e maximum shear stress developed
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Need to calculate maximum shear stress .
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- A hollow tube ABCDE constructed of monel metal is subjected to five torques acting in the directions shown in the figure. The magnitudes of the torques are T1= 1000 lb-in., T2= T4= 500 lb-in., and T3= T5= 800 lb-in. The tube has an outside diameter of d2= 1.0 in. The allowable shear stress is 12,000 psi and the allowable rate of twist is 2.0°/ft. Determine the maximum permissible inside diameter d1, of the tube.A solid brass bar of diameter d = 1.25 in. is subjected to torques T1as shown in part a of the figure. The allowable shear stress in the brass is 12 ksi. What is the maximum permissible value of the torques T1? If a hole of diameter 0.625 in. is drilled longitudinally through the bar, as shown in part b of the figure, what is the maximum permissible value of the torques T2? What is the percent decrease in torque and the percent decrease in weight due to the hole?A propeller shaft for a small yacht is made of a solid steel bar 104 mm in diameter. The allowable stress in shear is 48 MPa, and the allowable rate of twist is 2.0° in 3.5 meters. (a) Assuming that the shear modulus of elasticity is G = 80 GPa, determine the maximum torque that can be applied to the shaft. (b) Repeat part (a) if the shaft is now hollow with an inner diameter of 5d18. Compare values to corresponding values from part (a).
- A circular aluminum tube subjected to pure torsion by torques T(sec figure) has an outer radius r2equal to 1.5 times the inner radius r1. (a) If the maximum shear strain in the tube is measured as 400 × 10-6 rad, what is the shear strain y1at the inner surface? (b) If the maximum a1lo-abk rate of twist is 0.125 °/m and the maximum shear strain is to be kept at 400 × 10-6 rad by adjusting the torque T, that is the minimum required outer radius ( r2)Min?Two sections of steel drill pipe, joined by bolted flange plates at B, arc subjected to a concentrated torque 4000 kip-in. at x = 3 ft, and a uniformly distributed torque t0= 50 kip-ft/ft is applied on pipe BC. Let G = 11,800 ksi and assume that pipes AB and BC have the same inner diameter, d = 12 in. Pipe AB has a thickness tAB= 3/4 in., and pipe BC has a thickness tBC= 5/8 in. Find the reactive torques at A and C and the maximum shear stresses in each segment.Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 235 MPa. If the shaft diameter is doubled then the maximum shear (in MPa) developed corresponding to the same torque will be
- The hollow rod shown below is 2.99 meters long and is subjected to a torque of 988 Nm at end B. The diameter of the rod is 30 mm and has a wall thickness of 5 mm. The shear modulus of the material is 80 GPa, calculate the angle of twist in radians at B. TNm B metersTwo designs for a shaft are being considered. Both have an outside diameter of 80 m and are 100 cm long. One is solid but the other is hollow with an internal diameter of 50 mm. Both are made from steel (G=120 GPa). Compare the torsional shear stress, angle of twist of the two designs if they are subjected to a torque of T N. m. the missing Torque (T) = 199.A hollow shaft and a solid shaft constructed of the same material have the same length and the same outer radius. The inner radius R of the hollow shaft is 0.6R. Assuming that both shafts are subjected to the same torque, compare their shear stresses, angle of twist and weights.
- find the maximum shear stress induced in a solid circular shaft of diameter 150 mm when the shaft transmits 150 KW power at 150 RPM.A mild steel shaft of 60 mm diameter is subjected to a bending moment of 2200 Nm and a torque T. If the yield point of the steel in tension is 200 MPa, find the maximum value of this torque without causing yielding of the shaft according to:1. The maximum principal stress theory;2. The maximum shear stress theory; and3. The maximum distortion strain energy theory of yieldingA shaft consisting of an aluminum segment and a steel segment is acted upon by two torques. The shaft is attached to the wall at point C. The diameter of the aluminum segment is 0.062 m and the diameter of the steel section is 0.034 m. The shear elastic modulus for aluminum is G = 28 GPa and the shear elastic modulus for steel is G = 83 GPa. Calculate the angle of twist at point A if the value of T = 699 N m. Multiply your final answer by 104 and submit that as your input with four significant figures after the decimal point. For example, if your final answer is 0.00005555 then input 0.5555. Aluminum 900 mm 2T B HH Steel 600 mm T A