Mechanics of Materials Laboratory
Torsion of a Steel Bar
1
A. Introduction and Objective
The objectives of this experiment are to perform a torsion test of a steel bar, examine the
strain and stress distribution due to torsion in a bar with a circular cross-section,
experimentally determine the shearing modulus of steel, and observe the failure mode of
a ductile material in torsion.
B. Theory
1.0 Equilibrium
Shearing stresses develop in a material when torque is applied to a bar. Our goal
in this section is to derive an equation that relates applied toque to the shearing
stresses that develop in our bar. The torques and the shear stresses are related
through equilibrium.
Using a polar coordinate system in terms of (ρ, θ) centered at the center of the
circular cross-section, a small, or infinitesimal, amount of torque from the stress
at a point on the cross-section is given by
dT = dF = dA .
2
dA
The total torque is found by integrating (essentially adding up) the infinitesimal
torques over the cross-sectional area:
T = dT = A dA .
The preceding statement of equilibrium is valid whether or not the material
remains linear elastic. However, when the material is linear elastic, and shear
stress and strain are related through Hooke’s law for shear:
= G ,
which leads to
T = A G dA .
Unfortunately, the preceding equation is not very useful until we can find an
expression for the shear strain, γ.
The shearing strain in a twisted bar that has a circular cross-section can be
determined based on some simple observations and assumptions of elastic
deformation:
3
-the bar does not elongate,
-plane cross-sections remain plane, and
-radial lines on the cross-sections remain radial lines after deformation.
To illustrate these observations, consider a bar that is clamped on one end.
Before the torsional load is applied, points A and C lie on the outside radius at
the fixed end of the bar. Point B lies on the free end of the bar, and, on a side
view, the angle BAC creates a right angle.
T
A
B
C
After a torque is applied to the end of the bar, point B moves to B*. The
reduction in angle from the right angle is the shearing strain, γ.
2r
T
A
B
C
B
4
L
r
Geometrically,
tan =
BB *
.
L
Since the deformation of elastic metals are usually very small, the tangent of a
small angle and the angle (when expressed in radians) are approximately equal.
Additionally, the distance BB* is approximately equal to the arc length, rφ:
=
r
.
L
This equation gives the shearing strain on the outside edge of a bar of length L,
that has been twisted by an angle, φ, and its free end.
What is needed in the expression containing the integral, however, is the
shearing strain as a function of ρ and θ, the polar coordinates on the crosssection. Since our basic observation is that radial lines remain radial lines after
deformation, the shearing strain will not depend on its angular position, θ.
However, using the same concepts that were used to find the strain on the
outside of the bar, we can find the shearing strain as a function of position, ρ, an
arbitrary point radially outward from the center of the cross-section. We can
visualize this by examining a small length of the bar that has been removed from
the core of the bar, and examining the relative movement of points on each end
of the segment.
5
X
R
S
X
S’
R*
S*
R*
X
S’
S*
6
In the same way as presented previously,
=
SS *
=
,
x
x
which is the shearing stress over a length of the segment of the bar, x . To get
the strain at a point, we can take the limit as this length goes to zero:
= lim x =d .
x → 0
dx
d
is the twist rate of the bar, and is a constant. Then, the preceding
dx
equation implies that the shearing strain varies linearly with the distance from
the center of the bar (ρ). i.e. that the shearing strain is zero at the center of the
bar, and is maximum on the outer edge of the bar. For a linear elastic material
that obeys Hooke’s law in shear ( = G ), the shearing stress is also maximum
on the outside of the bar.
The term
An expression for the shearing stress in terms of the twist rate is
d
= G
.
dx
With an equation for the shearing strain determined in terms of the polar
coordinate variable, ρ, the torque on the bar can now be determined by
evaluating the integral
T = G dA = G
A
A
d
d
dA = G
A 2 dA .
dx
dx
The integral in the expression above contains only terms that involve the crosssection of the bar. It has a special name, and it is called the polar moment of
inertia, Ip.
This results in an equation for the torque, T, as
7
T =G
d
I.
p
dx
Eliminating the twist rate (which can be difficult to measure) by replacing it with
the expression containing the shearing stress results in:
T =G
I,
p
G
and finally upon solving for the shearing stress in terms of the torque results in
=
T
.
Ip
The polar moment of inertia can be quickly found by evaluating the integral, and
is an exercise typically assigned in Calculus class. For a solid circular crosssection, the polar moment of inertia is:
I p=
d 4,
32
where d is the diameter of the cross-section.
If the angle of twist of the bar, φ, is required, then torque-twist rate equation can
be separated and integrated:
d =
8
T
dx ,
I pG
=
T
L
0
I pG
dx =
TL
I pG
if the torque, shear modulus, and polar moment of inertia are constant over the
length of the bar for which the twist angle is determined. Note also that the units
of the twist angle are in radians.
Given any four of the five variables in the twist equation, the remaining unknown
term can be found.
