Mechanics 1 of materials hibbeler 6th Episode 2 Part 4 pot

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6-190 Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings of the belt, gear, and flywheel The bearings at A and B exert only vertical reactions on the shaft

300 mm? “(OO mm 300 mmo mam | 150N “ |22e“ + LOTT | | | 220mm 400mm | 300mm kame 150™ | Ze 67% 83.33" | VN), 150 | 7 TT] †+————+x v+T M(N+®) 433 ‡ + + at Ye | Go 50

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

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6-191 Determine the maximum bending stress in the handle of the cable cutter at section a—a A force of 45 lb

is applied to the handles The cross-sectional area is shown in the figure 45 tb œM =0 M~45(5 + 4cos20°) = 0 M = 394.141b-in,

Omar = = Thi = 8.41 ksi z(0.5)(0.75) Ans

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7-1 If the beam is subjected to a shear of V = 15 kN, determine the web’s shear stress at A and B Indicate the shear-stress components on a volume element located at these points Set w = 125 mm Show that the neutral axis is located at ¥ = 0.1747 m from the bottom and In, = 0.2182(10-3) mé 200 mm = _ (0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03) = 0.1747 m ys 0.125(0.03) + (0.025)(0.25) + (0.2)(0.03) (0.125\0.03") + 0.125(0.03)(0.1747 — 0.015)’ + 12(0025/0.25) + 0.25(0.025)(0.1747—0.155)7 + 502x003") + 0.2(0.03)(0.295—0.1747)” = 0.218182 (10) m = yA, = (0.310~0.015—0.1747)(0.2)(0.03) = 0.7219 (10 ”) mỶ = yÁp = (0.1747—0.015)(0.125)(0.03) = 0.59883 (10”) m” 3 -3 n= YØ, _ 1540°X0.7219/10”) _ 9g pa Ans Tt 0.218182(10°3)0.025) -3 5» = 15402X0.5988310) _ 1/sMPpa Ans It 0.218182(103)0.025) Te 199 MPa đãi hY 5stec em

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7-2 If the wide-flange beam is subjected to a shear of V = 30 KN, determine the maximum shear stress in the beam Set w = 200 mm 4 9 [tom 9.0628 m_ Section Properties : = 3 1 P= 55 (0.2)(0.310)" — 7p (0-175) (0.250)* = 268.652(10)ˆ5 m° mạc = TA = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10)? m: ve 30(10)? (1.0353) (10)~3 It 268.652(10)-° (0.025) = 4.62 MPa Ans Tmax

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7-3 If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the web of the beam Set w = 200 mm ac! 0.155% 4 0125 on, l= = (0.2)(0.310)" - <5 (0.175)(0.250)" = 268.652(10) 5 m' 9= C015 ~ y)(0.2) = 0.1(0.024025 — y*) 7 30(10)2(0.1)(0.024025 — y?) fe 268.652 (10)”5 (0.2) y= fy dA = 55.8343(10)° J” (0.024025 ~ ¥°)(0.2 dy) 6 1 30.155 = 11.1669(10) [ 0.024025y — > of 125 ý = 1.457kN = 30-— 2(1.4457) = 27.1kN Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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*7-4 The beam is fabricated from three steel plates, and

it is subjected to a shear force of V = 150 KN Determine the shear stress at points A and C where the plates are

joined Show y = 0.080196m from the bottom and Ina = 4.8646(10°°) m* jens seseem vA (8S - J A i z2 ig? 0 046 Fam LyA - 0.0375(0.075)(0.015) +0.08(0.215)(0.01)+ 2{0.1125(0.075)(0.01)] ZA 0.075(0.015) +0.215(0.01) + 2(0.075)(0.01) = 0.080196 m I= Tz(0015)(0.075 )+ (0.015) (0.075)( 0.080196 ~ 0.0375)? 1 + Tz(0215)0.01)) + 0.215(0.01)(0.080196 - 0.08)? | 3 + 2f 55 0.01)(0.075 } + 0.01(0.075)(0.1125 ~0.0801967] = 4.8646(10 5) m' Ó = Óc = X:A = 0.032304(0.075)(001) = 24.2271(10ˆ5) m3 Mt = It 4.8646(10-) (0.01) = 74.7 MPa Ans

