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4500_HW1_Sol

# 4500_HW1_Sol - MAE 4500 Home work 1 Solution Fall Semester...

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MAE 4500 Home work # 1 Solution Fall Semester 2006 ______________________________________________________________________ 1) A pin and a bushing assembly (figure 1) is to be used as a part of a locating fixture for machine tools. The pin is to have a class 5 fit with the reamed bushing hole into which it fits. Make a sketch of the pin and the hole, dimension it, and do an allowance analysis. 0.8500 " PIN BUSHING FIGURE. 1 Solution: Since a class 5 fit is specified and the pin is to be used as a location positioning device, an LC 5 fit will be used. The basic hole system (the nominal size is the basic size of the hole) will be used. Clearance Hole Shaft 0.3 +0.8 -0.3 1.6 0.0 -0.8 Smallest hole size = Nominal hole size = Basic hole size = 0.8500" Largest hole = Smallest hole + 0.0008 = 0.8508" Largest shaft = Nominal size - 0.0003 = 0.8497" Smallest shaft = Largest shaft -(-0.0003+0.0008) = 0.8492" Allowance Analysis Minimum clearance = Smallest hole - largest shaft = 0.0003" Maximum clearance = Largest hole - smallest shaft = 0.0016"

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Hole 0008 0 0000 0 8500 0 . . . + Shaft 0000 0 0005 0 8497 0 . . . + 2) A 2.000 key is to fit a slot in a fixture in order to locate a positioning block.
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4500_HW1_Sol - MAE 4500 Home work 1 Solution Fall Semester...

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