V_-_Center_of_Gravity_and_Centroid

V_-_Center_of_Gravity_and_Centroid - CENTER OF GRAVITY AND...

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Unformatted text preview: CENTER OF GRAVITY AND CENTROID Center of Gravity, Center of Mass and Centroid and LEARNING OBJECTIVES LEARNING Be able to understand the concepts of Be center of gravity, center of mass, and centroid centroid Be able to determine the locations of Be these points for a system of particles or a body body PRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurement Integrations over lines, areas and Integrations volumes volumes SYSTEM OF PARTICLES Center of gravity (CG) is a point which locates the resultant weight of a system of particles x1 ⋅ W1 + x2 ⋅ W2 + ... + xn ⋅ Wn x= W1 + W2 + ... + Wn y1 ⋅ W1 + y2 ⋅ W2 + ... + yn ⋅ Wn y= W1 + W2 + ... + Wn z1 ⋅ W1 + z 2 ⋅ W2 + ... + z n ⋅ Wn z= W1 + W2 + ... + Wn SYSTEM OF PARTICLES Center of mass (CM) is a point which locates the resultant mass of a system of particles x1 ⋅ m1 + x2 ⋅ m2 + ... + xn ⋅ mn x= m1 + m2 + ... + mn m1 m2 mn y1 ⋅ m1 + y2 ⋅ m2 + ... + yn ⋅ mn y= m1 + m2 + ... + mn z1 ⋅ m1 + z 2 ⋅ m2 + ... + z n ⋅ mn z= m1 + m2 + ... + mn CG and CM have the same location when the gravity, g is constant, which is not always true, e.g. planets. CENTRE OF GRAVITY OF A BODY Center of gravity (CG) is a point which locates the resultant weight of a body ~ dW ∫x x=V ∫ dW ~ dW ∫y ; V y=V ∫ dW V ~ dW ∫z ; z=V ∫ dW V If γ is the specific weight of the body, then dW = γ dV ~γ dV ∫x x=V ∫ γ dV V ~γ dV ∫y ; y=V ∫ γ dV V ~γ dV ∫z ; z =V ∫ γ dV V CENTRE OF MASS OF A BODY Center of mass (CM) is a point which locates the resultant mass of a body ~ dM ∫x x=V ∫ dM ~ dM ∫y ; y=V V ∫ dM ~ dM ∫z ; z=V V ∫ dM V if ρ is the density of the body, then dM = ρ dV, (remember that γ = ρg and hence, ρ = γ/g ) ~ρ dV ∫x x=V ∫ ρ dV V ~ρ dV ∫y ; y=V ∫ ρ dV V ~ρ dV ∫z ; z =V ∫ ρ dV V CENTROID OF A BODY The centroid of a body is a point which locates the geometric center of the body For a volume ~ dV ∫x x =V ∫ dV ; y =V V For an area ~ dA x ∫ x=A ∫ dA A ~ dV ∫y ∫ dV ;z = V V ~ dA ∫y ;y=A ∫ dA A ~ dV ∫z ∫ dV V ~ dA ∫z ;z= A ∫ dA A CENTROID OF A BODY-Continued For a line ~ dL ∫x x=L ∫ dL L ~ dL ∫y ;y=L ∫ dL L ~ dL ∫z ;z=L ∫ dL L Note: CM and centroid are the same if the body is homogeneous (constant density). If an object has an axis of symmetry, then the centroid of the object is located on that axis. CALCULATION STEPS FOR CG, CM AND CENTROID OF A BODY 1. Set up the appropriate boundary of integration, which is the same for both numerator and denominator Triple integration for a volume Double integration for an area Line integration for a line 2. Set up the integrand for numerator and denominator For CG, the integrands for both the numerator and the denominator contain g For CM, the integrands for both the numerator and