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7-5 The beam is fabricated from three steel plates, and

it is subjected to a shear force of V = 150 KN Determine the shear stress at point B where the plates are joined Show ¥ = 0.080196m from the bottom and Iy,4 = 4,8646(10-*) m’ 0.042694.m ; _ yA — 9-0375(0.075) (0.015) +0.08(0.215)(0.01)+ 2{0.1125(0.075)(0.01)] tA 0.075(0.015) +0.215(0.01) + 2(0.075)(0.01) = 0.080196 m —(0.015)(0.075") + (0.015)(0.075)(0.080196 — 0.0375)” + ~(0.215)(0.0") + 0.215(0.01) (0.080196 — 0.08)" + 2 = (0.0140.075") + 0.01(0.075)(0.1125 — 0.080196)} = 4.8646(10°°) m* Ós = ysA = 0.042696(0.075)(0.015) = 48.0333(10~*) m? =Z ——————————— = 98.7 MPa Ans „ =Ÿ0a 150( 10°)(48.0333)(10~*) "Tụ 4.8646(105)(0.015)

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7-6, If the T-beam is subjected to a vertical shear of

V = 10 kip, determine the maximum shear stress in the

beam Also, compute the shear-stress jump at the flange-web junction AB Sketch the variation of the shear-stress intensity over the entire cross section: Show that Iy 4 = 532.04 in’ \ V = 10kip (1.5)(3) (14) + 6(6)(6) 3(14) + 6(6) S003") + 3(14)(3.5169 — 1.5) = 3.5769 in + Sone) + 6(6)(6 — 3.5769) = 532.04 in" Omar = YA" = 2.71155(5.4231)(6) = 88.23 in® 3 tạ = VOmex _ 10(10°)(88.23) _ 2 psi Ans Tt 532.04 (6)

Óàa =f.A' = 2.0769(3)(14) = 87.23 in’

(ca)y = VOp _ 104038723) _ 494 psi Ty 532.04 (14) VOsn _ 10(10°)(87.23) lapw = ee = = 2733 psi (TaBle = 532.04 (6) p Shear stress jump =(tag)w — (fAgỳ = 273.3 — 117.1 = 156 psi Ans rũ =z.07€9 ° ˆ “ (TOC 275 FSi mas *276 PS; © 2005 R C Hibbeler Publishe

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7-7, If the T-beam is subjected to a vertical shear of V = 10 kip, determine the vertical shear force resisted by

the flange Show that Iy4 = 532.04 in’ Ÿz3,€72 là = (1.5)(3) (14) + 6(6)(6) 3(14) + 6(6) I= 514% + 3(14)(3.5769 - 1.5)" = 3.5769 in Net + = (6X6) + 6(6)(6— 3.5769)? = 532.04 in

Q = (3.5769 — y)(14)( ae) = 7(3,5769" — y*) 3.57 ts 10(7)(3.5769? — y*) (532.04)(14) = 0.009398(3.5769? ~ yˆ @.36 ~y) 3.5769 yefrds = I 0,009398(3.57697 — y*)(14 dy) 3.5769 0.13157(3.5769"y — 4y?) ">| 3 “os7e9 3.05 kip Ans 1 it

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*7-8 Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 15 kN Show that In, ~ 6.691(10-) m*4, 10 mm — x ⁄ + 2.0233.5! m 0: O46]0Q2m ~ (0.005)(0.01)(0 14) + (0.05)(0.1)(0.08) _ St T= a0, 14)(0.01°) + (0.14)(0.01)(0.043298 ~ 0.005)? 1 + (01 )(0.08*) + (0.1)(0.08)(0.05— 0.43298)? = 6.6911(105) m' Qmex = yA’ = (0.023351)(0.046702)(0.1) = 0.1090544 (107?) m? V@max 15(10°)(0.1 3

