Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
2024/2/7
In this course, we concentrate on the analysis of deformation and internal forces
of straight, slender members subjected to external loads. A member is slender if its
cross-section dimensions are much less than the axial length. A slender member is
called a rod if its primary function is to support axial loads, a shaft to transmit torque,
and a beam to support transverse loads.
Rod
Shaft
Beam
1
Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
2024/2/7
A rod is a slender, uniform structure member used to support axial loads. In this
lecture, we discuss how to characterize the material properties of a rod and how to
analyze the deformation and axial forces of a rod and a structure consisting of rods.
Spring Constant of a Rod
A rod subjected to an axial force 𝑃 will behave like a linear
spring if the change in length 𝛿 is much smaller than its original
(undeformed) length 𝑙 . Specifically, 𝑃 = 𝑘𝛿 if 𝛿/𝑙
1 . The
𝑙
𝛿
𝑃
proportional constant 𝑘 is called the equivalent spring constant of
the rod. The sign conventions used are: 𝑃 is positive for tensile and negative for
compressive loads; 𝛿 is positive for elongation and negative for contraction. A natural
question arises: is 𝑘 a good measure of the stiffness of the material of which the rod is
made?
Q 1 Match the three steel rods with the lines in the 𝑃-𝛿 plot.
𝑃


𝛿

Axial Stress and Strain
The rod's spring constant depends on the material's
stiffness and the rod's dimensions (i.e., original length 𝑙 and
𝑃/𝐴

cross-sectional area 𝐴). If we plot 𝑃/𝐴 vs 𝛿/𝑙, all three lines
coincide. Define the (normal or axial) strain 𝜀 and (normal or
𝛿/𝑙
Fig. 1
axial) stress 𝜎 as
𝜀 = 𝛿/𝑙 and 𝜎 = 𝑃/𝐴,
(1)
respectively. In the above equation, 𝜀 indicates the deformation per unit length
2
Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
2024/2/7
and 𝜎 the average internal axial force per unit cross-sectional area. The axial stress 𝜎 is
positive (negative) if the rod is in tension (compression). The axial strain 𝜀 is positive
(negative) if the rod is elongated (shortened). Note that 𝜀 is dimensionless and 𝜎 has the
dimension of pressure.
Tensile Test
see ptt
Hooke’s Law
With the definitions of stress and strain, the results shown in Fig. 1 can be
expressed as
E
E
=
.
S
𝜎 = 𝐸𝜀
(2)
where the proportionality constant 𝐸 is called Young’s modulus or modulus of elasticity.
Equation (2) is known as Hooke’s law. Young’s modulus measures a material's stiffness,
or ability to resist deformation.
Most engineering structures are designed to undergo relatively small deformations.
Within the range of the designated load, the structure obeys Hooke’s law. In this case,
the deformation is proportional to the applied load, and after removing the applied load,
the structure returns to its original configuration.
Ex 1 Determine the spring constant of a uniform rod.
Sol: Substitute Eq. (1) into Hooke’s law and rearrange terms to obtain
𝑃𝑙
𝐸𝐴
(3)
𝐸𝐴
𝛿
𝑙
(4)
𝛿=
Equivalently,
𝑃=
Hence, the spring constant of a uniform rod of length 𝑙 and cross-sectional area 𝐴 is
𝑘 = 𝐸𝐴/𝑙
3
Mechanics of Materials
NTU_ME CJLu
Ex 2
Note#2 Axial Loading 4+
3
=
𝐹
Determine the deformation of the rod shown
8
𝐴 ,𝑙
𝐴 ,𝑙
on the right.
2024/2/7
AN
𝐹
.
2
I
B
Sol: Divide the rod into two parts and draw the FBD’s.
