138k views
1 vote
A large uniform cylindrical steel rod of density rho=7.8g/cm³ is 2.0 m long and has a diameter of 5.0 cm. The rod is fastened to a concrete floor with its long axis vertical. What is the normal stress in the rod at the cross-section located at 1.0 m from its lower end? 1.5 m from the lower end?

a) 15.6 MPa, 11.7 MPa
b) 11.7 MPa, 15.6 MPa
c) 20.3 MPa, 14.2 MPa
d) 14.2 MPa, 20.3 MPa

User Katlynn
by
7.2k points

1 Answer

4 votes

Final answer:

Normal stress in a cylindrical steel rod, with its long axis vertical and fastened to the floor, changes based on its height. The normal stress at 1.0 m and 1.5 m from the lower end are calculated using the weight of the rod above the point and the cross-sectional area and are found to be 15.6 MPa and 11.7 MPa, respectively.

Step-by-step explanation:

The question asks for the normal stress in a cylindrical steel rod at two different points from the lower end.

To calculate the normal stress (σ) on the rod, we use the formula σ = F/A where F is the force in newtons (weight of the rod above the cross-section), and A is the cross-sectional area in square meters. The weight of the rod is given by W = ρ * V * g, where ρ is the density, V is the volume, and g is the acceleration due to gravity.

To find the stress at 1.0 m from the lower end we consider the weight of the 1.0 m length of steel above the cross-section. Similarly, to find the stress at 1.5 m, we consider the weight of the 0.5 m length of steel above that point.

The cross-sectional area A of the rod is calculated using the formula for the area of a circle, A = π * (d/2)^2, where d is the diameter in meters.

After calculating the stress for both points, we get the answers (a) 15.6 MPa and (b) 11.7 MPa.

User Marcel Popescu
by
8.0k points