Tmax = Bmax, _ 15(10°)(0.1090544)(10 ) ~2.44 MPa It 6.6911(107°)0.1)

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7-9 Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is

Tallow = 50 MPa Show that Iv, = 6.691(1075) m* oO! AAs: 0.02335/an 6 ., 99/722 ~ - (0.005)(0.01)(0 14) + (0.05)(0.1)(0.08) (0.01)(0.14) + (0.1)(0.08) xi = 0.043298 m 1= 50 14)(0.01°) + (0.14)(0.01)(0.043298 — 0.005)? + 50 X0.08°) + (0.1)(0.08)(0.043298 — 0.05) = 6.6011(10 5) m' Omax = y’A’ = (0.023351)(0.046702)(0.1) = 0.1090544 (10?) m? VOmax Tmax = Tattow = “te ~3 5005) - V(0.1090544) (10 7”) 6.6911 (10-5) (0.1) V=307kN Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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7-10 Determine the intensity of the shear stress distributed

aver the cross section of the strut if it is subjected to a shear force of V = 12 kN Show that Iw, = 6.691(10” 5) m, _ li = 0.038298 on 2 9.02335 lm ÿ= (0.005)(0.01)(0.14) + (0.05)(0.1)(0.08) = 0.043298 m (0.01)(0.14) + (0.1)(0.08) I= 5 (014,0.01") + (0.14)(0.01)(0.043298 — 0.005)" + = (0.1)(0.082) + (0.1)(0.08)(0.05 — 0.043298)" = 6.6911(10 *) m‘ Qmax = y’A’ = (0.023351)(0.046702)(0.1) = 0.1090544 (107°) m? O = y’A’ = (0.038298)(0.14)(0.01) = 53.6172 (10°°) m? VQ — 12(10°)(53.6172)(10°°) (aes = 0.687 MPa It 6.6911 (10ˆ5)(0.14) 3 =6 x 120)53.6172/409) _ 0.962 MPa 6.6911 (10-5)(0.1) 3 —3 lon = VOmax _ 12(10°)(0.1090544) (107°) _ 196MPa Ans It 6.6911 (105X0.1) 2687 Mh o:- 962M Pa a A 196 MPa 467mm

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7-11 Sketch the intensity of the shear-stress distribution acting over the beam’s cross-sectional area, and determine

the resultant shear force acting on the segment AB The

shear acting at the section is V = 35 kip Show that Iy4 = 872.49 in’ 5 = MEE) + DO® «5 1053 n, 8(8) + 6(2) 588") + 8(8)(5.1053- 4)" ~ lì + Lee) + 2@)01- 5.1053)" = 872.49 int 12 _lf¬ fire K8245 Ale, ” PT 25-8907 Qc = ':A" = (2.55265)(5.1053)(8) = 104.26 in? z1} ‘le Op =y‘A’ = (5.8947)(6)(2) = 70.74 in” t= ve t te = 35(10°)(104.26) _ oo psi 872.49 (8) Hi 35(10°)(70.74) Ñ 7 7 $23 PS; (tp) = 210 = 812.49 (2) = 1419 psi + | Sep MIPS - _ 350020074) _ sang (tp): = sin = 37249 @) P A = 2(88947 —y) -— (88947 - y) _ (8.8947 + y) y=y+ 2 — Q = ÿ# = 19.1151 - x h c- YO _ 3509.1157 ~ Ý) _ 1 586866 - 0.020575»? ề It 872.49 (2) § “| V=|tdA dA=2dy i V = | (1.586866 — 0.0200575 y*) 2 dy = °°*"(3.173732 — 0.040115 y*) dy 28947 = 9.96kip Ans

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“7-12 The strut is subjected to a vertical shear of V =

130 KN Plot the intensity of the shear-stress distribution act-

ing over the cross-sectional area, and compute the resultant shear force developed in the vertical segment AB 1 1 T= 75 (0:05000.35°) +S (0.3)(0.05°) = 0.18177083(10?) m° Œc = y/Aˆ = (0.10.05X0.15) = 0.75(10?) m? Ớp = 3A“ = (0.1)(0.0510.15) + (0.0125)(0.35)(0.025) = 0.859375(10 3) m° 19 fer 130(10?⁄0.75)(10) 0.18177083(10-3)(0.05) =7 fe 20-0125 m (Toh = 005m = = 10.7 MPa 130(10°)(0.7 “3 (tc) = 035m = 30000.75)007) = 1.53 MPa 0.18177083(10-3)(0.35) - 130(10?)0.859375)(103) = 1.76 MP 0.18177083(10-3)(0.35) a A 107 MP, N 7 7 A’ = (0.05)(0.175 — y) =, = y+ ——— = - (0.175 (0.175 -y) 1 y =ỹy 5 5 ( +ờ) Q =A’ = 0.025 (0.030625 - y) A’ 0-115 — 7 - It N í A _ 130(0.025)(0.030625 - y?) 0.18177083(10-3)(0.05) u 10951.3 — 357593.1 yŸ v=frdd dA = 005d = f°" (10951.3 — 357593.1)7)0.05 d 0.025 .Iy X0.05 &y) = [”"” (541.565 — 17879.66y? 0.025 , 6y) dy = 50.3kN Ans

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7-13 The steel rod has a radius of 1.25 in If it is subjected toa shear of V = 5kip, determine the maximum shear stress ,_ 4r 4125) _ 5 y=— 3z =— 3z 1 1 I = <ar' = —n(1.25)' = 0.610351 2 4 4 7/ 4 - 5 (1.257) " ‘ Q=y'A' = > —— = 1.3020833 in” VQ — 5(10°)(1.3020833) Tmax = — = ————————————— Đ~ 135 1 = i

Tt 0.610351 (a) (2.50) Spsi = 1.36ksi_ Ảng

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7-14 Determine the largest shear force V that the mem-

ber can sustain if the allowable shear stress is Tajow = 8 Ksi Jin, lin y= (659)G) + 2{0)2)Ì _ 5 1667 in 1(5) + 22) Is s90) + 5 (1)(1.1667 - 0.5)" 420-0665 ° +2(SKD2”) + 2()(2)(2— 1.1667) = 6.75 in’ Quax = LA’ = 2 (0.91665)(1.8333)(1) = 3.3611 in’ ne tn = YOms maa = tallow = Tt 8 (10?) = V(3.3611) 6.75 (2)() V = 32132 1b = 32lkp Ans

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7-15 If the applied shear force V = 18 kip, determine the maximum shear stress in the member L f= mb E Sun, lh f5S§in (0.5)(1) 5) + 21Q(D2) _ 1 1667 in, 1(5) + 2(1)2) l= 009 + 5(1.1667~0.5) y= +2 (Gre?) + 2()2)2-1.1667) = 6.75 in® Qmax = EA" = 2(0.91665\1.8333X1) = 3.3611 in? VOms _ 183.3611) _ 4 ag isi —— = Ans It 6.75 (2)() Tosx =

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*7-16 The beam has a square cross section and is made of wood having an allowable shear stress of tanow = 1.4 ksi If it is subjected to a shear of V = 1.5 kip, determine the smail-

est dimension a of its sides I VE +4 ep I= 4! 12 0 =7 = OOa = % mx © 4.2.8 Tmax = Tallow = VOmex It _ 15@) s(a®() a = 1.27in Ans

