Find the force required to permanently deform (stretch such that it snaps back to >0.2% longer than original length) a .04 m diameter bar if YS = 1000 MPa.
http://en.wikipedia.org/wiki/Tensile_strength
force/area=1000M N/m^2
Put in the area, and solve for force in N.
Yield Strength times area.
1000e6 x pi(.02m)squared = answer in newtons
about 1.26e6 newtons
To find the force required to permanently deform a bar, we need to consider a few factors. One of the key properties we need to know is the Young's modulus (Y), which represents the stiffness of the material. In this case, you mentioned that the yield strength (YS) of the material is 1000 MPa, but we still need to determine the Young's modulus to proceed.
Unfortunately, the information provided (diameter and yield strength) is not sufficient to directly calculate the Young's modulus of the material. Therefore, we require additional information such as the material type or any other physical properties of the bar.
Once we have the Young's modulus, we can proceed with the following steps to calculate the force required to permanently deform the bar:
1. Calculate the cross-sectional area (A) of the bar using the given diameter.
A = (π/4) * d^2
= (π/4) * (0.04)^2
2. Determine the strain (ε) required for permanent deformation. In this case, it is specified that the bar should be stretched so that it snaps back to more than 0.2% longer than its original length.
ε = 0.002
3. Calculate the stress (σ) required for this strain using Hooke's law, which states that stress is proportional to strain:
σ = Y * ε
4. Calculate the force (F) required to produce this stress by multiplying the stress by the cross-sectional area:
F = σ * A
However, since the Young's modulus (Y) is not provided, I cannot give you the exact value for the force required to permanently deform the bar. Please provide the missing information, and I will be happy to help you with the calculations.