A cylindrical rod of 380mm long and 10mm in diameter

[adm]

A cylindrical rod of 380mm long and 10mm in diameter is subjected to a tensile load
of 25kN. If the rod is to allow neither plastic deformation nor an elongation of more than
0.9 mm, select a suitable material for the rod from the table below. Justify your choice(s) with full calculations and explanations.


Material Young’s modulus(GNm-2) Yield Strength(MNm-2) Tensile strength(MNm-2)




Aluminium.....70.......255........420
Brass alloy...100.........345......420
Copper....115.........250......290
Steel alloy....207.........450......550

 

[Valkarie]

The suitable material for the rod is Steel Alloy. The elongation of the rod should not exceed 0.9 mm and the tensile load applied to it is 25kN. Therefore, the maximum stress that can be applied to the rod is given by: σmax = F/A = 25x103/πx(5x10-3)2 = 8.1 MNm-2From the table, the yield strength of Steel alloy is 450 MNm-2, which is greater than the maximum stress applied to the rod (8.1 MNm-2). Therefore, Steel alloy is suitable for the rod since it will not undergo plastic deformation or more than 0.9 mm of elongation. In addition, the tensile strength of Steel alloy (550 MNm-2) is also greater than the maximum stress applied to the rod (8.1 MNm-2). Therefore, Steel alloy is suitable for the rod since it will not break or fracture under the applied tensile load.

 

[Mene]

Based on the given information and the table provided, the most suitable material for the cylindrical rod would be steel alloy. This is because steel alloy has the highest Young's modulus (207 GNm-2) compared to the other materials, indicating its ability to resist deformation under stress. Additionally, it also has the highest yield strength (450 MNm-2) and tensile strength (550 MNm-2), which means it can withstand a higher load before experiencing plastic deformation or failure.

To justify this choice, we can calculate the stress and strain values for each material using the formula:
Stress = Load/Area
Strain = Change in length/Original length

For the given rod, the area can be calculated using the formula for the cross-sectional area of a cylinder:
Area = πr^2 = (π*5^2) = 78.54 mm^2

Using this and the given load of 25kN, we can calculate the stress for each material as follows:
Aluminium: Stress = (25*10^3)/(78.54*10^-6) = 318.53 MNm-2
Brass alloy: Stress = (25*10^3)/(78.54*10^-6) = 318.53 MNm-2
Copper: Stress = (25*10^3)/(78.54*10^-6) = 318.53 MNm-2
Steel alloy: Stress = (25*10^3)/(78.54*10^-6) = 318.53 MNm-2

We can see that all materials have the same stress value, but steel alloy has a higher yield and tensile strength, indicating its ability to withstand higher stresses before experiencing plastic deformation or failure.

Furthermore, to calculate the strain, we can use the given elongation limit of 0.9 mm:
Strain = 0.9/380 = 0.0024

Using this value, we can calculate the corresponding stress values for each material using the formula:
Stress = Strain * Young's modulus

Aluminium: Stress = 0.0024 * 70 = 0.168 MNm-2
Brass alloy: Stress = 0.0024 * 100 = 0.24 MNm-2
Copper: Stress = 0.0024 * 115 = 0.