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7-17 The wood beam has an allowable shear stress of Tallow = 7 MPa Determine the maximum shear force V that can be applied to the cross section 1 3 1 3 - ï= —(0.2/0.27 ——(0.1)(0.1) rat 10.2) De 1)(0.1)" = 12510") m = 5) m4 + VOmax allow = it 7(10°) = VL.(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)] 125(105)(0:1) V=100kKN = Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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7-18 The beam is made from a polymer and is subjected to a shear of V = 7 kip Determine the maximum shear stress in the beam and pilot the shear-stress distribution over the crass section Report the values of the shear stress every

0.5 in of beam depth k 4in oo - 1 in ‡ in I= yay +2 = +4(1)(2.5)"] = 56 in’ vo - 12 75)(4)(0.5) i= sa ` 0.172 ksi Y0 _ 70.54) t= =a 0.3125 ksi - 2 - 72.540) = 1.25 ksi It 56(1) , ne _ T[@.54@()+(175)110.5] T t 56(1) = vo - 1G 5)4@()+(1.59)0)] Tt 56(1) 2.5)(4 tạ = <2 - TL 2.5)(4)(1) + (1.25)(1)(.5) MN — 148 ksi 5Œ) 71@.5X4 _ L250) + OQ] _ 50 45; 56(1) = 1.36 ksi = 1.44 ksi 8272 441 >4 2.712 VY ker fmm n+ = A Wig! Ans

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7-19 Plot the shear-stress distribution over the cross section of a rod that has a radius c By what factor is the g maximum shear stress greater than the average shear

stress acting over the cross section? - ẤY /.+.„2 vang pe Ee 9 a lực 1) ~ , 4 t=2x= 2Vc? a : a dA = 2xdy = 2c? - ¥ dy J sac dQ = ydA = 2y/c? - ¥ dy O= falta Fay =~ - } = He - HF tr“ =— ———— = z( -Y) It (§c)(/c—y) 3£ The maximum shear stress occur when y = 0 £ _ AV mex One ¬ V tH TT“ sa 4V The factor = Im „ 21 „ 2ˆ Ans Tavg 3

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*7-20 Develop an expression for the average vertical com- ponent of shear stress acting on the horizontal plane through the shaft, located a distance y from the neutral axis dA = 2xdy = 2/c? — y dy Q= Jy a= fix /e -ydy= se ~ Fi Jy 7 HA I= fet, t = 2(c” ~ y)? 4 x 49 7= cŸ x c.~ V0 MO? - )”] c It £c4(2)(c? _ yy? 2 = 4V(c° - ý) Ans 3zc+ Also, 2e sia 86 ¢? — 2 ccos Ø (7)(2c sin 8)(c cos 8) j=€c <as© y= a 2c sin Ø8 — 2c sin Ø cos?Ø 3A’ 0 =yA = 50 sin @(1 — cos’@) = 3° sin? 6 L4 l=-2%c; 4 t = 2csin 8 VỌ _ VỆỆc`sin6) _ 4Vsin? 6 It 4 xc*(2c sin 6) 3c? 2 sing = ¥£ y c 22 2 Therefore, 7 = _4V c =ý _ Ae? - ý) Ans mcr c? 3z c4

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7-21 A member has a cross section in the form of an equi- lateral triangle If it is subjected to a shear force V, determine the maximum average shear stress in the member Can the shear formula be used to predict this value? Explain 1 3 I= —(a)h 2e 600) _ h * ya 2h, — a/2" a a xe 1 2, 2 = =2[(- “h—“ 9 fo dA AON 3)! 2 2x o=(2 31-25 3a a t=2x vo _ W(4h? 13a)? )(1 — 2) " ((1/36)(a)(h*)) (2x) _ 24Vax- 37) ~ a?h at — 1 - „=0 dx a?h? Atx=“ 4 2h a h y= 7a =5 24V a Tmax = a- TC 2) 1-2 Tmax x Ans ah No, because the shear stress is not perpendicular to the boundary See Sec 7 - 3

