Statik Atalet Momentinin Bulunması 8 - Atalet Momentinin Bulunması VECTOR MECHANICS FOR ENGINEERS: STATICS Seventh Edition Seventh Edition Seventh 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 ‘ ‘ ‘ in in in STATICs STATICs STATICs kitaplar kitaplar kitapları ı ından ndan ndan d d dü ü üzenlenmi zenlenmi zenlenmiş ş ştir tir tir. . . D D Dü ü üzenleyen zenleyen zenleyen: : : Dr. N. MEYDANLIK Dr. N. MEYDANLIK Dr. N. MEYDANLIK T T Trakya rakya rakya University University University 2007 2007 2007- - -08 ED 08 ED 08 EDİ İ İRNE RNE RNE CHAPTER © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . 9 10. HAFTA Atalet Momentleri Atalet Momentleri ( (Moments Moments of of Inertia Inertia) ) © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 2 2 2 Yap Yapı ı elemanlar elemanları ın nı ın mukavemeti b n mukavemeti bü üy yü ük oranda k oranda onlar onları ın kesit b n kesit bü üy yü ükl klü üklerine ve klerine ve ş şekline ba ekline bağ ğl lı ıd dı ır, r, ö özellikle de kesit alan zellikle de kesit alanı ın nı ın 2. momenti veya atalet n 2. momenti veya atalet momentine ba momentine bağ ğl lı ıd dı ır. r. G G İ İ R R İ İ© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 3 3 3 UYGULAMALAR UYGULAMALAR Kiriş ve kolon gibi bir çok yapı elemanı have cross sectional shapes like I, H, C, etc. Diğerleri içi dolu kare yada daire kesitlerden çok tüb şeklindedir. Why do they usually not have solid rectangular, square, or circular cross sectional areas? • Tasarımda bu elemanların hangi özellikleri daha etkendir ? • How can we calculate this property? © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 4 4 4 Atalet Momenti Atalet Momenti, Statik hesaplarında kullanılmaz ama hesaplanması ağırlık merkezi hesaplarına benzediği için statik dersi içinde gösterilir. Atalet momenti daha çok mukavemet (malzeme mekaniği), dinamik ve akışkanlar mekaniği derslerinde ve hesaplamalarında kullanılır. • Bir cismin atalet momenti atalet momenti onun dönmeye karşı direncinin bir ölçümüdür. Günlük tecrübelerimizden de biliriz ki dönen büyük bir tekerleği durdurmak veya dönmeye başlatmak küçük tekerlekten daha zordur. Matematiksel olarak da bu olayın büyük tekerleğin daha büyük atalet momentine sahip olması nedeniyle olduğu gösterilebilir. •Atalet momenti çeşitli mühendislik hesaplamalarında kullanılır; - Hidrostatik basınç kuvvetlerinin bileşkesinin yerini bulmak için, - Kirişlerde gerilme ve sehim hesapları için, - Dönen cisimlerin kütle atalet momentleri hesabı© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 5 5 5 MOMENTS OF INERTIA FOR AREAS Sıvı içine daldırılmış bir plağı göz önüne alalım. Yüzeyden z kadar aşağıdaki sıvı basıncı p=?z ile verilir. where ? is the specific weight of the liquid. Bu noktada dA alanına etki eden kuvvet dF = p dA = (? z) dA. Bu kuvvet nedeniyle x-eksenine göre moment z(dF) dir. The total moment is ? A z dF = ? A ? z 2 dA = ?? A (z 2 dA). This integral term is referred to as the moment of inertia of the area of the plate about an axis. © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 6 6 6 AB kirişi için 3 farklı kesit şekli ve alanı göz önüne alalım. Toplam alanlar eşit ve aynı malzemeden yapılmış, dolayısıyla birim uzunluk için kütleleri aynı. - Verilen yükleme hali için hangisini tercih edersiniz ? Niçin ? (daha az gerilme ve çökmeyi dikkate alın). Yanıt x-eksenine göre atalet momentine bağlıdır. x-ekseninden en uzak alanların çoğu (A) da olduğu için en büyük atalet momentine sahip olan (A) dır. Bu nedenle de en az gerilme ve çökmeyi (?) veren de A şıkkıdır. Atalet momenti arttıkça (?) ve gerilme düşer. 