırlık merkezini hesapla. © All rights reserved to bana. Eighth Edition 5 - 30 Example: Problem 5.5 (also see Sample Problem 5.1) Locate the centroid of the plane area shown. Yanıt : X=49,35 mm Y=93,83 mm© All rights reserved to bana. Eighth Edition 5 - 31 ( ) ( ) ydx y Q dA y A y ydx x Q dA x A x x el y el ? ? ? ? = = = = = = 2 ( ) [ ] ( ) [ ] dx x a y Q dA y A y dx x a x a Q dA x A x x el y el - = = = - + = = = ? ? ? ? 2 ? ? ? ? ? ? = = = ? ? ? ? ? ? = = = ? ? ? ? ? ? ? ? d r r Q dA y A y d r r Q dA x A x x el y el 2 2 2 1 sin 3 2 2 1 cos 3 2 ? ?? ? ? ?? ? = = = = = = dA y dy dx y dA y A y dA x dy dx x dA x A x el el • Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip. Determination of Centroids by Integration © All rights reserved to bana. Eighth Edition 5 - 32 Steps Steps for for determining determining area area centroid centroid© All rights reserved to bana. Eighth Edition 5 - 33 Determine by direct integration the location of the centroid of a parabolic spandrel. SOLUTION: • Determine the constant k. • Evaluate the total area. • Using either vertical or horizontal strips, perform a single integration to find the first moments. • Evaluate the centroid coordinates. Sample Problem 5.4 ; © All rights reserved to bana. Eighth Edition 5 - 34 SOLUTION: • Determine the constant k. 2 1 2 1 2 2 2 2 2 y b a x or x a b y a b k a k b x k y = = = ? = = • Evaluate the total area. 3 ab 3 x a b dx x a b dx y dA A a 0 3 2 a 0 2 2 = ? ? ? ? ? ? = = = = ? ? ? Sample Problem 5.4 ;© All rights reserved to bana. Eighth Edition 5 - 35 • Using vertical strips, perform a single integration to find the first moments. 4 b a 4 x a b dx x a b x dx xy dA x Q 2 a 0 4 2 a 0 2 2 el y = ? ? ? ? ? ? = ? ? ? ? ? ? = = = ? ? ? Sample Problem 5.4 ; 10 ab 5 x a 2 b dx x a b 2 1 dx y 2 y dA y Q 2 a 0 5 4 2 a 0 2 2 2 el x = ? ? ? ? ? ? = ? ? ? ? ? ? = = = ? ? ? © All rights reserved to bana. Eighth Edition 5 - 36 • Or, using horizontal strips, perform a single integration to find the first moments. ( ) ( ) 10 4 2 1 2 2 2 0 2 3 2 1 2 1 2 1 2 0 2 2 0 2 2 ab dy y b a ay dy y b a a y dy x a y dA y Q b a dy y b a a dy x a dy x a x a dA x Q b el x b b el y = ? ? ? ? ? ? - = ? ? ? ? ? ? - = - = = = ? ? ? ? ? ? ? ? - = - = - + = = ? ? ? ? ? ? ? ? Sample Problem 5.4 ;© All rights reserved to bana. Eighth Edition 5 - 37 • Evaluate the centroid coordinates. 4 3 2 b a ab x Q A x y = = a x 4 3 = 10 3 2 ab ab y Q A y x = = b y 10 3 = Sample Problem 5.4 Sample Problem 5.4 © All rights reserved to bana. Eighth Edition 5 - 38 ( ) ( ) y y y h h b x dy y h h b dA dy x dA = ? ? ? ? ? ? - = - = = ~ 2 1 ~ ( ) ( ) 3 2 1 6 1 ~ 2 0 0 h h b h b y dy y h h b dy y h h b y dA dA y y h h A A = = ? ? ? ? ? ? - ? ? ? ? ? ? - = = ? ? ? ? ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) h 0 A h A 0 2 1 b b h y h y dy x dA 2 h h x b dA h y dy h 1 b h b 6 x 1 3 b h 2 ? ?? ? ? ?? ? ? ?? ? ? ?? ? - - - - - - - - ? ?? ? ? ?? ? ? ?? ? ? ?? ? ? ?? ? ? ?? ? ? ?? ? ? ?? ? = = = = = = = = ? ? ? ? ? ? ? ? - - - - ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = = = = = = = = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? % % % % Dik Dik Üç Üçgenin a genin ağı ğırl rlı ık merkezi k merkezi© All rights reserved to bana. Eighth Edition 5 - 39 Ex. 1 :Boyalı alanın ağırlık merkezini bulunuz . © All rights reserved to bana. Eighth Edition 5 - 40© All rights reserved to bana. Eighth Edition 5 - 41 © All rights reserved to bana. Eighth Edition 5 - 42 Solution 1. Choose dA as a horizontal rectangular strip. (x 1, ,y) (x 2 ,y) 2. dA = ( x 2 – x 1 ) dy = ((2 – y) – y 2 ) dy 3. x = ( x 1 + x 2 ) / 2 = 0.5 (( 2 – y) + y 2 ) ~ EXAMPLE Given: The area as shown. Find: The x of the centroid.