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# 2.3-6 - SECTION 2.3 Changes in Lengths under Nonunifnrm...

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Unformatted text preview: SECTION 2.3 Changes in Lengths under Nonunifnrm Conditions 109 Salotion 2.3—6 Steel columns in a building ”—1—th1 = 400 RN L __ 1.. = length of each column“ "5 3.75 m ' E = 206 8P0 .lele = 11,000 mm1 A = 1’ Q 2 BC. .:.,00 mm [11) SI'IC.)RTEI\'ING 5].“; [JIA' THE TWO COLUMNS N—L- N. L N J. E l l = .‘IB + I“ Ei'fll' Eflf'l [1‘ EA HC 83H." V“ (t 120 \$0003.75 :11) (206 GPaltl 1,000 mnﬂ (400 kNl(3.75 m) + ____.___—._____1.. (206 GP;t)(3.900 mm") ll 18535111111 + 1.867] mm m 3.7206 mm 511C = 3.72 mm <— Problem 2.3-7 A steel bar 8.0 ft long has a circular cross section , of diameter all = 0.75 in. over one-half 01" its length and diameter d: z 0.5 in. over the other half (see ﬁgure). The modulus ol‘ elasticity E x 30 x H)“ psi. (0] How much will the bar elongate under a tensile load P = 5000 lb‘.’ (l1) ll‘ thcsame volume of material is made into a bur of constant diameter (1 and length 8.0 ft. what will he the elongation under the same load P? Sultition 2.3-7 Bar in tension (11 = 0.75 in. E}: = 0.50 in. P ="50001b t l l W40 ft——+p——-——-:l.0 ft ___. F = 500001 E 2 30 X 106 psi L=4~ftx48i11 (b) ADDITIONAL. LOAD P.} '.-\T png C (531C.}i]111x n 40 mm 50 2 additional shortening of the two columns due to the loud P0 50 = (5:13)...“ —— 5M" = 4.0 mm — 3.7106 mm = 0.2794 mm P L P L P L l. l 0 + 0 0 ( . ) EAAH EABC E A150, (30 7‘ Solve for P0: _ ﬁ< A213 ABC ) .- Pu W L AAB "l“ ABC SUBSTITUTE NUMERICAL VALUES: 1-: m 206 x 10" N/nﬁ 50 = 0.2794 >< 1023 m L = 3.75 m A” 2 11,000 >< 10'“ m2 AM- : 3,900 x' 10*ﬁ m2 PU : 44.200 N ; 44.2 kN <—— id. = 0.75 in. E51" 0.50 in. P =300011; : l‘wto rim—lev—to n A (a) ELONGATION OF NONPRISMATIC BAR __ V‘Nll‘l = EV}. — “5,01,- E “A. 5 _ (5000100801.) 30 x 10“ psi 5 ...
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