STRENGTH OF MATERIALS- Elongation of Tapered Bar


  • When either the axial force or the cross-sectional area varies continuously along the axis of the bar, then equation δ =  PL/AE is no longer suitable.
  • Instead the elongation can be found by considering a differential element of the bar and then integrating over the entire length of the bar.
  • Let us consider a uniformly tapering circular bar, subjected to an axial force (P), consider a section X-X of length δx and diameter di at a distance x from the end A.
  • In order to compute the value of diameter of bar at a chosen location let us determine the value of K, from similar triangles.

(d1 - d2)/2/L = K/x

K = (d1 - d2) x/2L

Diameter, di = d1 + 2K = d2 + (d1 - d2) x /L

Extension of the short length, δΔ = P . δx/2

Hence extension of the whole length of the rod is

δ = 0L 4 . P . δx/π . (d2 + 2K)2 E

δ = 4 . P/π . E  0L δx/{d + (d1 - d2) k/L}2

After carrying out the integration we get,

δ = -4 . P . L/π . E [ 1/d21/d1 ]

δ = 4 . P . L/π . E . d1 . d2


When d1 = d2 = d, then δ = 4 . P . L/π . E . d . d = 4 . P . L/π . E . d2 = P . L/A . E ,  same as previous one.

Elongation of a prismatic bar under axial load is equal to elongation of an identical tapered bar under the same axial load. When diameter of prismatic bar is equal to the geometric mean of diameters of the tapered bar i.e. d = √d1 . d2