Exam 3 Solutions

Exam 3 Solutions - PROBLEM 5.71 Owl m 0”] m 0.125 m...

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Unformatted text preview: PROBLEM 5.71 Owl m 0”] m 0.125 m -22L2$ N- In *5.71 Beam AB supports a xmifomfly distn'buted load of’Z kN/m and have concentrated loads P and Q” It has been experimentally determined that the normal stress due to bending on the bottom edge of the beam is —569 MPa at A and -—299 MPa at C Draw the shear and bending—moment diagrams for the beam and determine the magnitudes of the loads P and Q. SOLUTION 4 I = r‘a(\8)(scl‘ Gmsux/o‘m 'I s = ET- : 3.883on mg: 3.888x166m2 A+ A M. = 6" MA=(3.888XIO‘°)(‘5é.°1)=-221-25 N-m M c Mg: 56’c M5 = (afisleo‘CX—Z‘H): -ne.25 N~ m 9 2M. = 0 221.23 — (ammo) — 0.2 P - 0.325Q = o 0.11” + 0.325 G. —- I3/.ZS m +DML = 0 H625 ‘(o.orl(2ool- O.I P - 0.22312 = O.IP+ 0.22561 r 106-25 (2) O Sofivina (l) Avml (23 simuj'i‘avxeousjj P= 500 N: 4-: Q: 250 N -‘ gear/hon "Force a)" A PA— 400-- 500 — 250 = 0 RA -: I350 N'M VA = “50 N V9: 250 MA: 422.25 Nam M‘f- llé.2§ N-m MD 2 ‘ N‘v" 5.97 Determine the largest permissible uniformly distributed load w for the shown, knowing that the allowable normal stress is +12 ksi in tension and - 19,5 ksi in compression” PROBLEM 5.97 SOLUTION Qeaol‘ious: Q +C-36w=0 B=C=lgaw Shear: VA : o Vg‘: 0— SW = ' 8W VB+= —8w+1$w= IOw Va": 10w - 20w =—H>W Va ‘: -’.!C>w + L8 w = 8w V9: 8w — 3w = O AV‘EQS T A 1‘» B "(-&)(2}é§8wl r—32 w 8 +«» E (#)(Io)(to w) : 50 w Benohnfi mome'l‘S? MA = 0 Mg : O ‘ 1’ ‘32W M53—32W'f50vl 1' I8w Cen‘f‘woi'cl and momen‘l‘ o‘P [me/4‘42; D._7£_I 0-9492 O.7l_lmq>_ 2,5294 n," Bendinfi momen+ 19ml}: M f - (VI/j Ten$:'on nit 8 duel C - (l23( ‘fl_‘3706) 2 - RI'P- {h COMP. ad 8 va. c ‘ (415w- 1-4343 3 = — 17. €144 tar," in <\ Tension d E -(12Y— Limits): H.212 kip-l in <1 Compression at} E - (-17.5 )(2-3906 )3 44.6 ‘0"... iv; AWwaMe foul w B F c: — 32w : — 27, 4m W: 0.87% top /."n E l8 W: [7. 2’2 W: 0.754 /fn Smal’les'l’ w .- 0379 lap A». r to.“ («P IH. «I 5.99 Beams AB, BC, and CD have the cross section shown and are pin-connected at B PROBLEM 5-99 and C Knowing that the allowable normal stress is +110 MRI in tension and e 150 MPa in compression, determine (a) the largest'permissible value of’w if beam BC is not to be overstressed, (b) the corresponding maximum distance a for which the cantilever beams AB and CD are not overstressed 125 mm r290 mm—v-I i SOLUTION T MB 2 ML 3 O g___i Vs = - Vc. = (%)(7.z.)w= as w Area. 8 +0 E 0‘? shear Amara“ $8-08-: w) = 6.42 w ME: O+6J42Hu r 6-48mi Cen‘l’roicl anal momew+ o‘F {ner‘l’iq 3‘10625 344.82 140625 HC.‘!3 9.043;); 3.5l6xlo‘ 53:25:: 7.073 xzo‘ 3.5tasrb‘ = I2L'-I3 mm _ 534250 4375‘ I = EAJ‘ + 23:" r we. “0‘ m: Leea‘l‘iou .Yé'm) I/ (fir-3x6? afléo (Io"m3) 4”? LO‘H’OM Bending momen‘l‘ Anti“: Mt - SI/y Tension «+2; i, - (How/ofW—37Azmo‘é) s: 1422403 Mu. Wail E3 -— (-lSoxlo")(z$8-éx107‘») =.—‘ 38. Sat/03 Mm Tagabu A MD: —(.Hoxlo‘) (253.5160 242.4910" Nu Comp.¢+ A41): - (-Isarlo‘X—Szwxlo‘fl:flamxlo’ N- = 1.420103 N/M = 1.48: kN/m «- Alpavagje In“, W 6.48%! '-‘ iéaxlcs W {Shear 3:} A LV‘; = (a.+'3.é) W Area A +1: 13 of? shear disarm iaG/A + VB) 1‘ %a(a.+7.2)w Bending momewl A (415.: D) M, a(a+"7.2.) W 7inta+22mthsnafi = - |3,|:u x103 7; o.‘ 4. 3.6 a. e— 9‘83? = o a.’= L635 m 4 5.37 and 5.38 Draw the‘shear and bending-moment diagams for the beam and loading shown and determine the maximum normal stress due to bending 300 20 mm SOLUTION c Al D chc boob, EFGH 30mm Nah «wa ME = 0 Jun 7 4» king. QEME = O 0.: H-(o-2).C‘+o?-(O.4))(3oo) = O H = 213.33 N .1sz : o vg-4o~—-300+2:3-33=0 vE = I24 -€7 N Skew: E +0 F v = 124.57 u-m F b G V = 86.67 N-m 6+0 H V =-2I3.83 N-m Banal-M3 mud a F E Mr :0 D M; -(O.2)ClZé.€7) = o V Mp r N-m was? Bending! Momen+ on" G 7 M; .. DEMG=O x (a - M; + (0.2)(2I3..33) = 0 M5 = N-m 28.3% |2€.G7 Frye bed] A BC 0 E +9? Mta = o —o_e A +(o.4 )(3ao) +(O.2)(3oo) -(0.1)(I2‘c.c1) 2 o A: 257.7? N 4' M, = O ~(O.2)(3oo\ —» (0-4)(300) -(o-z)(12(..57) + 0.629 = O D = 468.8? N A 3 MB Bending Momen‘f 03‘ 8 fl 92MB=O Muir/I“: 5!-5C N-m —Co.2)(257.78) + We '—‘ 0 “5733" M3 = 51.56 MM 5 = gbb" = 2"(20‘X30)z Boo = 3 “03 rams = 3KIO-‘ M3 A B c Bending meme/21‘s.! C _ 1 +92 r)? = 0) Normal) shes: —(o.4 257.72 +(0.2)(3oo) 1577" V + Me, = 0 Stfla— = III“: ><lo‘ Pa. M; r 42.“ Mm 3““) : I7.I°r MP0. < MD Benchnj momen‘} a} D C D 5 +3 EH, -, o _ T ‘- M9 ~(O.2)(1l3.33)= 0 V Mo = — 25.33 N-m 5.97 Determipe the largest permissible uniformly distributed load w for the PROBLEM 5.97 shown, knowmg that the allowable normal stress is +12 ksi in tension and ~19,5 ksi in compression“ SOLUTION Wax/Hons: B+C-3éw=0 B=C=l§,w VA: 0 Vg-r O~ 8w = #8“! VB+= -8w+13~w—‘ 10w V5: jaw—20w=-10W Vc+=410w+ 13w =8w VD: 8w—8w‘: 0 Areas? A 1‘. B "(-%)(2)&8wl r—32 w B +0 E (MIon w) :50 w Bending M0m9w+s 3 MA = 0 MB: r ‘32“) ME ~‘ —35Lw +50 w = I8w Cen+v~o£cl anal momen+ J‘iner-h‘a Bendinrj MOM€A+ firm}: Ten-Jon «f 8 wt C ~ (my 3906) = — 22.;37 mp. {n C9mp- of} B and C - («FLSY— 1.4393] 2 " 17. 749 tarp in 4 Tension ~+ E —(12Y— 1.4393): 17.212 id,» in <\ Compression a} E -(~H.5 )(2.3q06 )2 44,6 klr. in Mooth Lao! w B as C — 32w = - 2?. 