as the following standard
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- Repeat Problem 11.2-3 assuming that R= 10 kN · m/rad and L = 2 m.A motor driving a solid circular steel shaft with diameter d = 1.5 in, transmits 50 hp to a gear at B, The allowable shear stress in the steel is 6000 psi. Calculate the required speed of rotation (number of revolutions per minute) so that the shear stress in the shaft does not exceed the allowable limit.Compare the angle of twist 1 for a thin-walled circular tube (see figure) calculated from the approximate theory for thin-walled bars with the angle of twist 2 calculated from the exact theory of torsion for circular bars, Express the ratio 12terms of the non-dimensional ratio ß = r/t. Calculate the ratio of angles of twist for ß = 5, 10, and 20. What conclusion about the accuracy of the approximate theory do you draw from these results?
- A shaft with a diameter d = 25 mm must transfer a torque of 23 Nm by using a key joint. The key has the following standard dimensions: b x h x l = 8 mm x 7 mm x 18 mm. Calculate the stresses.Design a square key for a 3 in diameter shaft to withstand a torque of 30000lb-in. The design stresses in shear and bearing can be safely taken as 11 ksi and 22 ksi, respectively. Find the length (inches) From Table AT 19, b=t=3/4 in4. What length of a square key is required for a 4-in diameter shaft transmitting 1000 hp at 1000 RPM? The allowable shear and compressive stresses in the key are 15 ksi and 30 ksi, respectively.
- A keyed sprocket delivers a torque of 778.8 N-m thru the shaft of 54 mm OD. The key thickness is 1.578 cm and the width is 1.11 cm. Compute the length of the same key. The permissible stress value of 60 MPa for shear and 90 MPa for tension. (Ans: 4.33 cm)Bearing stress 4. Please provide proper discussion and illustration also complete solution and clear solution please thank you Determine the length of a square key to be used for a 3.73 kW, 1800 rpm electric motor if the motor shaft diameter is 35 mm and the width of the key is approximately one-fourth of the shaft diameter. The allowable shearing stress on the key is 2.45 MPax Question: The diameter of shaft used in the below design is 90 mm. Identical pulleys are used in the design with a radius of 160 mm. The forces acting on the system are defined as follow: F₁ = 1P kN; F₂ = 8P kN; F3 = 2P kN; F₁ = 6P kN; F5 = 4P kN; F = 7P kN. Employing the maximum shear stress theory, calculate the maximum permissible P for a safe design experience. L₁ = 1.4m; L₂ = 1.8 m; L3 = 1.4m; L4 = 1.6 m; allow = 75 MPa (allowable strength, σs) A F₁ F₂ L2 m F4 B 3 F6 Fs
- Design a rigid flange coupling to transmit a torque of 250N-m between two coaxial shafts. The shaft is made of alloy steel, flanges out of cast iron and bolts out of steel. Four bolts are used to couple the flanges. The shafts are keyed to the flange hub. The permissible stresses are given below: Shear stress on shaft =100MPa Bearing or crushing stress on shaft =250MPa Shear stress on keys =100MPa Bearing stress on keys = 250MPa Shearing stress on cast iron = 200Mpa Shear stress on bolts =100MPa Torque = 257N-mDesign a rigid flange coupling to transmit a torque of 257N-m between two coaxial shafts. The shaft is made of alloy steel, flanges out of cast iron and bolts out of steel. Four bolts are used to couple the flanges. The shafts are keyed to the flange hub. The permissible stresses are given below: Shear stress on shaft =100MPa Bearing or crushing stress on shaft =250MPa Shear stress on keys =100MPa Bearing stress on keys = 250MPa Shearing stress on cast iron = 200Mpa Shear stress on bolts =100MPa Answer in Word pleaseCompute the torsional shear stress in a circular shaft with a diameter of 50 mm that is subjected to a torque of 800N-m.