MECN3048A – Mechanics of Solids I
Lecture 2
Axial Load
Mr R.T. Tebeta – MIA Engineering
Lecture Contents
4.2 Elastic deformation of an axially loaded member
4.4 Statically indeterminate axially loaded members
4.6 Thermal stress
4.8 Inelastic axial deformation
4.9 Residual stress
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.2 Elastic deformation of an axially loaded member
Consider the bar in the next figure:
It has a cross sectional area that gradually varies along its length 𝐿, and it is
made of a material that has a variable modulus of elasticity.
The bar is subjected to concentrated loads at its ends (𝑃1 and 𝑃2 ) and a variable
external load distributed along its length.
We want to find the relative displacement 𝛿 (delta) of one end of the bar with
respect to the other end as caused by the loading.
Consider a differential element of length 𝑑𝑥 and cross sectional area 𝐴(𝑥),
isolated from the bar at the arbitrary position 𝑥, where the modulus of elasticity is
𝐸(𝑥).
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.2 Elastic deformation of an axially loaded member
The free body diagram (FBD) of the section is represented
on the right. The section is subjected to a normal force 𝑁(𝑥)
function of the position 𝑥 along the bar. The load 𝑁(𝑥) will
deform the section as indicated, so:
𝑁 𝑥
𝜎=
𝐴 𝑥
𝑙𝑓𝑖𝑛 − 𝑙𝑖𝑛
𝑑𝑥 + 𝑑𝛿 − 𝑑𝑥 𝑑𝛿
𝜀=
=
=
𝑙𝑖𝑛
𝑑𝑥
𝑑𝑥
If the stress does not exceed the proportional limit, then Hooke’s law applies:
𝜎 = 𝐸(𝑥) ∙ 𝜀
𝑁 𝑥
𝑑𝛿
= 𝐸(𝑥)
𝐴 𝑥
𝑑𝑥
𝑑𝛿 =
𝑁(𝑥)
𝑑𝑥
𝐴 𝑥 𝐸(𝑥)
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.2 Elastic deformation of an axially loaded member
To calculate 𝛿, we must integrate the previous expression over the length 𝐿 of
the bar:
𝐿
𝑁(𝑥)
𝛿=න
𝑑𝑥
𝐴
𝑥
𝐸(𝑥)
0
If the bar has a constant cross sectional area 𝐴 and the material is
homogeneous (constant 𝐸). Also if the external forces applied to each ends are
constants (then the internal force 𝑁(𝑥) throughout the length of the bar is also
constant: 𝑁 𝑥 = 𝑃) then:
𝑃𝐿
𝛿=
𝐴𝐸
If the bar is subjected to several different axial forces along its length, then the
above equation applies to each segment of the bar where these quantities
remain constant:
𝑃𝑖 𝐿𝑖
𝛿=෍
𝐴𝑖 𝐸𝑖
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.2 Elastic deformation of an axially loaded member
Sign convention
We will consider both the force and the displacement to be positive if they cause
tension and elongation.
(+)
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.4 Statically indeterminate axially loaded members
Consider the bar shown in the figure (a) which is fixed supported
at both of its ends. From its FBD (b) there are two unknown
support reaction 𝐹𝐴 and 𝐹𝐵 :
෍ 𝐹 +↑ = 0
→
𝐹𝐴 + 𝐹𝐵 − 500𝑁 = 0
This problem is called statically indeterminate, since the
equilibrium equation is not sufficient to determine both reactions
on the bar.
In order to introduce an additional equation needed for solution, it
is necessary to consider how the points on the bar are displaced.
An equation that specifies the condition for displacement is
referred to as a compatibility condition. In our case, a suitable
compatibility condition would require the displacement of end 𝐴
of the bar with respect to end 𝐵 to be equal zero, since the end
supports are fixed.
𝛿𝐴/𝐵 = 0
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.4 Statically indeterminate axially loaded members
Realising that the internal force in segment 𝐴𝐶 is +𝐹𝐴 , and in
segment 𝐶𝐵 is −𝐹𝐵 , then the compatibility equation can be
written as:
𝐹𝐴 𝐿𝐴
−𝐹𝐵 𝐿𝐵
𝛿𝐴/𝐵 = 𝛿𝐴 + 𝛿𝐵 =
+
=0
𝐴𝐸
𝐴𝐸
𝐹𝐴 = 1.5 ∙ 𝐹𝐵
𝐴𝐶
𝐶𝐵
Finally, using the equilibrium condition, the reactions are:
𝐹𝐴 = 300 𝑁
𝐹𝐵 = 200 𝑁
Note: since both results are positive, the directions of the reactions chosen in
the FBD are correct.
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.6 Thermal stress
A change in temperature can cause a body to
change its dimensions. Generally, if the
temperature increases the body will expand,
whereas if the temperature decrease it will
contract.
