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lutik1710 [3]
2 years ago
14

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha

t is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm2 without plastic deformation
Engineering
1 answer:
Anton [14]2 years ago
3 0

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

\sigma=\frac{F}{A}

F=\sigma*A

Convert given values to base units:

275 MPa = 275*10^{6} Pa

325 mm^{2} = 0.000325 m^{2}

Substituting in given values:

F = (275*10^{6})*(0.000325)=89375 N

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7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21
Irina18 [472]

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

In this problem, the first 3 timing stages are given as:

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We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

g = (\frac{v}{s}) (\frac{C}{X})

Let's first find the fourth stage critical lane group ratio \frac{v}{s} , using the formula:

C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}

Solving for (\frac{v}{s}), we have:

(\frac{v}{s}) = 0.185

Let's also calculate the volume capacity ratio X,

X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

g_4 = (\frac{v}{s}) (\frac{C}{X})

Substituting figures in the equation, we now have:

g_4 = (0.185) (\frac{60}{0.704})

g_4 = 15.78 seconds

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

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