City University of Hong Kong
Dept. of Physics & Materials Science
AP2104 Mechanics of Solids
Laboratory Manual
Experiment 1
Pure Bending of a Beam
Experiment 2
Torsional Deformations
Experiment 3
Yield Criteria for Ductile Materials under Plane
Stresses
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Experiment 1
Pure Bending of a Beam
Objective
1. To examine the stresses at various positions of the beam under a constant load of pure
bending.
2. To determine the curvature of deflection of the beam.
Introduction
1. Pure Bending and Nonuniform Bending
When analyzing beams, it is often necessary to distinguish between pure bending and
nonuniform bending. Pure bending refers to flexure of a beam under a constant bending moment.
Therefore, pure bending occurs only in regions of a beam where the shear force is zero ( because
V = dM/dx ). In contrast, nonuniform bending refers to flexure in the presence of shear forces,
which means that the bending moment changes as we move along the axis of the beam.
As an example of pure bending, consider a simple beam AB loaded by two couples M1|
having the same magnitude hut acting in opposite directions (Fig. 1a). These loads produce a
constant bending moment M = M1 throughout the length of the beam. Note that the shear force V
is zero at all cross sections of the beam.
Fig. 1 (a) Simple beam in pure bending. (b) Cantilever beam in pure bending
Another illustration of pure bending is given in Fig. 1b, where the cantilever beam AB is
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subjected to a clockwise couple M2 at the free end. There are no shear forces in this beam, and the
bending moment M is constant throughout its length. The bending moment is negative (M = - M2).
The symmetrically loaded simple beam of Fig. 2a is an example of a beam that is partly in
pure bending and partly in nonuniform bending, as seen from the shear-force and
bending-moment diagrams (Figs. 2b and c). The central region of the beam is in pure bending
because the shear force is zero and the bending moment is constant. The parts of the beam near the
ends are in nonuniform bending because shear forces are present and the bending moments vary.
Fig. 2 Simple beam with central region in pure bending and end regions in
nonuniform bending
2. Deformations of a Beam in Pure Bending
Consider a portion ab of a beam in pure bending subjected to positive bending moments M
(Fig. 3a). We assume that the beam initially has a straight longitudinal axis (the x axis in the
figure) and that its cross section is symmetric about the y axis. Under the action of the bending
moments, the beam deflects in the xy plane (the plane of bending) and its longitudinal axis is bent
into a circular curve (curve ss in Fig. 3b). The beam is bent concave upward, which is positive in
curvature.
F
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Fig. 3 Deformation of a beam in pure bending
In this bending deformation, o’ is the center of curvature

where
1


d
dx
  curvature
  radius of curvarure
Cross sections mn and pq rotate with respect to each other about axes perpendicular to the xy
plane. Longitudinal lines on the convex (lower) part of the beam are elongated, whereas those on
the concave (upper) side are shortened. Thus, the lower part of the beam is in tension and the
upper part is in compression. There is a surface in which longitudinal lines do not change in
length. This surface, shown by the dashed line ss in Figs. 3 and, is called the neutral surface of
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the beam.
Fig. 4 Relationship between the bending moment and the normal stresses
3. Maximum Stresses at a Cross Section
The maximum tensile and compressive bending stresses acting at any given cross section
occur at points located farthest from the neutral surface ( Fig. 4 ). Let c1 and c2 denote the
distances from the neutral surface to the extreme elements in the positive and negative y
directions, respectively.
The corresponding maximum normal stresses σ1 and σ2 are
1  
where
Mc1
I
I = moment of inertia =
2  
and