If the relationship between temperature and
change in shape is linear and the material is
homogeneous
and
isotropic,
then
the
displacement of a member of length 𝐿 is:
𝛿𝑇 = 𝛼 ∆𝑇 𝐿
Where:
𝛼 is a property of the material referred to as the linear coefficient of thermal
expansion. The units are 1Τ°𝐶 (Celsius) or 1Τ𝐾 (Kelvin)
[Steel: 12 ∙ 10−6 1Τ°𝐶]
∆𝑇 is the change in temperature of the member. [∆𝑇 = 𝑇𝑓𝑖𝑛 − 𝑇𝑟𝑒𝑓 ]
𝐿 original length of the member
𝛿𝑇 change in length of the member
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.6 Thermal stress
The change in length of a statically determinate member can easily be calculated
with the previous formula, since the member is free to expand or contract under
a temperature change.
For a statically indeterminate member, these thermal displacement will be
constrained by the supports, thereby producing thermal stresses in the
member. These thermal stresses adds to the mechanical stresses and have to
be considered in design.
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.8 Inelastic axial deformation
Up to this point we have only considered loadings that cause the material to
behave elastically. We are now to consider a load that causes the material to
yield and thereby permanently deform.
Assuming a material such as low-carbon steel with a stress-strain diagram as
reported in figure (a), for non-excessive yielding this behaviour can be modelled
as shown in figure (b). A material that exhibits the behaviour in figure (b) is
referred to as being elastic perfectly plastic or elastoplastic.
(a)
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
(b)
4.8 Inelastic axial deformation
To illustrate physically how such material behaves, consider the
bar in figure (a) subjected to the axial load 𝑁. If the load causes
an elastic stress 𝜎 = 𝜎1 then the equilibrium requires 𝑁 =
‫𝜎 ׬‬1 𝑑𝐴 = 𝜎1 𝐴. This stress causes the bar to strain 𝜀1 as indicated
in the stress-strain diagram.
If 𝑁 is now increased to causing yielding of the material, then 𝜎 =
𝜎𝑌 . This load 𝑁𝑝 is called the plastic load.
For this case, the strain are not uniquely defined:
when 𝜎𝑌 is reached, the bar is subjected
to the yield strain 𝜀𝑌 . Then the bar will
continue to yield (or elongate) producing
the strains 𝜀2 , then 𝜀3 , etc.
This elongation is expected to continue
indefinitely in our perfectly plastic material,
while in the reality, strain hardening will
occur in the material.
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.8 Inelastic axial deformation
Now consider the case of a bar having a hole through it, as shown in figure (a).
When N is applied, a stress concentration occurs in the material at the edge of
the hole, on section 𝑎 − 𝑎. The stress here will reach a maximum value of
𝜎𝑚𝑎𝑥 = 𝜎1 , and having the corresponding elastic strain of 𝜀1 .
The stresses and corresponding strains at other point of the cross section will be
smaller, figure (c).
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.8 Inelastic axial deformation
Equilibrium again requires 𝑁 = ‫ 𝐴𝑑 𝜎 ׬‬which is geometrically equivalent to the
volume contained within the stress distribution.
If the load is further increased to 𝑁 ′ so that 𝜎𝑚𝑎𝑥 = 𝜎𝑌 , then the material will
begin to yield outward from the hole, until the equilibrium condition 𝑁 ′ = ‫ 𝐴𝑑 𝜎 ׬‬is
satisfied, figure (d).
A further increase in load will eventually cause the material over the entire cross
section to yield, figure (e).
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.8 Inelastic axial deformation
When this happens, no greater load can be sustained by the bar.
The plastic load 𝑁𝑝 is defined as:
𝑁𝑝 = න 𝜎𝑌 𝑑𝐴 = 𝜎𝑌 𝐴
𝐴
where 𝐴 is the bar’s cross sectional area at section 𝑎 − 𝑎.
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.9 Residual stress
Consider a prismatic member (figure (a)) made of elastoplastic material having
the stress-strain diagram shown in figure (b).
If an axial load produces a stress 𝜎𝑌 in the material and a corresponding strain
𝜀𝐶 , then when the load is removed the material will respond elastically and follow
the line 𝐶𝐷 in order to recover some of the strain.
A recovery to zero stress at point 𝑂′ will be possible if the member is statically
determinate, since then the support reactions for the member will be zero when
the load is removed. Under these circumstances the member will be
permanently deformed so that the permanent strain in the member is 𝜀𝑂′ .
(b)
(a)
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering
4.9 Residual stress
If the member is statically indeterminate, however, removal of the external load
will cause the support forces to respond to the elastic recovery 𝐶𝐷. Since these
forces will constrain the member from full recovery, they will introduce residual
stresses in the member.
To solve the problem, the complete cycle of loading and unloading can be
considered as the superposition of a positive load (loading) on a negative load
(unloading). The loading, 𝑂 to 𝐶, results in a plastic stress distribution, whereas
the unloading, along 𝐶𝐷, results only in an elastic stress. The results of the
superposition of these stresses are the residual stresses in the member.
(b)
(a)
Copyright University of the Witwatersrand, Johannesburg
School of Mechanical, Industrial and Aeronautical Engineering