Question

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200 lb 400 lb 4 ft - to2 ft2 ft → EI = 1.9 x 105 lb. ft2 FIGURE P16-22

Answer 1

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200lb 1 Depliction in a UH ME simply supported beam at C cui to zoolb load. = AC Using conjugate beam method. bending moment diagram. m diagram PL . Чао Ibtv 400 EI А ЧЕ+ c че в А чH с чH moment of M diagram at A ma098 Re - (4 x 400 xu) (+4) - (¿a yo x ) (4) - 8 R6 – 42,72 2026 20

Rp : 801 GI Now, digram a moment of indige of of real beam deflection Ac = Mc = R8x4 - (2 x 400 x 4) + 4/5 3204 Et - 10 66.67 = 2137.33 EI = 0.0112 ft 2137.33 109x105 2) Deplechon at o deu to 400 lb point load yoolb 25+ Using conjugate beam mothed. Pab: 60016 60 Et A GH eft & oft C 24 me=0 30 ) ) - 8 8A -( ( đa) ( ) ( 8 RA - 1840 (“) - Bayo

RA : 100.5 lb Depliction at ce moment at c. Ac= 0.021746 0.021063P+ 1000.5 x 4 = 4002 EI EI 3) 200 lb 400 lb Deflection at 6 due to both loads. t C DA 2 TRB YA mA = 0 Cfor real beam) 8 RB - 100 x6 - 2004 4 =0 RB = 4oolb RA = 600-400 = 2001 (.Etyao ) BMD for the beam, 80lb Mc = RAXY = 8oolb mo Rox 2 = 800 lb ,800 lb - - हDB - method m diagram (conjugate beam 터 800 X Cr .AL 4 ZH 2H

Now, for conjugate beam, (EU) pent] (***CO x2) MB=0 I ÇŹD LAB T RB R 2 x 4 Ra'C® ) - (+1800 xy) (444) - The K2)(2+1) - a * * x2) (3x2) -0 4H 24 24 ERA' - 1600 (s. 33 ) -4800 - 1066.67=0 EI C Et EI RA = 1800 lb ЕТ Now, moment at C = deplection at c. Ar PÅ(4) - (2 x 800 x4) x 4 = 7200 - 2133.34 = 5066.67 EI EI Е1 0.0267