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137_Mechanics Homework Mechanics of Materials Solution

# 137_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM 2.101 .2401 The 30'1“!“ Square has AB Pas a length L = 2.2 m; it is made ofa mild steel that is assumed to be eiastopiastlc With E = 200 GPa and av = 345 MPa. A force p is applied to the bar until end A has moved down by an amount (5,“. Determinc the maxtmum value of the force P and the permanent set of the bar after the force has been removed, knowing that (a) 4,3 = 4.5 mm, (b) 6m= 8 mm. SOLUTION A = (.30)(3°\ t ‘00 mm‘ = goggle-c en‘- 5, : LEV " ES: ' 2.2)csvs'ew : eves-um“ 23.7?5’mm E 200 V10" IF S». 2 Sf P... = A6} ~(Qoox|o")(3¢f5‘uzo‘) = 310.5365) Uniﬁed-“M3 S. = %El£' ﬁg; * 5v 1 3.775— MM- Sf I. Sun-5‘ (at 8.. = 4.5». > 5v P... . SID-5H0“ N = 310.: w a 3pm '4 4.5km '- 3.715- Mun " O.7D\$‘nu '1‘ (lo) '5“ = 8m. > 51- 19.! No.5 no" N = 'Slo..s’ tau 4 SP‘ 1' 8.0m - 3.79.5- wn a \$205- mm ‘ 1.102 The 30-mm Iquere bar AB hes a length L - 2.5 m; it is made ofmild steel that is assumed to be eieeto'pieetie with E - 200 Oh and at, = 345 MPa. A force P is applied to the her and then removed to give it a permanent set JP. Determine the maximum value oi‘the force P and the maximum amount 6," by which the bar should be stretched it‘the desired value of q, is (a) 3.5 mm, (b) 6.5 mm. PROBLEM 2.102 SOLUTION A = (301(39) = 9’00 MM'" = 700x10" um" sf = ﬁg, -. LEE: = ﬂail-(53521913 a H.3t15'xto’m = 4.3125 mm Zea no“ When 5... exceeds S.” :Hwe PY‘OC‘UGI.M‘J a. Fernaueni‘ 5+t~¢+olq o'P SP: 'Hue Motrin-tum “Fore-e rs P... :‘AG‘Y = (900x!0")(3%'x10‘)= 3io.\$'wio’l\l = 310.5 to .- 3.” s." s' = sm— 3, sh: SP+SY (qt 59’ 3.5”.“ 3...: 3.5”...“ 4.3125“ e 7.3: MM (hi 8, : 6.5th i S...‘ 6.5%“ \$3l25'mm = 10.8l mm ...
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