Final answer:
The diameter of the shaft made of AISI 1018 HR steel can be determined by using Hooke's Law for elongation, accounting for the truck's weight and the steel's modulus of elasticity. The formula must be applied correctly, and since the modulus of elasticity value is not provided in the question, a specific calculation cannot be made. The diameter is calculated from the cross-sectional area, which is obtained through the formula rearrangement.
Step-by-step explanation:
To determine the diameter of a shaft made of AISI 1018 HR steel that does not stretch more than 1.0 cm when a 2.5 × 104-kg truck passes over it, we need to consider the elastic properties of the steel and apply them in the formula derived from Hooke's Law for elongation:
δ = rac{PL}{AE}
Where:
- δ is the elongation
- P is the force applied (weight of the truck)
- L is the length of the shaft
- A is the cross-sectional area of the shaft
- E is the modulus of elasticity for AISI 1018 HR steel
To find the shaft's cross-sectional area A, we first calculate the force P by multiplying the mass of the truck by the acceleration due to gravity (g = 9.81 m/s2).
P = mass × g = 2.5 × 104 kg × 9.81 m/s2
Using the given allowable elongation (δ = 1.0 cm) and the known modulus of elasticity for AISI 1018 HR steel (E), we can rearrange the formula to solve for the cross-sectional area A. The diameter can then be obtained using the area of a circle (A = π(d/2)2).
Without the specific value for the modulus of elasticity E, we cannot definitively calculate the diameter; we must instead rely on this value to determine which of the provided options is correct. Assuming the necessary calculations are performed with the correct E value, one can use the process described here to find the appropriate diameter from the options provided.