ME 215 – Engineering Materials I and Shear (PART II)

ME 215 – Engineering Materials I
Chapter 4 – Properties in Bending
and Shear (PART II)
Mechanical Engineering
University of Gaziantep
Dr. A. Tolga Bozdana
Assistant Professor
Properties in Shear
Shear stress plays a dominant role in the failure of ductile materials,
and hence shear tests are usually conducted to obtain a measure of
“shear strength” for specific applications.
Shear stresses are categorized into two main types:
1. Direct or transverse shear in rivets (Fig. 7a) and beams (Fig. 7c).
2. Pure or torsional shear in a shaft subjected to pure torsion (Fig. 7b).
Figure 7a
Figure 7b
Figure 7c
Direct or transverse shear tests are employed to obtain a measure of
shear strength for specific applications while the torsion test is usually
employed to evaluate the shear behaviour and properties of a material.
1
Properties in Shear: Direct Shear
In direct shear test, a round or flat stock specimen is clamped in a
fixture and a shearing force is applied through a suitable shear tool.
In single and double shear test (Fig. 8),
direct shear stress (ττ) due to the shearing
force (F) applied on a cross-sectional area
(A) is expressed as:
Figure 8
τ = F A (for single shear)
τ = F 2 A (for double shear)
Transverse shear test gives only value of “ultimate shear strength”
of machine parts such as bolts, rivets and pins. However, the results of
such test may be approximate due to appreciable errors depending on
hardness, sharpness and correct setting of shearing tools. Errors may
also be introduced by bending stresses and friction between parts.
2
Properties in Shear: Direct Shear
The punch shear test (Fig. 9) is a form of
direct shear test of flat stock plastics. For
a given punch diameter (d) and material
thickness (h), the shear strength (ττ) is:
F
τ=
π ∗d ∗h
Figure 9
Instead of cross-sectional
area, the punched area is
taken into consideration.
This test is also amenable to experimental
errors discussed in the previous slide. In
fact, actual shear stress is not of interest.
Quite often, the relative force necessary to blank or shear the plates of
different thickness or material is of greater interest. So, this test is often
used to test fresh supplies of sheet material simply by comparison with
the previous results. This way, damage to expensive dies is avoided by
the detection of hard plates or materials that are too soft for production.
3
Torsion Test
Torsion tests are carried out by applying a twisting moment to one end
of a specimen while measuring angular deformation at the other end.
Torsion test is not used in material specifications to the same extent as
the tension test due to the reason that no uniform shear stress can be
generated. Magnitude of shear stress varies from zero at the axis of
symmetry to maximum at the surface. When the surface reaches to the
elastic limit, the interior will still be in the elastic range. Thus, beginning
of yield at the surface will not be detectable until a considerable amount
of plastic deformation has taken place.
To overcome this problem, thin walled specimens are used where shear
stress is assumed to be uniform along the cross section. However, there
is a danger of buckling when using hollow tubes if the length to diameter
and the diameter to thickness ratios are not kept within elastic limits.
4
Torsion Test
Nevertheless, the torsion test is still useful in determining the material
properties such as shear modulus of elasticity, torsional yield strength
and shear modulus of rupture. Tests can also be carried out on full sized
engineering parts to determine their behaviour under service conditions.
Torsion test offers certain advantages over tensile test. During torsional
test, no necking occurs in the specimen. So, the torque increases up to
the moment of failure. Plastic deformation is almost uniform over entire
length of specimen, which enables the determination of deformations
and stresses reliably for highly ductile materials (especially pure metals).
Brittle or low ductility materials, which are often difficult to test in tension,
undergo quite measurable deformation in torsional test, which enables
the determination of their mechanical properties.
In addition, torsional tests can easily be conducted at high strain rates.
5
Torsion Test
A torsion testing machine (Fig. 10) consists of loading (A) and indicating (B)
units. Unit A is permanently mounted on the bed and unit B is free to move
(which is necessary to permit quick adjustment of varying specimen lengths
and automatic compensation of changes in length).
Load is applied to by a twisting (straining) head. Measuring system may be
simple pendulum type, compund lever type, hydraulic type or fully electronic
type (preferred). Measured data is plotted to obtain torque vs. angle of twist.
Cylindrical specimens
with square or hexagon
shaped ends are usually
used in torsion tests.
Figure 10
6
Elastic Behaviour in Torsion
Consider a cylindrical bar that is subjected to equal and opposite torsional
moments (torque, T) as in Fig. 11. The bar is twisted through an angle (θ)
relative to the other end. This torsional deformation induces shear stresses
on cross-sectional planes of the bar, which resist externally applied torque.
Figure 11
The shear strain for small deformations (γ) and the shear stress (τ) are:
τ =T ∗R J
J = πR 4 2 (for solid bar)
γ = tan φ = θ ∗ R L
T : externally applied torque (kg*mm)
R : radius of the bar (mm)
J : polar moment of inertia (mm4)
θ : angle of twist (radians)
L : length of the bar (mm)
7
Elastic Behaviour in Torsion
During the torsion test, torque
(T) and angle of twist (θ) are
measured. After that, similar to
“Load-Extension” diagram of a
tensile test, “Torque-Twist”
diagram is constructed for
graphical representation of
material’s torsional behaviour.