the denominator contain ρ Integrands for numerator contain either x, y, or z 3. Execute the integration (calculator can be used in this step) EXAMPLE 1 EXAMPLE Determine the Determine center of mass of the homogeneous rod and the support reactions if the rod has a mass per unit length of 0.5 kg/m. length SOLUTION 1 SOLUTION For this case, the centroid, CG and CM are the same Bar length: The length of a differential element dL is equal to 2 dy dL = dx + dy = 1 + dx dx 2 2 dy 2 Given y = x yields 2 y = 3x dx 2 2 3 3 2 dy 3x dy 3 = but y = x then =x dx 2y dx 2 1 2 SOLUTION 1- continued continued 1m 9 L = ∫ dL = ∫ 1 + x dx 4 0 8 9 L= 1 + x 27 4 3 2 1m 0 = 1.4397 m SOLUTION 1 - continued continued Recall that x dL ∫ L x= dL ∫ y dL ∫ L and y = L dL ∫ L 1 9 ∫ x ⋅ 1 + 4 ⋅ x dx x=0 1 ∫ 0 9 1 + ⋅ x dx 4 = 0.546 m SOLUTION 1 - continued continued Recall that ∫ x dL L x= ∫ dL ∫ y dL L and y = ∫ dL L 1 ∫ y=0 L 3 2 ⋅ 1 + 9 ⋅ x dx x 4 1 ∫ 0 9 1 + ⋅ x dx 4 = 0.447 m SOLUTION 1 – continued continued Solve for support reactions: W = 0.5(g)(L) = 0.5(9.81)(1.4397) = 7.06 N ∑ MO = 0 MO – 0.546(7.06) = 0 ROX MO 0.546 m MO = 3.85 N-m ∑ FY = 0; ROY – 7.06 = 0 ROY W ROY = 7.06 N ∑ Fx = 0 , ROX = 0 N EXAMPLE 2 EXAMPLE Determine the centroid of the cross- section SOLUTION 2 SOLUTION Due to symmetry, the centroid, CG and CM are the same at x = aπ/2 2 aπ y= a⋅sin x a ∫ y dA ∫ ∫ y dy dx A ∫ dA A = 0 0 x a⋅sin aπ a ∫∫ 0 0 dy dx = x aπ a ⋅ sin a dx ∫2 0 aπ x ∫ a ⋅ sin a dx 0 SOLUTION 2 SOLUTION y= x aπ 2 1 − cos 2 ⋅ a a dx ∫2⋅ 2 0 aπ x a ⋅ − cos a 0 2 aπ a a x 4 ⋅ x + 8 ⋅ sin 2 ⋅ a 0 y= aπ x 2 a ⋅ − cos a 0 2 3 3 a ⋅π 4 = a ⋅π y= 2 8 2⋅a CYLINDRICAL COORDINATES (review) (review) For some problems, it is more convenient to set the integrals in terms of cylindrical coordinates. For cylindrical coordinates Z Z (X, Y, Ζ) z (Z, r, θ) z X X θ in Y θ ● rs Y= y X ● r X= θ c os r Y CYLINDRICAL COORDINATES (review) (review) For some problems, it is more convenient to set the integrals in terms of cylindrical coordinates. For cylindrical coordinates Z (z, r, θ) ● z Xy r θ X Y x = rcosθ y = rsinθ z=z r =x +y 2 2 2 EXAMPLE 3 EXAMPLE For cylindrical coordinate system: h ⋅y z= a 2 2 Let h = 20 in. and a = 10 in. Determine the centroid of the above Let volume. EXAMPLE 3 EXAMPLE Step 1 – set up the simplest possible integration equation. For this problem, any horizontal cross section (a section parallel to the x-y plane, is a circle.The radius of the circle is equal to the coordinate y or the coordinate x of any point on the surface of the solid whose coordinates are either (0., y, z) or (x, 0., z). Hence, at any z, the area of the circular cross section can be written as A = πy2 = πx2. The volume of the solid can then be obtained using the following integration equation: h h V = ∫ πy dz = ∫ πx dz; 2 0 h 2 0 z 2a but y = 2 substituting yields h z 4a 2 π 