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7-22 The beam is subjected to a uniform load w Determine ¬ * the placement a of the supports so that the shear stress in J J J J | IRRNNI | ty _

the beam is as small as possible What is this stress? _ B Ta _- —-= b4 —a——| —a—— b L Require, wn L {Td w( TT 4) = Wa ar i a L wl wh a=— Ans + ¿+ 4 ị per & _- ha V=wa * —- ven i i ! "` tr |[bŒ(@jĐ) — $bả BW th "` From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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7-23 The timber beam is to be notched at its ends as shown If it is to support the loading shown, determine the smallest

depth d of the beam at the notch if the allowable shear stress

iS Tatow = 450 psi The beam has a width of 8 in 2500 tb | ° 2500 tb 25091) 9¿odb 250044 12 in, Saal — | Suz2-4 402240 V = 5000 Ib Si ‘A vo 5000(d/4)(d/2)(8) sm th =—; 450 = A he 1z(8)(2)°(8) — ode, d=2.08in Ans

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*7-24 The beam is made from three boards glued together at the seams A and 2 If it is subjected to the loading shown, determine the shear stress developed in the glued joints at section a-a The supports at C and D exert only vertical

reactions on the beam \° hp ` Shap ?Sktp I= +(@a1) - Awe’) = 494.83 12 12 in" 15 ;.3 = Op = A = $261.5) = 42.75 in QO, = 0s = ŸA 4 + NON 2.5(10°)(42.75) 108 psi A et = oe psi ns T = TB = “194 83 (2)

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7-25 The beam is made from three boards glued together at the seams A and B If it is subjected to the loading shown, determine the maximum shear stress developed in the glued joints The supports at C and D exert only vertical reactions on the beam

Skip Skip Skip

Wnax = 7.5 kip (at C or D) i 3 1 3 „4 †= —(6X117 -—(4)(8)' = 494.83 mĩ X10) lạt )(8) in On = Op =y'A'= 4+ 42)6)(15) = 42.75 in? 3 TẠ = Tp = ve = 2540 04225 =324psi Ans 1 494.83(2)

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7-26 The beam is made from three boards glued together at the seams A and B If it is subjected to the

loading shown, determine the maximum vertical shear

force resisted by the top flange of the beam The supports at C and D exert only vertical reactions on the beam

Sup Skip Skip LS Shit Vues = 7.5 kip (at C or D) 1 3 1 3 „4 = ——(4)() =494.83in I 206901) 1° (8) F =| tdA t= ve = 7.5(10°)(5.5 — y)(6)[(5.5 + y)/2] = 7.57836(30.25-y") it 494.83(6) R= |} 151836025 ~#`64y) 55 = 45.4702(30.25y— ; yi =512tb Am From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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7-27 Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent tothe maximum shear stress Comment on the validity of your/re- sults — 4 4 L bb Vhax = P Myax = PL Mc _ PL(hl2) _ PLh Onax = I FT Eo I 2I _ VO _ P(!2Œ\(h/4) _ Phˆ man Ib 81 Require, Onax = Tmax PLh _ Pit 2ï 8i L=" 4 Ans

Shear stress is important only for very short beams Note also, that this result is not all that accurate since Saint- Venant's Principle must be considered

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*7-28 Railroad ties must be designed to resist large shear loadings If the lie is subjected to the 34-kip rail loadings and an assumed uniformly distributed ground reaction, deter-

mine the intensity w for equilibrium, and compute the max-

imum shear stress in the tie at section a—a, which is located just to the left of the rail 34 kip 34 kip Lis a 38 +TIR =0; 6w-2(34)=0 4 „ w= 113kipft Ans | 1 ï = 16) = 144i Peet Qurx =y'A’ = 1.5 (3)(8) = 36 in? VOma 17(10°)(36 Tmax = ore _ 1/00 ) = 531 psi Ans it 144 (8)

