Final answer:
The magnitude of the load P that generates a normal stress of 130 MPa in the link BD with a cross-sectional area of 1300 mm2 is 169 kN.
Step-by-step explanation:
To determine the magnitude of the load P that causes a normal stress of 130 MPa in the central portion of the link BD with a uniform cross-sectional area of 1300 mm2, we need to use the formula for stress:
Stress (σ) = Force (F) / Area (A)
Given the stress (σ) is 130 MPa and the area (A) is 1300 mm2, we can rearrange the formula to solve for the force (F) as follows:
F = σ * A
Converting the given stress from MPa to N/mm2 (since 1 MPa = 1 N/mm2) and using the given area, we get:
F = 130 N/mm2 * 1300 mm2
F = 169,000 N
Therefore, the magnitude of the load P is 169 kN. This calculation is based on the relationship between stress, force, and area.