𝐹 +𝐹
𝐹 +𝐹
𝐴 ,𝑙
𝐹 +𝐹
𝐹
𝐹
𝐹
↳
-
𝐹
𝐴 ,𝑙
The elongation of the rod is the sum of that of the two parts (using Eq. (3)):
𝐹 +𝐹 𝑙
𝐹𝑙
𝛿=
+
𝐸𝐴
𝐸𝐴
DOTE
Principle of Superposition
-
=
F+
/silk
+𝐹
𝑙
𝑙
+
𝐸𝐴
𝐸𝐴
S
=
e
(5)
Equation (5) can be rewritten as
𝛿=
𝐹𝑙
𝐸𝐴
The above result indicates that the deformation due to the combined loads 𝐹 and 𝐹 is
the sum of the deformation due to 𝐹 alone and that due to 𝐹 alone. This result implies
a general principle - the principle of superposition. This principle states that the effect
of a given combined loading on a structure can be obtained by determining the effects
of the various loads separately and combining the results, provided that the following
conditions are satisfied (Beer and Johnston):
1. Each effect is linearly related to the load that produces it.
2. The deformation resulting from any given load is small and does not affect the
conditions of application of the other loads.
We will use this principle repeatedly in this course.
Rods with Continuously Varying Loads or Dimensions
Ex 3 Determine the elongation of a uniform rod subjected to
a uniformly distributed load 𝑝 as shown on the right.
N(X)
4
jox /
𝐴, 𝑙, 𝐸
𝑝
Mechanics of Materials
NTU_ME CJLu
𝑝 Δ𝑥
sidependent
𝑁 𝑥
N(X + 0x)
Note#2 Axial Loading
11
2024/2/7
Sol: Consider a small element of length Δ𝑥 at a
𝑁 𝑥 − 𝑝 Δ𝑥
distance 𝑥 from the fixed end as shown on the left.
𝑥
Δ𝑥
The elongation of this element is
N(XX) While Ox A
𝑁 𝑥 Δ𝑥
Δ𝛿 =
𝐸𝐴
𝑝
𝑁 𝑥
X-0
𝑙−𝑥
where the axial force 𝑁 𝑥 is
𝑁 𝑥 =
:it
i
𝑝 𝑑𝑥 = 𝑝 𝑙 − 𝑥
Combining the above two equations yields
Δ𝛿 =
𝑝 𝑙 − 𝑥 Δ𝑥
𝐸𝐴
The total elongation 𝛿 is the sum of all Δ𝛿's
𝛿 = lim
𝑝 𝑙−𝑥
𝑝 𝑙
𝑑𝑥 =
𝐸𝐴
2𝐸𝐴
Δ𝛿 =
→
= 21x-Es ! =
Alternatively, we can use the principle of superposition to solve this problem. The
elongation Δ𝛿 of the entire rod due to a concentrated load of 𝑝 Δ𝑥 at 𝑥 is
𝑝 Δ𝑥
Δ𝛿 =
𝑥
𝑝 Δ𝑥 ⋅ 𝑥
𝐸𝐴
The total elongation due to all loads is
𝛿 = lim
→
Δ𝛿 =
𝑝 𝑥𝑑𝑥 𝑝 𝑙
=
𝐸𝐴
2𝐸𝐴
Ex 4 Determine the elongation of the rod of smoothly
varying cross-sectional area 𝐴 𝑥
due to a
𝑃
𝐸, 𝐴 𝑥
concentrated load 𝑃.
Sol: The rod has a constant axial force, i.e., 𝑁 𝑥 = 𝑃 . The elongation of a small
𝑁
𝑁
𝑥
element of length Δ𝑥 at a distance 𝑥 from the left end is
Δ𝑥
Δ𝛿 =
The total elongation is
5
𝑁Δ𝑥
𝑃Δ𝑥
=
𝐸𝐴 𝑥
𝐸𝐴 𝑥
Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
2024/2/7
𝑃𝑑𝑥
𝐸𝐴 𝑥
𝛿=
(6)
The above result is based on the formula 𝛿 = 𝑃𝐿/𝐸𝐴, which in turn is derived using
the assumption that the stress is uniform, i.e., 𝜎 = 𝑃/𝐴 over every cross-section. This
assumption is not valid for a tapered rod. Hence, Eq. (6) gives satisfactory results only
if the cross-sectional area changes smoothly.