52 π 2 V = ∫ π 4 dz = 4 z a = ha h 5h 5 0 SOLUTION 3 SOLUTION For this case, the centroid, CG and CM are the same and because of symmetry their coordinates x = y = 0.0 Step 2 – set After obtaining the volume of the solid, the z coordinate of the centroid can be obtained using the following equation: h πa 4 πa 2 6 ∫ z h 4 z dz 6h 4 z 5 z = 0h 2 = 2 =h πa 5 6 πa 4 z z dz 4 ∫ h4 5h 0 2 z = 16.67 inch SOLUTION 3 (alternative) (alternative) Once again, the centroid, CG and CM are the same and because of symmetry; x = y = 0.0 2π a z= ∫ zdV V ∫ dV V h ∫∫ zdzrdrdθ ∫ ∫∫ dzrdrdθ ∫ 00 = h2r a 2π a h 00 h2r a h2 ⋅ r z2 = a SOLUTION 3 (alternative) (alternative) For this case, the centroid, CG and CM are the same and because of symmetry; x = y = 0.0 2π 10 ∫∫ 00 = 20 zdzrdrdθ ∫ 20 2 r 10 2π 10 20 ∫∫ 00 dzrdrdθ ∫ 20 2 r 10 h2 ⋅ r z2 = a 20944.0 z= = 16.67 in. 1256.6 SOLUTION 3 (alternative) (alternative) For this case, the centroid, CG and CM are the same and because of symmetry; x = y = 0.0 2π 10 20 ∫ ∫ ∫ zdzrdrdθ 00 z= 2π 10 ∫∫ 00 2π 10 1 2 202 r 202 r ∫ ∫ 2 20 − 10 rdrdθ 10 = 0π0 = 20 2 10 2 20 − 20 r rdrdθ dzrdrdθ ∫2 ∫ ∫ 10 20 r 0 0 2π 10 ∫ ∫ ( 200r − 20r )drdθ 2 00 2π 10 3 0.5 2 ∫ ∫ 20r − 40 r drdθ 00 10 2π 10 20 2 ∫ ∫ 100r − 3 r3 dθ 0 z = 20π 10 = 5 2 10r 2 − 400.5 r 2 dθ ∫ ∫ 5 0 0 2π 20000 ∫ 10000 − 3 dθ 0 2π 5 2 0.5 2 ∫ 1000 − 5 40 10 dθ 0 2π ( 3333.33) 20944 z= = = 16.67 inch 2π ( 200) 1256.6 SOLUTION 3 (alternative) (alternative) In the ZY plane 2 za r=y= 2 h 2 z a h 2 ∫0 π h 2 zdz zdv ∫0 πr zdz ∫= zc = = h 2 2 2 ∫ dv ∫0 πr dz h π z a dz ∫0 h 2 h 2 SOLUTION 3 (alternative) (alternative) In the ZY plane 2 za r=y= 2 h 62 z 5a 2 π 4 dz π z a ∫0 h 4 =6 h zc = 42 h z a π z 5a 2 ∫0 π h 4 dz 5 h 4 h h 0 h 0 πh a = 62 πha 5 2 2 20 0 20 0 = 16.67 EXAMPLE 5 EXAMPLE In the ZY plane r = y = 2z zc = zdv ∫ dv ∫ ∫ πr zdz = ∫ π ( = ∫ πr dz ∫ π ( 2 2 0 2 0 2 0 2 2 0 ) 2 2 z zdz ) 2 2 z dz EXAMPLE 5 EXAMPLE In the ZY plane z r=y= a h 2 z 2 ∫0 π h a zdz zdv ∫0 πr zdz ∫= zc = = h 2 2 h z dv πr dz ∫ ∫0 ∫0 π h a dz h h EXAMPLE 5 EXAMPLE In the ZY plane y2 2 r = Z = a 1 − 2 b 2 a 2 1 − y ydy b ∫0 π b 2 2 ydv ∫0 πr ydy ∫= yc = = b 2 2 ∫ dv ∫0 πr dy b π a 2 1 − y 2 dy ∫0 b 2 b 2 EXAMPLE 5 EXAMPLE In the ZY plane a r = y = ( h − z) h 4r xc = y c = 3π 3 h a 4 ∫ ( h − z ) dz 0 1 h 2 4r ∫ xdv = 4 ∫0 πr 3π dz = h xc = yc = 2 1h 2 h a dv ∫ πr dz 3∫ π ( h − z ) dz 4 ∫0 0 h 2 a πr 2 zdz ∫0 π h ( h − z ) zdz ∫ zdv = ∫0 zc = = h 2 2 h a ∫ dv ∫0 πr dz ∫ π ( h − z ) dz 0 h h h EXAMPLE 5 EXAMPLE In the ZX plane r =z= a −x 2 c 2 ) xdx ∫ ∫ = = = dv πr dx π ( a − x ) dx ∫ ∫ ∫ ∫ xdv a 0 πr xdx a 0 2 2 a 0 π a 0 ( a −x 2 2 2 2 2 2 EXAMPLE 5 EXAMPLE In the ZY plane z r = y = 0.25 − 6.25 2 2 z 2 ∫0 π 0.25 − 6.25 zdz ∫ zdv = ∫0 πr zdz = zc = h h 2 z dv πr dz ∫ ∫0 ∫0 π 0.25 − 6.25 dz h h EXAMPLE 5 EXAMPLE In the ZY plane 2 y r=z= 100 yc = ∫ ydv ∫ dv 100 ∫ = ∫ πr ydy 50 100 50 2 πr dy 2 = 2 ∫ y ydy π 100 ∫ y dy π 100 100 50 100 50 2 2 2 QUESTION The _________ is the point defining the geometric center of an object . A) center of gravity B) center of mass C) centroid D) none of the above QUESTION To study problems concerned with the motion of matter under the influence of forces, i.e., dynamics, it is necessary to locate a point called ________. A) center of gravity B) center of mass C) centroid D) none of the above QUESTION The steel plate with known weight has non-uniform thickness and density and is supported as shown. Of the three parameters (CG, CM, and centroid), which one is needed for determining the support reactions? Are all three parameters located at the same point? A) B) C) D) (center of gravity, no) (center of gravity, yes) (centroid, yes) (centroid, no) QUESTION When determining the centroid of the area of the plate, which type of differential area element requires the least computational work? A) Vertical B) Horizontal C) Polar D) any one of the above QUESTION If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of _____ . A) x B) y C) z D) any of the above QUESTION If a vertical rectangular strip is chosen, then what are the values of x and y? A) (x , y) B) (x / 2 , y / 2) C) (x , 0) D) (x , y / 2) QUESTION y 9 m 12 m x Which of the following is the centroid for the given area? A) x = 4 m, y = 7.5 m B) x = 9 m, y = 6 m C) x = 3 m, y = 4 m D) x = 8 m, y = 3 m QUESTION y y 9 = x 12 or 9m 3 y = ⋅x 4 or 4 x = ⋅y 3 12 m Approach I, confirmed with 1/3 d from the base! 3 x 12 4 x= ∫ ∫x d y dx 0 0 3 x 12 4 ∫ ∫ d y dx 0 0 x 432 = =m 8 54 QUESTION y y 9 = x 12 or 3 y = ⋅x 4 9m or 4 x = ⋅y 3 3 x 12 4 y= ∫ ∫ y d y dx 00 3 x 12 4 ∫ ∫ d y dx 00 162 = =3 m 54 12 m x Approach I, confirmed with 1/3 d from the base! QUESTION y 9 = x 12 y 9m or 3 y = ⋅x 4 or 4 x = ⋅y 3 9 12 12 m ∫ ∫ x d x dy x Approach II, confirmed with 1/3 d from the base! x 04y =3 9 12 = 432 =8 m 54 = 162 =3 m 54 ∫ ∫ d x dy 04y 3 9 12 ∫ ∫ y d x dy y 04y =3 9 12 ∫ ∫ d x dy 04y 3 QUESTION Which of the following is the centroid for the given area? A) x = 0 m, y = 2.12 m y B) x = 0 m, y = 2.53 m C) x = 5 m, y = 2.12 m D) x = 5 m, y = 2.53 m r=5m x x QUESTION, Solution Approach I 52 − x 2 5 x= ∫ ∫ x d y dx −5 0 5 52 − x 2 −5 0 ∫ ∫ d y dx 5 y= 0 = =0 m 39.27 0 ∫ ∫ y d y dx 5 52 − x 2 −5 0 x r=5m This can also be obtained from symmetry 52 − x 2 −5 y ∫ ∫ d y dx 83.33 = = 2.122 m 39.27 QUESTION, Solution Approach II 5 52 − y 2 ∫ ∫ x d x dy x= 0 − 52 − y 2 5 52 − y 2 ∫ ∫ d x dy y 0 = =0 m 39.27 0 − 52 − y 2 5 52 − y 2 ∫ ∫ y d x dy y= 0 − 52 − y 2 5 52 − y 2 ∫ ∫ d x dy 0 − 52 − y 2 83.33 = = 2.122 m 39.27 x r=5m ...
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This note was uploaded on 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.

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