chapter 11 - stress, strain and deformation in...

2001, W. E. Haisler Chapter 13: Beam Bending 1Chapter 13 – Beam Bending

The Classical Beam Bending TheoryClassical bending of beams is characterized by the following assumptions/restrictions:Geometry - long and slender prismatic beam (symmetric about z axis) with x axis along its length:

2001, W. E. Haisler Chapter 13: Beam Bending 2

Applied Loading - Transverse forces or tractions normal to the x axis. Moments about the y or z axis.

Kinematics - Predominate deflection is normal to the x axis. Predominate strain is in the axial (x) direction. Assume small strain and rotations of beam so that “axial strains vary linearly over cross-section” or “plane sections remain plane”. We will explain these assumptions shortly.

In terms of the general elasticity problem, the above can be stated as follows:

2001, W. E. Haisler Chapter 13: Beam Bending 3 Equilibrium . If the cross-sectional dimensions are small

compared to the beam length, then applied transverse tractions (in y and z directions) will be small compared to the resultant internal stress in the x direction. --> small transverse loads produce large axial stresses. Thus we assume that the only major stress is xx (all other stresses are zero or negligible).

Equilibrium reduces to . Implies that xx = xx (y,z).

Stress-Strain . We assume a linear isotropic material so that and

Kinematics/Displacements . Includes strain-displacement

relation and the kinematic (displacement)

boundary conditions for .

2001, W. E. Haisler Chapter 13: Beam Bending 4 We will only consider “bending about the z-axis” in this course. External transverse loading must therefore be in the z direction (as distributed loads or point loads), and any applied moments must be about the z-axis.

Such applied loadings (distributed normal loads, point loads and moments) will produce “bending” of the beam and internal axial stress and shear stresses on the cross-section.

Consider for example, the cantilevered T-section problem with a 300 lbf transverse force applied at its end as shown below.

2001, W. E. Haisler Chapter 13: Beam Bending 6The idealized internal stresses at x=0 will be given the following.

Note that the axial bending stress (figure a) varies through out the cross-section and tends to be a maximum at either the top or bottom and is zero at the “neutral axis” (which is the centroid

xx xy

( )xy

2001, W. E. Haisler Chapter 13: Beam Bending 7of the cross-section for homogeneous cross-sections). In this case, it is tensile on top and compressive on the bottom. The shear stress (figure b) likewise varies over the cross-section, is zero at the top and bottom of the cross-section, and is a maximum at the neutral axis (centroid of the cross-section for homogeneous cross-sections). We will derive the equations with which you can predict this axial bending and shear stress in this chapter.

While one could work directly with the internal stresses ( and ) within the Conservation of Linear Momentum equations,

the approach can be simplified using classical beam bending theory. These stresses can be put in terms of an equivalent internal axial force, P, shear force, , and bending moment, , resultant that acts at the centroid of the cross section.


For axial stress distribution and shear stress distribution acting on the entire cross-sectional area (left picture above),

we need to determine an equivalent set of forces and moment that are equivalent to the stress distribution (right picture). Assume the bar has a cross-sectional area of A as below and the x-axis is along the centroid of the beam cross-section:


Equating the stress distribution over area A to forces and , and moment gives:

Internal axial force in x direction =

Internal moment about z axis =

Internal shear force in y direction =


Kinematic Observations/Assumptions - In order to obtain a “feel” for the kinematics (deformation) for a beam subjected to bending loads, it is informative to conduct some experiments. The following photograph shows a long beam with a square cross-section. Straight longitudinal lines have been scribed on the beam’s surface, which are parallel to the top and bottom surfaces (an thus parallel to a centroidally-placed x-axis). 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.


xy

z


After loading, we note that the top line has stretched (tension) and the bottom line has shortened (compression). If measured carefully, we see that the longitudinal line at the center has not changed length ( at centroid). 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 the same length ( ). 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.


The flat planes originally normal to the longitudinal axis remain 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) and perhaps also some shear stress (since opposite vertical lines of a square have rotated different amounts).

