This document discusses uniaxial stresses and strains in prismatic bars with varying cross-sectional areas. It provides equations to calculate the elongation of bars under axial loads when the cross-sectional area is constant or varies continuously along the axis. For a tapered bar with uniformly decreasing diameter, the total extension can be found by integrating the strain over the length of the bar. The extension is compared for a tapered bar versus a uniform cylindrical bar of equal mean radius.

1. Mechanics of Solids (NME-302) Uniaxial Stresses
Members In Uni – Axial State Of Stress:
For a prismatic bar loaded in tension by an axial force P, the elongation of the bar can be determined as
Suppose the bar is loaded at one or more intermediate positions, then equation (1) can be readily adapted to
handle this situation, i.e. we can determine the axial force in each part of the bar i.e. parts AB, BC, CD, and calculate
the elongation or shortening of each part separately, finally, these changes in lengths can be added algebraically to
obtain the total charge in length of the entire bar.
When either the axial force or the cross – sectional area varies continuously along the axis of the bar, then equation
(1) is no longer suitable. Instead, the elongation can be found by considering a deferential element of a bar and
then the equation (1) becomes
퐝휹 =
푷풙퐝풙
푬푨풙
풍
휹 = ∫
푷풙퐝풙
푬푨풙
ퟎ
i.e. the axial force Px and area of the cross – section Ax must be expressed as functions of x. If the expressions for Px
and Ax are not too complicated, the integral can be evaluated analytically, otherwise Numerical methods or
techniques can be used to evaluate these integrals.
Stresses In Non – Uniform Bars:
Consider a bar of varying cross section subjected to a tensile force P as shown below.
Let
a = cross sectional area of the bar at a chosen section XX
then
Stress σ = p / a
If E = Young's modulus of bar then the strain at the section XX can be calculated
ε = σ/ E
Then the extension of the short element δx. = ε. original length = σ / E. δx
=
푷
푬
휹풙
풂
Yatin Kumar Singh Page 1

2. Mechanics of Solids (NME-302) Uniaxial Stresses
Thus, the extension for the entire bar is
풍
휹 = ∫
푷풙
푬
ퟎ
훅풙
풂
푻풐풕풂풍 푬풙풕풆풏풔풊풐풏 =
푷
푬
풍
∫
훅풙
풂
ퟎ
Now let us for example take a case when the bar tapers uniformly from d at x = 0 to D at x = l
In order to compute the value of diameter of a bar at a chosen location let us determine the value of dimension k,
from similar triangles
(푫 − 풅)/ퟐ
풍
=
풌
풙
풌 =
(푫 − 풅)풙
ퟐ풍
therefore, the diameter 'y' at the X-section is
or y = d + 2k
풚 = 풅 +
(푫 − 풅)풙
풍
Hence the cross –section area at section X- X will be
푨풙 퐨퐫 풂 =
흅
ퟒ
풚ퟐ =
흅
ퟒ
[풅 + (푫 − 풅)
풙
풍
]
ퟐ
hence the total extension of the bar will be given by expression
퐓퐨퐭퐚퐥 퐄퐱퐭퐞퐧퐬퐢퐨퐧 =
푷
푬
풍
∫
훅풙
풂
ퟎ
=
흅
ퟒ
푷
푬
풍
∫
훅풙
흅
ퟒ
[풅 + (푫 − 풅)
풙
풍
ퟐ
]
ퟎ
=
ퟒ푷풍
흅푬
ퟏ
푫
[
−
ퟏ
풅
] =
ퟒ푷풍
흅푬푫풅
An interesting problem is to determine the shape of a bar which would have a uniform stress in it under the action
of its own weight and a load P.
let us consider such a bar as shown in the figure below:
The weight of the bar being supported under section XX is
풙
= ∫ 흆품풂퐝풙
ퟎ
where ρ is density of the bar. Thus the stress at XX is
흈 =
풙
ퟎ
풂
푷 + ∫ 흆품풂퐝풙
풙
흈. 풂 = 푷 + ∫ 흆품풂퐝풙
ퟎ
Differentiating the above equation with respect to x we get
흈.
퐝풂
퐝풙
= 흆. 품. 풂
Yatin Kumar Singh Page 2

