The Theory of Shrinkage And Forced Fits With Tabulated Data And Examples From P
The Theory of Shrinkage And Forced Fits With Tabulated Data And Examples From P
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234. Sub- A stituting in (38) 8 = 0. 0003 inch, which, owing to the lower tensile strength of cast iron, is about one-third of the shrinkage-allowance in Example 1, although the stress is two-thirds of the elastic limit. For a forced fit, good practice gives (see Table VI) a unit-allowance of 0. 0013 inch, or one-third greater than that of Example 1. The stresses which such an allowance would produce are, however, uncer- tain, as will be further discussed in the following chapter. Example 4. Wh...at is the radial pressure P x in the above examples?
Pi For Examples 1 and 2, we find from Table I that - = 0. 341. Hence, T 2 P l = 25, 000 X 0. 341 = 8525 pounds per square inch.
Pi For Example 3, we find from Table I that = 0. 351. In this case T 2 r a = 4000, hence, P x = 4000 X 0. 351 = 1404 pounds per square inch.
Example 5. What is the resistance to slip per inch of length of hub in Example 3?
In Equation (30), A = 10, L = l, and from Example 4 we have P t = 1404; / may be taken as 0. 2. Then Q = 8817 pounds, which is the total resistance of a ring of the hub, one inch in length.
What to read after The Theory of Shrinkage And Forced Fits With Tabulated Data And Examples From P?
You can find similar books in the "Read Also" column, or choose other free books by William Ledyard Cathcart to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
234. Sub- A stituting in (38) 8 = 0. 0003 inch, which, owing to the lower tensile strength of cast iron, is about one-third of the shrinkage-allowance in Example 1, although the stress is two-thirds of the elastic limit. For a forced fit, good practice gives (see Table VI) a unit-allowance of 0. 0013 inch, or one-third greater than that of Example 1. The stresses which such an allowance would produce are, however, uncer- tain, as will be further discussed in the following chapter. Example 4. Wh...at is the radial pressure P x in the above examples?
Pi For Examples 1 and 2, we find from Table I that - = 0. 341. Hence, T 2 P l = 25, 000 X 0. 341 = 8525 pounds per square inch.
Pi For Example 3, we find from Table I that = 0. 351. In this case T 2 r a = 4000, hence, P x = 4000 X 0. 351 = 1404 pounds per square inch.
Example 5. What is the resistance to slip per inch of length of hub in Example 3?
In Equation (30), A = 10, L = l, and from Example 4 we have P t = 1404; / may be taken as 0. 2. Then Q = 8817 pounds, which is the total resistance of a ring of the hub, one inch in length.
What to read after The Theory of Shrinkage And Forced Fits With Tabulated Data And Examples From P?
You can find similar books in the "Read Also" column, or choose other free books by William Ledyard Cathcart to read onlineMoreLess
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