Answer:
Pmax = 67.5 KN
Explanation:
We need to calculate the maximum allowable value of P for both aluminum and steel bars.
<u>FOR STEEL BARS</u>:
Since,
(σallow)st = (Pmax)st/A
where,
(σallow)st = maximum allowable stress of steel bar = 200 MPa = 2 x 10⁸ Pa
A = Cross-sectional area of steel bar = 450 mm² = 0.45 x 10⁻³ m²
(Pmax)st = Maximum allowable force for steel bar = ?
Therefore,
2 x 10⁸ Pa = (Pmax)st/0.45 x 10⁻³ m²
(Pmax)st = (2 x 10⁸ Pa)(0.45 x 10⁻³ m²)
(Pmax)st = 9 x 10⁴ N = 90 KN
<u>FOR Aluminum BARS</u>:
Since,
(σallow)al = (Pmax)al/A
where,
(σallow)al = maximum allowable stress of Al bar = 150 MPa = 1.5 x 10⁸ Pa
A = Cross-sectional area of Aluminum bar = 450 mm² = 0.45 x 10⁻³ m²
(Pmax)al = Maximum allowable force for Aluminum bar = ?
Therefore,
1.5 x 10⁸ Pa = (Pmax)al/0.45 x 10⁻³ m²
(Pmax)al = (1.5 x 10⁸ Pa)(0.45 x 10⁻³ m²)
(Pmax)al = 6.75 x 10⁴ N = 67.5 KN
Since,
(Pmax)al < (Pmax)st
Therefore,
The maximum allowable force will be:
Pmax = (Pmax)al
<u>Pmax = 67.5 KN</u>
