mimithurmond03
12.02.2020 •
Mathematics
A ½in diameter rod of 5in length is being considered as part of a mechanical linkagein which it can experience a tensile loadingand unloading during application.The material being considered is a standard aluminum bronze with a Young’s modulus of 15.5 Msi. What is the maximum load therod can take before itstarts to permanentlyelongate?Considering the rod is designed so thatit does not yield, what is the maximum energy that the rod would store? [Hint: You can look up someproperties of aluminum bronze in yourmachinedesign book].
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Ответ:
a. Maximum Load = Force = 27085.09 N
b. Maximum Energy = 3.440 Joules
Step-by-step explanation:
Given
Rod diameter = ½in = 0.5in
Length = 5in
Young's modulus = 15.5Msi
By applying the 0.2% offset rule,
The maximum load the rod can hold before it gets to breaking point is given as follows by taking the strain as 0.2%
Young Modulus = Stress/Strain Make Stress the Subject of Formula
Stress = Strain * Young Modulus
Stress = 0.2% * 15.5
Stress = 0.002 * 15.5
Stress = 0.031Msi
Calculating the area of the rod
Area = πr² or πd²/4
Area, A = 22/7 * 0.5^4 / 4
A = 22/7 * 0.25 / 4
A = 5.5/28
A = 0.1964in²
The maximum load that the rod would take before it starts to permanently elongate is given by
Force = Stress * Atea
Force = 0.031Msi * 0.1964in²
Force = 31Ksi * 0.1964in²
Force = 6.089Ksi in²
Force = 6.089 * 1000lbf
Force = 6089 lbf
1 lbf = 4.4482N
So, Force = 6089 * 4.4482N
Force = 27085.09 N
b.
Using Strain to Energy Formula
U = V×σ²/2·E
Where V = Volume
V = Length * Area
V = 5 in * 0.1964in²
V = 0.982in³
σ = Stress = 0.031Msi
= 0.031 * 1000Ksi
= 31Ksi
= 31 * 1000psi
= 31000psi
E = Young Modulus = 15.5Msi
= 15.5 * 1000Ksi
= 15.5 * 1000 * 1000psi
= 15500000psi
So,
Energy = 0.982 * 31000²/ ( 2 * 15500000)
Energy = 943,702,000/31000000
Energy = 30.442in³psi
Converted to ftlbf
Energy = 2.537 ftlbf
Converted to Joules
Energy = 3.440 Joules
Ответ:
Uh Hi
Step-by-step explanation: