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HW 5 - .PR EM.7 The prismatic bar is made of material with...

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Unformatted text preview: :..PR EM . .7 ', The prismatic bar is made of material with modulus of elasticity _-.E = 28 x 108 psi and coefficient of thermal expansion _ -.:0= 8 x 10“ “F1. The temperature of the unconstrained bar . 3 “‘- I'is increased by 50°F above its initial temperature T. What is E '" — ’ - the change in the bar's length? (b) What is the change in the i _s__ Q bar‘s diameter? (c) What is the normal stress on a plane per- I M . ' endicular to the bar’s axis after the increase in temperature? l 13 “1' W STRATEGY — To answer part (c), obtain a free-body diagram "by passing a plane through the bar. SOLUTION: _ ‘ The new (heated) length of the bar will be: (5) Al. (a) (AT) m (15 in.) (5 x 105 5F") (50°F) gyms L' = 0.006 in. (b) The change in the diameter of the bar is: I AD = D (0) (AT) = (2 in.) (s x 106 °F") (50°F) _ ANS AD = 0.0008 in. ” -(0) Because the bar is unconstrained, no forces are L—E 5 (m r exerted on the ends of the bar. The normaE stress r———-——; <._ _‘- on EVERY plane passed through the bar is ' 5w gN -1. l4; -. ANS o t 0 ‘ I ' " ' ass 1 Suppose that the temperature of the unconstrained bar in 9 Problem 3.5.7 is increased by 50°F above its initial temperature m__- » ' Tand the bar is also subjected to 30,000—ib tensile axiai forces E {O t i l J in. _- at the ends. What is the resulting change in the bar's length? fDeterrnine the change iniength assuming that (a) the temper» , b m____.,_: ature us first increased and then the anal forces are applied; 0) the axial forces are applied first and then the temperature i is increased. (SOLUTION: = Considering first the case in which the temperature changes first, then the load is applier: t _ :7 (ALh- = L 01 (AT) = (15 in.) (a x10*3 DF") (50°F) = 0.005 in. PL (30,000!b)(15.006m) AL) =—=-—.-~——~—--—=0.005tt'n. ( F AE mm)2(28x1051.5rm2) Total change in length in this scenario is: At. = 0.006 in. + 0.0051 in. ANS At. = 0.0111 in. _ Considering now the case in which the load is apptied first, then the temperature is increased: PL 30 0003 5‘ (AQF =.__= 2 0.005102. AE zilin) (28x10 lot in ) (AL),- = L (1 (AT) = (15.005 in.) (5 ><10Ja OF“) (50°F) = 0.006 in. Total change in length in this scenario is: At. = 0.0051 in. + 0.006 in. .ANS _ AL=0.0TH in. ' 147 . ..-_:'..,:JJ,;'{' -" prismatic be is made of material with modulus of elasticity i | j E = 28 x 106 psi and coefficient of thermal expansion 3 3 u = 8 x 10'6 °F‘. it is fixed to a rigid wall at the left. There is a l I‘— 15 in. 5—4 j } "gap b = 0.0002 in. between the bar's right end and the rigid wall. —»+: i—— b - if the temperature is increased by 50°F above the bar's initial tamperature T, what is the normal stress on a plane perpendicular , - the bare axis? . \ I';:«\;J.! I ' a rigid wall wiil exert an axial force on the bar sufficient to restrict the bar's elongation to 0.002 inches. AI. = 0.00th. = Lo:(AT) _ 5-3-9 = (15inxsx10'“°F")(so°F) — {————f”’ i ] 115 28110 lb/t'n" S g = - 7,467 mm2 NOTE: The {—0 sign indicates a compressive stress. 5- - a at. . . " A has a cross-sectional area of 0.04 m2. modulus of elasticity I B 70 GPa, and coefficient of thermal expansion at = 14 x 10'6 °C'1. B has a cross-sectional area of 0.01 m2. modulus of elasticity 1 m I! 1 m 1 120 GPa, and coefficient of thermal expansion at = 16 x 10‘6 °C“. 1 “ - e is a gp b fi= 0.4 mm between the ends of the bars. What E-wb " u imum increase in the temperature of the bars above their initial 7- erature T is necessary to cause them to come into contact? . _ N: ' - sum of the expansions of the two bars will be 0.0004 m. a, = L 4494M) + 1.30:3 (AT) = (1m)(14110_6 1° CHAT) + (1m)(1tsx10'r3 [0 (3m?) 5 AT = 133°C . 4-... . A l-I- E 3 u - temperature of the bars in Problem 3.5.13 is increased ‘3 40°C above their initial temperature T, what are the 4 . ‘:.- stresses in the bars? ' ihl m lr—I m «at—Ht .3" l 0 : - reaction at the walls will be sufficient to restrict the total elongation of the bars to 0.0004 m. L AL=LAaA(AT)+LnaB(AT)—UA“A—O'Bfl e relationship between the stress in the two bars is: PA = P3 —> as = 4 0A [2] '. bstituting equation [2] into equation [1]: b.0004m = (im)(14x1 0“5 1° C)(40°C) + (imxmxio‘t /° C)(40°C) — o'A u 40;. . 120x109N/m- 120x109N/m- 93 = - 67.2 MPa NOTE: The (-) indicates a compressive stress. ‘ s g=~1§.§MP§ 149 a 00 a1; des gning a bar with a solid circular cross section Jwith tIZ-tn. diameter that is to support 3 4000-10 tensile axiai toad, and you want the factor of safety to be at least 8 = 3. ‘ :Choose a structural steel from Appendix B that satisfies this requirement. required cross—sectional area for the circular bar: 0' = 5 _) J = 4,0000“, A 3 22111401)” ay =‘ 61,115 :brm2 ‘ ANS ASTM — A 5‘14 wiii suggort the toad. ‘FR BLEM 3.1.5. ' horizontal beam of length L = 2 m supports a load = 13:: 30-kN. The beam is supported by a pin support and 1 brace BC. The dimension h = 0.54 m. Sup-pose that you want to make the brace out of existing stock that has areas-sectionas area A == 0.001s m2 and yield stress I lay: 400 MPa. if you want the brace to have a factor :: 29f safety 3 = 1.5, whatéshould the angle 909? s fires and! Diagram: i - 30,000N -- sownon: __ The maximum allowabie force in the brace is: a 400x106N/mz F = —-"2A a — ( sch.“ S 5 [ L5 Summing moments abbut the pin connection at the wait: 2M: 0 = (30,000 N) (2 m) — F3: (sin e) (0.54 m) = (30,000 N) (2 m) - (426.7 x 103 N) (cos 9) (0.54m) 7 _ cos B = 0.2604. e = 74.9" )(O.0016m3) = 426.7x103N 165 ...
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