Engineering Mechanics Statics - Examples Part 16 ppsx

Engineering Mechanics - Statics Chapter 10 I max I x I y + 2 R+= I max 4.92 10 6 × mm 4 = I min I x I y + 2 R−= I min 1364444.44 mm 4 = Problem 10-88 Determine the directions of the principal axes with origin located at point O, and the principal moments of inertia for the area about these axes. Solve using Mohr's circle Given: a 4in= b 2in= c 2in= d 2in= r 1in= Solution: I x 1 3 cd+()ab+() 3 π r 4 4 π r 2 a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= I x 236.95 in 4 = I y 1 3 ab+()cd+() 3 π r 4 4 π r 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= I y 114.65 in 4 = I xy ab+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ dc+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab+()dc+()da π r 2 −= I xy 118.87 in 4 = RI x I x I y + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 I xy 2 += R 133.67 in 4 = I max I x I y + 2 R+= I max 309in 4 = I min I x I y + 2 R−= I min 42.1 in 4 = 1051 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 θ p1 1− 2 asin I xy R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ p1 31.39− deg= θ p2 θ p1 π 2 += θ p2 58.61 deg= Problem 10-89 The area of the cross section of an airplane wing has the listed properties about the x and y axes passing through the centroid C. Determine the orientation of the principal axes and the principal moments of inertia. Solve using Mohr's circle. Given: I x 450 in 4 = I y 1730 in 4 = I xy 138 in 4 = Solution: RI x I x I y + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 I xy 2 += R 654.71 in 4 = I max I x I y + 2 R+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I max 1.74 10 3 × in 4 = I min I x I y + 2 R− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I min 435in 4 = 1052 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 θ p1 1 2 asin I xy R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ p1 6.08 deg= θ p2 θ p1 90 deg+= θ p2 96.08 deg= Problem 10-90 The right circular cone is formed by revolving the shaded area around the x axis. Determine the moment of inertia l x and express the result in terms of the total mass m of the cone. The cone has a constant density ρ . Solution: m 0 h x ρπ rx h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⌡ d= 1 3 h ρπ r 2 = l x 3m π hr 2 0 h x 1 2 π rx h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4 ⌠ ⎮ ⎮ ⌡ d= 3 10 mr 2 = l x 3 10 mr 2 = Problem 10-91 Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m. 1053 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 Solution: m ρ 2 π R= ρ m 2 π R = I 0 2 π θ m 2 π R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ R 2 R ⌠ ⎮ ⎮ ⌡ d= mR 2 = ImR 2 = Problem 10-92 The solid is formed by revolving the shaded area around the y axis. Determine the radius of gyration k y . The specific weight of the material is γ . Given: a 3in= b 3in= γ 380 lb ft 3 = Solution: m 0 b y γπ a y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d= m 2.66lb= I y 0 b y γπ a y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 1 2 a y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d= I y 6.46 lb in 2 ⋅= k y I y m = k y 1.56 in= 1054 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 Problem 10-93 Determine the moment of inertia I x for the sphere and express the result in terms of the total mass m of the sphere. The sphere has a constant density ρ . Solution: m ρ 4 π r 3 3 = ρ 3m 4 π r 3 = I x r− r x 1 2 3m 4 π r 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ π r 2 x 2 − () r 2 x 2 − () ⌠ ⎮ ⎮ ⌡ d= 2 5 mr 2 = I x 2 5 mr 2 = Problem 10-94 Determine the radius of gyration k x of the paraboloid. The density of the material is ρ . Units Used: Mg 1000 kg= Given: ρ 5 Mg m 3 = a 200 mm= b 100 mm= 1055 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 Solution: m p 0 a x ρπ b 2 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d= m p 15.71 kg= I x 0 a x 1 2 ρπ b 2 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d= I x 52.36 10 3− × kg m 2 ⋅= k x I x m p = k x 57.7 mm= Problem 10-95 Determine the moment of inertia of the semi-ellipsoid with