10cm 10cm 1cm 1cm x 3cm 10cm 3cm A B F (C) (B) (A) MOMENTS OF INERTIA FOR AREAS© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 7 7 7 İ İntegrasyonla ntegrasyonla bir alan bir alanı ın atalet momentinin bulunmas n atalet momentinin bulunması ı : : • Herhangi bir alanın x ve y eksenlerine göre ikinci Momenti or atalet momenti, ? ? = = dA x I dA y I y x 2 2 • dA alanının koordinat eksenlerine parelel ince şerit halinde seçilmesi (I x için yatay, I y için düşey eleman) integral hesabını basitleştirir. dI y = x 2 dA dI x = y 2 dA © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 8 8 8 Moment of Inertia of an Area by Integration Moment of Inertia of an Area by Integration • For a rectangular area, 3 3 1 0 2 2 bh bdy y dA y I h x = = = ? ? • Dikdörtgen alan için bulunan ifade eksenlere parelel olarak seçilecek ince şeritlere de uygulanabilir, örneğin bir tek düşey şerit eleman seçilerek her iki eksene göre atalet momentleri dx y x dA x dI dx y dI y x 2 2 3 3 1 = = =© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 9 9 9 Polar Moment of Inertia Polar Moment of Inertia • Polar Atalet momenti, dönen silindirik millerin burulmasında önemli bir parametredir. ? = dA r J 2 0 • Polar atalet momenti ile dik atalet momentleri arasındaki ilişki, ( ) x y I I J dA y dA x dA y x dA r J + = + = + = = ? ? ? ? 0 2 2 2 2 2 0 • Atalet momentinin birimi uzunluğun 4. kuvvetidir (m 4 ). © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 10 10 10 Bir alan Bir alanı ın atalet yar n atalet yarı ıç çap apı ı • Atalet momeni I x olan bir alanı göz önüne alalım. Bu alanın yerine x-eksenine parelel ve atalet momenti yine I x ‘e eşdeğer bir dikdörtgen şerit düşünürsek, A I k A k I x x x x = = 2 k x = x eksenine göre atalet yarıçapı • Similarly, A J k A k J A I k A k I O O O O y y y y = = = = 2 2 2 2 2 y x O k k k + = • Atalet yarıçapı özellikle kolonların tasarımında önemlidir.© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 11 11 11 Sample Problem 9.1 Sample Problem 9.1 Üçgen alanın tabanına göre atalet momentini hesaplayınız. SOLUTION: • A differential strip parallel to the x axis is chosen for dA. dy l dA dA y dI x = = 2 • For similar triangles, dy h y h b dA h y h b l h y h b l - = - = - = • y = 0 dan y = h’ a kadar dI x ’ in integrasyonu , 12 3 bh I x = ( ) h h h x y y h h b dy y hy h b dy h y h b y dA y I 0 4 3 0 3 2 0 2 2 4 3 ? ? ? ? ? ? - = - = - = = ? ? ? © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 12 12 12 Sample Problem 9.2 Sample Problem 9.2 a) Dairesel bir alanın polar atalet momentini bulunuz. b) a şıkkında bulduğunuz sonucu kullanarak, dairesel alanın çapından geçen x-eksenine göre atalet momentini bulunuz. SOLUTION: • An annular differential area element is chosen, ( ) ? ? ? = = = = = r r O O O du u du u u dJ J du u dA dA u dJ 0 3 0 2 2 2 2 2 ? ? ? 4 2 r J O ? = • From symmetry, I x = I y , x x y x O I r I I I J 2 2 2 4 = = + = ? 4 4 r I I x diameter ? = =© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 13 13 13 Given: The shaded area shown in the figure. Find: The MoI of the area about the x- and y-axes. Solution I x = ? y 2 dA dA = (4 – x) dy = (4 – y 2 /4) dy I x = 0 ? y 2 (4 – y 2 /4) dy = [ (4/3) y 3 – (1/20) y 5 ] 0 = 34.1 cm 4 4 4 (x,y) 4 cm 4 cm Sample Problem 9. Sample Problem 9.3 3 © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 14 14 14 I y = ? x 2 dA = ? x 2 y.dx = ? x 2 (2 ? x) dx = 2 0 ? x 2.5 dx = [ (2/3.5) x 3.5 ] 0 = 73.1 cm 4 4 4 In the above example, it will be difficult to determine I y using a horizontal strip. However, I x in this example can be determined using a vertical strip. So, I x = ? (1/3) y 3 dx = ? (1/3) (2?x) 3 dx . y (x,y) 4 cm 4 cm© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 15 15 15 Given: The shaded area shown. Find: I x and I y of the area. Solution I x = ? (1/3) y 3 dx = 0 ? (1/3) x dx = [x 2 / 6 ] 0 = 10.7 in 4 8 8 (x,y) Sample Problem 9. Sample Problem 9.4 4 © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 16 16 16 I Y = ? x 2 dA = ? x 2 y dx = ? x 2 ( x (1/3) ) dx = 0 ? x (7/3) dx = [(3/10) x (10/3) ] 0 = 307.18 in 4 8 8 (x,y)© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 17 17 17 ATTENTION QUIZ 1. When determining the MoI of the element in the figure, dI y equals A) x 2 dy B) x 2 dx C) (1/3) y 3 dx D) x 2.5 dx 2. Similarly, dI x equals