Derive expression for prismatic bar of length and a and m
Answers
Consider a bar of circular cross section and uniform diameter throughout. Consider it to be suspended from a rigid support and its top end, such that it is in a hanging in a vertical position as shown in the figure.
Let,
A = Uniform cross sectional area of the bar
E = Young’s modulus for the bar
L = Length of the bar
ρ = Weight of the bar, per unit length, for the material of the bar
Consider an element of length ‘dy’ at a distance of ‘y’ from the bottom of the bar being elongated due to the force ‘P’, at section x-x, as shown in the figure.
Weight of the portion below x-x = P = ρ × A × y
Change in the length of the element ‘dy’ = PlAE=ρ×(A×y)×dyAEPlAE=ρ×(A×y)×dyAE
=ρ×y×dyEρ×y×dyE
For total change in the length of the bar, we need to integrate along the length
Total change in length = ∫L0ρy.dyE∫0Lρy.dyE
On integrating, we get,
δL=ρL22EδL=ρL22E
This is the expression for the elongation of a uniform bar under self weight
For a prismatic bar loaded in tension by an axial force P, the elongation of the ... Suppose the bar is loaded at one or more intermediate positions, then equation ( 1) ... be added algebraically to obtain the total charge in length of the entire bar.