Answer Happy

shaft length as 11.5 in
Similar solution:
w4=451lbf W3 = 481.6lbf W54 = 154.25lbf From Σ Mx finf the torques in the shaft between the gears T = W₂3 (¹/2) T 1408 3 (1¹/²) = = = 84481bf * in 2 Find reactions and and determine bending moment diagram as similar one (attached below) W23 ↓ Wr IN R₁ w/3 = 1408lbf Az RAY A W23 I 54 54 J K 1 I 1 L I I RB: RBY B 1

Gear 3 −1 -/-/- alt 4 Gear 5 Gear 4 2 IN

Solution Perform free body diagram analysis to get reaction forces at the bearings. RAZ 115.0 lbf = RAY = 356.7 lbf RBz = 1776.0 lbf RBY = 725.3 lbf From EM, find the torque in the shaft between the gears, T = W23(d3/2) = 540(12/2) 3240 lbf in. Generate shear-moment diagrams for two planes. Z RAZ x-z Plane y RAY A T V I 1 | 1 1 115 W23 W 23 1 13240 T 655 I T I T + I T T I I T T I I L I T T L T I T 1 I 1 WS4 W4 • J • K T I | 1 + I T I 1 I | 1 I I RBz T I -1776 RBY B 1 I T I 1 1 1 4 i T I T 1 I 1 I I I 1 1 T + ||

Combine orthogonal planes as vectors to get total moments, e.g., at J, V3996² + 1632² 4316 lbf in. = x-y Plane M M MTOT 357 1 + 230 V 1 1 I 1 1 T 1 1713 1 1 1 I 1 t 1 T 1 749 160 3341 T I 1472 1 1 1 3651 I I 13996 + 725 -1632 1 T I 14316 I 2220 907 T T I 2398