Trabalho de Resistencia dos Materiais 1

ZA/WD/AB = Wf D= W= SumC= 0 SumD= 0.250-D=0 SumW= 0 w= 24 + 26K w= 0.1m - 0.2m - 0.3m - 0.4m 0.4m 3 Sum mc = 0 TC = 500 NI/m TC = 500 NI/m A 0.9m a= 600 NI/m Summb = 0 TB + 350 NI/m - 500 NI/m =0 TB = 350 NI/m - 500 NI/m TB= 150 NI/m 4 80N Sub-Fx=0 Fx=Fcos15\nFx=80.cos15\nFx=113.8N SumFz=0 -Fx+NA=0 NA=910 NA= 77.3N Sum E = 0 -Ma MA - 89.875 = 3.30 -Ma = 33.3W Tg θ = 1,2 / 1,6 Tg σ = 2 / 1,6 d = 3,9m, 3,6,96° θ - ϕ = 14,48 mAB + ∑MA = 0 FBC 5m 14,48 - 5000 1,2 = 0 FBC 5m 14,48 = 6000 N/m FBC = 6000 N/m / 5m 14,48 FBC = 2399,5 N/m ℓ ΣME = 0 FBC m 14,418 9,6 - 5000 Sin 36,96 (9,6) - MC = 0 MC = 0 BC ΣFx = 0 -NE - FBC cos 14,48 - 5000 cos 36,96 = 0 NE = -2399,5 cos 14,48 + 5000 cos 36,96 NE = -2723,3 KW ZFy = 0 VE + FBC 5m 14,48 - 5000 cos 36,96 = 0 VE = -2399,5 5m 14,48 + 5000 cos 36,96 VE = 0 ME = 0 VE = 0 750KN/m 675KW FRA = 750KN/m 0,9m FRA = 675 KN €MA = 0 ED + FRA d + MA = 0 - 150N, 0,9m - 675 KN, 0,165m + MA = 0 MA, 1500N, 0,9 + 675 KN, 0,015m MA = 1653,7 KW/m ∑FyA = 0 - F - FRA + VA = 0 VA = F + FRA VA = 1500 + 675 VA = 2175 KW FRb: 750 KN/m, 3,3m £mb = 0 - b - Pd - Fbd, - mb = 0 - 1500 N, 3,3m - 2175 KN, 1,65m + mb mb: 1500N, 3,3m + 9,033 KIlom £ Fxb = 0 N = 0 FRDC = 750 KN, 1,5m FRDC = 1125 KN/m FRFC = 750 KN 2,9 m FRFC = 2925 KN/m ∑ MC = 0 £ Fr = 0 £ Fl = 0 VC z = 0 VC = 5550 KN N = 0 FRDC = 750 KN, 1,5m FRDC = 1125 KN/m FRFC = 750 KN 2,9 m £ MC = 0 £ Fz = 0 VC z = 0 VC = 5550 KN VC = 0 11 1,5 kN/m 4 kN A 1,5m D 1cm C 2,1m B 1,35 E 1,35m FR1 = W . d = 4,5 . 6 FR1 = 22 kN/m CM1 = d \triangulo\ CM1 = \frac{2}{3} = 3 m 2° passo: achar as reações C \vline l d = 3m FR1 = 22kN/m 19 kN FRe2 = 8,1 kN/m FR2: W . d = area triangular retângulo FR2 = \frac{6 . 1,7}{2} = 8,1 kN/m CM2 = d \triangulo\ CM2:\frac{2}{3} = 0,99 m \sum Fy = 0 - FR2 . 6,9 + 2b . 6 - FR1 . 6 = 0 - 8,1 kN/m . 6,9m - 22 kN/m = - Rb . 6 - 81 kN - 55,8 9 kN = - 6m . Rb - 136,8 9 kN /6m = Rb Rb = - 22,815 KN RA + Rb - FR1 - FR2= 0 RA = - 22,815 + 27 + 8,1 RA = 12, 255 kN 3° passo para carga E 6 kN 3kN/m B 1,35m E 1,35m E 1,35 FRe = W . d. triangulo retângulo FRe = 3 . (1,35 = 2,02 kN CMe = 1,35 = 0,45 m 2 015 FRe = 2,02 kN VE NE RA = ME = 0 -VF+ VE= 0 VE - FRe ( VE = 2,02KN ME = 0 FRe . 0,15 = ME = 0 0,90 = - Me ME = - 0,908 kN.m M 1,5 kN/m A 1,8m VD 1,2 235 KV/m D FR = 4,5 kN/m, 1,8 m FR = 8,1 KN CM = d \frac{2}{3} CM = \frac{1,9}{3} \approx 0,99 m \sum Fy = 0 - FR + RA + VD = 0 VD = + FR - KA VD = 9,9 21,295 VD= \frac{41,13}{5 KN} \sum Fx = 0 VD = ND = 0 \sum MD = 0 - Fre . d + KA . d - MD = 0 - 41,094 + 12,05 \ x 1.5 0 = - 2, 94 + = - MD 1232 - MD MA = - 4,823 kN/m \text{desenho trístup voluntheros da cate tesuda} 12 D 5m45 FR = 3,600N \frac{C}{BE} \frac{65}{65} 5: 2545 N B A c d e F \frac{600 N}{m} \text{BE} = 600N 3-2415kN Cm = \frac{b}{2} \sum MA = 0 NC + 2547.5m\cdot 450 = 0 NC = - 489kN b b= 2,54 kN \sum Fy = 0 252f.5 m . D - 244 + VC = 0 VC = - 211 - \frac{09}{m} \sum MC = 0 MC + 21100.2 - 254247.45m.450 = 0 MC = 4040kN 2-2567 2650 kN VC = 21 \sum Fy = 0 VC - 2, 45M\cdot 450 = 0 \sum MC = 0 MC + 2400 . 2 - 2547 + 451 = 0 MC = 4040KN 3-252500N/M \ 255m/45 MC 84 44 2jNC 900 kN/m 2m 1m 1m A B C Proporciona carga 900kN x 4m = x x 5m x = 900kN x 4m x = 720kN Cm = b/3 y/3 Cm = 1/3 Cm = 4/3 Fr = 720kN 4m - 1440 kN/m - = -Fe + Vb = 0 Vb = -Fe Vb = 1440 kN/m + Mb + Vb . b/3 = 0 Mb = -Vb . b/3 Mb = -1440 kN . 4/3 Mb = -1440 kN . 1.33m Mb = -1920 kN . 1m Nb = 0