A
Mc 2
I
y 2 dA
Apparatus
Square and rectangular polyacetal beams; four-points bending fixture; strain gauges and
strain gauge indicator; digital camera.
Procedure
1.
Setting up :
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a) Strain gauges are adhered at the mid points of the top and bottom faces of the sample beam
in order to record the compressive and tensile stresses arising from the pure bending.
2.
b)
The sample beam is placed securely on the four-points bending fixture which is mounted
on the cross-head of the tensile tester.
c)
The strain gauges are connected to the strain gauge indicator and its initial reading is
zeroed.
d)
The digital camera was placed on an adjustable platform and set at an appropriate height
to catch the front view of the beam.
Carrying out the experiment :
a)
Choose a suitable downward rate for the cross-head of the tensile tester and start the
bending of the sample beam.
b)
Take photos with the digital camera at two different bending positions ( note: the bending
should not exceed the yielding of the material ) and each time record the readings of the
compressive and tensile stresses from the strain gauge indictor.
Discussion
a)
Evaluate the curvature of the bending from the contour of the sample beam taken in the
photos.
b) Also calculate the curvature from the readings of the compressive and tensile stresses and
compare the result with the one obtained from (a).
c)
Show that the strains in a beam in pure bending vary linearly with their distance from the
neutral surface regardless of the shape of the stress-strain curve of the material.
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Experiment 2
Torsional Deformations
Objectives
1.
To carry out torsion tests for ductile materials.
2.
To measure the modulus of rigidity of the material.
Introduction
For a circular shaft, when subjecting to twisting couples or torques, there will be torsional
deformation occurring.
The distribution of stress in the cross section of a circular shaft is statically indeterminate.
However, every cross section remains plane and undistorted.
1. Strain :
The shearing strain in a small element with sides parallel and perpendicular to the axis of the
shaft and at a distance  from that axis can be expressed as :
 

L
where  is the angle of the twist for a length
L of the shaft. The above equation shows that the
shearing strain in a circular shaft varies linearly
with the distance from the axis of the shaft.
Therefore, the strain is maximum at the surface.
 max 
c
L
 
==>

c
 max
2. Stress :
Within the elastic range, by using the Hooke ' s law, the stress  is
  G 

c
G max     max
c
where G is the modulus of rigidity of the material. It shows that the shearing stress in a
circular shaft also varies linearly with the distance from the axis of the shaft. Therefore, the
stress is also maximum at the surface.
3. How to obtain the  max
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Equating the sum of the moments of the elementary forces exerted on any section of the shaft
to the magnitude T of the torque applied to the shaft, we have the elastic torsion formula :
 max 
where J 
Tc
J
and
T
J
 
1 4
c , the centroidal polar moment of inertia.
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4. Relation between torque and the angle of twist
Within the elastic range, the angle of the twist  of a circular shaft is proportional to the
torque T applied to it with the relationship of
 
