Engineering Mechanics - Statics Chapter 10 Solution: Moment and Product of Inertia about x and y Axes: Since the shaded area is symmetrical about the x axis, I xy 0in 4 = I x 1 12 2ac 3 1 12 b 2a() 3 += I x 10.75 in 4 = I y 1 12 2ab 3 2ab a b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1 12 c 2a() 3 += I y 30.75 in 4 = Moment of Inertia about the Inclined u and v Axes I u I x I y + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos 2 θ () + I xy sin 2 θ () −= I u 15.75 in 4 = I v I x I y + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos 2 θ () − I xy sin 2 θ () += I v 25.75 in 4 = Problem 10-78 Determine the directions of the principal axes with origin located at point O, and the principal moments of inertia for the rectangular area about these axes. Given: a 6in= b 3in= Solution: I x 1 3 ba 3 = I x 216in 4 = I y 1 3 ab 3 = I y 54in 4 = I xy a 2 b 2 ab= I xy 81in 4 = 1041 © 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 tan 2 θ () 2− I xy I x I y − = θ 1 2 atan 2 I xy I x − I y + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 22.5− deg= I max I x I y + 2 I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 ++= I max 250in 4 = I min I x I y + 2 I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 +−= I min 20.4 in 4 = Problem 10-79 Determine the moments of inertia I u , I v and the product of inertia I uv for the beam's cross-sectional area. Given: θ 45 deg= a 8in= b 2in= c 2in= d 16 in= Solution: I x 2 3 ab+()c 3 1 12 2bd 3 + 2bd d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I x 5.515 10 3 × in 4 = I y 1 12 2 ab+()[] 3 c 1 12 2b() 3 d+= I y 1.419 10 3 × in 4 = I xy 0in 4 = I u I x I y + 2 I x I y − 2 cos 2 θ () + I xy sin 2 θ () −= I u 3.47 10 3 × in 4 = I v I x I y + 2 I x I y − 2 cos 2 θ () − I xy sin 2 θ () += I v 3.47 10 3 × in 4 = 1042 © 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 I uv I x I y − 2 sin 2 θ () I xy cos 2 θ () += I uv 2.05 10 3 × in 4 = Problem 10-80 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. 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 = tan 2 θ p () I xy − I x I y − 2 = θ p 1 2 atan 2 I xy I x − I y + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ p 31.39− deg= θ p1 θ p = θ p1 31.39− deg= θ p2 90 deg θ p1 += θ p2 58.61 deg= I max I x I y + 2 I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 ++= I max 309in 4 = 1043 © 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 I min I x I y + 2 I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 +−= I min 42.1 in 4 = Problem 10-81 Determine the principal moments of inertia for the beam's cross-sectional area about the principal axes that have their origin located at the centroid C. Use the equations developed in Section 10.7. For the calculation, assume all corners to be square. Given: a 4in= b 4in= t 3 8 in= Solution: I x 2 1 12 at 3 at b t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 12 t 2b 2t−() 3 += I x 55.55 in 4 = I y 2 1 12 ta t−() 3 ta t−() at− 2 t 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 12 2bt 3 += I y 13.89 in 4 = I xy 2− at− 2 t 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ b t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ta t−()= I xy 20.73− in 4 = I max I x I y + 2 I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 ++= I max 64.1 in 4 = I min I x I y + 2 I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 +−= I min 5.33 in 4 = 1044 © 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 10-82 Determine the principal moments of inertia for the angle's cross-sectional area with respect to a set of principal axes that have their origin located at the centroid C. Use the equation developed in Section 10.7. For the calculation, assume all corners to be square. Given: a 100 mm= b 100 mm= t 20 mm= Solution: x c tb t 2 at−()tt at− 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + tb a t−()t+ = x c 32.22 mm= y c tb b 2 at−()t t 2 + tb a t−()t+ = y c 32.22 mm= I x 1 12 t 3 at−()ta t−()x c t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1 12 tb 3 + tb b 2 x c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I x 3.142 10 6 × mm 4 = I y 1 12 bt 3 bt x c t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1 12 ta t−() 3 + ta t−()t at− 2 + x c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I y 3.142 10 6 × mm 4 = I xy x c t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − b 