A steel rod, 150 mm diameter, is inserted in a copper rod of 300 mm diameter. The internal diameter of copper rod is 150 mm. The axial load acting on the rod is 75 kN and the lengt... A steel rod, 150 mm diameter, is inserted in a copper rod of 300 mm diameter. The internal diameter of copper rod is 150 mm. The axial load acting on the rod is 75 kN and the length of the rod is 1 m. Calculate stresses in the rod. Take E_steel = 200 GPa, E_copper = 120 GPa. Read more

Steps to Solve

1. Identify Given Values

First, we need to identify all the given values from the problem. For instance, the areas of the rods, the Young's modulus for steel ($E_s$) and copper ($E_c$), and the axial load ($P$).

2. Calculate the Stress in Each Rod

Stress ($\sigma$) is defined by the formula:

$$ \sigma = \frac{P}{A} $$

Where $P$ is the axial load and $A$ is the cross-sectional area. Calculate the stress in both the steel and copper rods using their respective areas.

3. Determine Strain using Young’s Modulus

Next, use Young's modulus to relate stress and strain. The relationship is given by:

$$ E = \frac{\sigma}{\epsilon} $$

Where $E$ is the Young's modulus, $\sigma$ is stress, and $\epsilon$ is strain. Rearranging gives us:

$$ \epsilon = \frac{\sigma}{E} $$

Calculate the strain for both rods.

4. Check Compatibility of Strain

For rods in series, strain in both materials must be equal:

$$ \epsilon_s = \epsilon_c $$

Substituting the expressions derived from Young's modulus will yield a relationship that relates the stresses in the two materials.

5. Solve for Individual Stresses

Utilize the compatibility equation to find the unknown stresses. Use the earlier formulas to express the stresses in terms of the known variables and each other.

6. Final Calculation

After solving the equations, substitute any known values into the final equations to calculate the stress for both materials.