Answer :
In a uniform cross-section bar, the load always be divided ( shared between the bar or beam). So, if there is X number of bars holding the external weight then bar weight is equals to X/2
i assume we are talking about longitudinal elongation under tensile / compressive force - equal to weight of rod.
Strain = stress / Y
ΔL / Length = W / Area Y
ΔL = W L / A Y
If the external load is Same as W, then total tensile/compressive force is 2 W.
Then ΔL will be twice to that of ΔL under its own weight = 2 ΔL.
So elongation due to external load W is ΔL and elongation due to its own weight W is ΔL.
Their ratio is 1 : 1
Strain = stress / Y
ΔL / Length = W / Area Y
ΔL = W L / A Y
If the external load is Same as W, then total tensile/compressive force is 2 W.
Then ΔL will be twice to that of ΔL under its own weight = 2 ΔL.
So elongation due to external load W is ΔL and elongation due to its own weight W is ΔL.
Their ratio is 1 : 1