respect to the x axis and express the result in terms of the mass m of the semiellipsoid. The material has a constant density ρ . Solution: m 0 a x ρπ b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⌠ ⎮ ⎮ ⎮ ⌡ d= 2 3 a ρπ b 2 = ρ 3m 2 π ab 2 = I x 0 a x 1 2 3m 2 π ab 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ π b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⌠ ⎮ ⎮ ⎮ ⌡ d= 2 5 mb 2 = I x 2 5 mb 2 = 1056 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 Problem 10-96 Determine the radius of gyration k x of the body. The specific weight of the material is γ . Given: γ 380 lb ft 3 = a 8in= b 2in= Solution: m b 0 a x γπ b 2 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 3 ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d= m b 13.26 lb= I x 0 a x 1 2 γπ b 2 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 3 b 2 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 3 ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d= I x 0.59 slug in 2 ⋅= k x I x m b = k x 1.20 in= Problem 10-97 Determine the moment of inertia for the ellipsoid with respect to the x axis and express the result in terms of the mass m of the ellipsoid. The material has a constant density ρ . 1057 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 Solution: m a− a x ρπ b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⌠ ⎮ ⎮ ⎮ ⌡ d= 4 3 a ρπ b 2 = ρ 3m 4 π ab 2 = I x a− a x 1 2 3m 4 π ab 2 π b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⌠ ⎮ ⎮ ⎮ ⌡ d= 2 5 mb 2 = I x 2 5 mb 2 = Problem 10-98 Determine the moment of inertia of the homogeneous pyramid of mass m with respect to the z axis. The density of the material is ρ . Suggestion: Use a rectangular plate element having a volume of dV = (2x)(2y) dz. Solution: V 0 h za 1 z h − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⌡ d= 1 3 ha 2 = ρ m V = 3m a 2 h = I z 3m a 2 h 0 h z 1 6 a 1 z h − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 4 ⌠ ⎮ ⎮ ⌡ d= 1 10 ma 2 = I z 1 10 ma 2 = 1058 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 Problem 10-99 The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of inertia I y . The specific weight of concrete is γ . Given: γ 150 lb ft 3 = a 6in= b 4in= c 8in= Solution: I y 1 2 γπ ab+() 2 ca b+() 2 0 c y 1 2 γπ a 2 y c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 y c ⌠ ⎮ ⎮ ⌡ d−= I y 2.25 slug ft 2 ⋅= Problem 10-100 Determine the moment of inertia of the thin plate about an axis perpendicular to the page and passing through the pin at O. The plate has a hole in its center. Its thickness is c, and the material has a density of ρ Given: a 1.40 m= c 50 mm= b 150 mm= ρ 50 kg m 3 = Solution: I G 1 12 ρ a 2 ca 2 a 2 + () 1 2 ρπ b 2 cb 2 −= I G 1.60 kg m 2 ⋅= I 0 I G md 2 += 1059 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 m ρ a 2 c ρπ b 2 c−= m 4.7233 kg= I 0 I G masin 45 deg()() 2 += I 0 6.23 kg m 2 ⋅= Problem 10-101 Determine the moment of inertia I z of the frustum of the cone which has a conical depression. The material has a density ρ . Given: ρ 200 kg m 3 = a 0.4 m= b 0.2 m= c 0.6 m= d 0.8 m= Solution: h da ab− = I z 3 10 ρ 1 3 π a 2 h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ a 2 3 10 ρ 1 3 π a 2 c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ a 2 − 3 10 ρ 1 3 π b 2 hd−() ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ b 2 −= I z 1.53 kg m 2 ⋅= Problem 10-102 Determine the moment of inertia for the assembly about an axis which is perpendicular to the page and passes through the center of mass G. The material has a specific weight γ . Given: a 0.5 ft= d 0.25 ft= 1060 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. [...]