Reaction at support and internal axial force resultant

Consider two straight bars of uniform cross section made of the same material. Bar 1 has an axial length 1m of and a square cross section 1mm with side length . Bar 2 has an axial length of 3 m and a round cross section with 2mm diameter. When subjected to 1KN 2 axial tension, bar 1 elongates by 5mm . What will be the elongation of bar 2, , if it is compressed axially with a load of 25KN ?

To find the elongation of bar 2, we can use Hooke's Law, which states that the elongation (ΔL) of a bar is directly proportional to the applied force (F) and inversely proportional to the cross-sectional area (A) and the Young's modulus (E) of the material.

First, let's calculate the cross-sectional area of both bars.

For Bar 1:
Given that the side length of the square cross section is 1 mm, we can calculate the area using the formula:
A1 = side length^2 = (1 mm)^2 = 1 mm^2

For Bar 2:
Given that the diameter of the round cross section is 2 mm, we can calculate the area using the formula:
A2 = π * (diameter/2)^2 = π * (2 mm/2)^2 = π mm^2 ≈ 3.14 mm^2

Now, let's find the elongation of Bar 2 using Hooke's Law equation:
ΔL2 = (F2 * L2) / (A2 * E)

Given:
F2 = 25 kN (compressive force)
L2 = 3 m (axial length)
A2 = 3.14 mm^2 (cross-sectional area)
E (Young's modulus) is not provided in the question, so we cannot calculate the exact value.

To find the elongation of Bar 2, we need to know the Young's modulus of the material. Young's modulus is a material property and can vary depending on the material. Once you have the value for Young's modulus (in N/m^2 or Pa), you can substitute it into the equation to calculate the elongation (ΔL2) of Bar 2.