A) (1/3) x 1.5 dx B) y 2 dA C) (1/12) x 3 dy D) (1/3) x 3 dx (x,y) y 2 = x © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Parelel Parelel eksenler Teoremi & Bile eksenler Teoremi & Bileş şik alanlar ik alanları ın atalet momentleri n atalet momentleri Yapı elemanlarının kesit alanları genellikle basit şekillerden yada ilkel basit şekillerin birleşmesinden oluşmuştur. Bu basit alanların atalet momentlerini bulmak için integrasyon yöntemi ile karşılaştırıldığında daha basit yöntemler var mıdır ? ??? EVET © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 19 19 19 PARALLEL-AXIS THEOREM FOR AN AREA Bu teorem bir alanın ağırlık merkezinden geçen eksenlere göre atalet momentinin yine bu eksenlere parelel başka eksenlere göre atalet momentleri ile ilgilidir. Bileşik alanların atalet momentlerinin bulunması için pratik bir yöntemdir. Gözönüne alınan alanın ağırlık merkezi C dir. x' and y' axes ağ. mer. C’ den geçmektedir. x‘-eksenine parelel ve d y kadar mesafede bir x-eksenine göre atalet momenti parelel eksen teoremi ile bulunur. © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 20 20 20 Ağırlık merkezinin tanımını kullanarak: y' = (? A y' dA) / (? A dA) . Now since C is at the origin of the x' – y' axes, y' = 0 , and hence ? A y' dA = 0 . Thus I X = I X ' + A d y 2 I X = ? A y 2 dA = ? A (y' + d y ) 2 dA = ? A y' 2 dA + 2 d y ? A y' dA + d y 2 ? A dA y = y‘ + d y Similarly, I Y = I Y ' + A d X 2 and J O = J C + A d 2 Parelel Parelel Eksen Teoremi Eksen Teoremi© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 21 21 21 © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 22 22 22© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 23 23 23 Parallel Axis T Parallel Axis Teoremi eoremi ö örnek rnek • Use the value of I for a circle from the table on the following page and the parallel-axis theorem to find I T , the moment of inertia about an axis tangent to the circle. ( ) 4 4 5 2 2 4 4 1 2 r r r r Ad I I T ? ? ? = + = + = • Use the value of I AA’ along the base of a triangle from the table on the following page and the parallel-axis theorem to find I BB’ , the moment of inertia along a parallel axis through the centroid of the triangle ( ) 3 36 1 2 3 1 2 1 3 12 1 2 2 bh h bh bh Ad I I Ad I I A A B B B B A A = - = - = + = ' ' ' ' © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 24 24 24 En En ç çok bilinen baz ok bilinen bazı ı ilkel kesitlerin atalet momentleri ilkel kesitlerin atalet momentleri© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 25 25 25 En En ç çok bilinen baz ok bilinen bazı ı ilkel kesitlerin atalet momentleri ilkel kesitlerin atalet momentleri © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 26 26 26© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 27 27 27 CONCEPT QUIZ 1. For the area A, we know the centroid’s (C) location, area, distances between the four parallel axes, and the MoI about axis 1. We can determine the MoI about axis 2 by applying the parallel axis theorem ___ . A) directly between the axes 1 and 2. B) between axes 1 and 3 and then between the axes 3 and 2. C) between axes 1 and 4 and then axes 4 and 2. D) None of the above. d 3 d 2 d 1 4 3 2 1 A • C Axis © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 28 28 28 2. For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number? A) Axis 1 B) Axis 2 C) Axis 3 D) Axis 4 E) Can not tell. d 3 d 2 d 1 4 3 2 1 A • C Axis CONCEPT QUIZ smallest MoI© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 29 29 29 ATTENTION QUIZ 1. For the given area, the moment of inertia about axis 1 is 200 cm 4 . What is the MoI about axis 3 (the centroidal axis)? A) 90 cm 4 B) 110 cm 4 C) 60 cm 4 D) 40 cm 4 A=10 cm 2 • C d 2 d 1 3 2 1 • C d 1 = d 2 = 2 cm I 1 = I 3 + Ad 2 I 3 = I 1 - Ad 2 = 200 - 10(4) 2 = 40 cm 4 © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 30 30 30 ATTENTION QUIZ 2. The moment of inertia of the rectangle about the x-axis equals A) 8 cm 4 B) 56 cm 4 C) 24 cm 4 D) 26 cm 4 2cm 2cm 3cm x I x = I x’ + Ad 2 I x = (bh 3 )/12 + Ad 2 = (3 . 