CE 323/ BES 222
MECHANICS OF DEFORMABLE BODIES
SUMMER 2020
CHAPTER 2 – STRAIN (Continuation)
D. SHEARING DEFORMATION
Shearing forces cause shearing deformation. An element subject to shear does not change in length but undergoes a
change in shape.
The change in angle at the corner of an original rectangular element is called the shear strain and is expressed as
The ratio of the shear stress τ and the shear strain γ is called the modulus of elasticity in shear or modulus of rigidity
and is denoted as G, in MPa.
The relationship between the shearing deformation and the applied shearing force is
where V is the shearing force acting over an area As.
E. GENERALIZED HOOKE’S LAW
E.1 Uniaxial loading; Poisson’s ratio
Experiments show that when a bar is stretched by an axial force, there is
a contraction in the transverse dimensions, as illustrated in Fig. 2.8. In 1811,
Sime´on D. Poisson showed that the ratio of the transverse strain to the axial
strain is constant for stresses within the proportional limit. This constant,
called Poisson’s ratio, is denoted by ν (lowercase Greek nu). For uniaxial
loading in the x-direction, Poisson’s ratio is ν = - ɛt /ɛx, where ɛt is the
transverse strain. The minus sign indicates that a positive strain (elongation)
in the axial direction causes a negative strain (contraction) in the transverse
directions. The transverse strain is uniform throughout the cross section and
is the same in any direction in the plane of the cross section.
Therefore, we have for uniaxial loading
ɛy = ɛz = - νɛx
Poisson’s ratio is a dimensionless quantity that ranges between 0.25 and 0.33 for metals.
Using σx = Eɛx yields the generalized Hooke’s law for uniaxial loading (σy = σz = 0):
ɛy = ɛz = - ν
E.2 Multi-axial Loading
CE 323/ BES 222
MECHANICS OF DEFORMABLE BODIES
SUMMER 2020
E.2.1 Biaxial Loading
Poisson’s ratio permits us to extend Hooke’s law for uniaxial loading to biaxial and triaxial
loadings. Consider an element of the material that is subjected simultaneously to normal stresses in the x-
and y-directions, as in the figure. The strains caused by σx alone are given as . Similarly, the
strains due to σy are ɛy = σy /E and ɛx = ɛz = - ν σy /E. Using superposition, we write the combined
effect of the two normal stresses as
(σx - νσy) (σy – νσx) (σx + σy)
which is Hooke’s law for biaxial loading in the xy-plane (σz = 0). The first two can be inverted to express the stresses
in terms of the strains:
( ) ( )
E.2.1 Tri-axial Loading
Hooke’s law for the tri-axial loading in Fig. 2.10 is obtained by adding the contribution
of σz, ɛz = σz /E and ɛx = ɛy = - ν σz /E, which yields
[σx – ν (σy + σz) [σy – ν (σz + σx) [σz – ν (σx + σy)
Sample Problems:
Problem 1: A rectangular steel block is 3 inches long in the x direction, 2 inches long in
the y direction, and 4 inches long in the z direction. The block is subjected to a triaxial
loading of three uniformly distributed forces as follows: 48 kips tension in the x
direction, 60 kips compression in the y direction, and 54 kips tension in the z direction. If
ν = 0.30 and E = 29 × 106 psi, determine the single uniformly distributed load in the x
direction that would produce the same deformation in the y direction as the original
loading.
Solution:
CE 323/ BES 222
MECHANICS OF DEFORMABLE BODIES
SUMMER 2020
Problem 2: A 2-in.-diameter steel tube with a wall thickness of 0.05 inch just fits in a rigid hole. Find the tangential
stress if an axial compressive load of 3140 lb is applied. Assume ν = 0.30 and neglect the possibility of buckling.
Solution:
𝜀𝑥 𝐸
(σx – ν σy) = 0
σx = ν σy
where σx = tangential stress
σy = Longitudinal stress
σy = Py / A = 3140/ [𝜋(2)(0.05)]
σy = 31400 / 𝜋 psi
σx = 0.30(31400 / 𝜋)
σx = 2298.5 psi
