Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_82

⚡ Mecademy AIENG정역학 · ch6  Problem Statement Determine the expression for the torque required to turn the shaft whose thrust is supported by a conical pivot bearing. The coefficient of friction is , and the bearing pressure is constant. Problem 6/82 (a) Torque Expression for Conical Pivot Bearing 1. Formula: We start by considering a differential area element on the conical bearing surface. For a conical pivot with vertex angle , an element at radius has a width along the slant . The area element is: The differential normal force is , where is the constant bearing pressure. The axial thrust is the integral of the vertical component of the normal force: The required torque is the integral of the moments of the differential friction forces : 2. Substitution: First, solve for the constant pressure in terms of the thrust and radii : Now substitute into the torque integral: 3. Calculation: Evaluate the integral and simplify the expression: ML μ dA αr ds= sin(α/2) dr dA=2πrds= sin(α/2) 2πrdr dN=pdAp L L= d N sin( α /2)= ∫ p sin( α /2)= ∫ r 1 r 2 ( sin(α/2) 2πrdr ) 2 π p r d r∫ r 1 r 2 M dF=μdN M=r d F = ∫ μ r d N = ∫ d r∫ r 1 r 2 sin(α/2) 2πμpr 2 pL r = 1 d /2,r = 12 d /2 2 L= 2 π p = [ 2 r 2 ] r 1 r 2 πp(r − 2 2 r )⟹ 1 2 p= π(r −r ) 2 2 1 2 L p M= r d r sin(α/2) 2πμ ( π(r −r ) 2 2 1 2 L )∫ r 1 r 2 2 Integral part: Torque expression: Convert radii to diameters (): Substitute back: 4. Result: ● Final Conclusion: The torque required to turn the shaft is . ✨ Final Answer Summary (a)