A13.0 Introduction

The bar AB in Fig. a is subjected to an axial compressive load $P$. If the compressive stresses are such that no buckling of the bar takes place, then bar sections such as 1-1 and 2-2 move parallel to each other as the bar shortens under the compressive stress.

In Fig. b the same bar is used as a simply supported beam with two applied loads $P$ as shown. The shear and bending moment diagrams for the given beam loading are also shown. The portion of the beam between sections 1-1 and 2-2 under the given loading are subjected to pure bending since the shear is zero in this region.

Experimental evidence for a beam segment $\Delta x$ taken in this beam region under pure bending shows that plane sections remain plane after bending but that the plane sections rotate with respect to each other as illustrated in Fig. c, where the dashed line represents the unstressed beam segment and the heavy section the shape after pure bending takes place. Thus the top fibers are shortened and subjected to compressive stresses and the lower fibers are elongated and subjected to tensile stresses. Therefore at some plane n-n on the cross-section, the fibers suffer no deformation and thus have zero stress. This location of zero stress under pure bending is referred to as the neutral axis.

A13.1 Location of Neutral Axis

Fig. d shows a cantilever beam subjected to a pure moment at its free end, and under this applied moment the beam takes the exaggerated deflected shape as shown.

The applied bending moment vector acts parallel to the Z axis, or in other words the applied bending moment acts in a plane perpendicular to the Z axis. Consider a beam segment of length L. Fig. d shows the distortion of this segment when plane sections remain plane after bending of the beam.

It will be assumed that the beam section is homogeneous, that is, made of the same material, and that the beam stresses are below the proportional limit stress of the material or in other words that Hooke's Law holds.

From the geometry of similar triangles, $$ \frac{e}{y} = \frac{e_c}{c} $$ or $$ e = \frac{y}{c} {e_c} $$       

We have from Young's Modulus, $E$, that $$ E = \frac{\text{unit stress}}{\text{unit strain}} = \frac{-\sigma _y}{e/L} $$ where $ {\sigma _y} $ is the bending stress under a deformation $e$ and since it is compression it will be given a minus sign.

Solving for $ {\sigma _y} $, $$ {\sigma _y} =  - E\frac{e_c}{c} \frac{y}{L} $$ The most remote fiber is at a distance $y = c$.

Hence $$ {\sigma _y} =  - E\frac{e_c}{L} $$ whence $$ {\sigma _y} =  - \frac{\sigma _c}{c} y $$

For equilibrium of the bending stress perpendicular to the beam cross-section or in the $x$ direction, we can write $\sum {F_x} = 0 $, or $$ \sum F_x  =  - \frac{\sigma _c}{c} \int {y\,da}  = 0 $$ however in this expression, the term $ \frac {\sigma _c}{c} $ is not zero, hence the term $ \int {y\,da} $ must equal zero and this can only be true if the neutral axis coincides with the centroidal axis of the beam cross-section.

The neutral axis does not pass through the beam section centroid when the beam is nonhomogeneous that is, the modulus of elasticity is not constant over the beam section and also when Hooke's Law does not apply or where the stress-strain relationship is non-linear. These beam conditions are described later in this Chapter.

A13.2 Equations for Bending Stress, Homogeneous Beams, Stresses Below Proportional Limit Stress

In the following derivations, it will be assumed that the plane of the external loads contain the flexural axis of the beam and hence, the beam is not subjected to torsional forces which, if present, would produce bending stresses if free warping of the beam sections was restrained, as occurs at points of support. The questions of flexural axes and torsional effects are taken up in later chapters.

Fig. A13.1 represents a cross-section of a straight cantilever beam with a constant cross-section, subjected to external loads which lie in a plane making an angle $ \theta $ with axis Y-Y through the centroid O. To simplify the figure, the flexural axis has been assumed to coincide with the centroidal axis, which in general is not true.

Let NN represent the neutral axis under the given loading, and let $\phi$ be the angle between the neutral axis and the axis X-X. The problem is to find the direction of the neutral axis and the bending stress $\sigma$ at any point on the section.

In the fundamental beam theory, it is assumed that the unit stress varies directly as the distance from the neutral axis, within the proportional limit of the material. Thus, Fig. A13.2 illustrates how the stress varies along a line such as mm perpendicular to the neutral axis N-N.

Let $\sigma $ represent unit bending stress at any point a distance ${y_n}$ from the neutral axis. Then the stress $\sigma $ on $da$ is $$\sigma = k{y_n}$$            

where $k$ is a constant. Since the position of the neutral axis is unknown, ${y_n}$ will be expressed for convenience in terms of rectangular coordinates with respect to the axes X-X and Y-Y.

