Solutions-Set 02 (1)

# Solutions-Set 02 (1) - l—Sii The arteher belt was pulled...

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Unformatted text preview: l—Sii. The arteher belt was pulled eut ef the eenerete wall and the failure surface termed part el’ a [mature and cylinder. This indicates a shear iaiILLre deeuITed aleng the cylinder BC and tensiert failure aleng the frusturn AB. II the shear and natural stresses aleng these 5L1 riaees have the magnitudea shewn. determine the force 1" that must have heert applied In the belt. Average Normal Sit-em: For the frusturrl. A = 2:1 = 2r{ﬂ.[l25 + may: '-._- ere-*- + ere!) = [11.12221 I'll] F1 [1.[12221 :r = —; 3[1[1“} = £1 = eemm Average Shear Stress: Per the ey'linder. A = ﬁ[[l.i}|{i}.[l3} = i}.[lltl4?l2 rnI _ _ L. - _L ' ' A' 4'21?) ' enema avg .F: = 21.21kN Equatlezn ei‘ Equlltnrium: +TEFF=EL P—2121—ﬁﬁ.645irt45"= P = 6-8.3 liN Ans. Ans: P = 53.3 EN 1—62. The crimping ieul is used In crimp lhe end nl' lhe lﬂEIN wire E. II a fume el III: N is applﬂed [e the handles. deten'nlne Ihe average shear stress in the pin at A_The pin is suhjeeleui in den hle shear and has a diameter el' 5 mm. Only a vertical fume 1's exeneti en lhe wire. Suppeﬂ Hamlin-M: Frern FBqua} C +EMD = n; 1Eﬂ123}—By{ﬁ}=[:l grams: 11 H. = e B; = n 5?- Frern FBquh) iZF = e: A = I: mm mm “'5 ‘ " a} 1am I; +EME = D: A}.{313}— SIHHHTJ} = [J A], = l lﬁﬁﬁ? N In“ “J Average Shear 51m Pin A is subjeeied In den hle shear. Hence I.“ [L F A VA=TA=Ty= 533.33}; Mm “ﬁlm If; -5H]le _ _ v.4. _ 33333 / {IA}nvg _ A—A _ 51:51.} = 29.?[E' MP3 ' Am“ ﬁ-Ilmm * l—ﬁﬁ. The pedestal in the shape at a fruslum {If a came '15 made tJI concrete having a speciﬁc 1Height tnl' 2400 kgan. Delermine the average annual stress acting in the pedestal at its mltlheight, z = 1.2 In. Him: The mlumeuf a tame tnl‘ radius r and height I! is V = %r.r2ﬂ. =—--_ ﬂ=T2ITI w = r [0.3151]. 5 — %{ :}{e.31}[4_s} {24th} = 1e34_31 ltg = HIS-13.1 N I +TEF,=:1; P—1te411 =n .r- = 1:13411 N a. T ﬂ‘ 3 P 10343.1 H w I m r, = _ = —I= 2343.35 Pa = 23.423m AM J" A rwjﬁ‘ﬁj ' 1'2 "" 0.3.75 ﬂ'b DASH-r l—T‘i The SU-kg “(menu-:11 ix suspended {mm wires AB and EC‘hhiCJ] have diameters et’ 1.5 mm 3.11:1 2 mm. respeuivel}: If [he wires have a nurrna] failure MISUSE el' r15“ = 351} MP3. determine IJ'Je Teeter efsai'ely‘ [ml each wire Lm L1H HPIHHHW Internal] Leading: The nerrnal [ere-e develeped in cables AB and BC can he delennined by mnsideﬁng the equilihrium ufjulnl 3. Fig.1 1—: EF‘ = [1; FE: (\$545" — Em mew" = U- {1) +TEﬁ. = [1; FM. min 3n" + Farm-ﬂ 4.” — SHEEN} = [:I {2) Sewing qu [1} and {2). F,” = 35911? N F3: = 4393?? N Menage Normal Stu-5w: The enmsrweiitma] area el wires AB and BC are A” =Ewm1if = name-F}. m1 and A3: = Emma-1} = 3.142(1t1'ﬁm1. FM 359m {r- } =— = — = EUGJQMPH 1 "a A” A” I_?ﬁ7‘{1ﬂ"*} Fm 439.7? r: = — = — = 139.98 MPa E “3}” Am 3.142(1n'5} ‘lﬂr'e email]. U'rzil 35E] _ [FE-JAR — UFHE}AH — EDS-19 — 1.12 KI“. - 350 (35.31,“- = A — — 2.31} Aux. ﬁrm)“ _ 139.93 = *l—tH. The frame is suhjeeted Ie the lead ef 4 kN 1.I.'hiI:'h acts on member All!) al f). Determine the required Ltiameler ef the pins al I) and C' if the alluwahle shear stress fer Ihe material is - - ﬁlli- MPH. Pin I: is suhjeeled ten ’1IIII'I' _ Liuuhle shear. whereas pin f) is suhjeeleri to single shear. Referring tn the FED of member Diff-j. Fig. :1. f, —EM;,- = ii: ﬂyflj] — Fm;- sin 45" {I} = I} L 21-; = n I’m-ens 45" — LL = n Referring to the FED of member A Hf). Fig. a. f, —EM3 = [1; 4:13.”:43" {3] + Fat-sin 45" {Lil - I}, [3} = [3 Solving Eqs {2] and {3}. F3: = lililllkN I}, = 1&5? khl Suhstilute the resull of Fm- inlu [1} 1}}. = 2.31%?- khl Thus. the feree acting en pin I) is h) e} + 9,3 = x 5.65?! + 2.26:3: = am] Hi Pin f.‘ is suhjeeted tn Ltuuhie shear w hile pin ﬂ is suhjeeleri In single shear. Referring Ie the FBDs [if pins C. and J'J in Fig e and 0'. respectively. J" an} j”: = T = 4mm V”. = F” = arm n: IVE: Fur pin C. vc r 4.1m: [till .- . =—: 4mm =_1 JJIIJI- AC s } rig =[i.l}l‘12ﬂ rn =1I.J mm For pin f). if” r E..[193{l[i3) .— . =—: 44 lti' =_—, JJIIJA' An N: J Edi); rig = [3.[11393m = I19 mm P1: 56.0% hi: (6%) ’ ...
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