The data that will be collected is the torque applied to the bar, and the angle of
twist between a section of the bar. The dial needle of the torsion machine will
indicate the value of torque. The initial angle of twist is determined by the use of
a device called a troptometer and then later by the angular indictor on the testing
machine. Neither of these two instruments are connected to a data acquisition
system, so you will carefully determine the value of torque at a specified
troptometer reading.
9
10
It is essential to understand how the troptometer operates in order to calculate the
relative angle of twist. The troptometer consists of a dial gage, wire, a left and
right side, and a connecting bar as shown in the figure. The troptometer is
installed on the testing machine with the connecting bar in place. This establishes
a precise 2.0 inch spacing between the left and right sides of the troptometer.
Prior to loading the specimen, the connecting bar is removed.
As the bar is loaded, both the left and right sides of the troptometer rotate with the
specimen. However, because of the deformation of the material, one side of the
troptometer rotates more than the other. This relative rotation (over the 2.0
inches) is measured by dial gage as a thin wire moves along the arc length of a
wheel that is precisely 2.5 inches in radius. If s is the arc length, and R is the
radius of the wheel (2.5 inches in this case), the relative angle of twist is
determine by the arc length relation:
s = R , or
= s / R.
Since the dial gage reads in thousandths of an inch, to determine the radians of
relative twist over 2.0 inches, you will use
(radians) =
(dial gage reading / 1000)in.
.
2.5in.
After a certain amount of twist, the amount of rotation will not be able to be
measured with the troptometer, and the dial gage will be disconnected. At this
point, the angle of twist will be given approximately by the angular degree
markings on the testing machine. The torque will also need to be recorded for the
corresponding values of angle of twist in degrees.
The angle of twist in degrees will also need to be converted into radians, by the
simple conversion factor that 180º = π radians, or
11
(radians) =
(degree reading) *
.
180.
C. Lab Procedure
The testing machine should be on 20-30 minutes prior to testing.
Verify the diameter of the steel bar. Place the troptometer onto the specimen. Mount the
specimen into the torsion machine, being careful to ensure that the flat of the grips align
with the flat ends of the specimen. Tighten the chuck on the specimen and remove the
chuck key prior to testing. Verify the operation of the troptometer and wire system.
(Note: the way the troptometer is set up, the dial gage decreases when the angle of twist
changes.)
Remove the troptometer bar. Apply the torque in the forward testing machine mode, at a
relatively slow rate. Call out the values of dial gage reading listed in the table and the
specimen is loaded. Use a dry-erase marker to mark the torque values from the testing
machine needle directly on the glass face of the testing machine indicator. When the
troptometer has reached it’s useful limit repeat this process for every 90º interval based
on the testing machine indicator, until the specimen fractures.
Record your test data on the following data sheet.
After the specimen fails, photograph or sketch the shape of the specimen and failure
surface.
12
A
Torque Reading
(in-lb)
B
C
D
E
F
Change in dial gage
reading (thousandths
of an inch)
angle of
twist from
testing
machine
(degrees)
Conversion
of column
B into
radians
Conversion
of column C
into radians
Divide columns
D-E by the gage
length to get
radians per inch;
write all values in
this column
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
*200
45
90
180
270
360
450
540
630
*release dial gage wire
13
D. Lab Report
Complete the data table columns D-F.
Construct a graph with torque values on the y-axis (column A) and radians per inch on
the x-axis (column F). These data points should be plotted as open symbols (Do not
connect the points with a line plot). Draw a best fit curve through all of the data.
Indicate the ultimate torque value on the graph.
Construct a second graph that has a maximum x-axis value of 0.04 radians/inch which
corresponds to the dial gage reading of 200 thousandths of an inch. This graph is
expected to have a relatively straight-line portion and some initial curving in it. The
deviation from a straight line indicates the yield torque of the material. Note this value
on the graph. Determine the slope, m, of the best fit line from the initial portion of the
data.
E. Questions
Determine the shear yield stress of the material from the yield torque and the shear stress
equation.
14
Is it valid to use the shear stress equation to determine ultimate shearing stress of the
material? Explain.
Sketch the elastic shearing stress distribution for an elastic bar with a round cross-section
subjected to torsion.
15
Determine the shear modulus for this material, by using the slope determined from the
second graph (Show Calculation).
Hints:
m=
T
( / L)
T
=
L
I pG
G=
T
I p ( / L)
Shear yield stress from measurements
Accepted value of shear stress
Percent difference
16
=
m
Ip
.
F. Discussion
17
Geage length = 2 inch
Une formula in page-1
A
Torque Reading
B
Change in dial gage
reading (thousandths
of an inch)
D
Conversion
of column
B into
radians
С
angle of
twist from
testing
machine
(degrees)
E
Conversion
of column C
into radians
(in-1b)
F
Divide columns
D-E by the gage
length to get
radians per inch;
write all values in
this column
O
480
850
1160
1280
1980
1960
2220
2400
2680
2660
2740
2760
2820
2860
2580
2920
2960
29.90
2985
3000
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
*200
45
90
180
270
360
450
540
630
*release dial gage wire
Purchase answer to see full
attachment