Mechanics 1 of materials hibbeler 6th Episode 2 Part 3 potx

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*6-140 A 100-mm-diameter circular rod is bent into an S$ shape If it is subjected to the applied moments M = 125 N-m at its ends, determine the maximum tensile and compressive stress developed in the rad M = 125 N-m 400 mm 400 mm M=125N-m 2 =2m(F-vdr?-c?) Ar =2z (0.45— 0.45? — 0.052 ) = 0.01750707495 m A= nc? =n (0.057) =2.5(10°)2 m? -3 Ra [a4 ~ 0.017507495 = 25007 _ 0 248606818 F-R = 0.45 —0.448606818 = 1.39318138(10 °)m Compressive stress at A and D : _ M(R-ra) | ~125(0.4486068 18 — 0.4) n= = =—1.39 MPa(max) Ans Ary(F~R) 2.5(103)Z (0.4)(1.39318138)02) M(R-rp) 125(0.4486068 18 — 0.5) p= = =—~1.17 MPa Arp(-R)_ 2.5(103)z (0.5)(1.39318138)(103) Tensile Stress at B and C : _ M(R-rp) _ ~125(0.448606818 —0.5) =1.17MPa ¬ Arg(F~R)_ 2.5(03)z (0.5)(1.39318138)(1072) _ M(R—rc) _ 125(0.448606818—0.4) c= = = 1.39 MPa Ans Arc(F—R) 2.5(107-7) x (0.4)(1.39318138)(1077)

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-141 The member has an elliptical cross section If it is subjected to a nroment of M = SON - m, determine the bend- ing stress at points A and 8, Is the stress at point A’, which is located on the member near the wall, the same as that at A? Explain dA _2nb' 5 J7? a2) Ar a = 2n(0.0375) 9 175 - 0.1752 — 0.0752 ) = 0.053049301 m 0.075 A= nab = m (0.075)(0.0375) = 2.8125(10 *)x -3 -„A_ „281200 25 _ 0 166556941 [,4¿ - 0053049301 F~R =0.175~0.166556941 = 0.0084430586 ơ.= M(R—ra) - 50(0.166556941 ~ 0.1) =446kPa(T) Ans ‘ Ara(F-R) 2.8125(10)% (0.1)(0.0084430586) on = M(R~rp) - 50(0.166556941 ~ 0.25) =224 kPa(C) Ảns , Arp(—R) 2.8125(10”2)zr (0.25)(0.0084430586) No, because of localized stress concentration at the wall Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-142 The member has an elliptical cross section If the

allowable bending stress iS Osjtow = 125 MPa, determine the maximum moment M that can be applied to the member a = 0.075 m; b = 0.0375 m A = 7(0.075)(0.0375) = 0.002825% m? dA - = {= -= 22 (y_ Vĩ - a3) - 2200215) (0115 — (0.175 — 0.075%) = 0.053049301 m P a 0.075 A _ 0002812 R = ThS 0.166556941 m 252 0053049301 r— R = 0.175 - 0.166556941 = 8.4430586(10”) m M(R - r) Ar(r — R) Assume tension failure M(0.166556941 — 0.1) 0.00281257{0.1)(8.4430586)(10?) 125(10°) = M = 14.0kN.m_ (controls) Ans Assume compression failure : M(0.166556941 ~ 0,25) 0.0028 1252(0.25)(8.4430586)( 107?) M = 27.9kN-m ~125(10°) =

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-143 The bar has 2 thickness of 0.25 in and is made of a material having an allowable bending stress of canow = 18 ksi Determine the maximum moment M that can be applied wi4i, r 0.25 _ 925 h 1 h 1 From Fig 6-48, K= 1.45 Omax = Ke A! 0.5 18(10°) = 1AM OS) (0.25)4) M=517T1b-in = 43.1 fb-ft Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall,

