Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_67

⚡ Mecademy AIENG정역학 · ch6  Problem Statement Circular disk is placed on top of disk and is subjected to a compressive force of . The diameters of and are and , respectively, and the pressure under each disk is constant over its surface. If the coefficient of friction between and is , determine the couple which will cause to slip on . Also, what is the minimum coefficient of friction between and the supporting surface which will prevent from rotating? Problem 6/67 (a) Couple to cause disk to slip on 1. Formula: For a circular area with uniform pressure , the resisting torque is given by: 2. Substitution: Substitute the known values for disk slipping on : 3. Calculation: Perform the arithmetic: 4. Result: ● Final Conclusion: The couple required to cause disk to slip on disk is AB80 lb AB9 in.12 in. AB0.40 MAB μBCB MAB pM M= μ⋅ ∫ 0 R p⋅rdA= μ r (2 πr d r )= ∫ 0 R ( πR 2 P ) μ PR 3 2 AB μ=μ = AB 0.40 P=80 lb R=R = A = 2 d A = 2 9 4.5 in. M= (0.40)(80 lb)(4.5 in.) 3 2 M= ⋅ 3 2 0.40⋅80⋅4.5 M=0.40⋅80⋅ ⋅4.5( 3 2 ) M=32⋅3=96 lb⋅in. M=96 lb⋅in. AB96 lb⋅in. (b) Minimum coefficient of friction between and surface 1. Formula: For disk to remain stationary, the maximum frictional torque between disk and surface must be at least equal to the torque transmitted from disk . Using the uniform pressure formula for the - interface (): 2. Substitution: Substitute the values for disk and the result from part (a): 3. Calculation: Solve for : 4. Result: ● Final Conclusion: The minimum coefficient o