Engineering

Engineering, 18.07.2019 02:20, jallison61

Considering the shrinkage only condition, if the two cylinders are 0.2 m in length and using the radial pressure you determined in part (b) above, calculate the longitudinal force 'f' required to separate the two cylinders. take usteel=0.15

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Engineering, 17.07.2019 20:30, vaeh41

Acircular steel cylinder of similar external and internal dimension to that in part (a) above, is now made by shrinking a cylinder of external diameter 0.2 m and internal diameter 0.1 m, onto an internal cylinder of similar material with external diameter 0.1 m and internal diameter 0.05 m. determine the radial pressure 'or' at the common surface due to shrinkage alone, so that when there is an internal pressure of 300 mn/m2, the maximum hoop stress σh in the inner cylinder is 150 mn/m2.

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Engineering, 18.07.2019 19:30, sethjohnson386pbnm3x

One steel cylinder is shrunk onto another. the compound cylinder has an inside diameter of 120mm, an outside diameter of 240mm and a diameter of 180mm at the surface of contact. the shrinkage produces a radial pressure p at the surfaces in contact. after the compound cylinder is subjected to an internal pressure pi, find the ratio of p to pi so that the maximum hoop stress in the two cylinders would be the same.

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Business, 30.09.2019 23:30, livleluve76

All of the following statements regarding inventory shrinkage are true except: a. inventory shrinkage refers to the loss of inventory. b. inventory shrinkage is determined by comparing a physical count of inventory with recorded inventory amounts. c. inventory shrinkage is recognized by debiting cost of goods sold. d. inventory shrinkage is recognized by debiting an operating expense. e. inventory shrinkage can be caused by theft or deterioration.

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Mathematics, 26.10.2019 01:43, angela6844

3. a machine part is manufactured from a block of iron with circular cylindrical slots. the block of iron has a width of 14 in., a height of 16 in., and a length of 20 in. the number of cylinders drilled out of the block is determined by the weight of the leftover block, which must be less than 1,000 lb. a. if iron has a weight of roughly 491 lb/ft3, how many cylinders with the same height as the block and with radius 2 in. must be drilled out of the block in order for the remaining solid to weigh less than 1,000 lb.? . if iron ore costs $115 per ton (1 ton=2200 lb.) and the price of each part is based solely on its weight of iron, how many parts can be purchased with $1,500? explain your answer.

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Considering the shrinkage only condition, if the two cylinders are 0.2 m in length and using the rad...

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