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*6-144 The bar has a thickness of 0.5 in and is subjected to amoment of 60 Ib « ft Determine the maximum bending

stress in the bar wo 4g ho La OF O25 hot From Fig 6-48, K = 1.45 Mc 60(12)(0.5) = Ke = 14522 —-]= 12.5ksi Ans đa = ẤT (Tosa

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-145 The bar is subjected to a moment of M = 15 N - m Determine the maximum bending stress in the bar and sketch, approximately, how the stress varies over the critical section M=\5Nm & 60 mm M=15N-m r U5 _ 9575 h 20 rls tt 813 tt Ww 270M Pe From Fig 6-48, K=l.2 Omax = Ki = 12g C200) = 27.0 MPa Ans I 75 0.01)(0.023)

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-146 The allowable bending stress for the bar is Cattow = 175 MPa, Determine the maximum moment M that can be applied to the bar M=15N-m = 1S 0575 20 From Fig 6-48, K = 1.2 Me M(OO1) = K*E; 135409= 12T sa Gstow = K = 06) Lê 0000022 M = 97.22 N-m Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-147 The bar is subjected to four couple moments If it is in equilibrium, determine the magnitudes of the largest moments M and M’ that can be applied without exceeding an allowabie bending stress of ơanow = 22 ksi 10Ib-R 100 tb-ft S ‘o in win u = in| II o an = tt N œ os cẻ — _k— or M ƒ ` From Fig 6-48, K = 1.2 \ J Orttow = a : 2209) = 1.2, —4007)_) /2/ƒ( 100 6-H 7;(0.5)(1.5)3 M = 3437.5 tb-in = 286.46 Ib- ft = 286 Ib- ft Ans GEM = 0; 286.46 - 10- 100~ M@’ = 0; M' = 1796lb-ft Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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*6-148 The bar is subjected to four couple moments If M = 180 1b - ft and M' = 70 1b - ft, determine the maximum bending stress developed in the bar sin, 43m 0.90 in 15in 101b- 100 Ib-ñ w 45 2 3 r 99 — 06 h 1.5 h 1.5 From Fig 6-48, K = 1.2 Onax = kM = 1.2 TC Ịz 13.8 ksi Ans I 7x(0.5)(1.5)?

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-149 Determine the maximum bending stress developed in the bar if it is subjected to the couples shown The bar has a thickness of 0.25 in

For the larger section : wi 45 Lis h 160tbin = 3-01 3 win From Fig 6-48, K = 1.755 Mc 160( 1.5) Omx = K-—— = 1.755[ ] = 749 psi (controls) Ans I 75(0.25)(3)? For the smaller section : wi 45 23 r_ L2 _ o1s h 1.5 h 1.5 From Fig 6-48, K = 1.15 Mc 60(0.75) = K—~= 115 } = 736 psi max I L #(025)(1.5)

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-150 Determine the length L of the center portion of the bar so that the maximum bending stress at A, B, and C is the same The bar has a thickness of 10 mm 350 N 259% 60 mm | 40 mm mm ° 2 mm | 7 mm 42m1 la c al — ` mm! +- E— mm te r— 3 200 mm —] †—Z22mm[ % 7 t2, 1220mm | (95 2 — Mg 217502) =35.4m w 60 r 7 —=—=Íil.5 -=—=0.175 h 40 b 40 ° 1758 From Fig 6-48, K=1.5 35)(0.02) (ØA)ma„ = hac =1.5 _7002 ¡_ 19.6875 MP ‘ I (To01ne.04)! MPa (OB)max = (F4)max = ae s, _ 175(0.2 + $)(0.03) =—— ] 19.6875(10)= “T00000 { My: I15(œ2† 4) 4 ris “20m^ fz L=0.95 m = 950 mm Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-151 The bar is subjected to a moment of M = 153N-m Determine the smallest radius r of the fillets so that an allowable bending stress of Tallow = 120 MPa is not exceeded Gaara = ce M4 =153Nm s\_ xr_ (153002) : 120(10") = KT #;(0.007)(0.04)3 ] Probs 6-151/6-152 K=1.46 From Fig 6-48, = =0.2 h r=0.2(40) = 8.0mm Ans

