why would it be inappropriate to dimension to a feature on a surface that is not perpendicular to the line of sight?

1 answer:

4 0

If you are drawing and dimensioning with a computer program the dimension will be inaccurate... If it is mechanical drawing then the fabricator would not have enough information to accurately measure the component. ie a circle turned a few degrees away from perp. would appear to be an ellipse. and may actually dimension that way

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True or false? Don't break or crush mercury-containing lamps because mercury powder may be released.

mars1129 [50]

Answer:true

Explanation:

7 0

4 years ago

A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not e

mash [69]

Answer:

The steel is a candidate.

Explanation:

Given

P = 27,500 N

d₀ = 10.0 mm = 0.01 m

Δd = 7.5×10 ⁻³ mm (maximum value)

This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is applied; this means that the stress corresponding to this load not exceed the yield strength of the material.

Upon computing the stress

σ = P/A₀ ⇒ σ = P/(π*d₀²/4) = 27,500 N/(π*(0.01 m)²/4) = 350*10⁶ N/m²

⇒ σ = 350 MPa

Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.

Relative to the second criterion, (i.e., that Δd be less than 7.5 × 10 ⁻³ mm), it is necessary to calculate the change in diameter Δd for these four alloys.

Then we use the equation υ = - εx / εz = - (Δd/d₀)/(σ/E)

⇒ υ = - (E*Δd)/(σ*d₀)

Now, solving for ∆d from this expression,

∆d = - υ*σ*d₀/E

For the Aluminum alloy

∆d = - (0.33)*(350 MPa)*(10 mm)/(70*10³MPa) = - 0.0165 mm

0.0165 mm > 7.5×10 ⁻³ mm

Hence, the Aluminum alloy is not a candidate.

For the Brass alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(101*10³MPa) = - 0.0118 mm

0.0118 mm > 7.5×10 ⁻³ mm

Hence, the Brass alloy is not a candidate.