Answer
Brass alloy and Steel alloy
Work Step by Step
Given: cylindrical rod 500 mm long and diameter of 12.7 mm load is 29,000 N no plastic nor elongation of 1.3 mm. (Refer to the given Table (aluminum alloy, brass alloy, copper and steel alloy)
Required: possible candidates for the given condition
Solution: Using Equation 6.1, determine the yield strength, $σ = \frac{F}{A_{0}} = \frac{29,000 N}{π (\frac{12.7 \times 10^{-3} m}{2})^2}= 228.93 MPa \approx 229 MPa$
Aluminum alloy, brass alloy and steel alloy have yield strength greater than 229 MPa, thus possible candidates. Copper does not satisfies this condition.
Using combination of Equations 6.2 and 6.5, compute the maximum elongation of the three remaining candidates,
$Δl = \frac{σ l_{0}}{E} $
For aluminum alloy, $Δl = \frac{(229 MPa)(500 mm)}{70 \times 10^{3} MPa} = 1.64 mm$ (not a candidate, Δl exceeded 1.3 mm)
For brass alloy, $Δl = \frac{(229 MPa)(500 mm)}{100 \times 10^{3} MPa} = 1.13 mm$ (a candidate, Δl not exceeded 1.3 mm)
For steel alloy, $Δl = \frac{(229 MPa)(500 mm)}{207 \times 10^{3} MPa} = 0.55 mm$ (a candidate, Δl not exceeded 1.3 mm)
Thus, brass alloy and steel alloy satisfy the given conditions and are possible candidates.