2 3 ) / 12 + 6 . 3 2 = 56cm 4 x’ d© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 31 31 31 EXAMPLE Given: The beam’s cross-sectional area. Find: The moment of inertia of the area about the y-axis and the radius of gyration k y . Solution 1. The cross-sectional area can be divided into three rectangles ( [1], [2], [3] ) as shown. 2. The centroids of these three rectangles are in their center. The distances from these centers to the y-axis are 0 mm, 87.5 mm, and 87.5 mm, respectively. [1] [2] [3] © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 32 32 32 3. From the inside back cover of the book, the MoI of a rectangle about its centroidal axis is (1/12) b h 3 . I y[1] = (1/12) (25mm) (300mm) 3 = 56.25 (10 6 ) mm 4 [1] [2] [3] Using the parallel-axis theorem, I Y[2] = I Y[3] = I Y’ + A (d X ) 2 = (1/12) (100) (25) 3 + (25) (100) ( 87.5 ) 2 = 19.27 (10 6 ) mm 4 EXAMPLE© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 33 33 33 4. I y = I y1 + I y2 + I y3 = 94.8 ( 10 6 ) mm 4 k y = ? ( I y / A) A = 300 (25) + 25 (100) + 25 (100) = 12,500 mm 2 k y = ? ( 94.79) (10 6 ) / (12500) = 87.1 mm EXAMPLE © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 34 34 34 MOMENT OF INERTIA FOR A COMPOSITE AREA A composite area is made by adding or subtracting a series of “simple” shaped areas like rectangles, triangles, and circles. For example, the area on the left can be made from a rectangle minus a triangle and circle. The MoI of these “simpler” shaped areas about their centroidal axes are found in most engineering handbooks as well as the inside back cover of the textbook. Using these data and the parallel-axis theorem, the MoI for a composite area can easily be calculated.© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 35 35 35 STEPS FOR ANALYSIS 1. Divide the given area into its simpler shaped parts. 2. Locate the centroid of each part and indicate the perpendicular distance from each centroid to the desired reference axis. 4. The MoI of the entire area about the reference axis is determined by performing an algebraic summation of the individual MoIs obtained in Step 3. (Please note that MoI of a hole is subtracted). 3. Determine the MoI of each “simpler” shaped part about the desired reference axis using the parallel-axis theorem ( I X = I X’ + A ( d y ) 2 ) . © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 36 36 36 READING QUIZ 1. The parallel-axis theorem for an area is applied between A) an axis passing through its centroid and any corresponding parallel axis. B) any two parallel axis. C) two horizontal axes only. D) two vertical axes only. 2. The moment of inertia of a composite area equals the ____ of the MoI of all of its parts. A) vector sum B) algebraic sum (addition or subtraction) C) addition D) product© © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 37 37 37 Given: The shaded area as shown in the figure. Find: The moment of inertia for the area about the x-axis and the radius of gyration k X . Solution 1. The given area can be obtained by subtracting both the circle (b) and triangle (c) from the rectangle (a). 2. Information about the centroids of the simple shapes can be obtained from the inside back cover of the book. The perpendicular distances of the centroids from the x-axis are: d a = 5 in , d b = 4 in, and d c = 8in. (a) (b) (c) EXAMPLE © © © All rights reserved All rights reserved All rights reserved to to to bana bana bana. . . Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics Vector Mechanics for Engineers: Statics 3 3 3 - - - 38 38 38 3. I Xa = (1/12) 6 (10) 3 + 6 (10)(5) 2 = 2000 in 4 I Xb = (1/4) ? (2) 4 + ? (2) 2 (4) 2 = 213.6 in 4 I Xc = (1 /36) (3) (6) 3 + (½) (3) (6) (8) 2 = 594 in 4 I X = I Xa – I Xb – I Xc = 1190 in 4 k X = ? ( I X / A ) A = 10 ( 6 ) – ? (2) 2 – (½) (3) (6) = 38.43 in 2 k X = ? (1192 / 38.43) = 5.57 in. (a) (b) (c) EXAMPLE