Wastewater flows into a _________ once it is released into a floor drain.

For a bronze alloy, the stress at which plastic deformation begins is 266 MPa and the modulus of elasticity is105 GPa.

pentagon [3]

Answer:

88750 N

Explanation:

given data:

plastic deformation σy=266 MPa=266*10^6 N/m^2

cross-sectional area Ao=333 mm^2=333*10^-6 m^2

solution:

To determine the maximum load that can be applied without

plastic deformation (Fy).

Fy=σy*Ao

=88750 N

7 0

3 years ago

In this assignment, you will demonstrate your ability to write simple shell scripts. This is a cumulative assignment that will c

nevsk [136]

Answer:

Explanation:

Usage: flip [-t|-u|-d|-m] filename[s]

Converts ASCII files between Unix, MS-DOS/Windows, or Macintosh newline formats

Options:

-u = convert file(s) to Unix newline format (newline)

-d = convert file(s) to MS-DOS/Windows newline format (linefeed + newline)

-m = convert file(s) to Macintosh newline format (linefeed)

-t = display current file type, no file modifications

8 0

3 years ago

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0

3 years ago

Engineering Careers Scavenger Hunt

emmasim [6.3K]

Answer:

c

Explanation:

it's the only engineering career

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3 years ago

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Anaircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa 1/2.It has been d

navik [9.2K]

Answer:

Yes, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40MPa

Explanation:

Given

Toughness, k = 40Mpa

Stress, σ = 300Mpa

Length, l = 4mm = 4 * 10^-3m

Under which fracture occurred (i.e., σ= 300 MPa and 2a= 4.0 mm), first we solve for parameter Y (The dimensionless parameter)

Y = k/(σπ√a)

Where a = ½ of the length in metres

Y = 40/(300 * π * √(4/2 * 10^-3))

Y = 1.68 ---- Approximated

To check if fracture will occur of not; we apply the same formula.

Y = k/(σπ√a)

Then we solve for k, where

σ = 260Mpa and a = ½ * 6 * 10^-3

So,.we have

1.68 = k/(260 * π * √(6*10^-3)/2)

k = 1.68 * (260 * π * (6*10^-3)/2)

k = 42.4 MPa --- Approximately

Therefore, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40 MPa

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3 years ago

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