First of all, there are not <u>just</u> two applications that are solely applicable to the electron beam welding process. There are MANY.
Please visit out website at the URL below and you can click the "View Application" button under each listed Industry segment to view case studies of commonly EB welded applications.
https://www.ptreb.com/electron-beam-welding-applications
And for more general information on our welding process, we have an informational section you can peruse as well:
https://www.ptreb.com/electron-beam-welding-information
Good luck with your assignment- we are glad to hear they are teaching about EBW in high school!!!
Answer:
3.34×10^-6m
Explanation:
The shear modulus can also be regarded as the rigidity. It is the ratio of shear stress and shear strain
can be expressed as
shear stress/(shear strain)
= (F/A)/(Lo/ . Δx)
Stress=Force/Area
The sheear stress can be expressed below as
F Lo /(A *Δx)
Where A=area of the disk= πd^2/4
F=shearing force force= 600N
Δx= distance
S= shear modulus= 1 x 109 N/m2
Lo= Lenght of the cylinder= 0.700 cm=7×10^-2m
If we make Δx subject of the formula we have
Δx= FLo/(SA)
If we substitute the Area A we have
Δx= FLo/[S(πd^2/4]
Δx=4FLo/(πd^2 *S)
If we input the values we have
(4×600×0.7×10^-2)/10^9 × 3.14 ×(4×10^-2)^2
= 3.35×10^-6m
Therefore, its shear deformation is 3.35×10^-6m
A=area of the disk= πd^2/4
= [3.142×(4×10^-2)^2]/4
Answer:
1.805 mm
Explanation:
Extension in the steel wire = WL_{steel}/AE_{steel}
Extension in the aluminium wire = WL_{Al}/AE_{Al}
Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})
we have:
W = mg
W = 5 × 9.8
W = 49 N
Area A = π/4 × (0.001)²
= 7.85398 × 10 ⁻⁷ m²
Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})
Total extension = 49/ 7.85398 × 10 ⁻⁷ ( (1.5/ 200×10⁹) + 1.5/ 70×10⁹))
Total extension = 0.0018048
Total extension = 1.805 mm
Thus, the total extension = the resulting change in the length of this composite wire = 1.805 mm
Newton's second law states that the force applied to an object is equal to the product between the mass m of the object and its acceleration a:
Using
and
, we can find the value of the force applied to the roller-blade to obtain this acceleration:
