15 P― A C 3000 lb B D
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12: Determine the smallest force P needed to lift the 3000-lb load. The coefficient of static friction between A and C and between B and D is μs = 0.30, and between A and B μs = 0.40. Neglect the weight of each wedge.
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- = It is required to design a split muff coupling to transmit 60 kW power at 140 rpm. The shafts, key 420 N/mm²). The yield strength in and clamping bolts are made of plain carbon steel (σyt compression is 160% of the tensile yield strength. The factor of safety for shafts, key and bolts is 5. The number of clamping bolts is 8. The coefficient of friction between sleeve halves and the shaft is 0.3. (i) Calculate the diameter of the shaft. (ii) Specify the length and outer diameter of the sleeve halves. (iii) Find out the diameter of clamping bolts. (iv) Specify the size of key and check the dimensions for shear and compression criteria6-4On what is the flexure formula based?
- ! Required information A helical coil compression spring is needed for food service machinery. The load varies from a minimum of Fmin 4 lbf to a maximum of Fmax 16 lbf. The spring rate k is to be 9.5 lbf/in. The outside diameter of the spring cannot exceed 2.5 in. The spring maker has available suitable dies for drawing 0.08-in-diameter wire. Use a fatigue design factor nfof 1.5 and the Gerber-Zimmerli fatigue-failure criterion. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the values for each of the following for a value of 0.08 in: The torsional modulus of rupture, Ssu, is 163.7 kpsi. Torsional yield strength, Ssy, is 85.5 kpsi. The endurance strength, assuming unpeened spring wire, Sse, is The value of the mean diameter, D, is 0.662 ✔ in. shear stress, Ta, is The fatigue factor of safety, nf, is The free length, Lo, is The critical length, LO(cr), is The alter 1.32 x in. in. kpsi. 39.5 kpsi.Describe uncontrolled spring action without a shock absorber.Doosan Man 6L60MC-C8 (2 stroke) Engine, piston stroke 2022 mm. Engine maximum power at 108 rpm is 21536 hp. Please calculate the MEP (bar) of engine.
- A ball bearing carries a load of 1000 N at 1760 rpm for 0.4 of the time and a load of2000 N at 880 rpm for the remainder of the time. Life is to be 10 years at 6 hours per day. Find the minimum value of the basic rating load C that the bearing must have. Loads are steady. (it is assumed that the bearing has but one row of balls. Aworking year will be taken as 250 days.) Ans. C = 16,430 NPlease show all stepsFull steps round 3 significant figures and use correct units plz