The left figure shows the stress
distribution over a cross-section of a
non-uniform rod. The stress over the
cross section is not uniform. In this
case, the average stress (𝜎ave = 𝑃/𝐴)
we used in this lecture may not be a
good indicator of the stress over the
cross section because 𝜎
much larger than 𝜎ave .
Force equilibrium
·
RB
Statically Indeterminate Problems E **
may be
P
+
RA
=
Constitutive
·
① S,
:
Law
Be
=
Ra
Ex 5 (B&J CA2.3) Determine the stresses in portions 𝐴𝐶 and 𝐵𝐶 due
D
:
=
to the application of a load 𝑃 at point 𝐶.
&
=
8,
Sol: (I) Assume that point 𝐶 moves downward a distance 𝑑 due to the
r
application of 𝑃. The strains of portions 𝐴𝐶 and 𝐵𝐶 are respectively
𝜀
=
𝑑
and 𝜀
𝐿
=−
𝑑
𝐿
(7)
Here, we adopt the sign convention that the normal strain is positive for elongation and
negative for compression. The corresponding stress is positive for tensile and negative
for compressive loads. The stresses in the two portions are determined by Hooke’s law,
and the corresponding tensile forces are
=>
=>
6
-
EA 8
Sc
+
p
=
E Sc
(E EN)
li 12
+
=
,
8
,
=
- co
P
Sc
Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
𝐹
= 𝐸𝜀 𝐴 =
𝐸𝑑𝐴
and 𝐹
𝐿
2024/2/7
= 𝐸𝜀 𝐴 = −
𝐸𝑑𝐴
𝐿
(8)
Equilibrium of point 𝐶 requires that
𝐹
−𝐹
=𝑃
(9)
Substitute Eq. (8) into Eq. (9) and solve for 𝑑 to obtain
𝑑=
𝑃𝐿 𝐿
𝐸𝐴𝐿
where 𝐿 = 𝐿 + 𝐿 . Carrying this value into Eq. (8), we obtain
𝜎
𝐹
𝑃𝐿
=
and 𝜎
𝐴
𝐴𝐿
=
=
𝐹
𝑃𝐿
=−
𝐴
𝐴𝐿
(II) Let 𝑅 be the reaction at pin 𝐴. Remove the pin joint and apply the equivalent force
𝑅 as show on the left. By the principle of superposition, the
SAL
displacement of point 𝐴 due to 𝑃 and 𝑅 is
xRB
𝑅 𝐿 𝑃𝐿
𝛿 =
−
𝐸𝐴
𝐸𝐴
xP
>
-
However, point 𝐴 is fixed, i.e., 𝛿 = 0, hence
𝑅 =
find
RB
so
compatibility
that
#2 & E
𝑃𝐿
𝐿
𝑃𝐿
𝐿
Dividing the reactions by 𝐴 yields the stress as
𝜎
=
𝑅
𝑃𝐿
=
and 𝜎
𝐴
𝐴𝐿
=
𝑅
𝑃𝐿
=−
𝐴
𝐴𝐿
Ex 6 (SP 2.2 B&J) The rigid castings 𝐴 and 𝐵 are connected by two 18-mm-diameter
Est
steel bolts 𝐶𝐷 and 𝐺𝐻 and are in contact with
the ends of a 38-mm-diameter aluminum
rod 𝐸𝐹 . Each bolt is single-threaded with a
Sal
sa
<
-
pitch of 2.5 mm, and after being snugly fitted,
the nuts at 𝐷 and 𝐻 are both tightened one-
p
#
7
=
0
H
. Ite ****,
eq
The reaction at point 𝐵 can be determined by the equilibrium equation as
𝑅 =𝑅 −𝑃 =−
SB
·
Force
FA
Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
·
=
2
equil
.
Est
Constitutive law
Ste
2024/2/7
=
e
quarter of a turn. Knowing that 𝐸 is 200 GPa for steel and 70 GPa for aluminum,
Sit
determine the normal stress in the rod.
