Deformation in a Transverse Cross Section
In this section, the deformation that occurs when a straight prismatic beam, made of
homogenous material, is subjected to bending. The discussion will be limited to beams
having cross sectional area that is symmetrical with respect to its axis and the bending
moment is applied about “z” axis perpendicular to its axis as shown in the figure 1.
Deformation in pure bending is a statistically-indeterminate problem. However, the steps
will be considered in a different order since at this stage there is insufficient information
about the stress distribution to enable an equilibrium equation to be formulated. The
approach initially will be considered a prismatic beam of symmetrical cross-section
subjected to pure bending, from which the geometry of the deformation can be studied
and strain distribution determined. The stress-strain relations will give the stress
distribution which can be related to the forces and moments through an equilibrium
condition.
Figure 1
Before connecting the analysis it is necessary to make some assumptions as follows:
1. Transverse sections of the beam which are plane before bending will remain plane
plane during bending.
2. From the consideration of symmetry during bending, transverse sections will be
perpendicular to the circular arcs having a common center of curvature.
3. The radius of curvature of the beam during bending is larger compared with the
transverse dimensions.
4. Longitudinal elements of the beam are subjected only to simple tension or compression.
5. Young’s modulus for the beam material has the same value for the tension and
compression.
.
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Figure 2
By using a highly deformable material such as rubber, we can illustrate what happens
when a straight prismatic member is subjected to a bending moment. Consider, for
example, the undeformed bar which has a square cross section and is marked with a
longitudinal and transverse grid lines. When a bending moment is applied, it tends to
distort these lines into the pattern shown in figure 2. Notice that the longitudinal lines
become curved and the vertical transverse lines remain straight and yet undergo a rotation.
Figure 3
In order to show how this distortion will strain the material, we will isolate a small
segment of the beam located a distance x along the beam’s length and having an
undeformed thickness Δx. Notice that any line segment, Δx , located at the neutral
surface, does not change its length, whereas any line segment Δs, located at the arbitrary
distance y above the neutral surface, will contract and become Δs’ after deformation. By
definition, the normal strain along Δs is determined, namely ,
Ɛx =
lim
Δs →0
Δs′ −Δs
Δs
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(a)
(b)
Figure 4
The strain in term of location y of the segment and the radius of curvature 𝑝 of the
longitudinal axis of the element axis of the element. Before deformation, Δ𝑠 = Δ𝑥,
(figure 4 (a), after deformation Δ𝑥 has a radius of curvature 𝑝 with the center of
curvature at point O’ , figure 4(b). Since Δθ defines the angle between the sides of the
element, Δ𝑥 = Δ𝑠 = 𝑝Δ𝜃 . In the same manner, the deformed length of the Δ𝑠 become
Δ𝑠 ′ = (p-y) Δ𝜃 , the following equation can be obtained.
Ɛ=-
𝑦
𝑝
This important result indicate that the longitudinal normal strain of any element within
the beam depends on its location y on the cross section and the radius of the curvature of
the beam’s longitudinal normal strain will vary linearly within y from the neutral axis. A
contraction ( - Ɛ ) will occur in fibers located above the neutral axis ( + y), whereas
elongation ( + Ɛ ) will occur in fibers below the axis ( - y). The variation in strain over the
cross section is shown in the figure 5.
Ɛ=-(
𝑦
𝑐
)Ɛ max
y
Δ𝑥
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Figure 5 : Normal strain distribution
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Here the maximum strain occurs at the outermost fiber, located a distance of y = c from
𝑦
the neutral axis. Applying Ɛ = , since Ɛ max = c / 𝑝 , then by division
𝑝
Ɛ
Ɛ max
=-(
𝑦/𝑝
𝑐/𝑝
)
so that,
Ɛ=-(
𝑦
𝑐
) Ɛ max
This normal strain depends only on the assumption made regards to the deformation.
When a moment is applied to the beam, therefore, it will only cause a normal stress in the
longitudinal or x- direction. All the other components of normal and shear stress will be
zero. It is uniaxial state of stress that cause material to have longitudinal normal strain
component Ɛ max . Furthermore, by Poisson’ ratio, there must also be associated strain
components Ɛ y = - v Ɛ x and Ɛ z = - v Ɛ x , which deform the plane of cross-sectional area,
although these deformation is usually neglected. Such deformation will, however, cause
the cross-sectional dimensions to become smaller below the neutral axis and larger above
the neutral axis. The illustration of it can be seen in figure
Figure 6
Consideration of the beam subjected to pure bending shown in Figure indicates that the
lower surface stretches and is therefore in tension and the upper surface shorten and thus
is in compression. Hence there must be an xz-plane in between in which longitudinal
deformation is zero. This is termed the neutral plane, and a transverse axis lying in the
neutral plane is the neutral axis. Consider the deformations between two sections AC and
BD , a distance δx apart, of an initially straight beam. A longitudinal EF at the distance y
below the neutral axis will have initially the same length as the GH at the neutral axis.
During bending EF stretches to become E’F’ but GH being at the neutral axis is
unstrained when it become G’H’. Therefore, if R is the radius curvature of G’H’,
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G’H’ = GH = δx = R δ θ
E’F’ = ( R + y ) δ θ
and the longitudinal strain in fibre E’F’ is
𝐸 ′ 𝐹′ −𝐸𝐹
Ɛx =
𝐸𝐹
but EF = GH = G’H’ = R δ θ ; therefore
Ɛx =
( R + y ) δ θ −( R ) δ θ
(R )δθ
Hence,
𝑦
Ɛx = 𝑅
and since R= dx / d θ , therefore also
𝑑θ
Ɛx = 𝑦 𝑑𝑥
Figure 7
Regarding deformation in the y- and z- directions, it is apparent that the changes in the
length of the beam will result in changes in the transverse dimensions. For example, the
fibers in compression will be associated with an increase in thickness, whereas the region
in tension will show a decrease in beam thickness. The transverse strains will be σx
Ɛz = Ɛy = -
𝑉σx
𝐸
= - VƐx
and a beam of initially rectangular cross-section will take up the shape shown in Figure .
This can be easily demonstrated by bending an eraser. The neutral surface, instead of
being plane, will be curved. This behavior is termed anticlastic curvature. The
deformations are extremely small and do not affect the solution for the longitudinal
strains.
Figure 8
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Figure 9
Anticlastic curvature =
1
𝑣
𝑝
𝑝
′=
The relation obtained show that the elements
located above the neutral surface ( y > 0 ),
will expand in both the y and z direction,
while the elements located below the neutral
surface ( y < 0 ) will contract. In the case of a
member of a rectangle cross section, the
expansion and contraction of various
elements in the vertical direction will
compensate, and no change in the vertical
dimension of cross section will be observed.
As concerned, however, the expansion of the
elements located above the neutral surface
and the corresponding contraction of the
elements located below that surface will
result in the various horizontal lines in section
being bent into arcs of circle. The situation
observed here is similar to that observed
earlier in a longitudinal cross section. We can
conclude that the neutral axis of transverse
section will be bent into a circle of radius 𝑝’ =
𝑝/v.
Note : The radius of curvature , ρ , captures the intensity with which a segment of a
beam has deformed in bending and ρ is defined as the radius of circle into which a
longitudinal line of the initially straight beam has deformed. The reciprocal , K = 1/ ρ , is
referred to as the curvature.
In conclusion , the section of study can be sum up by a brief table below.
Stress Produced by Each Load Individually
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Stress
Distributions
Stresses
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