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3 years ago

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What is the maximum value of the bending stress at the critical cross-section?

1 answer:

Greeley [361]3 years ago

8 0

Answer:

The bending stress increases linearly away from the neutral axis until the maximum values at the extreme fibers at the top and bottom of the beam. The maximum bending stress is given by: where c is the centroidal distance of the cross section (the distance from the centroid to the extreme fiber).

Explanation:

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Why is a crank-rocker mechanism more useful than a double-rocker mechanism?

klasskru [66]

Answer:

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Explanation:

Crank rocker mechanism is formed when the shorter link revolves and the others rock( oscilllates)

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Crank rocker mechanism can be used to convert rotary motion into oscillatory motion.

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3 years ago

Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stre

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3 years ago

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To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105

GenaCL600 [577]

Answer:

$62.14\ \text{miles}$

$6213727.37\ \text{miles}$

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = $10^5\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}$

$1\ \text{mile}=1609.34\ \text{m}$

$\dfrac{10^5}{1609.34}=62.14\ \text{miles}$

The chain would extend $62.14\ \text{miles}$

Dislocation density = $10^{10}\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}$

$\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}$

The chain would extend $6213727.37\ \text{miles}$

3 0

3 years ago

A round bar of chromium steel, (ρ= 7833 kg/m, k =48.9 W/m-K, c =0.115 KJ/kg-K, α=3.91 ×10^-6 m^2/s) emerges from a heat treatmen

Lerok [7]

Answer:

Q = 424523.22 kw

Explanation:

$\rho =7833 kg/m$

k = 48.9 W/m - K

c = 0.115 KJ/kg- K

$\alpha = 3.91*10^{-6} m^2/s$

$T_s = 285 degree celcius$

T_∞ = 35 degree celcius

velocity of air stream = 15 m/s

D = 40 cm

L = 200 cm

mass flow rate $\dot m = \rho AV = 7833 *\frac{\pi}{4} 0.4^2*15$

$\dot m = 14764.85 kg/s$

$A_s = \pi DL = \pi 0.4*2 = 2.513 m^2$

$Q = \dot m C \Delta T = h A_s \Delta T$

$\dot m C \Delta T = h A_s \Delta T$

solving for h

$h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}$

h = 675.6 kw/m^2K

$Q = h A_s\Delta T$

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

7 0

2 years ago