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# HW4 - Problem 2 2.77 The plastic block shown is bonded to a...

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Unformatted text preview: Problem 2]? 2.77 The plastic block shown is bonded to a rigid su ppm: and which a 240-kN load P is applied. I” Knowing that for Lhe was - rm“ MPZL determine the deﬂection of the plate. G r 3050;:{0‘ Pa H N .9. d f = ‘- 23-3“)me | I' 1'1 3' 50mm 7 O.C)§Om g : MN {o.owd(23.2:oxzo‘“ -3 = Mﬁowo m Lemma,“r w =<H§1= 0.1m . .; emails" _ - 3.52: ' 0-05"? llh z A G ( X j Ahﬂf .. L3} _. ..+ W __, {-3594 r5 .- ( P HM; 2.2a J (b) P=123 kilos“ (62‘: Few R'ﬂje'f .D - Li .3. a] - 13 5‘ m .12 2 ‘f. D .3 d ‘55- = LGO A... ‘ 0-337580 6—H“ : KENfP Kw” r A-“ 61., : (0.932;)05) P 7‘23 :- L’f‘r’f FM 12-: am, gem-'7 EKG-'74: (o.':'?1(2.s)=' 0,425 in 4 Problem 299 2.99 The aluminum test specimen shown is subjected to two equal and opposite centric axial forces of magnitude P. (:2) Knowing that E = To GPa and a." = 200 ME, determine the maximum allowable value of P and the corresponding total elongation of the Specimen. (b) Solve part 0. assuming that the specimen has been replaced by an aluminum bar of Lhe same length and a uniform 60 X 15-mm rectangular cross section. 6“” T 200 :4le lat; E : Tango“ 13¢ A,“ = {Got-«MHISMM‘J: ﬁoo MM" : CfoOxfo" 1,“; {0A Tgsi Eggg-‘mE-ﬁi D= 75 WM“ 5?: 60mm. w:gmm "6 2* 35 I. ‘3: .6. , _ (3—60—25 5150-110 RM I“? 2.9mm : 2.95 6;“: RE . r 4-} 06} P , Afr" = (“loo JJOqSCRoow : “Slowly” P = \$1.8 km 4: .zr \ . _ f ‘4. I I, -3 2- lee area. A ‘5 (7Smul(t'5w:~‘;' I'l.{£..m-. : ..l,{5¥l0 M ii." = E E .L_~‘ : 32'305KIDBFO-ISD + 0.300 + 9,150 RE E A; "'10 no? LLIQero‘3 nooxlo" unfwo'i : 79! no" a». 5. r o_'1r=1|,MM 42:. (lgl Unli-l-ﬁor‘m ban. 9 r A 6m = Howio“)1200~<r0‘lr 180x103N P: 120.0 m A - '5 \ S= PL : l—M D HOMO“ - firww-fm 5 = WWW A E {ﬁOOxl0"i(7oxlo‘l - '2 a n . . i bl 3 7 3. 7 The solid spindle AB is made of a steel with an allowable shearing slress of Pro em ' 12 ksi, and sleeve CD is made of a brass with an allowable shearing stress of T ksi. Determine (a) the largest torque T that can be applied at A if the allowable shearing stress is not to be exceeded in sleeve CD, (b) we corresponding required value of the diameter a", of spindle AB. 2. ._. TC: Tin-u» J J— . n \ na‘ '-" C1 :‘ W : H.113 k\P0LV\ FM? Equipulgr‘ﬁw) T: kﬁo-WH Id Said?! Gar-Hallie :i T = l9.2l3 kid-hf! —_- 13-; = 327' u“ 'm:3 C: 33;- = 3/(mc1q.213) : Loom in. T11 T101) 015: 2:: = (210.0064) a5=2.oi ;n.--a Problem 3_1 1 3.11 Under normal operating conditions, the electric motor exerts a lorque of 2.8 W - m on shaft AB. Knowing that each shah is solid, determine the maximum _ shearing slress in (a) shaft AB, (b) shall BC. (c) shaft CD. TH = 1.4 k.\' - Ill '1" = [Ml L'X v l'.| 113: IF— : 2_T = gunman _ 69 T whomx‘ ‘ 31-20%? J» ’l’nfSLQMPadi! (m SMH 8:: Tea- m w-M = HUM; Mum Dig-9!: 24“: 0.02% 21‘ , (210.4qu Ta; = we“ ‘ it (0.0%): G437XPOLPG Tea: 3‘4-5HP'1 4‘ 03‘ ﬂail CD: EH 05mm T 05*“?! N-M, vial- Hm: 0.024 m - 2T {ZWOJKIDE} 4:19- EE» ‘ T Emmi Pa ﬁo=?3.0M?« ma Problem 5 3.15 The allowable shearing Stress is 15 ksi in the steel rod AB and 8 ksi in the brass rod BC. Knowing that a torque of magnitude T= IO kjp‘ in is applied 31A, determine the required diameter of (a) rod AB, (b) rod BC. _ c: _1 3._ 2T {WV—jg J"2u C*m’.., T: [O kip-in 2L“: k5: C = 0.75:5“ d = 26. = LSD?) in. .4 Sim‘F‘i‘ 8C! T: {O Tim“ 2 8 Mai o3=—L—\$)CS) = 0.7057? m” 6.2 0.0126? in d 2c = LSS3 in. 4 Problem 325 3.25 The two solid shaﬁs are connected by gears as shown and are made of a steel for which the allowable shearing stress is TODD psi. Knowing the diameters of the two shafts are, respectively. dag = 1.6 in. and d5; = 1.25 in. determine the largest torque TC that can be applied at C. 11,.“ -= 7000 F53 2 7.0 ks; S‘na‘FT' BC: alga: M; in. c =§al= 0.81m Tc = J:““ : lg’fwcf‘ ’ 1,77 £903.81“ = 5.63 up m SLaH—EF c433? 1.25m c=%¢i: DéEFw- T: T J?“ = E 1142‘“ 8;; ale-lich T : -TF = i {2.68% r 430 lurks” MPOWLLJP’Q white 0-]: T; is +512 5?Aé/y?/, ...
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HW4 - Problem 2 2.77 The plastic block shown is bonded to a...

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