Question

1. For the W6x20 beam with the loading shown, calculate the deflection at point C. E = 29,000 ksi 2000 16 W-400 94 OB 4 ft 6 ft 4ft

Answer 1

sol 1760 lb.it Solnion' Grun, 2oooh 40011/4 c f هو 2 * 4A 681 467 Now, find the support readions Ema0 +Rox10-200076 +400x4*2=0 C 2:0 to lo = 88olb EV=0 -Re-2720 16 RA+ Rig - 2000 +40ory) 3600 lb. Now construct the Bending moment diagram, For section Ab M = 9Box - 2000 (x-2) For section ABC Mas 3808 -2000 (A-2) + 2720 CM-10)-400 (31-10) By substituing the values for head of 2000lb, point s' and c 1.0 Me 7200 lbst M: 4000 lb.ft Point o is where i=0 1786 1 4.42 B E 1570 4000 FF1 a 7200 E1 H BMD

Moment is o at M=0= 8801 - 2000 (0-2) = 3.57 from point A By Moment- Area method. deflection at t wat B •distance of tals ES Lof AED * 6+) is given by area of in diagram b | centacid from :&- face od *3)*(* arca et DEO * (44272x1,5 • - [area area of OBH * (41)] - [aeea of Beaf * 2 2) - [ area of FGH * (*) * - [ 2*960 * 2 *(6+3)+CH M260x1,57-*(4.43+ 2x957)] - [ ****, 434) - [ apporte 2 -[2-a oa hely 1 (11933.33 + 7566.56 – 23549.88 - 32000-4266.67) 12 40516:657) Ibatt? rvé indicater downwyd deflection E) E = 29000 ksi x 122 40516.657 x 24000x1000x 41.4x 2.4 1 91.4x10 mm J = 41.4x2.4 in 05 2.02 xlo?inch