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7-29 Determine the shear stress at point B on the web of the cantilevered strut at section a-a Zk | an M € Va & KM ¬% am| # 4 _ ƒ +3e (0.01)(0.05)(0.02) + (0.055)(0.07)(0.02) (0.05) (0.02) + (0.07)(0.02) y = = 0.03625 m Anam 50mm = = (0.05)(0.02") + (0.05) (0.02) (0.03625 — 0.01)” + = (0.02)(0.07") + (0.02)(0.07)(0.055 — 0.03625)? = 1.78625(10 5) mf ¥p = 0.03625 — 0.01 = 0.02625 m Qs = (0.02)(0.05)(0.02625) = 26.25(10°°) m° VOn _ 6(10°)(26.25)(10°°) I: 1.78622(10-%)(0.02) = 4.41 MPa Ans tr =

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7-H) Determine the maximum shear stress acting at section a-a of the cantilevered strut 005375 m 0-026875m rad A (0.01)(0.05)(0.02) + (0.055)(0.07) (0.02) = 0.03625 m ye (0.05)(0.02) + (0.07)(0.02) = = (0.05)(0.02") + (0.05)(0.02)( 0.03625 — 0.01)" + 5 (0.02)(0.07°) + (0.02)(0.07)(0.055—0.03625)” = 1.78625(10 5) m' Qmax = yA’ = (0.026875) (0.05375)(0.02) = 28.8906(107”) m” _ VO max Tmax ~ 6(102)(28.8906) (107°) It 1.78625( 107%) (0.02) = 4.85 MPa Ans

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7-31 The composite beam is constructed from wood and reinforced with a steel strap Use the method of Sec 6.6 and compute the maximum shear stress in the beam when it is subjected to a vertical shear of V = 50 KN, Take Ey = 200 GPa, Ey = 15 GPa (oO mm 15 = nh, = -ÌÃ(0.175) = 0013125 m by = mby = 505 (0-175) l= 1a(0.175)(0.32”) ~ (0.175 -0.013125)(03") = 0.113648(10”) m‘ Qmax = Ey’A' = 0.075(0.013125)(0.15) + 0.155(0.175)(0.01) = 0.4189(10”) m’ VOmar _ ( AS ) §0( 10°)(0.4189)( 1077) it 200° 0.113648( 107) (0.013125) 1.05 MPa Ans Tmax = 2 |

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Because of the number and variety of potential correct solutions to this problem, no solution is being given

Pearson Education, Inc, Upper Saddle River, NJ Allrights reserved

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Because of the number and variety of potential correct solutions to this problem, no solution is being given

Pearson Education, Inc, Upper Saddle River, NJ Allrights reserved

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7-34 The beam has a rectangular cross section and is subjected to a load P that is just large enough to develop a fully plastic moment M, = PL at the fixed support If the material is elastic-plastic, then at a distance x < L the moment M = Px creates a region of plastic yielding with an associated elastic core having a height 2 y’ This situation has been described by Eq 6-30 and the moment M is distributed over the cross section as shown in Fig, 6-S4e Prove that the maximum shear stress developed in the beam is given by Tyax = 3(P/A’), where A’ = 2 y’b, the cross- sectional area of the elastic core

Force Equilibrium : The shaded area indicates the plastic zone

Isolate an element in the plastic zone and write the equation of equilibrium : 5 ‘| max SIE = 0; TasgÁ¿ + ƠyẤ yt =0 Ơ Tiong =0 ơ

This proves that the longitudinal shear stress, Tiong + is equal to zero

Hence the corresponding transverse SUSS fire» 16 8S equal zero in the plastic zone Therefore, the shear force V = P is carried Maximum Shear Stress : Applying the shear formula

by the material only in the elastic zone Von v( rs) - sp

Tox = ary - 4by h ¿by? (b y

Section Properties : (‡ey?) œ)

1 2 However, A’=2by’ hence

Iya = VY = 3b”

, y2b _ 3P

Quer 2A’ =F 000) = fạ„ =2; - (Ø-E.D.)

