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NED University of Engineering & Technology Engineering
Department of Petroleum Engineering
Lecture #04
CE-212 – Mechanics of Solids
SHEAR STRESSES IN BEAMS
INTRODUCTION
When beams are acted upon by lateral loads, both bending moments and shear forces
act on their cross-sections, resulting in shear stresses
which will be studied here.
As shown below, if top and bottom surfaces of each board are smooth and not bonded
together, then application of load P will cause the boards to slide rel

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NED University of Engineering & Technology Engineering Department of Petroleum Engineering
1
By: Dr. Huma Khalid
Lecture #04 CE-212
–
Mechanics of Solids SHEAR STRESSES IN BEAMS
INTRODUCTION
When beams are acted upon by lateral loads, both bending moments and shear forces act on their cross-sections, resulting in shear stresses
which will be studied here.
As shown below, if top and bottom surfaces of each board are smooth and not bonded together, then application of load P will cause the boards to slide relative to one another.
However, if boards are bonded together, longitudinal shear stresses will develop and distort x-section in a complex manner.
As shown, when shear force V is applied, the non-uniform shear-strain distribution over cross-section will cause it to warp, i.e.,
plane section does not remain plane
.
Recall that the flexure formula assumes that cross-sections must remain plane and perpendicular to longitudinal axis of beam after deformation. This is violated when beam is subjected to both bending and shear; we assume that the warping is so small it can be neglected. This is true for a slender beam (small depth compared with its length).
A shear stress, denoted by
, is defined as a stress which is applied parallel or tangential to a face of a material, as opposed to a normal stress which is applied perpendicularly.
NED University of Engineering & Technology Engineering Department of Petroleum Engineering
2
By: Dr. Huma Khalid
We call the deformation created by shear stress as shear s
train, given the symbol γ
(gamma). It is defined as the change in angle of the element, it is a non-dimensional quantity.
Similar to normal stress , there is a linear relationship (for most engineering materials) between the shear stress and shear strain, as shown in figure This relationship is called Hooke's law for Shear and is represented by equation
where:
G
=
Shear Modulus of Elasticity
(for short,
Shear Modulus
) or
Modulus of Rigidity
The calculation of shear stresses is tackled below.
CALCULATION OF SHEAR STRESSES
The development of a general shear stress relation for beams is based on static equilibrium
of forces acting on the beam’s cross
-section.
We start by assuming that the shear stresses
act parallel to the direction of the shear force V, i.e. in the vertical direction.
It is also assumed that the shear stress distribution is uniform across the width of the cross-section (i.e. in the z direction).
Let’s consider the shear stresses on an infinitesimal element of area
at distance
from the z-axis, as shown in the following sketch.
From the equilibrium of vertical forces, the shear stresses
on one vertical side of the element must be accompanied by an equal shear stress
acting on the other vertical side in the opposite direction.
NED University of Engineering & Technology Engineering Department of Petroleum Engineering
3
By: Dr. Huma Khalid
Similarly, from the equilibrium of moments about the centre of the element, the shear stress pair acting on the two vertical sides must be accompanied by equal pair acting on the two horizontal sides. The latter is known as the complementary shear stress.
Now in order to evaluate the distribution of shear stresses on the cross-section, consider the free body diagram shown below. It refers to a section of a beam subject to bending and shear. The cross-section is symmetrical about the y-axis, but otherwise of arbitrary shape.
Considering the shaded element shown on the beam’s side view at distance x from the
y-axis. The bottom face of the element is parallel to the N.A. and is at an arbitrary distance
from that axis.
The bottom face is acted upon by the horizontal shear
. The right and left sides of the element are subject to normal bending stresses
and
, respectively.
The sides are also affected by complimentary vertical shear stresses
, however these do not enter into the equilibrium equation of forces in the horizontal direction.
The beam is under non-uniform (i.e. variable) bending moment, which is denoted by
on one side of the element and
on the other side.
Note: if the moment does not vary, i.e. pure bending, then there will be no shear stresses to consider!
Considering an element of area
on the cross-section at distance
from the N.A., the horizontal force acting on this area is equal to
, and this is equal to
Therefore, the total horizontal force acting on the shaded area on the cross-section (at a distance x from the y axis) is equal to
NED University of Engineering & Technology Engineering Department of Petroleum Engineering
4
By: Dr. Huma Khalid
(1)
Similarly, the total horizontal force acting on the shaded area on the cross-section (at a distance
from the y axis) is equal to
(2)
F
inally, the horizontal force acting on the bottom of the shaded area on the beam’s
side view is
(3) where
is the area of the bottom surface of the shaded area.
The equilibrium of forces in the horizontal direction yields:
Therefore:
(4) Note that
is constant at the section thus it was taken out of the integration.
The above equation can be rewritten as follows:
()
(5) However, by definition: V
(6) where V is the applied shear force at the section.
Note also that the integral,
∫
is the first moment of area of the portion of the cross-sectional area with respect to the N.A.
Denoting this moment of area as Q, i.e.
∫
, and substituting Q and V , the formula for the horizontal / longitudinal shear stress is:
()
(7)
The above equation is also known as the shear formula and can be used to determine the shear stress at any point on the cross-section.
Note that V and I are constants, while Q and t are variables.

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