Fundamental Mechanics
ENGG 1300
Axially Loaded Members
Ir Dr. Ray Su
Faculty of Engineering
THE UNIVERSITY OF HONG KONG
All rights reserved
1
Course Outline
Axially Loaded Members
Statics Systems
Statically Determinate Structures
Reactions of statics systems, e.g. beams and frames
Internal forces in pin-jointed trusses and
cable structures
Course Outcomes
1. Determine stress and strain of axially loaded
members using Hooke’s law.
2. Understand statics, restraints and redundancy
3. Construct equilibrium equations to solve
statically determinate problems.
2
Reference Books
1.
Hibbeler, R.C., Engineering Mechanics: Static & Dynamics, 13th
Edition, Pearson, 2013
2.
Beer, F.P., Johnston, E.R. and DeWolf, J.T., Mechanics of Materials,
6th Edition, McGraw Hill, 2013.
3.
Gere, J.M. and Goodno, B.J., Mechanics of Materials, 8th Edition,
Cengage Learning, 2013.
4.
Hannah, J. and Hillier, M.J., Applied Mechanics, Pitman, 1995.
5.
Riley, W.F., Sturges, L.D. and Morris, D.H., Statics and Mechanics of
Materials, Wiley, 1995.
3
Axially Loaded Members
4
Axial Stress
Consider a bar with a cross sectional area A being subjected to
equal and opposite forces F pulling at the ends so the bar is under
tension.
F
F
axial stress
in the bar
A
In the bar, the material is experiencing an axial stress defined
to be the ratio of the force to the cross sectional area of the
bar.
F
=
Note: F is normal to A
A
The unit of stress is N/m2 or Pa (Pascal).
1 N/m2 = 1 Pa
1,000 N/m2 = 1 kPa
1,000,000 N/m2 = 1 MPa
Mega
1N
1,000,000N
1MN
Note that 1 MPa = 1 N/mm2, since 1mm 2 = 1,000,000m m 2 = 1m 2 = 1MPa
5
Example
Determine the axial stress in the following prismatic bars.
100 kN
1250 mm2
F 100 1000
= =
= 80 N/mm 2
A
1250
100 kN
15 kN
120 mm2
15 kN
=
F 15 1000
=
= 125 N/mm 2
A
120
6
Saint Venant’s Principle
For elastic analysis, the difference between the effects of two
different but statically equivalent loads becomes very small at
sufficiently large distances from the loads.
Two bars subjected to the same total force F
Internal axial stress distribution
F
h
F/2
q
q
F/2
>h
Hence we do not normally concern about the actual applied load
distribution provided that the centroid of the load is coincident
7
with the centroid of the section in the elastic analysis.
For information
Centroid
CL
CL
Centroid
Centroid
CL
Square section
Circle section
E.g.
xo
y xo=a/2
b

=
x
x1
Δx
a
0
A = Total area
xbdx
ab
xo A =  xdA
a 2b 2 a
=
=
ab
2
x1
x2
x
Centroid = xo

=
x2
x1
b
xdA
A
x
a
y x =2a/3
o
x2
y1(x)
h
Equilateral triangular section
ΔA = (y2-y1)Δx
xo
CL
h/3
Centroid by Integration
y
Centroid
CL
CL
y2(x)
CL
xo =

