The largest load P that can be applied to the rod is approximately 0.00791 MN (meg
To determine the largest load P that can be applied to the steel rod, we need to consider the shearing stress limits for both the steel rod and the aluminum plate.
First, let's calculate the area of the steel rod's flange. The diameter of the flange is given as 40 mm, so the radius would be half of that, which is 20 mm (0.02 m). Using the formula for the area of a circle (A = πr^2), we find:
A_steel = π * (0.02)^2 = 0.00126 m^2
Next, we need to calculate the area of the aluminum plate's hole. The diameter of the hole is given as 12 mm, so the radius would be half of that, which is 6 mm (0.006 m). Again using the formula for the area of a circle, we get:
A_aluminum = π * (0.006)^2 = 0.000113 m^2
Now, we can calculate the maximum load P that can be applied to the steel rod. The shearing stress for the steel rod must not exceed 176 MPa. The shearing stress is given by the formula:
τ = P / A
Rearranging the formula, we can solve for P:
P = τ * A
Substituting the given shearing stress and the area of the steel rod, we have:
P_steel = 176 MPa * 0.00126 m^2 = 0.22176 MN (mega newtons)
Next, let's calculate the maximum load P that can be applied to the aluminum plate. The shearing stress for the aluminum plate must not exceed 70 MPa. Using the same formula as before, we have:
P_aluminum = 70 MPa * 0.000113 m^2 = 0.00791 MN
Finally, to find the largest load P that can be applied to the rod, we need to consider the smaller of the two values:
P = min(P_steel, P_aluminum) = min(0.22176 MN, 0.00791 MN) = 0.00791 MN