© All rights reserved to bana. Eighth Edition 5 - 43 4. x = ( ? A x dA ) / ( ? A dA ) ~ ? A dA = 0 ? ( 2 – y – y 2 ) dy [ 2 y – y 2 / 2 – y 3 / 3] 1 = 1.167 m 2 1 0 ? A x dA = 0 ? 0.5 ( 2 – y + y 2 ) ( 2 – y – y 2 ) dy = 0.5 0 ? ( 4 – 4 y + y 2 – y 4 ) dy = 0.5 [ 4 y – 4 y 2 / 2 + y 3 / 3 – y 5 / 5 ] 1 = 1.067 m 3 0 1 1 ~ x = 1.067 / 1.167 = 0.914 m © All rights reserved to bana. Eighth Edition 5 - 44 ( ) ? ? ? ? ? ? ? ? = - = x x ~ dx y y dA 1 2 ( ) ( ) ( ) ( ) ft dx x x dx x x x x dx y y dx y y x dA dA x x A A 5 . 0 6 1 12 1 ~ 1 0 2 1 0 2 1 0 1 2 1 0 1 2 = = - - = - - = = ? ? ? ? ? ? Ö Örnek:2 rnek:2 ; ; Düşey diferensiyel eleman ile ( ) ( ) ( ) ? ? ? ? ? ? ? ? ? ? + = - + = - = 2 2 ~ 1 2 1 2 1 1 2 y y y y y y dx y y dA ( ) ( ) ( ) ( ) ( ) ( ) ft y dx x x dx x x x x y dx y y dx y y y y dA dA y y A A 4 . 0 6 1 60 4 2 2 ~ 1 0 2 1 0 2 2 1 0 1 2 1 0 1 2 1 2 = = - - ? ? ? ? ? ? ? ? + = - - ? ? ? ? ? ? + = = ? ? ? ? ? ?© All rights reserved to bana. Eighth Edition 5 - 45 Yatay diferensiyel eleman ile ( ) ( ) ( ) ? ? ? ? ? ? ? ? ? ? + = - + = - = 2 2 ~ 2 1 2 1 2 2 1 x x x x x x dy x x dA ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 5 . 0 2 2 2 2 ~ 1 0 1 0 2 1 0 1 0 2 2 2 1 1 0 2 1 1 0 2 1 2 1 1 0 2 1 1 0 2 1 2 1 = - - = - ? ? ? ? ? ? - = = = - - = - - + = = ? ? ? ? ? ? ? ? ? ? dy y y dy y y dy y y dy y y x y x y x dy x x dy x x dy x x dy x x x x dA dA x x A A ( ) ( ) ( ) ? ? ? ? ? ? ? ? ? ? + = - + = - = 2 2 ~ 1 2 1 2 1 1 2 y y y y y y dx y y dA ( ) ( ) ( ) ( ) ( ) ( ) ft y dx x x dx x x x x y dx y y dx y y y y dA dA y y A A 4 . 0 6 1 60 4 2 2 ~ 1 0 2 1 0 2 2 1 0 1 2 1 0 1 2 1 2 = = - - ? ? ? ? ? ? ? ? + = - - ? ? ? ? ? ? + = = ? ? ? ? ? ? © All rights reserved to bana. Eighth Edition 5 - 46 Theorems of Pappus-Guldinus • Surface of revolution is generated by rotating a plane curve about a fixed axis. • Bir dönel yüzeyin alanı, kendisini meydana getiren eğrinin boyu ile yüzeyin meydana gelmesi sırasında eğrinin ağırlık merkezinin katettiği uzaklığın çarpımına eşittir. L y A ? 2 =© All rights reserved to bana. Eighth Edition 5 - 47 Theorems of Pappus-Guldinus • Body of revolution is generated by rotating a plane area about a fixed axis. • Bir dönel cismin hacmi, kendisini meydana getiren alan ile cismin meydana gelmesi sırasında alanın ağırlık merkezinin katettiği uzaklığın çarpımına eşittir. A y V ? 2 = © All rights reserved to bana. Eighth Edition 5 - 48 Sample Problem 5.7 The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as shown. Knowing that the pulley is made of steel and that the density of steel is determine the mass and weight of the rim. 3 3 m kg 10 85 . 7 × = ? SOLUTION: • Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section. • Multiply by density and acceleration to get the mass and acceleration.© All rights reserved to bana. Eighth Edition 5 - 49 Sample Problem 5.7 SOLUTION: • Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section. ( )( ) ? ? ? ? ? ? × × = = - 3 3 9 3 6 3 3 mm m 10 mm 10 65 . 7 m kg 10 85 . 7 V m ? kg 0 . 60 = m ( )( ) 2 s m 81 . 