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*6-152 The bar is subjected to a moment of M= 17.5 N-m If r = 6mm determine the maximum bending stress in the material

From Fig 6-48, K = 1.555

Cmax = KE = 1555-00) 7 70.007)(0.04) _) =< ta6MPa Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-153 If the radius of each notch on the plate is r = 0.5 in determine the largest moment that can be applied The

allowable bending stress for the material is Osnow = 18 ksi b= 145-125 = 1.0m 2 14.5 in Lin , b 1 40 795 Loos r 05 h 125 From Fig 6-50: K = 2.60 Omax -K= 18(10°) = 2.60[ 2829) #p(1)(12.5)

M = 180288 ib-in = 15.0 kip-ft Ans

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6-154 The symmetric notched plate is subjected to bend-

ing If the radius of each notch is r = 0.5 in and the applied moment is M = 10 kip ft, determine the maximum bend-

ing stress in the plate 14.5 in, ny, r _ ĐỖ _ nga 0.5 h 12.5 From Fig 6-50: K=2.60 Cmax = KE =2.60iC0002)62”1 —12oksi Ans 1 œ02.5°

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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@155 The simply supported notched bar is subjected to two forces P Determine the largest magnitude of P that can be applied without causing the material to yield The material is A-36 steel Each notch has a radius of r = 0.125 in | mzs-E) _ 95 in * “H1 1.25 ia, 1.75 in, | i _L ———— ms-Ls-Lms—Las~l Z2/n - ^“27n dain P Pr px 15 L2 _ 02s P b = 025 = 2; : = 0.125 = 0.1 r 0.125 h 125 | } mazop From Fig 6-50, K = 1.92 Ss P(0.625 Oy = rie 36= 1.927 ZOOS) 9 7 4(0.5)(1.25) P = 122 Ib Ans

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"6-156 The simply supported notched bar is subjected to the two loads’ each having a magnitude of P= 100 tb Determine the maximum bending stress developed in the

bar, and sketch the bending-stress distribution acting over the cross section at the center of the bar Each natch has a radius of r = 0.125 in | 10016 10 tb 1,25 in 1.75 in [a mm ao in, "m n— 1.75—1 b= L=125 _ O05 2 b 0.25 -~=—— =?; r, 0125 _ on r 0.125 h 1.25 From Fig 6-50, K = 1.92 2000(0.625) 1z(0.5)(1.25)? Mc Omx = K—- = 19 ] = 29.5ksi Ans 008 lb £200 hin -2074 186 1b 3o-7 &c 30-7 fs

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6-157 A bar having a width of 3 in and height of 2 in is made of an elastic plastic materiai for which oy = 36 ksi

Determine the moment applied about the horizontal axis that will cause half the bar to yield GyseKst 38,098) :27%P S| fet ma tua IS z sạ Xip 3 Ì z@sG ) sel = 54(1.5) +27(0.667) = 99.0 kip-in = 8.25kip-ft Ans Also; Myc Cy = — rel My() 36(10)=———— « 1z)(2?) My = 72 000 Ib- in Yr M=2M[1~5 CÔ Half of the bar will yield, yy = h/4 11 M=- My = ¬ 02000) = 99000 lb -in.=8.25kip-ftƠ Ans Note: The above equation is valid only for rectangular sections

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6-158 Determine the plastic section modulus and the shape factor for the wide-flange beam L= = (0.2)(0.23)° - = (0.18)(0.2)° = 82.78333(10°°)m* C¡ =T¡ = ø(0.2)(0.015) = 0.00307 Cp = T = øy(0.1)(0.02) = 0.0020y M, = 0.00307 (0.215) + 0.0020y(0.1) = 0.000845 oy M, Ởy = —— l7 z- 00908422Y — g45(105m Ans oy Myc oy = — mT _ Ør(82.78333)105) = 0.000719855 oy 0.115 K = Me 0.0008450y _1ị? Ang My 0.0007198550y

