Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_79

⚡ Mecademy AIENG정역학 · ch6  Problem Statement For the flat sanding disk of radius , the pressure developed between the disk and the sanded surface decreases linearly with from a value at the center to at . If the coefficient of friction is , derive the expression for the torque required to turn the shaft under an axial force . Problem 6/79 (a) Expression for the torque 1. Present Final Formula: The total axial force is the integral of the pressure over the contact area: The torque required to overcome friction is given by the integral of the differential friction moment: 2. Substitute Values: The pressure varies linearly from at to at : Substitute this into the expression for : Substitute into the expression for : ap rp 0 p /2 0 r=a μM L M L L= p( r )⋅ ∫ 0 a 2πrdr M M= μ⋅ ∫ 0 a p(r)⋅r⋅(2πrdr) p(r)p 0 r=0p /2 0 r=a p(r)=p + 0 r = ( a−0 p /2−p 00 )p 1− 0 ( 2a r ) L L= 2 πp 1− r d r 0 ∫ 0 a ( 2a r ) M M= 2 π μ p 1− r d r 0 ∫ 0 a ( 2a r ) 2 3. Partial Operations: Evaluate the integral for the axial force : From this, we find the central pressure in terms of : Now evaluate the integral for the torque : 4. Final Calculation: Substitute the expression for into the equation for : ● Final Conclusion: The torque required to turn the shaft is . ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 6/79 L L= 2 πp r− d r = 0 ∫ 0 a ( 2a r 2 ) 2 πp − = 0 [ 2 r 2 6a r 3 ] 0 a 2 πp − = 0 ( 2 a 2 6 a 2 ) p 0 L p = 0 2πa 2 3L M M= 2 π μ p r− d r = 0 ∫ 0 a ( 2 2a r 3 ) 2 π μ p − = 0 [ 3 r