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# Stresses in beams

When a beam having an arbitrary cross section is subjected to a transverse loads the beam will bend. In addition to bending the other effects such as twisting and buckling may occur, and to investigate a problem that includes all the combined effects of bending, twisting and buckling could become a complicated one.

Stresses in beams

Preamble:

When a beam having an arbitrary cross section is subjected to a transverse loads the beam will bend. In addition to bending the other effects such as twisting and buckling may occur, and to investigate a problem that includes all the combined effects of bending, twisting and buckling could become a complicated one. Thus we are interested to investigate the bending effects alone, in order to do so, we have to put certain constraints on the geometry of the beam and the manner of loading.

Assumptions:

The constraints put on the geometry would form the assumptions:

1. Beam is initially straight , and has a constant cross-section.

2.    Beam is made of homogeneous material and the beam has a longitudinal plane of symmetry.

3. Resultant of the applied loads lies in the plane of symmetry.

4. The geometry of the overall member is such that bending not buckling is the primary cause of failure.

5. Elastic limit is nowhere exceeded and 'E' is same in tension and compression.

6. Plane cross - sections remains plane before and after bending.

Let us consider a beam initially unstressed as shown in fig 1(a). Now the beam is subjected to a constant bending moment (i.e. „Zero Shearing Force') along its length as would be obtained by

applying equal couples at each end. The beam will bend to the radius R as shown in Fig 1(b)

As a result of this bending, the top fibers of the beam will be subjected to tension and the bottom to compression it is reasonable to suppose, therefore, that some where between the two there are points at which the stress is zero. The locus of all such points is known as neutral axis . The radius of curvature R is then measured to this axis. For symmetrical sections the N. A. is the axis of symmetry but what ever the section N. A. will always pass through the centre of the area or centroid.

As we are aware of the fact internal reactions developed on any cross-section of a beam may consists of a resultant normal force, a resultant shear force and a resultant couple. In order to ensure that the bending effects alone are investigated, we shall put a constraint on the loading such that the resultant normal and the resultant shear forces are zero on any cross-section perpendicular to the longitudinal axis of the member, That means F = 0 since or M = constant.

Thus, the zero shear force means that the bending moment is constant or the bending is same at every cross-section of the beam. Such a situation may be visualized or envisaged when the beam

or some portion of the beam, as been loaded only by pure couples at its ends. It must be recalled that the couples are assumed to be loaded in the plane of symmetry.

When a member is loaded in such a fashion it is said to be in pure bending. The examples of pure bending have been indicated in EX 1and EX 2 as shown below :

When a beam is subjected to pure bending are loaded by the couples at the ends, certain cross- section gets deformed and we shall have to make out the conclusion that,

1.  Plane sections originally perpendicular to longitudinal axis of the beam remain plane and perpendicular to the longitudinal axis even after bending , i.e. the cross-section A'E', B'F' ( refer Fig 1(a) ) do not get warped or curved.

2. In the deformed section, the planes of this cross-section have a common intersection i.e. any time originally parallel to the longitudinal axis of the beam becomes an arc of circle.

We know that when a beam is under bending the fibres at the top will be lengthened while at the bottom will be shortened provided the bending moment M acts at the ends. In between these there are some fibres which remain unchanged in length that is they are not strained, that is they do not carry any stress. The plane containing such fibres is called neutral surface.

The line of intersection between the neutral surface and the transverse exploratory section is called the neutral axisNeutral axis (N A) .

Bending Stresses in Beams or Derivation of Elastic Flexural formula :

In order to compute the value of bending stresses developed in a loaded beam, let us consider the two cross-sections of a beam HE and GF , originally parallel as shown in fig 1(a).when the beam

is to bend it is assumed that these sections remain parallel i.e. H'E' and G'F' , the final position of the sections, are still straight lines, they then subtend some angle q.

Consider now fiber AB in the material, at adistance y from the N.A, when the beam bends this will stretch to A'B'

Since CD and C'D' are on the neutral axis and it is assumed that the Stress on the neutral axis zero. Therefore, there won't be any strain on the neutral axis

Consider any arbitrary a cross-section of beam, as shown above now the strain on a fibre at a distance „y' from the N.A, is given by the expression

Now the termis the property of the material and is called as a second moment of area of the cross-section and is denoted by a symbol I.

Therefore  M/I = sigma/y = E/R

This equation is known as the Bending Theory Equation.The above proof has involved the assumption of pure bending without any shear force being present. Therefore this termed as the pure bending equation. This equation gives distribution of stresses which are normal to cross-section i.e. in x-direction.

Stress variation along the length and in the beam section

Bending Stress and Deflection Equation

In this section, we consider the case of pure bending; i.e., where only bending stresses exist as a

result of applied bending moments. To develop the theory, we will take the phenomenological approach to develop what is called the 'Euler-Bernoulli theory of beam bending.' Geometry:

Consider a long slender straight beam of length L and cross-sectional area A. We assume the beam is prismatic or nearly so. The length dimension is large compared to the dimensions of the cross-section. While the cross-section may be any shape, we will assume that it is symmetric about the y axis

Loading: For our purposes, we will consider shear forces or distributed loads that are applied in the y direction only (on the surface of the beam) and moments about the z-axis. We have consider examples of such loading in ENGR 211 previously and some examples are shown below:

Kinematic Observations: In order to obtain a 'feel' for the kinematics (deformation) of a beam

subjected to pure bending loads, it is informative to conduct an experiment. Consider a rectangular lines have been scribed on the beam's surface, which are parallel to the top and

bottom surfaces (and thus parallel to a centroidally placed x-axis along the length of the beam). Lines are also scribed around the circumference of the beam so that they are perpendicular to the longitudinals (these circumferential lines form flat planes as shown). The longitudinal and circumferential lines form a square grid on the surface. The beam is now bent by moments at each end as shown in the lower photograph. After loading, we note that the top line has stretched and the bottom line has shortened (implies that there is strain exx). If measured carefully, we see that the longitudinal line at the center has not changed length (implies that exx = 0 at y = 0). The longitudinal lines now appear to form concentric circular lines.

We also note that the vertical lines originally perpendicular to the longitudinal lines remain straight

and perpendicular to the longitudinal lines. If measured carefully, we will see that the vertical lines remain approximately the same length (implies eyy = 0). Each of the vertical lines (as well as the planes they form) has rotated and, if extended downward, they will pass through a common point that forms the center of the concentric longitudinal lines (with some radius ?). The flat planes originally normal to the longitudinal axis remain essentially flat planes and remain normal to the deformed longitudinal lines. The squares on the surface are now quadrilaterals and each appears to have tension (or compression) stress in the longitudinal direction (since the horizontal lines of a square have changed length). However, in pure bending we make the assumption that. If the x-axis is along the length of beam and the y-axis is normal to the beam, this suggests that we have an axial normal stress sxx that is tension above the x-axis and compression below the y-axis. The remaining normal stresses syy and szz will generally be negligible for pure bending about the z-axis. For pure bending, all shear stresses are assumed to be zero. Consequently, for pure bending, the stress matrix reduces to zero.

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