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6-159 The beam is made of an elastic plastic material for

which oy = 250 MPa Determine the residual stress in the

beam at its top and bottom after the plastic moment M, is applied and then released I= 30202 _ = (0.18)(0.2) = 82.78333(10°°)m* Cy =F, = o(0.2)(0 015)= 0.0030, 2293-5 MP 42 SMR C2 = he = dy(0.1)(0.02) = 0.00207 250M Pa, v29 _ 4 ce M, = 0.0030y (0.215) + 0.00207 (0.1) = 0.000845 oy =e = 0.000845(250)(10”) = 211.25 kN -m Ä 3# TP tp ty _ Mpc _ 211.25(10°)(0.115) ———T† ¡ 8278333405 „7440 5 y Os = 0.09796 m = 98.0 250 293.5” yew’ m= 4v mm Orop = Obotom = 293.5— 250 = 43.5 MPa Ans

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*6§-160 Determine the shape factor for the cross section of the H-beam ed T & = —(0.2)(0.02") + 200202) = 26.8(10%)m‘* ei 12 C, = J = oy(2)(0.09)(0.02) = 0.00360, €2 = B = oy(0.01)(0.24) = 0.00249, 1, = 0.0036ơy (0.11) + 0.0024Øy(0.01) = 0.00042ơy ~ Ør(6.8X10”%) My 0.1 = 0.002680, M4, _ 0000420y K = sale My 0.000268a, =1.57 Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-161 The H-beam is made of an elastic-plastic material

for which ay = 250 MPa Determine the residual stress in the top and bottom of the beam after the plastic

moment M, is applied and then released k= =5(0.2K002") + = 10.02)(02) =26.8(105)m' Cy = Tj = ay(2)(0.09)(0.02) = 0.00360, Cp = T = 6y(0.01)(0.24) = 0.00240, My = 0.0036ay (0.11) + 0.002407(0.01) = 0.00042ay M, = 0.00042(250)(10°) = 105 kN - m ; AĐMh - »3904/4/ 2/14 os Myc = 105(10ˆX0.1) =392 MPa } I” 268q0% = od 52+ FG ) : Ể By cả: 230 392 7 = 0.0638 = 63.8 mm ¬h P j 222 đr = đp =392-250= 142MPA = Ans #228 mm

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6-162 The rod has a circular cross section If it is made of an elastic plastic material, determine the shape factor and the plastic section modulus Z Plastic moment : nr 2 Ca T= OS) = % ur oo 1 I=—mr 4 My = —- = te = oy r € r 4 Shape factor M, ‘oy — 16 = 2 = J = — = 1.70 Ans Plastic section modulus 4r3 3 Z= M = = Or = ar’ Ans Oy Oy 3

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6-163 The rod has a circular cross section If it is mace or an elastic plastic material, determine the maximum elastic moment and plastic moment that can be applied to the cross

section, Take r = 3 in., oy = 36 ksi Elastic moment : 1= lay 4 My = OH 2 orm’) _ ary € r 4 3 = (36) = 163.4 kịp - in = 63.6 kíp - ft Ans ác 47 Plastic moment : + —— mr xr 37 \ € Ca T= oS) = =—+T1- có T mr? 89rẻ 4n 4 TT M, = ——~oy(—) = —oy= -(3°)\(36 Ẫ 5 >) 3 3 (36) = 1296 kip-in = 108 kip: ft Ans

