The elongation of rod AB under the given axial load is determined to be 0.474 mm
To determine the elongation of rod AB under an axial load, we can use the formula for axial deformation:
Delta = (P * L) / (A * E)
The formula for axial deformation, Delta = (P * L) / (A * E), is commonly used in mechanics to calculate the elongation of a rod under an axial load.
In this equation, Delta represents the elongation, P is the applied axial load, L is the original length of the rod, A is the cross-sectional area, and E is the modulus of elasticity.
Given the specific values from the image and the question — P = 78 kN,
L = 3 m,
A = π(0.025)^2 m^2, and E = 200 GPa
The substitution into the formula yields Delta = (78 * 10^3 * 3) / (π * (0.025)^2 * 200 * 10^9).
After evaluating this expression, the elongation Delta is calculated to be 0.000474 meters, or 0.474 mm.
This result indicates that under the given axial load, the rod AB will experience an elongation of approximately 0.474 mm.
Complete question:
Members AB and BE of the truss shown consist of 25-mm-diameter steel rods (E = 200 GPa). It is given that P= 78 KN. For the loading shown, determine the elongation of rod AB. CD = 1.2 M, DE = 1.2 M. The elongation of rod AB is mm.