2 y c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ bt at− 2 t+ x c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ y c t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ at−()t−= I xy 1.778− 10 6 × mm 4 = I max I x I y + 2 I x I y − 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ I xy −= I max 4.92 10 6 × mm 4 = I min 2.22 10 6 × mm 4 = I min I x I y + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + I xy += 1045 © 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 10-83 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. Given: I x 450 in 4 = I y 1730 in 4 = I xy 138 in 4 = Solution: tan 2 θ () 2− I xy I x I y − = θ 1 2 atan 2 I xy I x − I y + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 6.08 deg= I max I x I y + 2 I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 ++= I max 1745in 4 = I min I x I y + 2 I x I y − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 +−= I min 435in 4 = Problem 10-84 Using Mohr’s circle, determine the principal moments of inertia for the triangular area and the orientation of the principal axes of inertia having an origin at point O. Given: a 30 mm= b 40 mm= 1046 © 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: Moment of inertia I x and I y : I x 1 12 ba 3 = I x 90 10 3 × mm 4 = I y 1 12 ab 3 = I y 160 10 3 × mm 4 = Product of inertia I xy : I xy 0 b x x 2 a a b x− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⌡ d= I xy 60 10 3 × mm 4 = Mohr's circle : OA I x I y + 2 I x − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 += OA 69.462 10 3 × mm 4 = I max I x I y + 2 OA+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I max 194.462 10 3 × mm 4 = I min I x I y + 2 OA− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = I min 55.5 10 3 × mm 4 = tan 2 θ () I xy I x I y + 2 I x − = θ 1 2 atan 2 I xy I x − I y + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 29.9 deg= 1047 © 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 10-85 Determine the directions of the principal axes with origin located at point O, and the principal moments of inertia for the rectangular area about these axes. Solve using Mohr's circle. Given: a 6in= b 3in= Solution: I x 1 3 ba 3 = I x 216in 4 = I y 1 3 ab 3 = I y 54in 4 = I xy a 2 b 2 ab= I xy 81in 4 = RI x I x I y + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 I xy 2 += R 114.55 in 4 = I max I x I y + 2 R+= I max 250in 4 = I min I x I y + 2 R−= I min 20.4 in 4 = θ p1 1− 2 asin I xy R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ p1 22.50− deg= θ p2 θ p1 90 deg+= θ p2 67.50 deg= 1048 © 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 10-86 Determine the principal moments of inertia for the beam's cross-sectional area about the principal axes that have their origin located at the centroid C. For the calculation, assume all corners to be square. Solve using Mohr's circle. Given: a 4in= b 4in= t 3 8 in= Solution: I x 2 1 12 at 3 at b t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 12 t 2b 2t−() 3 += I x 55.55 in 4 = I y 2 1 12 ta t−() 3 ta t−() at− 2 t 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 12 2bt 3 += I y 13.89 in 4 = I xy 2− at− 2 t 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ b t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ta t−()= I xy 20.73− in 4 = RI x I x I y + 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 += R 29.39 in 4 = I max I x I y + 2 R+= I max 64.1 in 4 = I min 20.45 in 4 = I min I x I y + 2 R−= Problem 10-87 Determine the principal moments of inertia for the angle's cross-sectional area with respect to a set of principal axes that have their origin located at the centroid C. For the calculation, assume all corners to be square. Solve using Mohr's ciricle. 