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The hydraulic cylinder CF,which partially controls the position of rod DE, has been locked in the position shown. Member BD is 15 mm thick and is connected to the vertical rod by a 9-mm-diameter bo t. Knowing that P = 2 kN and theta = 75 Degree. Determine (a) the average shearing stress in the bolt. (b) the bearing stress at C in member BD.
The hydraulic cylinder CF,which partially controls the position of rod DE, has been locked in the position shown. Member BD is 15 mm thick and is connected to the vertical rod by a 9-mm-diameter bo t. Knowing that P = 2 kN and theta = 75 Degree. Determine (a) the average shearing stress in the bolt. (b) the bearing stress at C in member BD.
Answer: The average shearing stress in the bolt is 31.48 MPa and the bearing stress at C in member BD is 17.3 MPa.Thehydraulic cylinderCF, which partially controls the position of rod DE, has been locked in the position shown. Member BD is 15 mm thick and is connected to the vertical rod by a 9-mm-diameter bolt.Knowing that P = 2 kN and theta = 75 Degree, we can determine(a) the average shearing stress in the bolt, and(b) the bearing stress at C in member BD.For (a), we can use the equation for average shearing stress:τavg= P/Awhere A is the cross-sectional area of the bolt, which is A = π * (d/2)2 = π * (9/2)2 = 63.6 mm2.Plugging in our values, we get:τavg= 2 kN/63.6 mm2= 31.48 MPaFor (b), we can use the equation for bearing stress:σb= P*sin(theta)/twhere t is the thickness of member BD, which is 15 mm.Plugging in our values, we get:σb= (2 kN)(sin 75°)/15 mm = 17.3 MPaTherefore, the average shearing stress in the bolt is 31.48 MPa and the bearing stress at C in member BD is17.3 MPa.Learn more aboutshearing stresshere:brainly.com/question/13034525##SPJ11...