ME211solution_7 - PL-5(103(8 = = =-3.64(10-6 AE(0.42...

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3 6 3.64(10 ) = 22 9 5(10 )(8) (0.4 0.3 )200(10 ) 4 PL AE δ π == ote: Negative sign means contraction. N
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0; (12) 5(6) 0 AE MT =− = i p = = i p = Tk 2.5 E 0; 2.5(2) (10) 0 DC 0.5 C / 23 2.5(1.5)(12) 0.0316 (0.25) (29)(10 ) 4 BE PL in AE δ π == = 0.5(2)(12) (0.25) (29)(10 ) 4 C PL AE 0.00843 in = From similar triangle 22 (0.00843) 0.00169 in = 10 10 EC δδ BEB E δδδ =+ = + / 0.00169 0.0316 0.0333 in =
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0; 60 0 ys t c o n FP P =+ = ------(1) s tc o δδ = n ; {} 33 (8)(12) 18(29)(10 ) 9(16) 18 (4.2)(10 ) st con PP = 0.98639 s t P = c o n P i p i p ------(2) Solving Eqs. (1) and (2) yields 29.795 st Pk = , 30.205 con = 29.795 18 st st st P A σ == 1.66 ksi = 30.205 9(16) 18 con con con P A 0.24 ksi = Either the concrete or steel can be used for the deflection calculation 3 29.795(8)(12) 18(29)(10 ) st st PL AE δ 0.0055 in =
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Equation of equilibrium: 0; 25 25 0 xb s FP P =+ = ------(1) 50 bs PP += Compatibility: δδ = 29 2 2 (220) (200) (0.02) (200)(10 ) (0.05 0.04 )(200)(10 ) 44 ππ = 9 s P N N 0.40404 b P = ------(2) Solving Eqs. (1) and (2) yields 35.61 s Pk = , 14.4 b =
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Note: Since adjacent rails expand, each rail will be required to expand δ/2 on each end, or δfor the entire rail. { } 6 6.6(10 ) 90 ( 20) (40)(12) 0.348 TL in δα = −− = 0.34848 TF δδ =− {} 6 3 (40)(12)
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ME211solution_7 - PL-5(103(8 = = =-3.64(10-6 AE(0.42...

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