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daser333 [38]
2 years ago
10

A part has been tested to have Sut = 530 MPa, f = 0.9, and a fully corrected Se = 210 MPa. The design requirements call for the

part to be loaded at three different fully-reversed loads and cycled at each one for a set number of cycles. First, it will be loaded at ±350 MPa for 5,000 cycles. Then, it will be loaded at ±260 MPa for 50,000 cycles. Finally, it will be loaded at ±225 MPa until it fails. How many cycles do we expect the part to last at the final loading? Use Miner's method. :g

Engineering
1 answer:
kobusy [5.1K]2 years ago
8 0

Answer:

126984 cycles

Explanation:

Given data :

Sut = 530 MPa

f = 0.9

fully corrected Se = 210 MPa

using Miner's method attached below is the detailed solution of the given problem

when loaded with ± 225 MPa the number of cycles before it fails will be

≈  126984

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13- Convert the following numbers to the indicated bases. List all intermediate steps.
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Answer:

Following are the conversion to this question:

Explanation:

In point (a):

\to \frac{36459080}{8} = 4557385   + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673   + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \  \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\

\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}

In point (b):

\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \  \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}

In point (c):

\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\

\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}

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0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O

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