Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_96

⚡ Mecademy AIENG정역학 · ch6  Problem Statement Repeat Prob. 6/95, only now the coefficients of static friction are as follows: 0.60 at A and C and 0.20 at B. Plot as a function of for , where is the common radius of all three shafts. What are the limiting values of for and for large values of ? (Note: From Prob. 6/95, the positions of shafts and are fixed at a horizontal distance of apart, while shaft is located midway between them and its vertical position can be varied. All shafts are fixed against rotation.) Problem 6/96 (a) Analysis of the Limiting Values and General Expression for 1. Formula: The tension required to initiate motion is given by the belt friction formula: where is the coefficient of static friction and is the wrap angle in radians for each shaft. 2. Substitution: From the geometric arrangement: Shafts and are separated horizontally by . Shaft is midway, so the horizontal distance from to (and to ) is . Let be the angle the straight rope segment makes with the horizontal. Considering the finite radius , the tangency condition for a rope passing under and over yields: The wrap angles are determined as follows: Shaft (under): Shaft (over): Shaft (under): (by symmetry) T/mgy0≤y≤10rr T/mgy=0 yAC 6rBy T/mg = mg T e = μ β ∑ ii e μ β +μ β +μ β AABBCC μ i β i AC6rB ABBCL=3r αAB rA B 3rsinα−ycosα=2r⟹3sinα− cos α = r y 2 Aβ = A + 2 π α Bβ = B 2α Cβ = C + 2 π α Substituting friction coefficients and : 3. Calculation: For : Substitute into the geometric relation: For la