Question

The two aluminum rods AB and BC have diameters of 25 and 20 mm, respectively. Determine the largest vertical force P that can be supported. The failure stress for the aluminum is o fail = 400 MPa. Use F.S. = 2.

Answer 1

The largest vertical force that the aluminum rods AB (25 mm diameter) and BC (20 mm diameter) can support is 62,800 N, with the smaller rod determining the maximum force due to its smaller cross-sectional area when using a factor of safety of 2 and a failure stress of 400 MPa.

Step-by-step explanation:

To determine the largest vertical force (P) that two aluminum rods AB and BC can support, given that they have diameters of 25 mm and 20 mm respectively and an aluminum failure stress (σfail) of 400 MPa with a factor of safety (F.S.) of 2, we must consider the weakest rod (with the smaller diameter and cross-sectional area) because it will fail first under the applied load. The cross-sectional area (A) of a rod with diameter (d) is calculated using the formula A = π(d/2)2, ensuring to convert millimeters to meters as the stress units are in megapascals (MPa). The allowable stress (σallow) is found by dividing the failure stress by the factor of safety: σallow = σfail / F.S. Then, the largest vertical force (P) the rod can support is calculated by multiplying the allowable stress by the cross-sectional area: P = σallow × A.

Step-by-step calculation:

1. Calculate the area of the smaller rod (d = 20 mm): A = π(0.02/2)2 = 3.14 × (0.01)2 = 3.14 × 0.0001 m2 = 3.14 × 10-4 m2.
2. Calculate the allowable stress: σallow = 400 MPa / 2 = 200 MPa = 200 × 106 N/m2.
3. Compute the largest vertical force (P): P = σallow × A = 200 × 106 N/m2 × 3.14 × 10-4 m2 = 62,800 N.

Therefore, the largest vertical force P that can be supported by the aluminum rods is 62,800 N