... publisher Engineering Mechanics - Statics Chapter 10 a ⌠ 2 2 ⎮ 1 ⎛ 3m ⎞ ⎛ b x ⎞ ⎛ b x + b⎞ dx = 93 m b2 Ix = ⎮ ⎟ ⎜ ⎟ π ⎜ + b⎟ ⎜ a 70 ⎠ ⎝ ⎠ ⎮ 2 ⎝ 7π a b2 ⎠ ⎝ a ⌡0 Ix = 93 2 mb 70 Problem 1 0-1 16 Determine the product of inertia for the shaded area with respect to the x and y axes Given: a = 1m b = 1m Solution: ⌠ ⎮ ⎮ Ixy = ⎮ ⎮ ⌡ b 1 1 ⎛ y⎞ ya⎜ ⎟ 2 ⎝ b⎠ 3 1 3 ⎛ y⎞ dy ⎟ ⎝ b⎠ a⎜ Ixy = 0.1875 m 4 0 Problem 1 0-1 17... writing from the publisher Engineering Mechanics - Statics Chapter 11 Problem 1 1-9 A force P is applied to the end of the lever Determine the horizontal force F on the piston for equilibrium Solution: δs = 2 l δθ x = 2 l cos ( θ ) δx = −2 l sin ( θ ) δθ δ U = − P δs − F δ x = 0 −P 2lδθ + F2l sin ( θ ) δθ = 0 F = P csc ( θ ) Problem 1 1-1 0 The mechanism consists of the four pin-connected bars and three... in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 10 Problem 1 0-1 14 Determine the moment of inertia for the shaded area about the y axis Given: a = 2 in b = 8 in Solution: a ⌠ 3 ⎮ 2⎡ ⎛x⎞ ⎤ Iy = ⎮ x ⎢b − b ⎜ ⎟ ⎥ dx ⎣ ⎝ a⎠ ⎦ ⌡ 0 Iy = 10.67 in 4 Problem 1 0-1 15 Determine the mass moment of inertia Ix of the body and express the result... permission in writing from the publisher Engineering Mechanics - Statics Chapter 10 Problem 1 0-1 06 Each of the three rods has a mass m Determine the moment of inertia for the assembly about an axis which is perpendicular to the page and passes through the center point O Solution: ⎡1 I O = 3⎢ ⎣12 IO = 2 2⎤ ⎛ a sin ( 60 deg) ⎞ ⎥ ⎟ 3 ⎝ ⎠⎦ ma + m⎜ 1 2 ma 2 Problem 1 0-1 07 The slender rods have weight density... copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 11 Problem 1 1-5 Each member of the pin-connected mechanism has mass m1 If the spring is unstretched when θ = 0ο , determine the required stiffness k so that the mechanism is in equilibrium when θ = θ0... publisher Engineering Mechanics - Statics Chapter 10 Problem 1 0-1 11 Determine the moment of inertia for the solid steel assembly about the x axis Steel has a specific weight γst Given: a = 2 ft b = 3 ft c = 0.5 ft d = 0.25 ft lb γ st = 490 ft 3 Solution: h = ca c−d ⎡ 2 ⎛ c2 ⎞ π 2 ⎛ 3c2 ⎞ π 2 ⎛ 3d2 ⎞⎤ ⎢π c b⎜ ⎟ + c h⎜ ⎟ − d ( h − a) ⎜ ⎟⎥ Ix = γ st ⎣ ⎝ 2 ⎠ 3 ⎝ 10 ⎠ 3 ⎝ 10 ⎠⎦ Ix = 5.64 slug⋅ ft 2 Problem 1 0-1 12... publisher Engineering Mechanics - Statics Chapter 10 c = 0.1 m d = 0.3 m Solution: b bρ r 2 yc = IG = 2 2 + π d ρ s( b + d) − π c ρ s( b + d) 2 yc = 0.888 m 2 bρ r + π d ρ s − π c ρ s + ρ r2a 1 1 2 2 2 2aρ r ( 2a) + 2aρ r yc + bρ r b 12 12 2 1 2 2 2 2 ⎞ ⎛b + bρ r ⎜ − yc⎟ + π d ρ s d + π d ρ s ( b + d − yc) 2 ⎝2 ⎠ + 1 2 π c ρ s c − π c ρ s ( b + d − yc) 2 IG = 5.61 kg⋅ m 2 2 2 2 Problem 1 0-1 13 Determine.. .Engineering Mechanics - Statics b = 2 ft c = 1 ft Chapter 10 e = 1 ft γ = 90 lb ft 3 Solution: IG = 1 2 2 1 2 2 1 2 2 γ π ( a + b) e ( a + b) − γ π b ( e − d) b − γ π c d c 2 2 2 IG = 118 slug⋅ ft 2 Problem 1 0-1 03 Determine the moment of inertia for the assembly about an axis which is perpendicular to the... form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 10 Given: Wp = 12 lb a = 1 ft Wr = 4 lb b = 1 ft c = 3 ft d = 2 ft Solution: ( 2 ) 1 2 ⎛ c + d − c⎞ + 1 W a2 + b2 + W ⎛c + b ⎞ I0 = W r ( c + d) + W r ⎜ ⎟ ⎟ p p⎜ 12 12 ⎝ 2 ⎠ ⎝ 2⎠ k0 = I0 Wp + Wr 2 k0 = 3.15 ft Problem 1 0-1 09 Determine the moment of inertia for the overhung crank about the x... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 10 Solution: 4 Ix = πr Iy = Ix 8 4 Ixy = 0 mm Iu = Iv = Iuv = Ix + Iy Ix − Iy cos ( 2θ ) − Ixy sin ( 2θ ) Iu = 5.09 × 10 mm Ix + Iy Ix − Iy − cos ( 2θ ) − Ixy sin ( 2θ ) 2 2 Iv = 5.09 × 10 mm 2 + 2 Ix − Iy sin ( 2θ ) + Ixy cos ( 2θ ) 2 6 6 Iuv = 0 m 4 4 4 Problem 1 0-1 18 Determine the moment of inertia for the shaded . publisher. Engineering Mechanics - Statics Chapter 10 I x 0 a x 1 2 3m 7 π ab 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ π bx a b+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 bx a b+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d= 93 70 mb 2 = I x 93 70 mb 2 = Problem 1 0-1 16 Determine. writing from the publisher. Engineering Mechanics - Statics Chapter 10 θ p1 1− 2 asin I xy R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ p1 31.39− deg= θ p2 θ p1 π 2 += θ p2 58.61 deg= Problem 1 0-8 9 The area of the cross. writing from the publisher. Engineering Mechanics - Statics Chapter 10 θ p1 1 2 asin I xy R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ p1 6.08 deg= θ p2 θ p1 90 deg+= θ p2 96.08 deg= Problem 1 0-9 0 The right circular cone