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*6-164 The T-beam is made of an elastic-plastic material Determine the maximum elastic moment and the plastic moment that can be applied to the cross section o, = 36 ksi Elastic Analysis : " 8(22Œ) + 102/0) _ 4 339 in 82) + 102) I= 12") + 8(2)(3.33337) + = (2)(10") + 10(2)(2.6667?) = 492 in’ M y = 226) _ 2310 kip-in = 193kip ft Ans 7.667 Plastic analysis : 2 [g4A=0 C+Q-h =0 4 36(8)(2) + 36(2)(10—đ) - 36(2)(4) = 0 d d=9in< 10in, OK M, = 36(8)(2)(2) + 36(2)(1)(0.5) + 36(2)(9)(4.5) 4104 kip-in = 342kip-ft Ans H From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-165 Determine the plastic section modulus and the shape

factor for the beam Elastic Analysis : 8 (2)(1) + 10207) _ 4 333 in 8(2) + 10(2) ‘ ‘ la = 62") + 8(2)(3.33337) + = (210°) + 10(2)(2.66677) = 492 in‘ y= Myc Ởy = —— tt v= 1923) _ 2310kip-in 7.667 Plastic analysis : 2 Joad=0; CG +Q-F =0 4 36(8)(2) + 36(2(10-d) - 36(2)(d) = 0 d d = 9in.< 10 in OK M, = 36(8)(2)(2) + 36(2)(1)(0.5) + 36(2)(9)(4.5) = 4104 kip- in M, oy = — 1y ze Me 2 AN iain? Ans Oy 36 cam My K= 4104 = 1.78 Ans 231

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6-166 Determine the plastic section modutus and the shape factor for the cross section of the beam 1 3 1 3 ~6, 4 = —(0 7 (0-2)(0.2 ) .22)° — —(0.185)(0.2)° 7h } (0.2) = 54.133(10 (10) m C, = oy(0.01)(0.2) = (0.002) oy Cz = oy(0.1)(0.015) = (0.0015) ay M, = 0.002Øy(0.21) + 0.0015Øy(0.1) = 0.000570y M Ơy = —— l7 z- 0000510 _ s10 15) mộ Ans Oy Myc oy = — rT -6 My = oy(54.133)(10) = 0.00056, 0.11 x = Me „ 000060 _ 1 16 Ans My 00005ơy

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6-167 Determine the plastic moment M, that can be supported by a beam having the cross section shown oy = 30ksi J øaA =0 Cc + Cy - Tñ =0 +(? ~ 1⁄30) + (10 ~ 4140) - d(1)(30) = 0 3z + 10 — 2d = 0 d=97124in.< 10in, OK M, = (2° ~ 1°)(30)(2.2876) + (0.2876)(1)(30)(0.1438) + (9.7124)(1)(30)(4.8562) = 2063 kip - in = 172kip-ft Ans

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*6-168 The thick-walled tube is made from an elastic- plastic material Determine the shape factor and the plastic section modulus Z Plastic analysis : Location of centroid C, - 4 y= XyA = 3x (172) — (Fr? _ Ar} - 7?) 2A Fr} ~ r?) 3#(rỆ — r‡) 7z T=C= sứ? - roy _- #22 2 8(rý — rỉ) 4 Mẹ = ~Œ¿ — rị)Øy[——————~] lp 5! 2 1) ru = ry = -(rì - r 3” rj )Oy Elastic analysis : T4 4 l=~Œy —r 4°” 1) ol | rt ~ rh 4 _ r4 My = T _ 772 TỦ -Ö *ZỨứ; Ty c Tạ 4n Shape factor 4/„3 3 rahe le -Ö l6r;(rỶ - rị) Ans rh¬ứr My i Y gy 3z(rÿ — r‡) Plastic section modulus : 47,3 Mẹ _ x72 -ri)oy_ 4 = — r 3.43 _ Oy oy 3! 2 — r) Ans Z=

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-169 Determine the shape factor and the plastic section modulus for the member Plastic analysis : 1 h bh = C= —(b)(- = T=C= 53! MG) er 7 Oy bh_oh bh M, = ——=Ơy(—-) = ——Ơ: b= Ons) = Ty Elastic analysis : 1 ha bh} I= 2[—(b)(-)] = — (3G) ] 48 My= —— = ——— = — c x 22 7 Shape factor M, bh oy K= —P = ity =2 Ans M, bh? Y 3a Ơy Plastic section modulus : bh? Mẹ _ TrƠy _ bw