a
0
a
ab
x
b

x xdx 
a
 = 3 = 2a
ab / 2
ab / 2 3
2
8
Axial Force from Distributed Load
Rectangular section (ab) subjected to the following distributed
loads which are expressed in terms of the axial stress.
Case 1
Uniformly
distributed
load
σ
2
 (in N/m )
Total downward force
=volume of the stress block
=σab (unit in N)
a
b
Case 2
Triangularly
distributed
load
σ
Total downward force
=volume of the stress block
=(σa/2)b = σab /2
a
b
9
Example
Determine the vertical stress distribution at the base of a column
with dimensions bd under (a) an axially loaded vertical load and
(b) an eccentrically loaded vertical load (e = eccentricity).
(Assumptions: a linear elastic material (e.g. p17) and plane section remaining
plane, more details will be given in latter parts of this course)
(b1)
(b2)
F
F
(a)
b
(b)
section
F
F
e
=
d
F
F
e
F
=
F M = Fe
e
+
d
Solution
σ
support
σ+σf
σ-σf
(a) Vertical stress = σ = F/A = F/(bd)
σf
σ
2d/3= d - d/6- d/6
1d
32
2d/3
Resultant force = σfb(d/2)/2 σ
f
(b2) Moment equilibrium: [σfb(d/2)/2]2d/3=Fe; σf = 6Fe/(bd2)
-σf
d
6
d/2
Combining cases (b1) and (b2), the vertical stresses at the left
and right sides of the stress block are σ+σf and σ-σf respectively.
10
For information
centroid
a/3
Now a = d/2,
hence 4a/3 = 2d/3
centroid
2a/3
a
22a/3
=2d/3
11
Hydrostatic pressure
Water exerts force in all directions, and its water pressure
acting on any surface is always normal to that surface. The
magnitude of hydrostatic pressure increases with the depth of
water.
Example
12
Example
A cantilever wall of a water tank is shown below. Determine the
horizontal hydrostatic pressure acting on the wall.
water
ρ = 1,000 kg/m3
Solution
h
ρgh
Hydrostatic
pressure
(1) Consider a water column with a plan area of 1 m  1 m
and a depth of h from the water surface.
(2) The mass of water column is m = ρh(11) = ρh.
h
1m1m
(3) The weight of water column is w = mg = ρgh.
(4) The vertical stress (or pressure) at the base of water column is
w/A = ρgh/(11) = ρgh.
(5) The hydrostatic pressures in any directions are the same, hence the
horizontal pressure is also ρgh.
13
Example
A 10 mm thick cylindrical pressure vessel with an internal
diameter of 1.5 m is subjected to a uniform internal pressure of
200 kPa (1 Pascal = 1 N/m2). Considering a unit length (i.e. 1 m)
of the vessel, determine the internal force and internal stress
1000 mm
in the vessel.
F
F
0.01 m
1.5 m
Solution
200
kPa
10
200
kPa
F
F
F
F
Free body diagrams
3-D view
Horizontal force equilibrium: 2F = 2001.51 = 300 kN
F = 150 kN
The internal stress in the circumferential direction:
σ = F/A = 150,000/(101000) = 15 N/mm2 or 15 MPa.
This stress acting along the circumferential direction is called
hoop stress.
14
Material strength
Material strength is defined as the maximum stress that can be
resisted by the material.
Tensile and compressive strengths of selected materials
Material
Tensile Strength
(MPa)
Compressive Strength
(MPa)
Steel
400 – 700
400 – 700
Wood
~40
~35
Glass
50
50
Concrete
2–4
20 – 80
Brick
0.28
7
Granite
4.8
130
15
Strain
A bar with an original length Lo is subjected to equal and opposite
forces F pulling at the ends. The change in length of the bar
under tension is ΔL.
F
F
Lo
ΔL
L
The deformation of the bar can be
=
Lo
quantified by the strain which is defined as
Strain is dimensionless and therefore has no unit.
Typical values for strain are less than 0.005 and are often
expressed in microstrain units:
e.g. Microstrain = strain  10-6
SI prefix: μ = 10-6
When the applied forces have been removed and the bar can
restore its initial dimensions and shape, the material is elastic.
Otherwise the material is inelastic.
16
Young’s modulus
A
F
Lo
stress
F
ΔL
strain
In 19th-Century, a British scientist Thomas Young found that
within elastic range, axial stress (=F/A) is proportional to axial
strain (=ΔL/Lo). The proportional constant is called Young’s
Modulus (or elastic modulus or modulus of elasticity).
Hooke’s law:
Axial stress
Axial strain
= constant (Young’s Modulus)
The unit of Young’s modulus is N/m2 or Pa (Pascal).
1,000,000,000 N/m2 = 109 Pa = 1 GPa
(G = giga)
Linear materials mean the materials obey Hooke’s Law.
17
Young’s modulus
Young’s moduli of selected materials
Material
Young’s Modulus
(GPa)
Steel
200
Wood
1 – 10
Glass
50 – 90
Concrete
20 – 30
Brick
2 – 12
Granite
26 – 70
For practical engineering applications, it is generally accepted
that the Young’s moduli in tension and compression are the
same.
18
Example
A steel circular wire with Lo = 3 m long and dia = 2 mm in
diameter is fixed to the ceiling and a weight with 400 N is
attached to the end. Calculate the strain and extension of the
wire, given that the Young’s modulus of the wire is E = 200 GPa.
Solution
Sectional area of the wire = 22π/4 = 3.1416 mm2
Axial stress = 400/3.1416 = 127.3 N/mm2 (or MPa)
Hooke’s law E =


Axial strain = 127.3 MPa/ 200,000 MPa
= 0.000637 or 637μ
Axial extension = 0.0006373
= 0.00191 m (or 1.91 mm)
Dia =
2 mm
400 N
Lo =
3m
?
19
Poisson’s ratio
In early 1800s, a French scientist S.D. Poisson realized that
within elastic range the ratio of lateral strain and axial strain is a
constant. This constant is called Poisson’s ratio ν.
A circular bar under tension
Lo
L
After
Before
deformation
d do
d
L L − Lo
d − do
=
=
, Axial strain =
Lateral strain =
Lo
Lo
do
do
Poisson’s ratio = -
Lateral strain
Axial strain
Poisson’s ratio is dimensionless and therefore has no unit.
20
Poisson’s ratio
Poisson’s ratios of selected materials
Material
Poisson’s ratio
Steel
0.3
Wood
0.3 – 0.48
Glass
0.2 – 0.25
Concrete
0.2
Masonry
0.2 – 0.25
Granite
0.2 – 0.3
21
Ended
22
Open-ended questions
▪How to measure the axial deformation, axial strain
and axial stress?
▪What are flexural and shear stresses?
▪Why stresses are important in engineering?
▪Besides linear elastic materials, do you know other
materials?
23
For information
Centroid
A physical way to find the centriod
Irregular shape
Centroid
Determine the centriod of
the irregular plane.
Moments on both sides will
automatically balance.
The centroid is at the
interaction of two lines.
24