Question

how do i get all the values in the afd sfd bmd?

Frictionless pulley 200 mm in diameter 1.1 m 3.5 m 2 m D 30 kg mass wkN/m (max) b

Answer 1

T 200mm = 0.2 m. lolm 130 kg 30x9.8= 294 N Free body diagram +2941 294N 1.1m = 1 m. A 2 B D 3.5m 2 m, 294 294 - 294 294 1299 -294 294

AF - 294 N. AF Axial force (AF): at point E - - 294 N. at point (just above) AF at point D (guer below) - – 294 + 294 20. In member BD AF at point (in member BD) ? - 29 4 N. member DE. - 294 M. AF at point B (Just right) AF at point B (Just left) 294+ 294 ZO. In member BC AF at point e =o. at point B B a O. AF In member AB AF at point Bzo. AF at point A=0

Shear force (SF) - 294 Member DEN E 294 N. SF at SF at D (Just above) 2 294 N. ud 294 Member BCI 294 SF at point c- - 294 N. SF at point B (Just above) =~294 N. K 294 Member ABD : 294 of B SF at Do - 294 N. SF at B 294 N. SF at 294 N. A moment calculation: Bending BM at E = 0. Bm at D = - 294 & 1+29 4 x 0 = - 294 N-m. Bm at mid of BD = - 294 x 1+29 4 x1 =0. +29 4 Nom. B m at B in BD) ? - 294 x 1+29 4x2 2

Bm at C = o. B in BC) 294% 1.1 2 323.4 N-m. Bm at Bm at B (in AB) = – 29.4 x 1 + 294 × 2 + 294x1.1 617.4 N-m. BM at A (in AB) = -294 x 1 + 294 * (2+3.5) + 294+1+1 16 46.4 N-m.