PROBLEM Area of Problem I = 471,040 mm xy 2(471,040) 252,757 1,752,789 = or θ m = and = (1,002,773 ± 885,665) mm

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1 PROBLEM 9.88 F the area indicated, determine the ientation of the principal aes at the igin the cresponding values of the moments of inertia. Area of Problem From Problem 9.8: Problem 9.75: = 5,757 mm = 1,75,789 mm = 71,00 mm Now Eq. (9.5): tan θ m = (71,00) = 5,757 1,75,789 = Then θ m = θ m = Also Eq. (9.7): Then ma, min = ± + ma, min 5, , 75, 789 5, 757 1, 75, 789 = ± = (1,00,773 ± 885,665) mm B inspection, the a ais cresponds to min the b ais cresponds to ma. ma 6 = mm min 6 = mm PROPRETARY MATERAL. 013 The McGraw-Hill Companies, nc. All rights reserved. No part of this Manual ma be displaed, reproduced distributed in an fm b an means, without the pri written permission of the publisher, used beond the limited distribution to teachers educats permitted b McGraw-Hill f their individual course preparation. f ou are a student using this Manual, 1511

2 PROBLEM 9.9 Using Mohr s circle, determine the moments of inertia the product of inertia of the area of Problem 9.75 with respect to new centroidal aes obtained b rotating the aes 5 clockwise. From Problem 9.8: = 5,757 mm = 1,75,789 mm Problem 9.75: = 71,00 mm The Mohr s circle is defined b the diameter XY, where X (5,757; 71,00) Y (1,75,789; 71,00). Now 1 1 ave = ( + ) = (5, ,75,789) = 1,00,773 mm The Mohr s circle is then drawn as shown. 1 R= ( ) + 1 = (5,757 1,75,789) 71,00 + = 885,665 mm tan θ m = θ m = (71,00) = 5,757 1,75,789 = Then α = 180 ( ) = PROPRETARY MATERAL. 013 The McGraw-Hill Companies, nc. All rights reserved. No part of this Manual ma be displaed, reproduced distributed in an fm b an means, without the pri written permission of the publisher, used beond the limited distribution to teachers educats permitted b McGraw-Hill f their individual course preparation. f ou are a student using this Manual, 150

3 PROBLEM 9.9 (Continued) Then = ave + Rcosα = 1, 00, , 665cos = mm = ave Rcosα = 1,00, ,665cos = mm = Rsinα = 885,665sin mm = PROPRETARY MATERAL. 013 The McGraw-Hill Companies, nc. All rights reserved. No part of this Manual ma be displaed, reproduced distributed in an fm b an means, without the pri written permission of the publisher, used beond the limited distribution to teachers educats permitted b McGraw-Hill f their individual course preparation. f ou are a student using this Manual, 151

4 PROBLEM 9.10 Using Mohr s circle, determine f the area indicated the ientation of the principal centroidal aes the cresponding values of the moments of inertia. Area of Problem 9.77 (The moments of inertia of the area of Problem 9.10 were determined in Problem 9.). From Problem 9.: Problem 9.77: = in =.5080 in =.530 in The Mohr s circle is defined b the diameter XY, where X(18.18,.530) Y (.5080,.530). Now 1 1 ave = ( + ) = ( ) = in The Mohr s circle is then drawn as shown. 1 R = ( ) + 1 = ( ) (.530) + = in tan θ m = θ m = θ m = (.530) = = 0.65 The principal aes are obtained b rotating the aes through counterclockwise about C. PROPRETARY MATERAL. 013 The McGraw-Hill Companies, nc. All rights reserved. No part of this Manual ma be displaed, reproduced distributed in an fm b an means, without the pri written permission of the publisher, used beond the limited distribution to teachers educats permitted b McGraw-Hill f their individual course preparation. f ou are a student using this Manual, 1536

5 PROBLEM 9.10 (Continued) Now ma,min = ave ± R = ± = in ma = 3.9 in min From the Mohr s circle it is seen that the a ais cresponds to ma the b ais cresponds to min. PROPRETARY MATERAL. 013 The McGraw-Hill Companies, nc. All rights reserved. No part of this Manual ma be displaed, reproduced distributed in an fm b an means, without the pri written permission of the publisher, used beond the limited distribution to teachers educats permitted b McGraw-Hill f their individual course preparation. f ou are a student using this Manual, 1537

6 PROBLEM Using Mohr s circle, prove that the epression is independent of the ientation of the aes, where,, represent the moments product of inertia, respectivel, of a given area with respect to a pair of rectangular aes through a given Point O. Also show that the given epression is equal to the square of the length of the tangent drawn from the igin of the codinate sstem to Mohr s circle. First observe that f a given area A igin O of a rectangular codinate sstem, the values of ave R are the same f all ientations of the codinate aes. Shown below is a Mohr s circle, with the moments of inertia,, the product of inertia,, having been computed f an arbitrar ientation of the aes. From the Mohr s circle Then, fming the epression = + Rcos θ ave = Rcos θ ave = Rsin θ = ( ave + cos θ)( ave cos θ) ( sin θ) R R R ( ave θ) = R cos ( R sin θ) ave R = which is a constant is independent of the ientation of the codinate aes Q.E.D. Shown is a Mohr s circle, with line OA, of length L, the required tangent. Noting that OAC is a right angle, it follows that = ave L R = L Q.E.D. PROPRETARY MATERAL. 013 The McGraw-Hill Companies, nc. All rights reserved. No part of this Manual ma be displaed, reproduced distributed in an fm b an means, without the pri written permission of the publisher, used beond the limited distribution to teachers educats permitted b McGraw-Hill f their individual course preparation. f ou are a student using this Manual, 157