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4 years ago

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A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of

24 kpsi. After some calculation, the designer has selected a standard 1/2 in diameter (d) bar. The factor of safety (n) is ____. Round the answers to three significant digits.

1 answer:

schepotkina [342]4 years ago

5 0

Answer:

$n = 2.36$

Explanation:

The stress experimented by the circular bar is:

$\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)$

$\sigma = 10.186\,kpsi$

The safety factor is:

$n = \frac{24\,kpsi}{10.186\,kpsi}$

$n = 2.36$

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Answer:

A) ν = 0.292

B) ν = 0.381

Explanation:

Poisson's ratio = - (Strain in the direction of the load)/(strain in the direction at right angle to the load)

In axial tension, the direction of the load is in the length's direction and the direction at right angle to the load is the side length

Strain = change in length/original length = (Δy)/y or (Δx)/x or (ΔL/L)

A) Strain in the direction of the load = (2.49946 - 2.5)/2.5 = - 0.000216

Strain in the direction at right angle to the load = (7.20532 - 7.2)/7.2 = 0.0007389

Poisson's ratio = - (-0.000216)/(0.0007389) = 0.292

B) Strain in the direction of the load = (2.09929 - 2.1)/2.1 = - 0.0003381

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Poisson's ratio = - (-0.0003381)/(0.0008868) = 0.381

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3 years ago

Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of th

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Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, $\sigma$ is given by

$\sigma=\frac {F}{A}$ where F is force and A is area and area is given by $\frac {\pi d^{2}}{4}$ hence

$\sigma=\frac {4F}{\pi d^{2}}$ where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

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From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

$\epsilon=\frac {\triangle l}{l}$ where $\epsilon$ is the strain, $\triangle l$ is elongation and l is original length and making elongation the subject

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(b)

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