Thus, $$\begin{aligned} {c}{y_n} &= (y - x\tan \phi )\cos \phi \\ &= y\cos \phi  - x\sin \phi \end{aligned}$$

Then Eq. 1 becomes $$\sigma  = k\left( {y\cos \phi  - x\sin \phi } \right)$$

This equation contains three unknowns, $\sigma $, $k$ and $\phi $. For solution, two additional equations are furnished by conditions of equilibrium namely, that the sum of the moments of the external forces that lie on one side of the section ABCD about each of the rectangular axes X-X and Y-Y must be equal and opposite, respectively, to the sum of the moments of the internal stresses on the section about the same axes.

Let $M$ represent the bending moment in the plane of the loads; then the moment about axis X-X and Y-Y is $ {M_x} = M\cos \phi $ and $ {M_y} = M\sin \phi $. The moment of the stresses on the beam section about axis X-X is $\int {\sigma \,da\,y} $. Hence, taking moments about axis X-X, we obtain for equilibrium, $$ \begin{aligned} M \cos \theta &= \int {\sigma \, da \, y} \\ &= \int {k\left( {\cos \phi \,{y^2}\,da - \sin \phi \,xy\,da} \right)} \\ &= k\cos \phi \int {y^2}\,da  - k\sin \phi \int {xy\,da} \end{aligned} $$

In similar manner, taking moments about the Y-Y axis $$ M\sin \theta  = \int {\sigma \,da\,x} $$ whence $$ M\sin \theta  =  - k\sin \phi \int {x^2}\,da + k\cos \phi \int {xy\,da} $$

A13.3 Method 1. Stresses for Moments About the Principal Axes

In equation 4, the term $$ \int {y^2}\,da $$ is the moment of inertia of the cross-sectional area about axis X-X, which we will denote by ${I_x}$, and the term $$\int {xy\,da} $$ represents the product of inertia about axes X-X and Y-Y. We know, however, that the product of inertia with respect to the principal axes is zero. Therefore, if we select XX and YY in such a way as to make them coincide with the principal axes, we can write equation 4: $$M\cos \theta  =  k\cos \phi \,{I_{x_p}}$$

In like manner, from equation 4a $$M\sin \theta  =  - k\sin \phi \,{I_{{y_p}}}$$

To find the unit stress ${\sigma}$ at any point on the cross-section, we solve equation 5 for $\cos \phi $ and equation 6 for $\sin \phi $, and then substituting these values in 3, we obtain the following expression, giving $\sigma $ the subscript $b$ to represent bending stress: $$ {\sigma _b} = \frac {M\cos \theta \, y_p}{I_{x_p}} + \frac{M\sin \theta \, x_p}{I_{y_p}}$$

Let the resolved bending moment $M\cos \theta $ and $M\sin \theta $ about the principal axes be given the symbols ${M_{{x_p}}}$ and ${M_{{y_p}}}$. Then we can write $$ {\sigma _b} =  - \frac{ M_{x_p} y_p }{I_{x_p}} - \frac{ M_{y_p} x_p }{I_{y_p}} $$

The minus signs have been placed before each term in order to give a negative value for ${\sigma _b}$ when we have a positive bending moment, or ${M_{x_p}}$ is the moment of a couple acting about ${X_p} - {X_p}$, positive when it produces compression in the upper right hand quadrant.  ${M_{{y_p}}}$ is the moment about the ${Y_p} - {Y_p}$ axis, and is also positive when it produces compression in the upper right hand quadrant.

BENDING STRESS EQUATION FOR SYMMETRICAL BEAM SECTIONS

Since symmetrical axes are principal axes (term $\int {xy\,da}  = 0$), the bending stress equation for bending about the symmetrical XX and YY axes is obviously, $$ {\sigma _b} =  - \frac{ {M_x}y }{I_x} - \frac{ {M_y}x }{I_y} $$

A13.4 Method 2. Stresses by use of Neutral Axis for Given Plane of Loading.

The direction of the neutral axis NN, measured from the XXp principal axis is given by dividing equation 6 by 5. $$\tan a =  - \frac{I_{x_p} \tan \theta } {I_{y_p}}$$

The negative sign arises from the fact that $\theta $ is measured from one principal axis and $\phi $ is measured in the same direction from the other principal axis.

Since equation 8 gives us the location of the neutral axis for a particular plane of loading, the stress at any point can be found by resolving the external moment into a plane perpendicular to the neutral axis N-N and using the moment of inertia about the neutral axis, hence $${ \sigma _b} = \frac{ \left[ {M\cos \left( {\theta  - a} \right)} \right] {y_n}} {I_n} = \frac{M_n y_n}{I_n} $$

${I_n}$ can be determined from the relationship expressed in Chapter A3, namely, $$ {I_n} = {I_{x_p}}{\cos ^2}a + {I_{y_p}}{\sin ^2}a $$

A13.5 Method 3. Stresses from Moments, Section Properties and Distances Referred to any Pair of Rectangular Axes through the Centroid of the Section.