3. Mechanics of Solids (NME-302) Uniaxial Stresses
integrating
∫
퐝풂
풂
= ∫
흆.품
흈
퐝풙
퐥퐨퐠풆 풂 =
흆. 품. 풙
흈
+ 퐜퐨퐧퐬퐭퐚퐧퐭
applying boundary conditions
at x = 0; a = a0 thus constant = 퐥퐨퐠풆 풂풃
퐥퐨퐠풆 풂 =
흆. 품. 풙
흈
+ 퐥퐨퐠풆 풂ퟎ
풂
풂ퟎ
퐥퐨퐠풆 (
) =
흆. 품. 풙
흈
풆
흆.품.풙
흈 =
풂
풂ퟎ
also at x = 0
흈 =
푷
풂ퟎ
Thus
풂
풂ퟎ
= 풆
흆.품.풙풂ퟎ
푷
Illustrative Problem: A round bar, of length L, tapers uniformly from radius r1 at one end to radius r2at the other.
Show that the extension produced by a tensile axial load P is
푃퐿
2휋퐸푟1
2. If r2 = 2r1 , compare this extension with that of a
uniform cylindrical bar having a radius equal to the mean radius of the tapered bar.
Solution:
Consider the above figure let r1 be the radius at the smaller end. Then at a X cross-section XX located at a distance
x from the smaller end, the value of radius is equal to
= 풓ퟏ +
풙
푳
(풓ퟐ − 풓ퟏ) = 풓ퟏ(ퟏ + 풌풙)
퐰퐡퐞퐫퐞 풌 = (
풓ퟐ − 풓ퟏ
푳
) .
ퟏ
풓ퟏ
퐒퐭퐫퐞퐬퐬 퐚퐭 퐬퐞퐜퐭퐢퐨퐧 퐗퐗 =
퐥퐨퐚퐝
퐚퐫퐞퐚
=
푷
흅풓ퟏퟐ
(ퟏ + 풌풙)ퟐ
퐒퐭퐫퐚퐢퐧 퐚퐭 퐭퐡퐢퐬 퐬퐞퐜퐭퐢퐨퐧 =
퐬퐭퐫퐞퐬퐬
퐄
=
푷
푬. 흅풓ퟏퟐ
(ퟏ + 풌풙)ퟐ
Yatin Kumar Singh Page 3

4. Mechanics of Solids (NME-302) Uniaxial Stresses
Thus for a small length dx of the bar at this section the extension is
푷. 퐝풙
푬흅풓ퟏퟐ
(ퟏ + 풌풙)ퟐ
integrating
푳
퐄퐱퐭퐞퐧퐬퐢퐨퐧 = ∫
푷. 풅풙
푬흅풓ퟏퟐ
(ퟏ + 풌풙)ퟐ
ퟎ
=
푷
푬흅풓ퟏퟐ
푳
∫(ퟏ + 풌풙)−ퟐ
ퟎ
퐝풙 =
푷
푬흅풓ퟏퟐ
[
(ퟏ + 풌풙)−ퟏ
−풌
푳
=
]
ퟎ
푷
푬흅풓ퟏퟐ
[
(ퟏ + 풌풙)−ퟏ
−풌
−
ퟏ
−풌
]
=
푷
푬흅풓ퟏퟐ
[ퟏ −
ퟏ
ퟏ + 풌푳
] =
푷푳
푬흅풓ퟏퟐ
(ퟏ + 풌푳)
since 풌 = (
풓ퟐ−풓ퟏ
풓ퟏ.푳
)
Thus ퟏ + 풌푳 =
풓ퟐ
풓ퟏ
퐓퐡퐮퐬, 퐄퐱퐭퐞퐧퐬퐢퐨퐧 =
푷푳
푬흅풓ퟏ풓ퟐ
Comparing of Extensions:
For the case when r2 = 2.r1, the value of computed extension as above becomes equal to
푷
ퟐ푬흅풓ퟏퟐ
The mean radius of taper bar = 1 / 2(r1 + r2) = 1 / 2(r1 +2 r2) = 3 / 2 .r1
Therefore, the extension of uniform bar = Original length. Strain
= 푳
흈
푬
=
푳
푬
.
푷
흅 (
ퟑ
ퟐ
풓ퟏ)
ퟐ =
ퟒ푷푳
품흅푬풓ퟏퟐ
퐄퐱퐭퐞퐧퐬퐢퐨퐧 퐨퐟 퐮퐧퐢퐟퐨퐫퐦
퐄퐱퐭퐞퐧퐬퐢퐨퐧 퐨퐟 퐭퐚퐩퐞퐫퐞퐝
=
ퟒ푷푳
품흅푬풓ퟏퟐ
푷
ퟐ푬흅풓ퟏퟐ
=
ퟖ
품
Yatin Kumar Singh Page 4