9 kg 0 . 60 = = mg W N 589 = W • Multiply by density and acceleration to get the mass and acceleration. © All rights reserved to bana. Eighth Edition 5 - 50 Distributed Loads on Beams • A distributed load is represented by plotting the load per unit length, w (N/m) . The total load is equal to the area under the load curve. ? ? = = = A dA dx w W L 0 ( ) ( ) A x dA x A OP dW x W OP L = = = ? ? 0 • A distributed load can be replace by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the area centroid.© All rights reserved to bana. Eighth Edition 5 - 51 Distributed Loads on Beams Illustration: The lumber on the rack shown below distributes the weight evenly across the supporting beam. This uniform loading is represented by a load curve with equal length lines. The total weight (resultant) equals the area under the load curve and it acts at the centroid of the load curve. (Ref: Statics, 9 th Edition, by Hibbeler) © All rights reserved to bana. Eighth Edition 5 - 52 Sample Problem 5.9 A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports. SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. • The line of action of the concentrated load passes through the centroid of the area under the curve. • Determine the support reactions by summing moments about the beam ends.© All rights reserved to bana. Eighth Edition 5 - 53 Sample Problem 5.9 SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. kN 0 . 18 = F • The line of action of the concentrated load passes through the centroid of the area under the curve. kN 18 m kN 63 · = X m 5 . 3 = X © All rights reserved to bana. Eighth Edition 5 - 54 Sample Problem 5.9 • Determine the support reactions by summing moments about the beam ends. ( ) ( )( ) 0 m .5 3 kN 18 m 6 : 0 = - = ? y A B M kN 5 . 10 = y B ( ) ( )( ) 0 m .5 3 m 6 kN 18 m 6 : 0 = - + - = ? y B A M kN 5 . 7 = y A© All rights reserved to bana. Eighth Edition 5 - 55 Center of Gravity of a 3D Body: Centroid of a Volume • Center of gravity G ( ) ? ? - = - j W j W r r ( ) ( ) [ ] ( ) ( ) ( ) j W r j W r j W r j W r G G r r r r r r r r - × ? = - × ? - × = - × ? ? ? ? = = dW r W r dW W G r r • Results are independent of body orientation, ? ? ? = = = zdW W z ydW W y xdW W x ? ? ? = = = zdV V z ydV V y xdV V x dV dW V W ? ? = = and • For homogeneous bodies, © All rights reserved to bana. Eighth Edition 5 - 56 Centroids of Common 3D Shapes© All rights reserved to bana. Eighth Edition 5 - 57 Composite 3D Bodies • Moment of the total weight concentrated at the center of gravity G is equal to the sum of the moments of the weights of the component parts. ? ? ? ? ? ? = = = W z W Z W y W Y W x W X • For homogeneous bodies, ? ? ? ? ? ? = = = V z V Z V y V Y V x V X © All rights reserved to bana. Eighth Edition 5 - 58 Sample Problem 5.12 Locate the center of gravity of the steel machine element. The diameter of each hole is 1 in. SOLUTION: • Form the machine element from a rectangular parallelepiped and a quarter cylinder and then subtracting two 1-in. diameter cylinders.© All rights reserved to bana. Eighth Edition 5 - 59 Sample Problem 5.12 © All rights reserved to bana. Eighth Edition 5 - 60 Sample Problem 5.12 ( ) ( ) 3 4 in .286 5 in 08 . 3 = = ? ? V V x X ( ) ( ) 3 4 in .286 5 in 5.047 - = = ? ? V V y Y ( ) ( ) 3 4 in .286 5 in .618 1 = = ? ? V V z Z in. 577 . 0 = X in. 577 . 0 = Y in. 577 . 0 = Z