= se Ast
Sol: Casting 𝐴 can be treated as fixed. Because of symmetry, the
𝐹
two steel rods 𝐶𝐷 and 𝐺𝐻 are subjected to the same internal force
𝐹
and undergo the same deformation. Let 𝐹 be the compressive
·
Compatibilitfye
𝐹
axial force in the steel rods and 𝐹 the tensile axial force in the
aluminum rod 𝐶𝐷. The equilibrium of casting 𝐵 requires that
𝐹 − 2𝐹 = 0
(10)
這是一個靜不定的問題，必需考
慮鋁桿和螺栓的變形。因為結構
(a)
的完整性，鋁桿和螺栓的變形滿
足特定的關係式–匹配方程式。
參考左圖：
（a）代表初始不受力
(b)
狀態。最後的平衡狀態（c）可以
𝛿
𝑝/4
𝛿
想像是先將鋁桿移除，旋轉右方
鑄件的螺帽1/4圈，因此右方鑄
(c)
件向左移動𝑝/4（圖（b）
，𝑝代表
螺距）；接著將右方鑄件向右移
動𝛿 （這個位移量即是螺栓的伸長量）
，同時將鋁桿壓短𝛿 後，放入兩鑄件間（圖
（c））。
𝛿 + 𝛿 = 𝑝/4
(11)
The load-displacement relations are
𝛿 =
𝐹 𝑙
𝐸 𝐴
and 𝛿 =
𝐹 𝑙
𝐸 𝐴
We now solve Eqs. (10)-(12) simultaneously and obtain the axial forces as
𝐹 =
𝑝 𝑙
2𝑙
+
4 𝐸 𝐴
𝐸 𝐴
8
and 𝐹 = 2𝐹
(12)
Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
2024/2/7
Thermal Effects
Changes in temperature produce expansion or
contraction of structural materials, resulting in
thermal deformations. Consider a homogeneous rod
𝐴𝐵 resting freely on a smooth horizontal surface
(Fig. 2.30). The elongation 𝛿
of the rod is
proportional to both the temperature rise Δ𝑇 and the
length 𝐿 of the rod. Define thermal strain as
𝜀 = 𝛿 /𝐿
Then
𝜀 = 𝛼 Δ𝑇
(13)
where 𝛼 is a constant characteristic of the material, called the coefficient of thermal
expansion.
If the rod is subjected to both axial stress and temperature change, the strain in the
rod is the sum of that due to the axial stress and thermal strain; that is
𝜎
(14)
+ 𝛼 Δ𝑇
𝐸
ET
Er
Ex 7 Consider a rod of length 𝐿 , Young’s modulus 𝐸 , and coefficient of thermal
𝜀=
expansion 𝛼. Determine the strain and stress induced by a temperature rise Δ𝑇 for
the following two cases.
(a) The left end of the rod is fixed while the right end is free.
(b) Both ends of the rod are fixed.
Sol: (a) The rod can expand freely. There is no stress in the rod. The deformation is due
to the thermal expansion. Hence, 𝜎 = 0 and 𝜀 = 𝜀 = 𝛼Δ𝑇
(b) Because both ends are fixed, the total strain of the rod is zero. By Eq. (14),
𝜎 = −𝐸𝛼Δ𝑇
9
E
o
+
=
20T
-
=
0
ExoT
Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
2024/2/7
where the negative sign indicates a compressive stress. This kind of stress due to a
temperature change is called thermal stress.
Alternatively, one can solve this problem by the principle
of superposition (see the figure on the left). Because the
right end of the rod cannot move, the thermal
elongation 𝛿 due to the temperature change must be
pushed back by the reaction 𝑃. Obviously,
𝛼 Δ𝑇 𝐿 =
𝑃𝐿
𝐸𝐴
Hence, the compressive stress is
𝜎 = 𝑃/𝐴 = 𝐸𝛼Δ𝑇
(15)
It is worth noting that
(i)
For a structure which can deform freely, a uniform temperature change
throughout the structure will induce thermal strains but no thermal stresses. On
the other hand, for a structure that has supports preventing free deformation,
even a uniform temperature change will produce thermal stresses.