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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6-170 The member is made from an clastic-plastic mater- ial Determine the maximum elastic moment and the plastic moment that can be applied to the cross section Take h 4in.,h - Gin, oy = 36 ksi Elastic analysis : f= 2 (4)(3)"] = 18 in‘ ØyI _ 36(18) My = TT = —— =216kip- in = lồkp: ft Ans e TA Plastic analysis : a T=C= 2(9((6) = 216 kíp FT, He r 6 ca Mẹ = 216C) = 432kip- in = 36 kip - R Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 32

6-171 The wide-flange member is made from an elastic- plastic material Determine the shape factor and the plastic section modulus Z Plastic analysis : h-2t R= C, = oybt; TF = Cz = ay( yt h-2t h-2t 2 )@(——) oy[b Kh—2) + Hư 7] Mẹ; = oyb th—1t) + oy( Elastic analysis : 143 1 3 I= —bh ~ —(b-)(h—-28 12 mạ )( ) = -rppŠ lbh — (b—Ð(h-209] (m— ad —2 n3 My = —— e i bh? - (b—Ð(h—20) =———————ơd; 6h Shape factor : Ke M _ {b (h—Ð) + ‡(h~20?]0y bh3 ~ (b—t)(R—203 My — set 0y _ 3h 4b (h—) + t(h~— 207 Ans 2° bR- (b~Ð(h-203 Plastic section modulus : z« My „ ØrbẤh—9 + z@—20?] Ơy Oy = bÁh~0 + ;ứ—20” Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 33

*6-172 The beam ts made oi an elastic-piastic material tor which ‘oy = 200 MPa If the largest moment in the beam occurs within the center section a—a, determine the magni- tude of each force P that causes this moment to be (a) the largest elastic moment and (b) the largest plastic moment M=2P q) , a) Elastic moment | 3 | 1 I=— 13 (0100.2 ) = 66.667(10°°) m4 3) 6 #=—— a o, = Me I P P M, = 200(10°)(66.667)(10°) P 0.1 = 133.33kN-m }M From Eq (1) 2m 133.33 =2P P P = 66.7kN Ans b) Plastic moment bh’ M = a7 _1 obs 0.1(0.2? af ph = —— ) (200)(10°) — tere, bh = 200kN-m “7 é From Eq (1) 200 = 2P = 100 kN Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall,

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Trang 34

6-173 The beam is made from an elastic-plastic material

for which oy = 200 MPa Determine the magnitude of force P that causes this moment to be (a) the largest

elastic moment and (b) the largest plastic moment 8) _ Elastic analysis : 2 + 5 P My = 5l „ 200009)03(015(022) a MBP “foam € 01 q M= 8P = 200; P = 250kN Ans — b) Plastic analysis : — se T = C = 200(10°)(0.15)(0.1) = 3000kN rên [om M, = 3000(0.1) = 300kN - m teed i 7 I M = 8P = 300; P= 37.5KN Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 35

6-174 The box beam is made from an elastic-plastic material for which oy = 25 ksi Determine the intensity of the distributed load wo that will cause the moment to be (a) the largest elastic moment and (b) the largest plastic moment Elastic analysis : 1 3 1 ï= —(8)(16) mĩ )(16) - — (62) 3) = 1866.67 in in’ Moss = si 2?wa(12) = = Wo = 18.0kip/ft Ans Plastic analysis : Cy = T = 25(8)(2) = 400kip C2 = F = 25(6)(2) = 300kip " M, = 400(14) + 300(6) = 7400 kip - in 279 (12) = 7400 Wo = 22.8kip/ft Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 36

6-175 The beam is made of an elastic plastic materiai for

which oy = 30 ksi If the largest moment in the beam occurs

at the center section a-a, determine the intensity of the dis- tributed load w that causes this moment to be (a) the largest elastic moment and (b) the largest plastic moment tal — + 4 ———un † Wn | 10° (0° /0w iow 10wW) M = 50w (1) a) Elastic moment J= 28/8) = 341.33 in’ Myc 1 _ 3064133) ~ 4 = 2560 kip: in = 213.33 kip: ft From Eq (1), 213.33 = 50w w = 4.27 kip/ft Ans b) Plastic moment 1€ C = T =30(8)(4) = 960 kíp | La +” M, = 960(4) = 3840kip-in = 320 kíp -ft LI T From Eg (1) 320 = 50w w= 640kipfL Ans oy = My

