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VikaD [51]
3 years ago
13

g A plane stress element has components sigma x = 160 MPa, tau xy = 100 MPa (CW). Determine the two values pf sigma y for which

the maximum shear stress is 73 MPa.

Engineering
1 answer:
Volgvan3 years ago
6 0

Answer:

The question is mentioned in the attachment.

Explanation:

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A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu
Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

The characteristic length of the strip is given by;

L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m

The Biot number is given as;

B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952

B_i < 0.1,  thus apply lumped system approximation to determine the constant time for the process;

\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s

The time for the heating process is given as;

t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0
3 years ago
Most licensed architects are members of which association?
olga2289 [7]

Answer:

The ALA, or Association of Licensed Architects. Hope this helps.

Explanation:

7 0
3 years ago
Read 2 more answers
In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini
storchak [24]

Answer:

attached below

Explanation:

8 0
3 years ago
I need the answer, step by step
Fynjy0 [20]

Answer:

HFOUGTYSLDÑÑYHÑ.HDug hyfñi755+464654795

Explanation:

+10000000iq

6 0
4 years ago
A room in the lower level of a cruise ship has a 30 cm diameter circular window If the midpoint of the window is 4 m below the w
AleksAgata [21]

Answer:

F = 2840.3 N

Explanation:

Given:

- Diameter of window D = 0.3 m

- Midpoint of window from sea level h = 4 m

- Specific gravity of sea water S.G = 1.024

- Density of water p = 1000 kg/m^3

Find:

The hydro-static force F_r acting on the mid-point of the window.

Solution:

- The average pressure P acting on the midpoint of the window:

                               P = S.G p*g*h

                               P = 1.024*1000*9.81*4

                               P = 40181.76

- The hydro-static force F_r acting on the mid-point of the window:

                               F = P*A = P*pi*D^2 / 4

                               F = 40181.76*pi*0.3^2 / 4

                               F = 2840.3 N

7 0
3 years ago
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