Answer: It does make sense, because I've been involved in these careers and have a long family line of them. And other questions?
Explanation:
Answer: both mm and inches on each dimension in a sketch (with the main dimension in one format and the other in brackets below it), in the way you can have dual dimensions shown when detailing an idw view.
personally think it would look a mess/cluttered with even more text all over the sketch environment, but everyone's differenent.
If it's any help - you know you can enter dimensions in either format? If you're working in mm you can still dimension a line and type "2in" and vice-versa. Probably know this already, but no harm saying it, just in case.
You can enter the units directly in or mm and Inventor will convert to current document settings (which you can change - maybe someone can come up with a simple toggle icon to toggle the document settings). Tools>Document Settings>Units
Unlike SolidWorks when you edit the dimension the original entry shows in the dialog box so it makes it easy to keep track of different units even if they aren't always displayed. (SWx does the conversion or equation and then that is what you get.)
I work quite a bit in inch and metric and combination (ex metric frame motor on inch machine) and it doesn't seem to be a real difficulty to me.
Answer:
β = 
= 0.7071 ≈ 1 ( damping condition )
closed-form expression for the response is attached below
Explanation:
Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s
computing a solution :
M = 1,
c = 2,
k = 2,
Wn = 
= 
next we determine the damping condition using the damping formula
β = = 0.7071 ≈ 1
from the condition above it can be said that the damping condition indicates underdamping
attached below is the closed form expression for the response
Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
Answer:
Pressure = 11.38 psi
Force = 13.981 Ibf
Explanation:
Step by step solution is in the attached document.
