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1244_Mechanics Homework Mechanics of Materials Solution

# 1244_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM 10 57 10.57 Using allowable stress design, determine the allowable centric load for a ' column of 6.5-!!! eﬁ‘ective length that is made ﬁ'om the following rolled—steel shape: (:1) W250 x 49.1, (b) W250 x 80. Use ay= 250 MP9. and E = 200 GPa. SOLUTION . 212' (Zoom?) = — 5+“). 32/2; ————m—./ 250nm 125.69% (51.) wzsoxtmn : czao x10 m rm: 49.2,:10'3 w; Le _ as 7 ‘ ms? 132.” > C: _ 1725 ‘ Tr‘(:¢‘oo><lo“) _ ‘ 6““ ‘ l.‘t2(L/r‘)" ' (mlesan‘ " 55m ”’0 P‘“ PM = A6,}. = (czsom-‘)c\$2.wo‘) = sear/0’ N r 368 kN «4 (b) W250>< 80 A =' 107.00: 10“m" mﬁ £50qu 1-; _ ii... _ L/v‘ Y‘ ' 95(5):]0'3 ' loo < Cc Cr. =' 0.79577 F25. =- =§ + §(o.7as77)-%(o.7757?}3 = Lemma _ £31» [ L/r ]_ 25m 0‘ L a _ l 5‘” _ F33 1-306: ) ‘ﬂ—ﬁazon [1- 2(0.7¢is773]- 89.82on Pa P.» = A 61.... :- ([0200x10‘5)(8‘7.82»!0") = mnrogw = 9m 1w .. 10.58 A W8 x 3! rolled-steel shape is used to form a column of 21-11 eﬁ‘ective PROBLEM 10-53 length. Using allowable stress design, determine the allowable centric load If the yield ﬁmgth of the grade of steel used' 15 (a) a,= 36 ksi, (1:) 0": 50 ksi. Use E— ‘* 29 X 1 p31 SOLUTION sired: E = zqooo In: W s new A = 9.13 in‘ m“ = 2.02 m L5 2! H = 252 .‘n zLem- )rzu 75 1 L./v~< cc 55/! = 0.93%; ES. = g + i-(omﬁsz) - %(0-‘38ﬁ52l3 '-“ MUCH! 5;”: %[I-§(l‘g)]= %m[a—é(o.aznszli] = 9.59 ks; Pa =- GZuA = (9.59X9J3) = 87.6 kg,» .... (bl 6} =' 50 ksf Cc : .Zﬁ_(_‘l__.)_‘ 2 °°° = 107.00 9.59 ks; Put! 9 GauA = (15930133) -' 31\$ kiPs ...
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