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 37

*6-176 The beam has a rectangular cross section and is made of an elastic-plastic material having a stress strain diagram as shown Determine the magnitude of the moment M that must be applied to the beam in order to create a max- imum strain in its outer fibers of emax ~ 0.008 to (MPa) 200 aan 0.004 2008 C, = F = 200(10°)(0.1)(0.2) = 4000 kN Œ =1 = 2(200105/0.1)/02) = 2000 kN M = 4000(0.3) + 2000(0.1333) = 1467kKN-m=14.7MN-m Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 38

6-177 A beam is made from polypropylene plastic and has

a stress-strain diagram that can be approximated by the curve shown If the beam is subjected to a maximum tensile and compressive strain of € = 0.02 mm/mm, determine the maximum moment M +ơ(Pa) M ự 100 ae ‘ o= 10(106) £ 1⁄4 30 mm €(mnv/mm) Chay? 2.22 tL Ze 0.05m™ /] # Emax = 0.02 + o = 10(10°)(0.02)'* = 3.761 MPa 002 _c 05 y c=04y ơ= 10(1051(0.4)1⁄4y1⁄4 đ=7.9527(105)y!4 M= J yodA =2 pe 9(7.9527)(10°)y"”*(0.03)dy 005 M = 0.4771 6405 [ y dy = 047716(10°N(=) (0.05 M=903kN.m Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 39

6-178 The bar is made of an aluminum alloy having a stress-strain didgram that can be approximated by the straight line segments shown Assuming that this diagram is the same for both tension and compression, determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is emạ„ = 0.03 otksi) „ ” si 80 eo L.o.05 3in, tị c(in.in.} à 0.006 0025 0.0: o-80 _— 9-80 —————— = ——; Ø = 82ksi 0.03 - 0.025 0.05 - 0.025 r 1 CQ = = 2(0.3333)(80 + 82)(3) = 81 kip <c.z.1 — 26k: woe” \Í Ni % Co = B = 5(1.2666)(60 + 80)(3) = 266 kip 1+ 3) RES .- Te 1 \ Enas 20-03 7 T Cs = B = 5(0.4)(60)(3) = 36 kip #*L&1” cons; BRI BHF: M = 81(3.6680) + 266(2.1270) + 36(0.5333) = 882.09 kip - in = 73.5kip - ft Ans

Note : The centroid of a trapezodial area was used in calculation of moment areas

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 40

6-179 The bar is made of an aluminum alloy having a stress-strain diagram that can be approximated by the straight line segments shown Assuming that this diagram is the same for both tension and compression, determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is ¢max = 0.05 a(ksi) 90 = 80 4 ứ 40 =| 3 % €(in./in.} 0.006 0.025 0.05 ơi = -6_z = 10(10°)e 0.006 ơ -60 _ 80~ 60 e— 0006 0.025 - 0.006 Ø; = 1052.63£ + 53.684 © é 4 3 - 80 _ _0- 80 , Ø; = 400 + 70 vei e - 0.025 0.05 — 0.025 2.6 ? ' 2-2 0.05 02£ ce= 9 = 0.025y Substitute € into Ø expression : O, = 250y Os y < 0.24 in Ø; = 26.315y + 53.684 0.24 <y< lin 6, = 10y + 70 lin.< y S$ 2ïn dM = yodA = yoa(3 dy) 0.24

M = 213 y 250y? dy +3 fj, (2631597 + 53.684y) dy + 3 luw + 70y) dyÌ

= 980.588 kip.in = 81.7kip-ft Ans

Also, the solution can be obtained from stress blocks as in Prob : 6-178

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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