The fiber stresses can be found without resort to principal axes or to the neutral axis.

Equation 4can be written: $${M_x} = k \cos \phi \, {I_x} - k \sin \phi \, {I_{xy}}$$ where ${I_x} = \int {y^2} da $ and $ I_{xy} = \int {xy\,da} $ and ${M_x} = M\cos \theta $.

In like manner $${M_y} =  - k\sin \phi \,{I_y} + k\cos \phi \,{I_{xy}}$$

Solving equations 11 and 12 for $\sin \phi$ and $\cos \phi$ and substituting their values in equation 3, we obtain the following expression for the fiber stress ${\sigma _b}$ $${\sigma _b} = - \frac{({M_y}{I_x} - {M_x}{I_{xy}})} { {I_x}{I_y} - I_{xy}^2 } x \, - \frac{({M_x}{I_y} - {M_y}{I_{xy}})} { {I_x}{I_y} - I_{xy}^2 } y $$

For simplification, let $$\begin{aligned} {K_1} &= I_{xy} \left( {I_x}{I_y} - I_{xy}^2 \right) \\ {K_2} &= I_{y} \left( {I_x}{I_y} - I_{xy}^2 \right) \\ {K_3} &= I_{x} \left( {I_x}{I_y} - I_{xy}^2 \right) \end{aligned}$$

Substituting these values in Equation 13 $${\sigma _b} = - \left( K_3 M_y - K_1 M_x \right) x - \left( K_2 M_x - K_1 M_y \right)y$$

In Method 2, equation 8 was used to find the position of the neutral axis for a given plane of loading. The location of the neutral axis can also be found relative to any pair of rectangular centroidal axes X and Y as follows:-

Since the stress at any point on the neutral axis must be zero, we can write from equation 14 that:- $$\left( K_3 M_y - K_1 M_x \right) x =  - \left( K_2 M_x - K_1 M_y \right) y$$ for all points located on the neutral axis. From Fig. A13.1 $\tan \phi  = \frac{y}{x}$

Thus $$\tan \phi  =  - \frac{ {\left( { {K_3}{M_y} - {K_1}{M_x} } \right)} }{ {\left( { {K_2}{M_x} - {K_1}{M_y} } \right)} }$$

It frequently happens that the plane of the bending moment coincides with either the X-X or the Y-Y axis, thus making either ${M_x}$ or ${M_y}$ equal to zero. In this case, equation 15 can be simplified. For example, if ${M_y} = 0$ $$\tan \phi  = \frac{I_{xy}}{I_y}$$ and if ${M_x} = 0$ $$\tan \phi  = \frac{I_x}{I_{xy}}$$

A13.6 Advantages and Disadvantages of the Three Methods.

Method 2 (bending about the neutral axis for a given plane of loading) no doubt gives a better picture of the true action of the beam relative to its bending as a whole. The point of maximum fiber stress is easily determined by placing a scale perpendicular to the neutral axis and moving it along the neutral axis to find the point on the beam section farthest away from the neutral axis. In airplane design, there are many design conditions, which change the direction of the plane of loading, thus, several neutral axes must be computed for each beam section, which is a disadvantage as compared to the other two methods.

In determining the shears and moments on airplane structures, it is common practice to resolve air and landing forces parallel to the airplane XYZ axes and these results can be used directly in method 3, whereas method 1 requires a further resolution with respect to the principal axes. Methods 1 and 3 are more widely used than method 2.

Since bending moments about one principal axis produces no bending about the other principal axis, the principal axes are convenient axes to use when calculating internal shear flow distribution.

A13.7 Deflections

The deflection can be found by using the beam section properties about the neutral axis for the given plane of loading and the bending moment resolved in a plane normal to the neutral axis. The deflection can also be found by resolving the bending moment into the two principal planes and then using the properties about the principal axes. The resultant deflection is the vector sum of the deflections in the direction of the principal planes.

A13.8 Illustrative Problems. Example Problem 1

Fig. A13.3 shows a unsymmetrical one cell box beam with four corner flange members a, b, c and d. Let it be required to determine the bending axial stress in the four corner members due to the loads Px and Py acting 50" from the section abcd.

In this example solution the sheet connecting the corner members will be considered ineffective in bending. The stresses will be determined by each of the three methods as presented in this chapter.

SOLUTION

The first step common to all three methods is the calculation of the moments of inertia about the centroidal X and Y axes. Table A13.1 gives the detailed calculations. The properties are first calculated about the reference axes x'x' and y'y' and then transferred to the parallel centroidal axes.

TABLE A13.1

etc