BeerMOM_GE_C07-P001-P050BeerMOM_GE_C07-P051-P100BeerMOM_GE_C07-P101-P169CCHHAAPPTTEERR 77PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.1 For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 7.2. SOLUTION 0: 18 cos15 sin 15 45 cos15 cos15 27 sin 15 sin 15 + 18 sin 15° cos15 0FAA A A AσΣ=+ °°+ °°− °° °= 2218 cos 15 sin 15 45 cos 15 27 sin 15 18 sin 15 cos 15σ=− ° °− °+ °− °° 49.2 MPaσ=−  0: 18 cos15 cos 15 45 cos15 sin 15 27 sin 15 cos 15 18 sin 15° sin 15 0FAA A A AτΣ=+ °°− °°− °°− °= 2218(cos 15 sin 15 ) (45 27)cos 15 sin 15 τ=− °− ° + + ° ° 2.41 MPaτ= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.2 For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 7.2. SOLUTION 0FΣ= 90 sin 30 cos30 90 cos30 sin30 60 cos30 cos30 0AA A Aσ−°°−°°+ °°= 2180sin 30 cos30 60cos 30 32.9 MPaσ=°°−°=  0FΣ= 90 sin 30 sin 30 90 cos30 cos30 60 cos30 sin 300AA A Aτ+°°−°°− °°= 2290(cos 30 sin 30 ) 60cos30 sin 30 71.0 MPaτ=°−°+°°= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.3 For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 7.2. SOLUTION    0:Σ=F 120 cos20 cos 20 45 cos 20 sin 20 45 sin 20 cos 20 60 sin 20 sin 20 0AA AAAσ− ° °− ° °− ° °− ° °= 22120cos 20 45cos 20 sin 20 45sin 20 cos 20 60sin 20σ=°+°°+°°+° 14.19 MPaσ=  0:Σ=F 120 cos 20 sin 20 45 cos 20 cos20 45 sin 20 sin 20 60 sin 20 cos 200AA A A Aτ+ ° °− ° °+ ° °− ° °= 22120cos 20 sin 20 45(cos 20 sin 20 ) 60sin 20 cos20τ=− ° °+ °− ° + ° ° 15.19 MPaτ= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.4 For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 7.2. SOLUTION 084 cos55 cos55 42 sin 55 sin 55 0FAA AσΣ=−°°+°°= 2284cos 55 42sin 55 0.546 MPaσ=°−°=−  084 cos55 sin 55 42 sin 55 cos55FAA AτΣ=−°°−°° 126 cos55 sin 55 59.2 MPaτ=°°= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.5 For the given state of stress, determine (a) the principal planes, (b) the principal stresses. SOLUTION 60 MPa 40 MPa 35 MPaxyxyσστ=− =− = (a) 2(2)(35)tan 2 3.5060 40xypxyτθσσ== =−−−+ 2 74.05pθ=− ° 37.0 , 53.0pθ=− ° °  (b) 22max,min222260 40 60 40(35)2250 36.4 MPaxy xyxyσσ σσστ+−=± +−− −+=± +=− ± max13.60 MPaσ=−  min86.4 MPaσ=− PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.6 For the given state of stress, determine (a) the principal planes, (b) the principal stresses. SOLUTION 28 MPa 140 MPa 42 MPaxy xyσστ== =− (a) 2(2)( 42)tan 2 = 0.75028 140xypxyτθσσ−==−− 2 36.87pθ=° 18.43 ,108.43pθ=° ° (b) 22max, min22+2228 140 28 140(42)2284 70xy xyxyσσ σσστ+−=±+−=± +−=±max154 MPaσ= min14 MPaσ= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.