A = p 4 (0.05)2 = (103 ) m 2. J = p 2 (0.025)4 = (104 ) m 4. s = P A = 2(10 3 ) (103 = MPa. t = Tc J = 500(0.


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1 014 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he publisher The shor concree cylinder having a diameer of 50 mm is subjeced o a orque of 500 N # m and an axial compressive force of kn. Deermine if i fails according o he maximumnormalsress heory. The ulimae sress of he concree is s ul = 8 MPa. kn 500 N m A = p 4 (0.05) = (103 ) m 500 N m J = p (0.05)4 = (104 ) m 4 s = P A = (10 3 ) (103 = MPa ) = Tc J = 500(0.05) = 0.37 MPa (106 ) s x = 0 s y = MPa xy = 0.37 MPa kn s 1, = s x + s y s 1, = ; A a s x  s y b + xy ; A a 0  (1.019) b s 1 = MPa s = MPa Failure crieria: s 1 6 s al = 8 MPa s 6 s al = 8 MPa No. OK OK No. 1033
2 014 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he publisher If he in.diameer shaf is made from brile maerial having an ulimae srengh of s ul = 50 ksi, for boh ension and compression, deermine if he shaf fails according o he maximumnormalsress heory. Use a facor of safey of 1.5 agains rupure. Normal Sress and Shear Sresses. The crosssecional area and polar momen of ineria of he shaf s crosssecion are 4 kip f 30 kip A = pa1 B = pin J = p A14 B = p in4 The normal sress is caused by axial sress. s = N A =  30 p = ksi The shear sress is conribued by orsional shear sress. = Tc J = 4(1)(1) = ksi p The sae of sress a he poins on he surface of he shaf is represened on he elemen shown in Fig. a. InPlane Principal Sress. s x = ksi, s y = 0 and xy = ksi. We have s 1, = s x + s y ; A a s x  s y b + xy = ; A a b + (30.56) = ( ; 30.99) ksi s 1 = 6.15 ksi s = ksi Maximum NormalSress Theory. s allow = s ul = 50 = ksi F.S. 1.5 s 1 = 6.15 ksi 6 s allow = ksi s = ksi 7 s allow = ksi (O.K.) (N.G.) Based on hese resuls, he maerial fails according o he maximum normalsress heory. Yes. 1037
3 014 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he publisher If he in.diameer shaf is made from cas iron having ensile and compressive ulimae srenghs of 1s ul = 50 ksi and 1s ul c = 75 ksi, respecively, deermine if he shaf fails in accordance wih Mohr s failure crierion. Normal Sress and Shear Sresses. The crosssecional area and polar momen of ineria of he shaf s crosssecion are A = pa1 B = p in J = p A14 B = p in4 4 kip f 30 kip The normal sress is conribued by axial sress. s = N A =  30 p = ksi The shear sress is conribued by orsional shear sress. = Tc J = 4(1)(1) = ksi p The sae of sress a he poins on he surface of he shaf is represened on he elemen shown in Fig. a. InPlane Principal Sress. s x = ksi, s y = 0, and xy = ksi. We have s 1, = s x + s y ; A a s x  s y b + xy = ; A a b + (30.56) = ( ; 30.99) ksi s 1 = 6.15 ksi s = ksi Mohr s Failure Crieria. As shown in Fig. b, he coordinaes of poin A, which represen he principal sresses, are locaed inside he shaded region. Therefore, he maerial does no fail according o Mohr s failure crieria. No. 1038
4 014 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he publisher Coninued Maximum Disorion Energy Theory. s 1  s 1 s + s = s allow ( ) + ( ) = s allow s allow = MPa Thus, he facor of safey is F.S. = s Y s allow = = 1.64 F.S. =
5 014 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he publisher The yield sress for heareaed beryllium copper is s Y = 130 ksi. If his maerial is subjeced o plane sress and elasic failure occurs when one principal sress is 145 ksi, wha is he smalles magniude of he oher principal sress? Use he maximumdisorionenergy heory. Maximum Disorion Energy Theory: Wih s 1 = 145 ksi, s 1  s 1 s + s = s Y s + s = 130 s  145s = 0 s = (145) ; (145)  4(1)(415) (1) = 7.5 ; Choose he smaller roo, s = 38.9 ksi s = 38.9 ksi 1045
6 014 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he publisher The sae of sress acing a a criical poin on he sea frame of an auomobile during a crash is shown in he figure. Deermine he smalles yield sress for a seel ha can be seleced for he member, based on he maximumshearsress heory. Normal and Shear Sress: In accordance wih he sign convenion. s x = 80 ksi s y = 0 xy = 5 ksi InPlane Principal Sress: Applying Eq ksi 80 ksi s 1, = s x + s y ; a s x  s y b + xy A = ; a 800 b + 5 A = 40 ; s 1 = ksi s = ksi Maximum Shear Sress Theory: and have opposie signs so s 1  s = s Y s 1 s (7.170) = s Y s Y = 94.3 ksi s Y = 94.3 ksi 1048
7 014 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he publisher The sae of sress acing a a criical poin on a machine elemen is shown in he figure. Deermine he smalles yield sress for a seel ha migh be seleced for he par, based on he maximumshearsress heory. The Principal Sresses: s 1, = s x + s y ; a s x  s y b + xy A 10 ksi 4 ksi 8 ksi = 810 ; a 8  (10) b + 4 A s 1 = ksi s = ksi Maximum Shear Sress Theory: Boh principal sresses have opposie sign. hence, s 1  s = s Y ( ) = s Y s Y = 19.7 ksi s Y = 19.7 ksi 1051
8 014 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he publisher The gas ank is made from A36 seel and has an inner diameer of 1.50 m. If he ank is designed o wihsand a pressure of 5 MPa, deermine he required minimum wall hickness o he neares millimeer using (a) he maximumshearsress heory, and (b) maximumdisorionenergy heory. Apply a facor of safey of 1.5 agains yielding. (a) Normal Sress. Assuming ha hinwall analysis is valid, we have s 1 = s h = pr = 5A106 B(0.75) = 3.75A106 B s = s long = pr = 5A10 6 B(0.75) = 1.875A106 B Maximum Shear Sress Theory. s allow = s Y FS. = 50A106 B 1.5 = A10 6 BPa s 1 and s have he same sign. Thus, s 1 = s allow 3.75A10 6 B = A10 6 B = 0.05 m =.5 mm Since = , hinwall analysis is valid. (b) Maximum Disorion Energy Theory. Thus, r = s allow = s 1  s 1 s + s = s allow s Y = 50A106 B F.S. 1.5 = A10 6 BPa C 3.75A106 B S  C 3.75A106 B SC 1.875A106 B S + C 1.875A106 B S = c166.67a10 6 B d 3.476A10 6 B = A10 6 B = m = 19.5 mm Since r = = , hinwall analysis is valid. (a) =.5 mm, (b) = 19.5 mm 1056
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