Final answer:
The largest load P that can be applied to the steel bar, ensuring a factor of safety of 2.0 with respect to the weld material's tensile yield strength, is 32.5 kN. This is calculated based on the allowable tensile strength of the weld after applying the factor of safety and the cross-sectional area of the steel bar.
Step-by-step explanation:
The student's question relates to the structural engineering concept of material strength and loading. The maximum load P that can be safely applied to a welded steel bar can be determined by considering the tensile and shear yield strengths of the two materials involved, i.e., the weld material and the bar material. Using the given factor of safety of 2.0, the tensile strength of the weld and the bar, and the cross section of the bar, we can calculate the allowable tensile stress and then the maximum load P by dividing the allowable tensile stress by the area of the cross section.
To begin with, we find the allowable tensile strength by dividing the weld material's yield strength (which is the weaker material concerning tensile strength) by the factor of safety:
- Allowable tensile strength for weld = 325 MPa / 2 = 162.5 MPa
Next, we need to calculate the area of the cross-section of the bar:
- Area (A) = 20 mm x 10 mm = 200 mm² = 200 x 10⁻⁶ m²
Now, we convert the units of the allowable tensile strength to N/m² (because 1 MPa = 1 x 10⁶ N/m²) and calculate the maximum tensile load (Ptensile) that the weld can sustain:
- Allowable tensile strength for weld in N/m² = 162.5 x 10⁶ N/m²
- Ptensile = Allowable tensile strength for weld x Area
- Ptensile = (162.5 x 10⁶ N/m²) x (200 x 10⁻⁶ m²)
- Ptensile = 32.5 x 10³ N = 32.5 kN
Thus, the largest load P that can be applied to the steel bar to satisfy the safety criteria is 32.5 kN.