Problem 1: If the parts of the shaft below are a 1" diameter solid steel rod and a 2" OD diameter steel pipe with a 0.25" wall thickness, what is the torsional stress in each section of the shaft? B 6 in. 15 lb 8 in. 15 lb
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- 3) A 25 mm solid round shaft is supported by self-aligning bearings at A and B (the bearings do not support axial loads). Two chain sprockets are attached to the shaft as shown. a. Find the force, F, on the 125 mm diameter sprockets. b. Construct the V-M-N-T diagrams for the shaft. c. Determine the shaft location with the most severe state of stress. d. Sketch the Mohr's circle and find the principal stresses at that location. A 125 mm dia. 50-mm dia. 75 mm 25mm dia. 5000 N 150 mm 75 mmx 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 Fs2.A lever secured to a 2 inches round shaft by a steel tapered pin (d = 3/8”) has a pull of 50 lbs at a 30 inch radius from shaft center. Find the unit working stress of the pin.
- please box in the following answers: normal stresses for beam AB and BC (mid section) normal stresses for beam AB and BC (at connections A and C) shear stress for pin A, B and C bearing stress along member AB and BC at pin section for A and C 600mm Flat end 40 mm FRONT VIEW d-25 mm d-25 mm 50 mm 125 mm 30 mm 25mm ,20 mm TOP VIEW OF ROD BC -800 mm Q-30 kN TOP VIEW OF BOOM AB d=25mm Flat end Q-30KN 20mm 20 mm END VIEW 600 mm 50 mm 20 mm 800 mm 30 AN 1A flanged bolt coupling is used to connect a solid shaft 90 mm in diameter to a hollow shaft 100 mm in outsude diameter and 90 mm in inside diameter. If the allowable aring stress in the shafts and the bolts is 60 MPa, how many 10 mm diameter steel bolts must be used on a 200 mm diameter bolt circle so that the coupling will be as strong as the weaker shaft?100 kip 10 kip -ft The 4-in. diameter shaft shown above is made from a ductile brass alloy. Which of the following are closest to the principal stresses in the shaft? O01 = 31.8 ksi 02 = 76.4 ksi O 01 = 94.0 ksi 02=-62.1 ksi O 01 = 14.3 ksi %3D 02 = -6.37 ksi O o1 = 7.96 ksi !! 02 = 9.55 ksi
- 100 kip 10 kip ft The 4-in. diameter shaft shown above is made from a ductile brass alloy. Which of the following are closest to the principal stresses in the shaft? 01 = 31.8 ksi 02 = 76.4 ksi = 94.0 ksi 02 = -62.1 ksi O0g - 14.3 ksi 02 = -6.37 ksi Oo= 7.96 ksi 02 = 9.55 ksi500 lb in. 1200 lb - in. 400 lb - in. dep-0.9 in. dec =0.75 in. B dan = 0.6 in. Specifications: Shafts AB, BC, CD are solid material. 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.The maximum torsional shear stress of a shaft with diameter D can be found at what distance from the center of the shaft?
- A sleeve made of AISI 1040 CD steel has inner and outer diameters 40mm and 50mm, respectively. It is press fit on a solid shaft of same material with diameter 42mm such that the nominal diameter between sleeve and shaft became 41mm. a. Find the pressure developed between shaft and sleeve. b. Draw the tangential, radial, and total stress as a function of radius, for both sleeve and shaft. c. Find the maximum shaft diameter that can be used before failure.4. As shown below, the prosthesis is made of two parts a core (Ge=150 ksi) and a tube (G=85 ksi). The prostheses is under torsional torque of 15 lbf.in due to foot movement. a) Determine the maximum torsional stresses in the core and tube. b) Determine the angle of twist 15 Ibf.in 15 lbf.in 15in Diameter 1.5 in Diameter 1in 15 Ibf.in (Answer: a)t, = 23.13 psi; t = 19.67 psi b) = 0.0046 rad)FIGURE 12 - 7 in.- Refer to figure 12. The constant wall thickness of a steel tube with the cross section shown is 3/8 in. If a 5,625 lb-ft torque is applied to the tube, find the shear stress in the wall of the tube. Neglect 3 in. 5 in. stress concentrations at the corners. 4 in. *Dimensions are measured on-center