Figure 12
Normalised mild steel
(D = 6.35 mm, L = 152.4 mm)
Fig. 12 shows typical torque twist diagrams for a ductile
(normalised mild steel) and a
brittle (cast iron) material.
Cast iron
(D = 6.35 mm, L = 76.2 mm)
8
Elastic Behaviour in Torsion: Stiffness
A material has high stiffness in torsion
when its deformation in elastic range
is relatively small. Stiffness in torsion
is measured by “shear modulus” or
“modulus of rigidity” denoted by G:
G = τ γ = (T ∗ L ) ( J ∗θ )
It can also be obtained from Young’s
Modulus (E) provided that Poisson’s
Ratio (µ) of that material is known:
G = E [2(1 + µ )]
Poisson’s ratio and modulus of rigidity
of various materials are shown in the
table (µ
µ is 0.3 for most materials).
9
Elastic Behaviour in Torsion: Elastic Shear Strength
Elastic shear strength is measured by the maximum stress in specimen
corresponding to torque during the transition from elastic to plastic range.
The first onset of yielding is usually not apparent
in most materials, hence the “offset yield point”
is commonly employed in torsion test (i.e. point
P in Fig. 13). The offset angle of twist is usually
taken as 4*10-5 radian/mm of gauge length. So,
the elastic shear strength (Ssy) is defined by:
S sy = Tsy ∗ R J
Tsy : torque at proportional limit
(i.e. at offset angle of twist)
Figure 13
T
P
Tsy
//
O offset
θ
Tubular specimens are also used to precisely determine the torsional elastic
limit or yield strength. Using a specimen having a thin walled circular crosssection, shear stress along the wall is uniform for all practical purposes as
such specimens do not benefit from “strengthening effect” of inner fibers.
10
Elastic Behaviour in Torsion: Elastic Shear Strength
Strengthening effect of inner fibers is illustrated in Fig. 14 for aluminum alloy
6061-T4. Theoretically speaking, thinner is the wall thickness, more reliable
is the measurement of elastic shear strength since all fibers are at about the
same stress. However, if a thin walled tube is subjected to torsion, it would
first fail by buckling before the shear strength of material is reached. Thus, a
tubular specimen with a ratio of length to diameter of at least 10 and a ratio
of diameter to wall thickness of about 10 has been recommended for yield
strength or modulus of rigidity determinations without buckling. For a tubular
specimen, the elastic shear strength is defined by:
S sy =
Figure 14
Tsy
t = (do - di)/2
2
2π ∗ R ∗ t
R = (do + di)/2
Consequently, the torque - twist
diagram can be converted to the
shear stress - shear strain diagram
using equations mentioned so far.
11
Elastic Behaviour in Torsion: Resilience
As in tension, the area under the torque - twist diagram represents the total
work in stressing the specimen to the proportional limit. Assuming that the
total work is absorbed by the gauge length of the specimen, the modulus
of resilience (Us)is expressed as:
2
T
∗
θ
S
1 sy sy
sy
Us = ∗
=
2 A∗ L
4G
Tsy : torque at proportional limit or yield
θsy : corresponding angle of twist
A : cross-sectional area of the specimen
L : gauge length of the specimen
Ssy : elastic shear strength
G : modulus of rigidity
12
Plastic Behaviour in Torsion: Ultimate Strength and Toughness
1. Plastic Shear Strength: approximate
solid specimen:τ u
definition of ultimate shear strength
tubular specimen:τ u
or modulus of rupture:
= Tu ∗ R J
= Tu (2π ∗ R 2 ∗ t )
For accurate measure of ultimate shear strength, tubular specimens with
short reduced sections are recommended (i.e. a ratio of reduced section
length to diameter of 0.5 and a ratio of diameter to wall thickness of 10).
According to Nadai, it can be calculated as:
τ a = (12Tu ) (πD 3 )
2. Toughness: its definition using torque - twist diagram is also approximate.
Toughness index number in torsion (To) can be defined as:
τ o = (Tu ∗θ f ) ( A ∗ L )
θf : fracture angle of twist
13
Failure Types in Torsion
The torsional fracture is quite distinct from either tension or compression
fracture. There is almost no localized reduction or area (i.e. no necking).
A ductile material fails by shear along one of the planes of maximum shear
as illustrated in Fig. 15a. The fracture is usually silky in texture and the axis
about which the final twisting took place may usually be observed.
Rupture of a brittle material (Fig. 15b)
for which tensile strength is less than
shearing strength occurs by separation
in tension along a helicoidal surface.
Cast iron is a good example that its
shearing and compressive strengths
are greater than its tensile strength.
(a)
Figure 15
(b)
(c)
Fig. 15c illustrates the typical buckling
failure of a thin walled tubular specimen.
14
Failure Types in Torsion
15