TL
JG
Apparatus
Torsion testing tools; ductile rods (e.g. metallic); a ruler.
Procedure
Set up the test as shown in the right diagram.
A circular shaft is fixed at one end. A force exerts on
the other end of the shaft to the magnitude T of the
torque applied to the shaft.
Measure the length and radius of each testing
shaft. And prepare a table for these parameters and the
 and T recorded in the testing.
1.
Maximum strain :
 max 
2.
Maximum stress :
 max 
where J 
3.
c
L
Tc
J
1 4
c , the centroidal moment of inertia.
2
Modulus of rigidity of the material :
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G 
4.
 max
 max
Relation between torque and the angle of twist
Plot the angles against the applied tongues for at least two samples and illustrate the
relationship between them.
Discussion
1.
Why in a torsion test ductile materials, which generally fail in shear, will break along a plane
perpendicular to the axis of the specimen, while brittle materials, which are weaker in
tension than in shear, will break along surfaces forming 45° angle with that axis? What have
you observed in the experiment?
2.
Give a brief discussion on the modulus of rigidity of the different materials you have tested.
3.
Could you obtain the modulus of rigidity of the materials from the plot of  vs T ? If yes,
compare the values with those obtained above. Which procedure do you suggest to use in
practice? Why?
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Experiment 3
Yield Criteria for Ductile Materials under Plane Stresses
Objectives
1.
To investigate the principal stresses σa and σb at any given point of a structural element or
machine component when it is in a state of plane stress.
2.
To evaluate the two different yield criteria ( Tresca and Von Mises ) for ductile materials
under plane stresses.
Introduction
Structural elements and machine components made of a ductile material arc usually designed
so that the material will not yield under the expected loading conditions. When the element or
component is under uniaxial stress (Fig. 1), the value of the normal stress σx, which will cause the
material to yield, may be obtained readily from a tensile test conducted on a specimen of the same
material, since the test specimen and the structural element or machine component are in the same
state of stress. Thus, regardless of the actual mechanism which causes the material to yield, we
may state that the element or component will be safe as long as σx<σY, where σY is the yield
strength of the test specimen.
On the other hand, when a structural element or machine component is in a state of plane
stress (Fig. 2a), the principal stresses σa and σb at any given point can be determined (Fig. 2b). The
material may then be regarded as being in a state of biaxial stress at that point. Since this state is
different from the state of uniaxial stress found in a specimen subjected to a tensile test, it is
clearly not possible to predict directly from such a test whether or not the structural element or
machine component under investigation will fail. Some criteria regarding the actual mechanism
of failure of the material must be established.
Maximum-Shearing-Stress Criterion.
According to this criterion, a given structural component is safe as long as the maximum
value τmax of the shearing stress in that component remains smaller than the corresponding value
of the shearing stress in a tensile-test specimen of the same material as the specimen starts to
yield.
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The maximum value of the shearing stress under a centric axial load is equal to half the value
of the corresponding normal axial stress, hence the maximum shearing stress in a tensile-test
specimen is ½σY as the specimen starts to yield. On the other hand, for plane stress, the maximum
value τmax of the shearing stress is equal to ½|σmax| if the principal stresses are either both positive
or both negative, and to ½|σmax – σmin| if the maximum stress is positive and the minimum stress
negative. Thus, if the principal stresses σa and σb have the same sign, the
maximum-shearing-stress criterion gives
|σa| < σY
|σb| < σY
If the principal stresses σa and σb have opposite signs, the maximum-shearing-stress criterion
yields
|σa – σb| < σY
The relations obtained have been represented graphically in Fig. 3. Any given state of stress
will be represented in that figure by a point of coordinates σa and σb, where σa and σb are the two
principal stresses. If this point falls within the area shown in the figure, the structural component
is safe. If it falls outside this area, the component will fail as a result of yield in the material. The
hexagon associated with the initiation of yield in the material is known as Tresca's hexagon.
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Maxirnum-Distortion-Enoryy Criterion
According to this criterion, known as the von Mises criterion, a given structural component
is safe as long as the maximum value of the distortion energy per unit volume in that material
remains smaller than (ud)Y, the distortion energy per unit volume required to cause yield in a
tensile-test specimen of the same material. If the distortion energy per unit volume in an isotropic
material under plane stress is ud, the maximum-distortion-energy criterion indicates that the
structural component is safe as long as ud < (ud)Y, i.e., as long as the point of coordinates σa and σb
falls within the area shown in Fig. 4. This area is bounded by the ellipse of equation
σa2 – σaσb + σb2 = σY2
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Apparatus
Tensile tester, Fixture for performing the shear yield test, polypropylene (pp) strip
specimens for shear yield tests, pp dumbbell specimens for tensile yield tests.
Procedure
1. Perform the tensile yield test with the pp dumbbell specimens and evaluate the yield
point from the resulting stress-strain curve.
2. Mount the fixture on the tensile tester and perform the shear yield test with the pp strip
specimens.
3. Construct the corresponding Tresca’s hexagon and the von Mises’ ellipse and
demonstrate the validity of these yield criteria.
Discussion
Explain the principle of deriving the two yield criteria (Tresca and von Mises)
and their main differences, and how they can be applied in the design of structural
components.
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