Thus, to begin the theoretical development, we make some kinematic assumptions based on the experimental observation. We assume that the predominate deflection is normal to the x axis, i.e., . Predominate strain is in the axial (x) direction, i.e., . Assume small strain and


rotations of beam so that “axial strains vary linearly over cross-section” or “plane sections remain plane”.

In terms of the general elasticity problem, we make the following assumptions for the case of pure bending of a beam:

1. Conservation of Linear Momentum. If the cross-sectional dimensions are small compared to the beam length, then applied transverse tractions (in y direction) will be small compared to the resultant internal stress in the x direction. --> small transverse loads produce large axial stresses and is small. For pure bending, we assume that the shear stress is negligible. Thus we


assume that the only major stress is (all other stresses are zero or negligible). The stress tensor reduces to

. Equilibrium (Conservation of Linear

Momentum) reduces to . (neglecting body forces)

2. Stress-Strain. We assume a linear elastic isotropic material so that stress and strain are linearly related to each other:


3. Strain-Displacement. and

4. Kinematic assumptions: Since , this implies that . Let be the transverse

displacement of the centroidal axis in the y direction (the zero subscript means the transverse displacement is measured at y=0). Hence,

Draw a sketch of the undeformed and deformed beam and consider the geometry in light of assuming that a normal to centroidal axis remains normal and straight after bending:


After bending, the transverse displacement of the centroidal x-axis will be defined by as shown below. The subscript “0” means that uy is measured at y=0 (i.e,, at the centroidal axis position). The rotation of the beam at any point x is given by the derivative of the transverse displacement with respect to x:

Since me make the assumption that a normal to the centroidal axis remains straight and normal, then the normal will also rotate by this same amount . For a point “A” located at some position y above the centroidal axis, we note that point A will have moved to the left as shown on


the sketch. From geometry, the displacement in the x direction can be written as (note that for +, is negative):

Above means that the axial displacement can be written entirely in terms of the transverse displacement of the centroidal axis and that the displacement is linear with transverse position y. Substituting the above into the axial strain, we obtain


(1)

The above satisfies the assumption that the axial strain at y=0 (the centroidal axis) is zero. Because the strain is zero along the x-axis passing through the centroid, it is sometimes referred to as the neutral axis.

We can now rewrite the internal bending moment in terms of displacements by substituting the strain-displacement equation into the stress-strain equation and that result into the moment equation to obtain


Note that we must integrate over the cross-section A which lies in the y-z plane. Assume that Young’s modulus E is a constant over the cross-section, i.e., E=E(x). Hence, we write:

The integral term is a geometrical property of the cross-section and can be easily integrated:


where is called the moment of inertia of the cross-section about the z axis. Note: We will assume that the cross-section is symmetric about the y axis. If the cross-section is not symmetric about the y axis, a transverse load may produce twisting of the cross-section which we have not considered here. Note that bending is occurring about the z axis since bending moments are about the z axis.

With this definition of the moment of inertia, the bending moment equation becomes


or

(2)

The last equation is an ordinary, second order differential equation that defines the transverse displacement in terms of the bending moment. Note that the bending moment will in general be a function of x. In addition, E and may be functions of x.


The stress may now be written in terms of the bending moment. Substitute (2) into the strain equation (1) to obtain:

Substitute strain into Hooke's Law to obtain:

Thus,


Note that varies linearly with y (i.e., linearly from top to bottom surface of the beam) and is zero at the centroidal axis. Since the bending moment , then the stress also varies with position along the length of the beam. For a particularly location x, the bending stress will obviously be a maximum at the maximum value of y (i.e., either the top or bottom of the beam).


The sketch above shows the centroid as if it were halfway between the top and bottom surface. In this case, the bending stress on the top and bottom surface would be equal in magnitude but opposite in sign (i.e., one compression, the other tension).

Important: For cross-sections like the inverted T section shown earlier wherein the centroid is not an equal distance from the top and bottom surface, the bending stress will have different magnitudes at the top and bottom. For these types of problems, the surface with maximum y value will produce the largest magnitude bending stress. We will work some problems shortly, which show this type of behavior.

chapter 11 - stress, strain and deformation in...

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