Question 8.3: A simple beam AB with span length L = 6 ft supports a concen...

We begin by using the flexure and shear formulas to calculate the stresses acting on cross section mn. Once those stresses are known, we can determine the principal stresses and maximum shear stresses from the equations of plane stress. Finally, we can plot graphs of these stresses to show how they vary over the height of the beam.
As a preliminary matter, we note that the reaction of the beam at support A is R_{A} = P/3 = 3600 lb, and therefore the bending moment and shear force at section mn are

Normal stresses on cross section mn. These stresses are found from the flexure formula (Eq. 8-17a), as follows:

\sigma_{x}=-\frac{My}{I}=-\frac{12My}{bh^{3}}=-\frac{12(32,400  lb-in.)y}{(2  in.)(6  in.)^{3}}=-900y                (a)

in which y has units of inches (in.) and \sigma_{x} has units of pounds per square inch (psi). The stresses calculated from Eq. (a) are positive when in tension and negative when in compression. For instance, note that a positive value of y (upper half of the beam) gives a negative stress, as expected.
A stress element cut from the side of the beam at cross section mn (Fig. 8-17) is shown in Fig. 8-18. For reference purposes, a set of xy axes is associated with the element. The normal stress \sigma_{x} and the shear stress \tau_{xy} are shown acting on the element in their positive directions. (Note that in this example there is no normal stress \sigma_{y} acting on the element.)
Shear stresses on cross section mn. The shear stresses are given by the shear formula (Eq. 8-17b) in which the first moment Q for a rectangular cross section is

Q=b\left(\frac{h}{2}-y\right)\left(y+\frac{h/2-y}{2}\right)=\frac{b}{2}\left(\frac{h^{2}}{4}-y^{2}\right)                  (8-18)

Thus, the shear formula becomes

\tau=\frac{VQ}{Ib}=\frac{12V}{(bh^{3})(b)}\left(\frac{b}{2}\right)\left(\frac{h^{2}}{4}-y^{2}\right)=\frac{6V}{bh^{3}} \left(\frac{h^{2}}{4}-y^{2}\right)            (8-19)

The shear stresses \tau_{xy} acting on the x face of the stress element (Fig. 8-18) are positive upward, whereas the actual shear stresses τ (Eq. 8-19) act downward. Therefore, the shear stresses \tau_{xy} are given by the following formula:

\tau_{xy}=-\frac{6V}{bh^{3}}\left(\frac{h^{2}}{4}-y^{2}\right)                          (8-20)

Substituting numerical values into this equation gives

\tau_{xy}=-\frac{6(3600  lb)}{(2  in.)(6  in.)^{3}}\left(\frac{(6  in.)^{2}}{4}-y^{2}\right)=-50(9-y^{2})                        (b)

in which y has units of inches (in.) and \tau_{xy} has units of pounds per square inch (psi).
Calculation of stresses. For the purpose of calculating the stresses at cross section mn, let us divide the height of the beam into six equal intervals and label the corresponding points from A to G, as shown in the side view of the beam (Fig. 8-19a on the next page). The y coordinates of these points are listed in column 2 of Table 8-1 on the next page, and the corresponding stresses \sigma_{x} and \tau_{xy} (calculated from Eqs. a and b, respectively) are listed in columns 3 and 4. These stresses are plotted in Figs. 8-19b and 8-19c. The normal stresses vary linearly from a compressive stress of -2700 psi at the top of the beam (point A) to a tensile stress of 2700 psi at the bottom of the beam (point G). The shear stresses have a parabolic distribution with the maximum stress at the neutral axis (point D).

TABLE 8-1 STRESSES AT CROSS SECTION mn IN THE BEAM OF FIG. 8-17

Principal stresses and maximum shear stresses. The principal stresses at each of the seven points A through G may be determined from Eq. (7-17):

\sigma_{1,2}=\frac{\sigma_{x}+\sigma_{y}}{2}\pm\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}}              (8-21)

Since there is no normal stress in the y direction (Fig. 8-18), this equation simplifies to

\sigma_{1,2}=\frac{\sigma_{x}}{2}\pm\sqrt{\left(\frac{\sigma_{x}}{2}\right)^{2}+\tau^{2}_{xy}}                    (8-22)

Also, the maximum shear stresses (from Eq. 7-25) are

\tau_{\max}=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}}                          (8-23)

which simplifies to

\tau_{\max}=\sqrt{\left(\frac{\sigma_{x}}{2}\right)^{2}+\tau^{2}_{xy}}                      (8-24)

Thus, by substituting the values of \sigma_{x} and \tau_{xy} (from Table 8-1) into Eqs. (8-22) and (8-24), we can calculate the principal stresses \sigma_{1} and \sigma_{2} and the maximum shear stress \tau_{\max}. These quantities are listed in the last three columns of Table 8-1 and are plotted in Figs. 8-19d, e, and f.
The tensile principal stresses \sigma_{1} increase from zero at the top of the beam to a maximum of 2700 psi at the bottom (Fig. 8-19d). The directions of the stresses also change, varying from vertical at the top to horizontal at the bottom. At midheight, the stress \sigma_{1} acts on a 45° plane. Similar comments apply to the compressive principal stress \sigma_{2}, except in reverse. For instance, the stress is largest at the top of the beam and zero at the bottom (Fig. 8-19e).
The maximum shear stresses at cross section mn occur on 45° planes at the top and bottom of the beam. These stresses are equal to one-half of the normal stresses \sigma_{x} at the same points. At the neutral axis, where the normal stress \sigma_{x} is zero, the maximum shear stresses occur on the horizontal and vertical planes.
Note 1: If we consider other cross sections of the beam, the maximum normal and shear stresses will be different from those shown in Fig. 8-19. For instance, at a cross section between section mn and the concentrated load (Fig. 8-17), the normal stresses \sigma_{x} are larger than shown in Fig. 8-19b because the bending moment is larger. However, the shear stresses \tau_{xy} are the same as those shown in Fig. 8-19c because the shear force doesn’t change in that region of the beam. Consequently, the principal stresses \sigma_{1} and \sigma_{2} and maximum shear stresses \tau_{\max} will vary in the same general manner as shown in Figs. 8-19d, e, and f but with different numerical values.
The largest tensile stress anywhere in the beam is the normal stress at the bottom of the beam at the cross section of maximum bending moment. This stress is

(\sigma_{tens})_{\max} = 14,400 psi

The largest compressive stress has the same numerical value and occurs at the top of the beam at the same cross section.
The largest shear stress \tau_{xy} acting on a cross section of the beam occurs to the right of the load P (Fig. 8-17) because the shear force is larger in that region of the beam (V = R_{B} = 7200 lb). Therefore, the largest value of \tau_{xy}, which occurs at the neutral axis, is

(\tau_{xy})_{\max} = 900 psi

The largest shear stress anywhere in the beam occurs on 45° planes at either the top or bottom of the beam at the cross section of maximum bending moment:

\tau_{\max}=\frac{14,400  psi}{2} = 7200 psi

Note 2: In the practical design of ordinary beams, the principal stresses and maximum shear stresses are rarely calculated. Instead, the tensile and compressive stresses to be used in design are calculated from the flexure formula at the cross section of maximum bending moment, and the shear stress to be used in design is calculated from the shear formula at the cross section of maximum shear force.