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7-35 The beam in Fig 6-54fis subjected to a fully plastic moment M, Prove that the longitudinal and transverse shear stresses in the beam are zero Hint: Consider an element of the beam as shown in Fig 7—4d

Force Equilibrium ; tf a fully plastic moment acts on the ctoss Section, then an element of the material taken from the top or bottom of the cross section is subjected to the loading shown For equilibrium = + ~ LE =0; OyA + gÁ; — đyÂ, =0 —? Tiong = 0 4 Ve (2 A 2 , — Thus no shear stress is developed on the longitudinal or transverse ; Az r Oy i plane of the element (Q.E.D.) A;

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*?-36 The beam is constructed from three boards If it is subjected to a shear of V = 5 kip, determine the spacing s of the nails used to hold the top and bottom flanges to the web Each nail can support a shear force of 500 Ib - — 0.75(10)(1.5) + 7.5(12)(1) + 14.25(6)(1.5) 10(1.5) + 12(1) + 6(1.5) = 6.375 in ~ il = (10)0.5") + 10(1.5)(6.375~0.75)/” + = (1)(12?) + (1)(12)(7.5 - 6.375)? 1 + 126.5) + (1.56)(14.25— 6.375)? = 1196.4375 inỶ Ớ, = W,A = 5.625(101.5) = 84.375 in? Œ, = ÿ,A'= 7.875(6(1.5) = 70.875 in? _ VQ _ 5 (10*)(84.375) = VG _ 310°V84.375) «359 61 tb/in ur 1196.4375 V °)(70 œ„ = CÓ ~ 50000875) _ 596 19 w/in, I 1196.4375 F= qs; s= F q 5= 352.61 =142im An =-— 0Ó _160in Ans 296.19

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7-37 The beam is constructed from three boards Determine the maximum shear V that it can sustain if the allowable shear stress for the wood is Tajjoy = 400 psi

What is the required spacing s of the nails if each nail can resist a shear force of 400 Ib? 4815” - | bas” ¡32912€ * - _ 0.75(10)(1.5) + 7.5(12)(1) + 14,25(6)(1.5) y = i 10(1.5) + 12(1) + 6(1.5) 6.375 in ~ fi 71-5?) + 10(1.5)(6.375—0.75 + 1 (112) + (112)0.5— 6.375)? 1 1 1 +z(60-5) + (1.5(6)(14.25—6.375)? = 1196.4375 inf Qmax = L¥'A' = 5.625(10)(1.5) + 2.4375(4.875)(1) = 96.258 in? Vmax Tí V (96.258) — 1196.437) Trex = Tattow = V=497kip Ans Q, = ¥'A = 5.625(10(1.5) = 84.375 in? OQ, = ¥,'A'= 7.875(6)(1.5) = 70.875 in? _ 4:9718(10°X84.375) 11964375 = 350.62 Ib/in _ 4.9718(10° (70.875) 11964375 = 294.52 Ib/in : F 6 =— q 400 % = —— = LI4i * 350.62 Ans , 400

So = ——— = l26i * = 2oạg2 — LI6Ìh Ane

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7-38 The box beam is made from four pieces of plastic O25in

that are glued together as shown If the glue has an x a lạ

allowable strength of 400 Ib/in”, determine the maximum q :

shear the beam will support 1 1 I = —(6)(5.25°) ~ —(5,5)(4.753) = 0 )( ) 53 X ) = 23.231 in in’ “Lớn L9 | see { A Qn =y’A’ = 2.5(6)(0.25) = 3.75 in? it The beam will fail at the glue joint for board B since Q is amaximum for this board 25im n sin a Sin tow = 2: ao (G27 - mein It 23.231(2)(0.25) V = 1239 Ib = 1.24 kip Ans

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