1049 © 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 Given: a 100 mm= b 100 mm= t 20 mm= Solution: x c tb t 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ at−()tt at− 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + tb a t−()t+ = x c 32.22 mm= y c tb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ at−()t t 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + tb a t−()t+ = y c 32.22 mm= I x 1 12 t 3 at−()ta t−()x c t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1 12 tb 3 + tb b 2 x c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I x 3.142 10 6 × mm 4 = I y 1 12 bt 3 bt x c t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1 12 ta t−() 3 + ta t−()t at− 2 + x c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += I y 3.142 10 6 × mm 4 = I xy x c t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − b 2 y c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ bt at− 2 t+ x c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ y c t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ at−()t−= I xy 1.778− 10 6 × mm 4 = RI x I x I y + 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 I xy 2 += R 1.78 10 6 × mm 4 = 1050 © 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. [...]... writing from the publisher Engineering Mechanics - Statics Chapter 10 Problem 1 0-9 3 Determine the moment of inertia Ix 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: 3 4π r m=ρ 3 ρ = r 3m 3 4π r ⌠ 1 3m 2 2 ⎞ 2 2 2 2 ⎛ Ix = ⎮ ⎜ 3 ⎟ π r − x r − x dx = 5 m r ⎮ 2 4π r ⎠ ⌡− r ⎝ ( )( ) Ix = 2 2 mr 5 Problem 1 0-9 4 Determine the radius... publisher Engineering Mechanics - Statics Chapter 10 Problem 1 0-9 6 Determine the radius of gyration kx of the body The specific weight of the material is γ Given: lb γ = 38 0 ft 3 a = 8 in b = 2 in Solution: a ⌠ 2 ⎮ ⎮ 3 ⎞ ⎮ γ π b2 ⎛ x ⎟ dx mb = ⎜ ⎮ ⎝ a⎠ ⌡ mb = 13. 26 lb 0 ⌠ ⎮ ⎮ Ix = ⎮ ⎮ ⌡ a 2 1 2 x⎞ γ πb ⎛ ⎟ ⎜ 2 ⎝ a⎠ 3 2 2⎛x⎞ b 3 ⎜ ⎟ dx ⎝ a⎠ Ix = 0.59 slug⋅ in 2 0 kx = Ix mb kx = 1.20 in Problem 1 0-9 7 Determine... publisher Engineering Mechanics - Statics 2 Chapter 10 2 m = ρa c − ρ πb c m = 4 .7 233 kg I0 = IG + m ( a sin ( 45 deg) ) I0 = 6. 23 kg⋅ m 2 2 Problem 1 0-1 01 Determine the moment of inertia Iz of the frustum of the cone which has a conical depression The material has a density ρ Given: kg ρ = 200 m 3 a = 0.4 m b = 0.2 m c = 0.6 m d = 0.8 m Solution: h = da a−b Iz = 3 ⎡ ⎛ 1 2 ⎞⎤ 2 3 ⎡ ⎛ 1 2 ⎞⎤ 2 3 ⎡ ⎡1 2... publisher Engineering Mechanics - Statics Chapter 10 b 1 ⌠ ⎮ y2a yc = A⎮ ⌡0 Ix' = Ix − A yc y b yc = 120.0 mm dy 6 2 4 Ix' = 146 × 10 mm Problem 1 0-1 21 Determine the area moment of inertia for the triangular area about (a) the x axis, and (b) the centroidal x' axis Solution: h ⌠ 2b 1 3 Ix = ⎮ y ( h − y) d y = ⋅h ⋅b ⎮ h 12 ⌡0 3 Ix' = bh 12 − 1 2 bh⎛ ⎜ h⎞ 2 Ix = 1 3 ⎟ = ⋅h ⋅b 3 36 ⎝ Ix' = 1 12 1 36 bh 3 3 bh... 10 a ⌠ 2 2 ⎮ 1 ⎛ 3m ⎞ ⎛ b x ⎞ ⎛ b x + b⎞ dx = 93 m b2 Ix = ⎮ ⎟ ⎜ ⎟ π ⎜ + b⎟ ⎜ a 70 ⎠ ⎝ ⎠ ⎮ 2 ⎝ 7 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 = 0.1 875 m 4 0 Problem 1 0-1 17 Determine the area moments of inertia Iu and... the publisher Engineering Mechanics - Statics Chapter 10 ⎡1 ⎛ a ⎞ 1 ⎡ 2⎤ 2 2 Ix = 2⎢ m ⎜ ⎟ + m ( d − e) ⎥ + M⎣( 2d) + e ⎤ ⎦ 12 2 ⎝ 2⎠ ⎣ ⎦ 2 Ix = 3. 25 × 10 3 kg⋅ m 2 Problem 1 0-1 10 Determine the moment of inertia for the overhung crank about the x' axis The material is steel having density ρ Units used: Mg = 1000 kg Given: ρ = 7. 85 Mg m 3 a = 20 mm b = 20 mm c = 50 mm d = 90 mm e = 30 mm Solution:... writing from the publisher Engineering Mechanics - Statics Chapter 10 Solution: m = ρ 2π R 2π ⌠ I=⎮ ⎮ ⌡0 ρ = m 2π R ⎛ m ⎞ R 2 R dθ = m R 2 ⎜ ⎟ ⎝ 2π R ⎠ I=mR 2 Problem 1 0-9 2 The solid is formed by revolving the shaded area around the y axis Determine the radius of gyration ky The specific weight of the material is γ Given: a = 3 in b = 3 in lb γ = 38 0 ft 3 Solution: b ⌠ 2 ⎮ ⎡ ⎛ y ⎞ 3 m = ⎮ γ π ⎢a ⎜ ⎟ ⎥... c + d) ( a + b) − ⎜ + πr a ⎟ 3 ⎝ 4 ⎠ Ix = 236 .95 in 4 Iy = 1 3 ⎛ πr 2 2⎞ ( a + b) ( c + d) − ⎜ + πr d ⎟ 3 ⎝ 4 ⎠ Iy = 114.65 in 4 Ixy = ⎛ a + b ⎞ ⎛ d + c ⎞ ( a + b) ( d + c) − d aπ r2 ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠ Ixy = 118. 87 in R = ⎡ ⎛ Ix + Iy ⎞⎤ 2 ⎢Ix − ⎜ ⎟⎥ + Ixy ⎣ ⎝ 2 ⎠⎦ 4 4 4 2 Imax = Imin = R = 133 . 67 in 2 −R 4 Imax = 30 9 in Ix + Iy +R 2 Ix + Iy 4 4 Imin = 42.1 in 4 1051 © 20 07 R C Hibbeler Published by Pearson... 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... ⌠ 1 ⎮ ⎮ 3 2 ⎛ y⎞ Ix = ⎮ y a ⎜ ⎟ d y ⎮ ⎝ b⎠ ⌡ 0 Ix = 30 7 in 4 10 67 © 20 07 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 . writing from the publisher. Engineering Mechanics - Statics Chapter 10 m ρ a 2 c ρπ b 2 c−= m 4 .7 233 kg= I 0 I G masin 45 deg()() 2 += I 0 6. 23 kg m 2 ⋅= Problem 1 0-1 01 Determine the moment of. from the publisher. Engineering Mechanics - Statics Chapter 10 Solution: Moment of inertia I x and I y : I x 1 12 ba 3 = I x 90 10 3 × mm 4 = I y 1 12 ab 3 = I y 160 10 3 × mm 4 = Product of. 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