Mechanics of Materials(ME-294)
Lecture 7:
Principal Stresses and Strains
(chapter 7 of R. S. Khurmi)
Introduction
• Transformations of Stress and Strain
( chapter 7 of Mechanics Of Materials 6TH Edition, by
Ferdinand P. Beer ,E. Russell Johnston, Jr. John T.
Dewolf, David F. Mazurek)
• Combined Stress
(chapter 3 Strength Of Materials Fifth Edition by William
A. Nash, C. Potter,Schaum’s Outline Series)
• Principal Stresses and Strains
(chapter 7 of R. S. Khurmi)
Introduction
We have studies
▫ Normal/Direct Stresses (Tensile and Compressive)arising in bars
subject to axial loading
 1-D (In one direction)
 3-D
▫ as well as shearing stresses
• But frequently bars are simultaneously subject to several kinds
of loadings, and it is required to determine the state of stress
under these conditions.
• It is the purpose of this chapter to investigate the state of stress
on an arbitrary plane through an element in a body subject to
both normal and shearing stresses.
• There are two methods to find the state of stress.
▫ Analytical
▫ Graphical
Stress Elements
• The most useful way of
representing the stresses is to
isolate a small element of
material, such as the element
labeled C in Fig( c) , and then
show the stresses acting on all
faces of this element.
• An element of this kind is called
a stress element. The stress
element at point C is a small
rectangular block (it doesn’t
matter whether it is a cube or a
rectangular parallelepiped) with
its right-hand face lying in cross
section mn.
• The dimensions of a stress element
are assumed to be infinitesimally
small, but for clarity we draw the
element to a large scale, as in Fig.a.
• In this case, the edges of the element
are parallel to the x, y, and z axes,
and the only stresses are the normal
stresses σx acting on the x faces
(recall that the x faces have their
normals parallel to the x axis).
• Because it is more convenient, we
usually draw a two-dimensional view
of the element (Fig.b) instead of a
three-dimensional view.
Stresses on Inclined Sections
• Larger stresses may occur on inclined sections. Therefore, we
will begin our analysis of stresses and strains by discussing
methods for finding the normal and shear stresses acting on
inclined sections cut through a member
• To obtain a more complete picture, we need to investigate the
stresses acting on inclined sections, such as the section cut by
the inclined plane pq in Fig. a.
Stresses on Inclined Sections
• Because the stresses are the
same throughout the entire bar,
the stresses acting over the
inclined section must be
uniformly distributed, as
pictured in the free body
diagrams of Fig. b (threedimensional view) and Fig. c
(two-dimensional view).
• From the equilibrium of the free
body we know that the resultant
of the stresses must be a
horizontal force P.
• The resultant is drawn with a
dashed line in Figs. b and c.
Let us now return to the task of finding the stresses acting on section pq . As already
mentioned, the resultant of these stresses is a force P acting in the x-direction.
• This resultant may be resolved into two components, a normal force N that is perpendicular to
the inclined plane pq and a shear force V that is tangential to it.
• These force components are
N =P cosθ
V =P sinθ
• Associated with the forces N and V are normal and shear
stresses that are uniformly distributed over the inclined section
(Figs. c and d).
• The normal stress is equal to the normal force N divided by the
area of the section, and the shear stress is equal to the shear
force V divided by the area of the section. Thus, the stresses are
Sign convention for stresses acting on an
inclined section. (Normal stresses are positive
when in tension and shear stresses are
positive when they tend to produce
counterclockwise rotation.)
For a bar in tension, the normal force N produces positive normal stresses σθ
and the shear force V produces negative shear stresses τθ. These stresses are
given by the following equations :
Following expressions for the normal and shear stresses
These equations give the stresses acting on an inclined section oriented at an angle
u to the x axis .
It is important to recognize that above Eqs. were derived only from statics, and
therefore they are independent of the material.
Thus, these equations are valid for any material, whether it behaves linearly or
nonlinearly, elastically or inelastically.
Maximum Normal and Shear Stresses
The maximum normal stress occurs at θ = 0 and is
The shear stress τ is zero on cross
sections of the bar (θ =0) as well as on
longitudinal sections (θ =90°).
Between these extremes, the stress
varies as shown on the graph,
reaching the largest positive value
when θ= 45° and the largest negative
value when θ =-45°.
These maximum shear stresses have
the same magnitude
• Even though the maximum shear stress
in an axially loaded bar is only one-half
the maximum normal stress, the shear
stress may cause failure if the material
is much weaker in shear than in
tension.
• An example of a shear failure is
pictured in Fig. , which shows a block of
wood that was loaded in compression
and failed by shearing along a 45°
plane.
• A similar type of behavior occurs in
mild steel loaded in tension
General Case of Plane Stress
• In general if a plane element is removed from a body it will be
subject to the normal stresses σx and σy together with the
shearing stress τxy , as shown in Fig. 3-1. We assume that no
stresses act on the element in the z-direction.
For normal stresses,
tensile stress is
considered to be
positive, compressive
stress negative. For
shearing stresses, the
positive sense* is that
illustrated in Fig. 3-1.
If moments are taken about
the center of the face
shown, we would see that
τxy = τyx since the normal
stresses pass through the
center. We shall assume
that the stresses σx , σy
and τxy are known.
Frequently it is desirable
to investigate the state of
stress on a plane inclined
at an angle θ, measured
counterclockwise to the
x-axis, as shown in Fig. 31. The normal and
shearing stresses on such
a plane are denoted by σ
and τ and appear as in
Fig. 3-2.
Principal Stresses and Maximum Shearing Stress
Mohr’s Circle-Graphical Form
• All the information contained in the above
equations may be presented in a convenient
graphical form known as Mohr’s circle.
• In this representation normal stresses are plotted
along the horizontal axis and shearing stresses along
the vertical axis.
• The stresses σx,σy, and τxy are plotted and a circle is
drawn through these points having its center on the
horizontal axis.
Mohr’s Circle
Figure 3-3 shows Mohr’s
circle for an element
subject to the general case
of plane stress, shown in
Fig. 3-1.
The ends of diameter BCD
are the points (σy, τyx) at B
and (σx, –τxy) at D. The
center C is at [(σx + σy)/2,
0)].
,, and
For applications see
Problems 3.4, 3.5, 3.6, 3.7,
3.9, and 3.11. of Strength
Of Materials Fifth Edition