In this figure (page 259) explain why the EMG and torque curves look the way that they do. **Be detailed
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- In this figure (page 259) explain why the EMG and torque curves look the way that they do. **Be detailed.
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- A round shaft, made of cold-finished AISI 1020 steel, is subjected to a variable torque whose maximum value is 7238 cm-kg. For N 5 on the Soderberg criterion, determine the diameter if (a) the torque is reversed, (b) the torque varies from zero to a maximum, (c) the torque varies from 3141 in-lb to maximum.understand the remaing of the answers but why he done 120 - 0.6 = 119.4 m why he subtracted 120 - 0.6 ( can you explain )to me whyFor the given question and solution, Why is taux'y' after the 25 degree rotation = positive Rsin(phi) = 1.05MA instead of negative Rsin(phi) ? how can i determine the sign convention after rotation? i get mixed up with the sign convention of the shear stress after rotation.
- Falcon company Nizwa is conducting a campus interview to you. You are asked to design the propeller shaft with the following specification. (i) Power transmission =60 kW,(ii) Shaft speed = 160 rpm,(iii)The angle of twist / m length = 1.1º(iv)Maximum twisting moment is 30% greater than its mean. (v) Take Modulus of rigidity as 105 Gpa. (i) Mean torque of the shaft (unit in Nm) = (ii)Maximum torque of shaft (unit in Nm) = (iii) The diameter of the shaft (unit in mm) = iv) Maximum shear stress in the shaft (Unit in MPa) isA steel rim, with a mean diameter of 1 m, rotates at 1 000 r/min. Determine thecircumferential stress-induced if the density of steel is 7.8 Mg/m3ᴄᴏᴍᴘᴜᴛᴇ ᴛʜᴇ ɴᴏᴍɪɴᴀʟ ꜱʜᴇᴀʀ ꜱᴛʀᴇꜱꜱ ᴀᴛ ᴛʜᴇ ꜱᴜʀꜰᴀᴄᴇ, ɪɴ ᴍᴘᴀ, ꜰᴏʀ ᴀ 50 ᴍᴍ ᴅɪᴀᴍᴇᴛᴇʀ ꜱʜᴀꜰᴛ ᴛʜᴀᴛ ɪꜱ ꜱᴜʙᴊᴇᴄᴛᴇᴅ ᴛᴏ ᴀ ᴛᴏʀQᴜᴇ ᴏꜰ 0.48 ᴋɴ-ᴍ
- Determine the diameter of a solid steel shaft that will transfer 30 MW at 1500 RPM with a 1 degree twist angle for every 20 diameters of length. G=80GN/m2. Give me the solution of this please. Ans. D = 3.03 mFalcon company Nizwa is conducting a campus interview to you. You are asked to design the propeller shaft with the following specification. ) power tranmission = 270 kw (1) speed 65 rpm, () Maximum shear stress = 55 MPa (iv) diameter ratio = 1.78 (v) The maximum value of the twisting moment is 40% greater than its mean. Find, (i)The mean torque is (Unit is in Nm)=, (ii)The maximum torque is (Unit is in Nm)=. (iii) The outer diameter of the shaft (unit in mm) (iv) The inner diameter of the shaft (unit in mm)Please don't copy Find the torque and power transmitted by a 100 mm diameter shaft, if maximum angle of twist and length of shaft are 1.2° and 2 m. Take G = 70 GPa, speed of shaft = 230 r.p.m.
- Define the term Shear-Flow Resultants?A vertical shaft 25 mm diameter and 0.75 m long is mounted in long bearings and carries a pulleyof mass 10 kg midway between the bearings. The centre of pulley is 0.5 mm from the axis of theshaft. Find (a) the whirling speed, and (b) the bending stress in the shaft, when it is rotating at 1700r.p.m. Neglect the mass of the shaft and E = 200 GN/m2.A pressure difference of 35 pa is available to force -10c air. through a circular sheet metalduct 400 mm in diameter and 20m long. Find thevelocity in m/s. A.11.64 B.38.20 C.12.81 D.20.60 SHOW DETAILED SOLUTIONS