A flanged bolt coupling consists of eight d-mm diameter steel bolts on a bolt circle 267 mm in diameter, and four d-mm diameter steel bolts on a concentric bolt circle 190 mm in diameter. If the torque that can be applied is 8166 Nm without exceeding a shearing stress of 43.32 MPa in the bolts, what is the diameter of the bolts? Round off the final answer to two decimal places.

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A motor supplies 173 hp at 650 rpm to flange A of the shaft. Gear B transfers 75 hp of power to operating machinery in the factory, and the remaining power in the shaft is transferred by gear D. Shafts (1) and (2) are solid aluminum [G = 4400 ksi] shafts that have the same diameter and an allowable shear stress of T = 5.8 ksi. Shaft (3) is a solid steel [G=9100 ksi] shaft with an allowable shear stress of T = 10.2 ksi. Assume that L₁=2.3 ft, L₂=1.2 ft, L3-2.9 ft. Determine: (a) the minimum permissible diameter d₁=d₂ for aluminum shafts (1) and (2). (b) the minimum permissible diameter d3 for steel shaft (3). (c) the rotation angle D/A of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b). Answers: (a) d₁=d₂ = i (b) d3= i (c) & DIA = i (1) L₁ TB. B (2) L2 in. in. o' (3) L3 Tp Mommu D

A coupling must transmit 30 hp at a speed of 175 r/min. 3/8 inch diameter bolts are arranged in a single 6 inch diameter circle. The shear stress in the bolts should not exceed 5,000 psi. What is the minimum number of bolts that should be used? Answer: seven bolts

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An extension spring made from A313 stainless steel has a mean coil diameter of 10.00 mm and spring wire diameter of 1.38 mm with 72 body coils. If the spring has full twisted loop ends (hook and body have the same diameter), determine the factor of safety for end hook bending under a 21.5 N service load. Round your answer to 3 decimal places.

The answer is 1290N, but I am not entirely sure how they got there. However, I noticed that when I calculated the torques for the person and that of the ladder, my answer would be correct if I multiplied by sin(60) or cos (30) instead. Using the sum of these values, I would divide it by the length of the ladder and sin(30), which would give the correct answer, but I'm a bit confused as to why I need to take sin(60) or cos(30) to calculate the torques for the weight of the ladder and the human, but then use sin(30) when calculating the torque for the perpendicular force. I would appreciate a work through of the problem and some explanation for my difficulties above, thank you.

Please solve, include the diagrams, and do not copy from other bartleby answers.  Handwritten solution is more preferred than typed. A flanged bolt coupling connects two shafts. The bolt circle diameter is 250 mm in which there are 12 bolts. The allowable shear stress is 56 MPa in the shafts and 22 MPa in the bolts. One shaft is hollow with outside diameter of 100 mm, while the other one is solid of diameter 50 mm. Determine the ratio of the outside and inside diameter of the hollow shaft and the bolt diameter so that both shafts and the flanged bolt coupling are having the same strength in torsion.