(ii)
The stress in a structure member should not exceed the designated allowable
stress, 𝜎allow , which is usually much less than Young’s modulus. To ensure the
thermal stress considered in case (b) to be less than 𝜎
𝛼Δ𝑇
Take stainless steel, with 𝛼 = 17
, by Eq. (15),
𝜎allow
𝐸
10
°C
and 𝜎allow /𝐸 = 10
, as an
example. The allowed temperature change is Δ𝑇 = 59 °C. Note that a relatively
modest change in temperature can produce significant thermal stresses. Hence,
temperature effects can be important in engineering design.
10

Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
2024/2/7
Q 2 For a (linear) statically determinate structure,
uniform temperature changes in the members
produce thermal strains but no thermal stresses. True
or False
Gere M of M, Fig. 2-21
Q 3 For a (linear) statically indeterminate structure,
uniform temperature changes in the members
may or may not produce thermal stresses. True
or False
Gere M of M, Fig. 2-22
Ex 8 (B&J 2.4) The rigid bar 𝐶𝐷𝐸 is attached to a pin support at 𝐸 and rests on the 30mm-diameter brass cylinder 𝐵𝐷. A 22-mm-diameter steel rod 𝐴𝐶 passes through
a hole in the bar and is secured by a nut which is snugly fitted when the temperature
of the entire assembly is 20 °C. The temperature of the brass cylinder is then raised
-
to 50 °C while the steel rod remains at 20 °C. Assuming
-
that no stresses were present before the temperature
T- T + OT
change, determine the stress in the cylinder.
Rod 𝐴𝐶 : steel, 𝐸 = 200 GPa , 𝛼 = 11.7
10 /°C ,
Cylinder 𝐵𝐷: brass 𝐸 = 105 GPa, 𝛼 = 20.9
10 /°C
Sol: Let the elongations of rod 𝐴𝐶 and cylinder 𝐵𝐷 be 𝛿
Because bar 𝐶𝐷𝐸 is rigid, 𝛿
is related to 𝛿
𝛿
by
= 2.5𝛿
and 𝛿
, respectively.
-
↑
-
it wit
↓
#
z
(16)
The deflection of cylinder 𝐵𝐷 consists of two parts: the thermal expansion due to the
temperature change and the contraction due to the compressive axial force 𝐹
Specifically,
· NO
·
Tan Or O
Soc (d
=
Sp .
#
,
Force
& ME
+
dz) O
dzO
&Ac
11
(
equil
-
0
=
(d
,
+
dz) + FBDdz
=
0
Constitutive law
·
8ac
Sop
=
:
+
LBP e
e
.
↑
eq
.
= I
Mechanics of Materials
NTU_ME CJLu
Note#2 Axial Loading
𝛿
The elongation 𝛿
Compatibility
·
=𝛼
Δ𝑇 𝐿
−
2024/2/7
𝐹 𝐿
𝐸 𝐴
(17)
is purely caused by the tensile axial force 𝐹 ,
𝛿
=
𝐹 𝐿
𝐸 𝐴
(18)
Moment equilibrium about point 𝐸 requires that
0.75𝐹
− 0.3𝐹
= 0 or 𝐹
= 0.4𝐹
(19)
Substituting the above equation into Eq. (18) gives
𝛿
=
0.4𝐹 𝐿
𝐸 𝐴
(20)
Substitution of Eqs. (17) and (20) into Eq. (16) results in an equation for 𝐹
0.4𝐹 𝐿
𝐸 𝐴
= 2.5 𝛼
Δ𝑇 𝐿
−
𝐹 𝐿
𝐸 𝐴
Rearranging terms gives
𝐹
=𝛼
Δ𝑇 𝐸 𝐴
0.16
𝐿
𝐿
𝐸 𝐴
𝐸 𝐴
+1
Carrying the values of the parameters into the above equation yields 𝐹
and the resulting compressive stress
𝜎
=
𝐹
𝐴
= 44.8 MPa
12
= 31.7 kN