Shear Stress; Id no.: 10.01.03.033Presentation Transcript
Prestress Concrete Design Sessional
Ms. Sabreena Nasrin
Mr. Galib Muktadir Ratul
Department of Civil Engineering
A shear stress, denoted
(Greek: tau), is defined as the
component of stress coplanar
with a material cross section.
Shear stress arises from the
force vector component
parallel to the cross section.
It is the form of stress that
subjects an object to which
force is applied to
skew, tending to cause shear
Shear Stress Parallel to the
Cross Section (Horizontal)
Shear Stress in 2D
Shear Stress Parallel to the
Cross Section (Inclined)
Shear Stress in 3D
A shear stress between
two objects occurs when a
force pulls the object along
the same plane as the face of
the object abutting another
object that is being pulled in
the opposite direction.
A shear stress within an
object will occur when a
force parallel to the plane
causes one plane of the
material to want to slip
against another, thus
deforming the material.
If a fluid is placed between two parallel plates spaced 1.0
cm apart, and a force of 1.0 dyne is applied to each square
centimeter of the surface of the upper plate to keep it in
motion, the shear stress in the fluid is 1 dyne/cm2 at any
point between the two plates.
The formula to calculate average shear stress is:
= the shear stress;
= the force applied;
= the cross-sectional area of material with area parallel to
the applied force vector.
Measure the area, say value A, of the material over which the
force is applied. The area of a simple rectangular or squareshaped cross section is obtained by multiplying the length by the
height. The area of a circular cross section is calculated by the
equation A= pi*r^2. The area of a circle is equal to the value of
pi (3.14159) multiplied by the squared radius of the circle.
Measure the force that is to be applied over the area, say value F.
Simple forces of weight can be measured with a scale that
displays results in pounds. Substitute the values obtained in the
above steps as the following formula: T=F/A; where T = the
shear stress, F = the force applied and A = the cross-sectional
area over which the force was applied at first. Divide the
numerical value for F by the value for A and the resulting
number is the calculated shear stress.
It is the shear component of an
applied tensile (or compressive)
stress resolved along a slip plane
that is other than perpendicular or
parallel to the stress axis.
τ = σ cos Φ cos λ
It is the value of resolved shear
stress at which yielding begins; it is
a property of the material.
τ =σ (cosΦ cosλ)max
It is the stress on the mechanical
elements of that surface - something
like the stress in a bolt that is
connecting two pieces of metal. If
the bolt cracks straight across, if
failed due to the shear.
It is the stress when something
lands on a surface - something like
when a person falls off a bike and
skids across the ground. The shear
stress tears their skin.
Some Shear Testing Machines
Force: ΣF = 0 (V = RA
in this case)
Stress: fv = V/A
The circle is centered at
the average stress
value, and has a radius R
equal to the maximum
shear stress, as shown in the
The maximum shear
stress is equal to one-half
the difference between the
two principal stresses,
Ɵs is an important angle
where the maximum
shear stress occurs.
The shear stress equals
the maximum shear stress
when the stress element is
rotated 45 away from the
The transformation to the
maximum shear stress
direction can be
Horizontal & Vertical Shear Stress
Let us begin by
examining a beam of
rectangular cross section.
We can reasonably
assume that the shear
stresses τ act parallel to
the shear force V.
Let us also assume that
the distribution of shear
stresses is uniform across
the width of the beam.
Shear stresses on one side of an element are
accompanied by shear stresses of equal
magnitude acting on perpendicular faces of an
Thus, there will be horizontal shear stresses
between horizontal layers (fibers) of the
beam, as well as, transverse shear stresses on
the vertical cross section.
At any point within the beam these
complementary shear stresses are equal in
The existence of
stresses in a beam can
be demonstrated as
A single bar of depth 2h is much
stiffer that two separate bars each
of depth h.
Shown below is a rectangular
beam in pure bending.
Let Q = First moment of area =∫ydA
Where: V = transverse shear force
Q = first moment of area (section above area of interest)
I = moment of inertia
b = width of section
For the rectangular section shown above:
τ =V/2I(h² /4) – y1²)
As shown above, shear stresses vary quadratic ally with
the distance y1 from the neutral axis. The maximum shear
stress occurs at the neutral axis and is zero at both the top
and bottom surface of the beam.
For a rectangular cross section, the maximum
shear stress is obtained as follows:
Q = (bh/2)(h/4) = bh²/8
I = bh²/12
Ʈmax = 3V/2A
For a circular cross section:
Ʈmax = 4V/3A
Steel is affected by the compression component of
Shear. In case of tension there is no problem.
Concrete is affected by the tensile component of
principal shear stress. In case of compression there is
Distribution in a
Distribution in a
Shear Stress Distribution in a Wide
a T Section