6. A 2.5 in.-diameter steel propeller shaft of a boat transmits 2800 hp @ 1800 rpm. Bending and axial loads are negligible. Clearly draw and label necessary schematic diagrams and list necessary assumptions. (a) What is the shear stress at the surface of shaft? (b) If a hollow shaft of inside diameter 0.88 times outside diameter is used, what outside diameter would be required to give the same outer surface stress? (c) Compare weights of the solid and hollow shafts. What is % saving in weight of shaft material if hollow shaft is used? (d) As a design engineer, what important lesson have you learned from solving this this? (e) What is Polar Moment of Inertia?

6. A 2.5 in.-diameter steel propeller shaft of a boat transmits 2800 hp @ 1800 rpm. Bending and axial loads are negligible. Clearly draw and label necessary schematic diagrams and list necessary assumptions. (a) What is the shear stress at the surface of shaft? (b) If a hollow shaft of inside diameter 0.88 times outside diameter is used, what outside diameter would be required to give the same outer surface stress? (c) Compare weights of the solid and hollow shafts. What is % saving in weight of shaft material if hollow shaft is used? (d) As a design engineer, what important lesson have you learned from solving this this? (e) What is Polar Moment of Inertia?

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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8.9 (A/B). A solid steel shaft of 200 mm diameter transmits 5 MW at 500 rev/min. It is proposed to alter the horsepower to 7 MW and the speed to 440 rev/min and to replace the solid shaft by a hollow shaft made of the same type of steel but having only 80% of the weight of the solid shaft. The length of both shafts is the same and the hollow shaft is to have the same maximum shear stress as the solid shaft. Find: (a) the ratio between the torque per unit angle of twist per metre for the two shafts; (b) the external and internal diameters for the hollow shaft. [L.Mech.E.] [2.085; 261, 190 mm.]
A steel shaft with a diameter 45 mm and length 1.30 m transmits 55.50 kW from an electric motor to a compressor. The allowable shear stress is 51.75 MPa and the allowable angle of twist is 2.1 degrees. G=83x10^3 psi Determine the polar moment of inertia (in mm^4) of the steel shaft. Determine the required torque (in N-m) as limited by stress. Determine the required torque (in N-m) as limited by the angle of twist.
4a) A thin cylindrical tank is subjected to an internal pressure (p) of 15 MPa and at the same time, is subjected to a twisting torque (T) of 900 Nm. Outside diameter (D) of the cylinder is 100 mm and the wall thickness (t) is 5 mm. What is the normal stress and shear stress at the inner wall C on a plane formed by a cut AB, inclinecd 30" to the axis of the tank, as shown in Figure 1. Choose a convenient coordinate system and show it over the diagram. Consider ll the stress components developed due to the loadings and calculate/identify their values by appropriate symbols. No derivation is necessary. Figure 1
A hollow steel shafts transmits 200kW of power at 150 rpm. The total angle of twist in a length of 5 m of the shaft is 3Â°. Find the inner and outer diameter of the shaft. If the permissible shear stress is 60 MPa. Modulus of rigidity (C) =80 GPa
-10 108 (mm) Figure 5.41 5.21 Determine the St Venant torsion constant for the thin tapered section in Fig. 5.41. Hence derive an expression for the shear stress acting along the longer edges in terms of the torque T when the thickness 10 st s 1. If T = 15 Nm estimate the position and magnitude of the maximum shear stress from a consideration of contours of constant shear stress
(2) The shaft consists of a 3-in diameter aluminum segment that is rigidly joined to a 2-in diameter steel segment. The ends of the shaft are attached to rigid supports, Calculate the angle of twist developed at the interface of the two segments when the torque T = 50 kip in is applied. Use G = 4×106 psi for aluminum and G = 12×106 psi for steel. | Aluminum 3-in. diameter 6 ft T Steel 2-in. diameter 3 ft D
Please present drawings/FBD’s illustrating your solution.
400 lb-in. 1200 lb-in.. Specifications: Shafts AB, BC, CD are solid material. 500 lb-in., B AB=0.6 in. C dsc-0.75 in. dcp = 0.9 in. Determine: a) The shaft section in which the maximum shear stress occurs and the magnitude of the stress. Now consider that the shaft is tubular with a 0.300 inner diameter. Determine: b) The shaft section in which the maximum shear stress occurs and the magnitude of the stress.
Steel shaft can be produced as solid body or circular tube with circular tube. The shaft must bear a torque of 1200 Nm. Since the shear strength for the shaft material is 40 MPa, the maximum allowable unit torsion angle of the bar is 0.75 ^ 0 / m and the shear modulus of steel is G = 78 GPa;a) Find the required diameter d0 for a solid circle shaft.b) Find the required outside diameter d2 for a ring-section shaft whose thickness is one tenth of the outer diameter.c) Find the diameter ratio (d2 / d0 ratio) and weight ratio for ring cross section and solid section shafts
Q2(c) The steel alloy shaft having a diameter of 38 mm is used to transmit torque, T as shown in Figure Q2(c). If the allowable shear stress of the shaft is 80kPa, i)Determine the maximum torque T that can be transmitted by the shaft. ii)What would be the maximum torque T' if a 25 mm diameter hole is bored through the shaft? Figure Q2(c.) |Q2 c) (i) T c The torsion formula :7 = J Т (0.019) T (0.019)^4 ; Therefore, the max torque T = x10 Nm 80 2 Q2 c) (ii) For point lying on the outside suurface of the shaft: x10 ="(0.019) * (0.019*-0.0125 ) ; Therefore the max torque T' Nm 80 2 Q2 c) (iii) The shear stress at point located at inner surface of shaft, T'c T = -= J ) (0.0125) = 7 (0.019*-0.01254) T'c kPa 4. 2 2.
A drilling riser is used to drill a 4km deep hole into the seabed. The riser can be approximated as a hollow steel shaft of 185mm and 160mm outer and inner diameter respectively. If the power exerted is 230kW and the speed of rotation of the drill is 75RPM, calculate the maximum shear stress in the shaft and the twist of one end relative to the other in degrees.(Note the angle of twist of a 4km drill string is going to be >> 360o ) G = 89 kN/mm2