Question

The 16.2-ft steel beam weighs 420 lb and is hoisted from rest where the tension in each of the cables is 210 lb. If the hoisting drums are given initial angular accelerations a1 = 4.8 rad/sec2 and a2 = 8.3 rad/sec, calculate the corresponding tensions TA and Ts in the cables. The beam may be treated as a slender bar. 16" 16" 5.4' 5.4' 5.4 Answers: lb lb

Answer 1

Free-body diagram 16" 16" a beam2 αι α2 T A 5.4' 5.4 5,4

Solution: Length of beam (L) = 16.2 ft Weight of beam (W) 420 lb Mass of beam (m) = 420 / 32.174 = 13.054 slugs α1-4.8 rad/sec, α2-8.3 rad/sec'. Radius of pulleys (r) 8 inches 8/12-0.6667 ft. LAB 5.4 ft Accelerations a1-r * a1 0.6667 * 4.8 3.2002 ft/sec2. a2 ra2 0.6667 * 8.3-5.5336 ft/sec2. aG - (a1 + a2) 2 (3.20025.5336) 2 4.3669 ft/sec2. abeam- (a2a) LAB-(5.5336 - 3.2002)/16.2 0.144 rad/sec2. Area moment of inertia: lG (1/12) ml Using, FBD of the entire beam Sum of forces in y-direction -maG (1/12) 13.054 (16.2A2) 285.491 lb-ft-sec2 TA + TB = (13.054 * 4.3669) + 420 = 477.0055 TA + TB-477.0055 Sum of moment about G lGQbeam (To-Ta) * (5.4 / 2)-285.491 * 0.144 To-Ta 15.226 After solving Equation 1 and 2 we get, TA-230.889 lb, TB 246.115 lb. Answer. Equation-1 Equation-2