449 w: 0.27% k7, A». E W 7 I7. w t O. /[M Smfles‘r w = 0.879 lap/n. r 10m up /H. «a CEN 302 — Mechanics of Materials Section Exam #2-A 30w 710K352. Name ................................................................................................. .. Problem 1. (30 points} The concrete post in Figure (a) is reinforced axially with four symmetrically placed bars, each with a cross sectional area of lSOOmmz. The moduli of elasticity are 150GPa for steel and 15 GPa for concrete. (a) Compute the stress in each material when thefil 000 kN load is applied. (10 points), (b) If the initial length of the column is L = 10 fi, how much does the column shorten when the load is applied? (.5 points) (0) If the allowable stress in the post is IOOMPa for steel and 4MPa for concrete, compute the maximum safe axial load P that may be applied. (5 points) Cw Wje (d) If instead of using steel and concrete we only used concrete determine the stress inWand the final length of the column (after the application of the load).(5 points) (e) What are the longitudinal and lateral strains in this case? What are the new cross—sectional dimensions (average value)? Assume that the Poisson’s Ratio is v = -0.33. (5 points) PCor-b n. at” o Mk; E4343 M A}? 31:23:31“ 2 Generate. ---------- u“ mm 7333mm 000 “$13 Sammnam Cid-5mg jam qooo \ {a v : 58%qlo‘ggflm , W / fl/ 5’ ’ 6,00a x 529 (C) 55% g E— -— #X 147/ (ywa : 33%. 29 100W 5 P ' ggs’ylé‘g :14 W M (a w a 4 » C 30 W" ngg ,— 5.36/6N «x» P5; : 359w; Wrow < 60000 on: o w to 494w? «2m Era=336+wo : 255m a J 6000” A N : 300x30Q—5 $0 5% £60 wk Xi‘okgm : 000:”; ; gnaggggfl ’75; W Labia? m éizw: .. (mamas? j W :D fl 500nm”: (.0 meat?) JM 0 Q73 Mn? 5 E,“ M X, floggm/ M'menjz’mm ’O [Egggggjfiga mg m f mm; mmh-memmmmeM/ “mu...” mm“ fiWngv. w. w" , ‘m. , ,wwmmwl M Problem 4 (15 points) A) Derive the expressions for the shear force and bending moment for each segment of the beam (1) and (3) (7.5 for each). Express shear and moment functions in terms of X. For segment 1, x should be taken as shown below (starting at the left support), While x in the segment 2 should be taken from the right support. 400 lb/ft 200 lb/ft / 50% 0‘7 7L 200% 2.. ‘Mflx *QOUX, Mi—f- ISD Q Problem 2 (35 Points) The box beam shown in the figure below supports the concentrated loads 8001b and 20001b. Calculate the shear and moment diagrams (10 points each). Compute the maximum tensile and maximum compressive stress due to the bending moment (10 points). Show the stress distribution in the section shown below (5 points). 6in 8001b 20001b <—————» [V] ~600‘ [M1 Section — Draw Stress ‘ Distribution Co wright 020905 it 4 2000x4x€ I {jinx-xfldg/i) 5’ 2' :43‘qi‘zl’fi‘ réxfg-j‘” «M33 {gm/1’37?! /L [2, \\ H b lé— ciOmm—av , k Fg L :QOmm g v 40mm % 4‘— G R GUM ‘te‘l MW‘R'WM COMPNS‘SWQ m4 4€n§31¢ SHESWS- Mm: ...
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Exam 3 Solutions - PROBLEM 5.71 Owl m 0”] m 0.125 m...

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