1.6. TORSIONAL STRESS AND DEFORMATIONSThis is possible because of defects called in the grain structure which can move
through the crystal structure. These dislocations or slips in the grain structure
allow the overall change in shape of the metal. Each grain can have a very large
number of dislocations which only visible under a powerful microscope. During
hammering and rolling, crystal grains break up into smaller units that increases
their elastic properties. Therefore, Metal samples with smaller crystal grains are
more rigid than those with larger grains. Cold working increases the yield strength
and elastic moduli.1.6 Torsional stress and Deformations
Torsion is a variation of pure shear, wherein a structural member is twisted in the manner
of Figure1.16. Shear modulus in the context of torsion is calledRigidity Modulus. Con-
sider a shaft rigidly clamped at one end and twisted at the other end by a torqueT=F.d
applied in a plane perpendicular to the axis of the bar (Figure1.16)^2. Such a shaft is said
to be in torsion. Therefore, torsion refers to the twisting of a structural member when it
is loaded by couples that produce rotation about its longitudinal axis[ 3 ].
F dFT= F.dshould be at least four times this diameter; 60 mm is common. Gauge length
is used in ductility computations, as discussed in Section 6.6; the standard value is
50 mm (2.0 in.). The specimen is mounted by its ends into the holding grips of the
testing apparatus (Figure 6.3). The tensile testing machine is designed to elongate
the specimen at a constant rate and to continuously and simultaneously measure the
instantaneous applied load (with a load cell) and the resulting elongations (using an
extensometer). A stress–strain test typically takes several minutes to perform and is
destructive; that is, the test specimen is permanently deformed and usually fractured.
[The (a) chapter-opening photograph for this chapter is of a modern tensile-testing
apparatus.]
The output of such a tensile test is recorded (usually on a computer) as load
or force versus elongation. These load–deformation characteristics are dependent
on the specimen size. For example, it will require twice the load to produce the same
elongation if the cross-sectional area of the specimen is doubled. To minimize these1214 in. 26.2 Concepts of Stress and Strain• 153TTFFFFFFFA 0A 0A 0(a) (b)(c) (d)!"l l 0 l 0 lFigure 6.
A standard tensile
specimen with
circular cross
section.Figure 6.
(a) Schematic
illustration of how a
tensile load produces
an elongation and
positive linear strain.
Dashed lines
represent the shape
before deformation;
solid lines, after
deformation.
(b) Schematic
illustration of how a
compressive load
produces contraction
and a negative linear
strain. (c) Schematic
representation of
shear strain , where
!tan.
(d) Schematic
representation of
torsional
deformation (i.e.,
angle of twist )
produced by an
applied torque T.fuggGauge length2"Reduced section
2 "
"Diameter
"(^14)
(^34)
(^38) Radius
0.505" Diameter
JWCL187_ch06_150-196.qxd 11/5/09 9:36 AM Page 153
d
A A^0 > A
d 0 > d , Δd = (d 0 - d)
l 0 < l , Δl = (l - l 0 )
longitudinal strain = Δl/l 0
d 0 lateral strain = -^ Δd/d^0
z
x
Poisson s ratio = ( l/ld/d 00 )
=dl^00 dl
T
Figure 1.16: A shaft subjected to torsion.
Twisting Couple: The two equal and opposite forces (with magnitudeF) acting at two
di erent points at a distancedon the shaft as shown in Figure1.16form the twisting
coupleT (also see Figures1.17(a) and (b)). E ects of a torsional load applied to a bar
F
a b
b’
T
T
L
γ a
b
b’
T
T
L
θ c
(a) Shear strain (γ) = <bab’ (b) Angle of twist (θ) =<bcb’
Figure 1.17: Shear strain(“)and twist angle(◊)of a shaft under torsion.(^2) The shaft is a machine element which is used to transmit power in machines. Acoupleis a pair of
forces, equal in magnitude, oppositely directed, displaced by perpendicular distance and do not share a
line of action. Torque(or moment of the force) is the product of the magnitude of the force and the
perpendicular distance of the line of action of force from the axis of rotation.
PH8151 19 LICET