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3 years ago

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For a rod of annealed AISI 1018 steel with a cross sectional area of 0.65 in^2?; what is the maximum tensile load Pmax that shou

ld be placed on it given a design factor of 3 to avoid yielding?

1 answer:

lawyer [7]3 years ago

7 0

Answer:

maximum tensile load Pmax is 11.91 ksi

Explanation:

given data

area = 0.65 in²

design factor of safety = 3

to find out

what is the maximum tensile load Pmax

solution

we know here area is 0.65 in² and FOS = 3

so by steel table for rod of annealed AISI 1018 steel table we know σy = 55 ksi

so

we use here design factor formula that is

$\frac{ \sigma y}{FOS} = \frac{Pmax}{area}$ .............1

put here all these value we get Pmax in equation 1

$\frac{55}{3} = \frac{Pmax}{0.65}$

Pmax = 11.91 ksi

so maximum tensile load Pmax is 11.91 ksi

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From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

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Part 2

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$250 = 70+85+40 +t_{INT}$

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A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

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