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# hw5-10sol - \N*5 — soumoNS‘ 3-2 Data taken from a...

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Unformatted text preview: \.\\N *5 — soumoNS‘ 3-2. Data taken from a stress—strain test for a ceramic are given in the table. The curve is linear between the origin and ‘ (in/in) the ﬁrst point. Plot the diagram, and determine the modulus 0 of elasticity and the modulus of resilience. 0.0006. 0.0010 Modulus of Elasticity: From the stress—strain diagram - . 0.0022 33.2 — O . . I E — — 55.3(103) kSI Ans. Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stress—strain diagram (shown shaded). u, = -:-(33.2)(103)<i1n—bz)(0.0006 = 9.96 "3;" Ans. ‘— Clﬂlo T"; 35.2. Emu/m] onto: one» “or: w mm: 3—3. Data taken from a stress—strain test for a ceramic are given in the table. The curve is linear between the origin and the ﬁrst point. Plot the diagram, and determine approximately the modulus of toughness. The rupture stress is a, = 53.4 ksi. Modulus of Toughness: The'modulus of toughness is equal to the area under the stress—strain diagram (shown shaded). 1n lb in. + 45.5(1-o3)<ifrl—2>(0.0012)(E) 1 . lb in. + 5(7.90)(103)(F>(0.0012)(i;)_ 1 ‘ lb ' . + 5(12.3)(103)(i—;2—)(0.0004)<:—:) (14,)“,me = %(33.2)(1o3)(7132>(0.0004 + 0.0010)(%) in =85.0i“'3lb L ’- <65.0?<,I Ans. 1‘3 Ram 3—10 mm) _J(_ assm GWEN“. AT Q=5°°Lb;\ L=\‘llu 4:04;“ ww DWI/9 Funk Y‘:SO° Lb ‘\'o V1=l<g00w‘ Hth E &M\UH.:'~N1 LW Mi Mimi L; 0* a cg v3=1300 w (Msumlqj ‘AMAY I—u‘hﬁﬂ War) scum”): A“: véxizwkoégy: o'msgga‘okz @ VF 500L151 (5—; : z 332\$“,L-‘L 7— : - _ P1 _ \‘toous e ‘ q . @ 131’ “600%: G"). ' 'A: ' b'nSb‘o “A: —' \4'314 (19. - \4314 lay 0,0 CK"; ‘ we” ’9’500‘0 M47 Qﬂ‘woua: ta-e‘) ~'— '93: ‘~ T2“. =o,ooo‘(5% ¢ “4 01-01 , .314Lﬁ~’6,‘\1q - I “VODSLMM; E 1 ': L__.__.___v\__._\(:_ .; \u" Gire‘ ‘ 0.00015 ‘ q». E = \5\‘600 6"; 1)» - “00¢ w ‘ ‘ 2 . O,\1SBLI\QL \ _ , 0341 _ 11.13%; 44.314 L4 WA!“ \M'. (TL) ’ E - = o. 000511 “A W» V = Who «J Raucous; AL: L¢ Le} 43 = \’L.; (0 1300577 = 0.00 may L; = \Th 7‘ 0.00% + gown : \1.0\5°\1.\ 3-14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire 80. If the wire has a diameter of 0.25 in., determine how much it stretches when a load of P = 600 lb acts on the pipe. Here, we are only interested in determining the force in wire BD. Referring 4ft I to the FBD in Fig a §+2MA = o;_ FBD(§)(3) — 600(6) = o FED = 1500 lb The normal stress developed in the wire is FED _ 1500 =-—— = .6 ‘= . ' a'BD ABD %(0252) 305 (103)p51 3056k51 Since 030 < 0-,, = 36 ksi, Hooke’s Law can be applied to determine the strain in the wire. O'BD = E630; = 630 = 1.054(10‘3) in./in. The unstretched length of the wire is LBD = V32 + 42 = 5ft = 60 in. Thus, the wire stretches - 5313 = 630 LED = = 0.0632 in PE» 101 Stress-Strain Diagrams Modulus of Elasticity The ﬁgure below plots four stress-strain curves for four different types of material. A Rank these situations, from greatest to least, on the basis of the modulus of elasticity. Q Q Q Q Q 0 col.— 4;»— oou.» ton—- OOIUI as. com 0 Q Greatest 1 B 2 C " 3 A 4 :9 Least ' ' Or, the modulus of elasticity is the same for every material. Please carefully explain your reasoning. A 3/156 2/ 0" E5: 7:; 3 E: =39?) How sure are you of your ranking? (Circle one) Basically Guessed. Sure Very Sure = 1 , 2 3 4 5 ' 6 7 - I 8 9 10 23 Stress-Strain Diagrams Modulus of Resilience The ﬁgure below plots four stress-strain curves for four diﬁ‘erent types of material. ' Rank these situations, from greatest to least, on the basis of the modulus of resilience. QAQQ Q Q con—I m— oou.» NlI-I 0010- Alt» OON Q Q Q Greatest 1 A 2 ID 3 ’B 4 Least ) Or, the modulus of resilience is the same for every material. Please carefully‘explajn your reasoning. WAANS VW\W.: (:yu. LW V75» a; “up Um Kigali: 6) = a ‘6 - ‘3 15 \. - — 6'6 \ I av . ' Urf a LKWXY 4 W How sure are you of your ranking? (Circle one) Basically Guessed . Sure Very Sure 1 2' 3 4 5 3'6‘ 7 8 9 10 24 ...
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hw5-10sol - \N*5 — soumoNS‘ 3-2 Data taken from a...

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