Question 5.9: A tapered cantilever beam AB of solid circular cross section...

If the angle of taper of the beam is small, the bending stresses obtained from the flexure formula will differ only slightly from the exact values. As a guideline concerning accuracy, we note that if the angle between line AB (Fig. 5-24) and the longitudinal axis of the beam is about 20° , the error in calculating the normal stresses from the flexure formula is about 10%. Of course, as the angle of taper decreases, the error becomes smaller.
Section modulus. The section modulus at any cross section of the beam can be expressed as a function of the distance x measured along the axis of the beam. Since the section modulus depends upon the diameter, we first must express the diameter in terms of x, as follows:

{d}_{x}=d_{A}+\left(d_{B}-d_{A}\right) \frac{x}{L}                              (5-30)

in which d_{x} is the diameter at distance x from the free end. Therefore, the section modulus at distance x from the end (Eq. 5-19b) is

S_{x}=\frac{\pi d_{x}^{3}}{32}=\frac{\pi}{32}\left[d_{A}+\left(d_{B}-d_{A}\right) \frac{x}{L}\right]^{3}                    (5-31)

Bending stresses. Since the bending moment equals Px, the maximum normal stress at any cross section is given by the equation

\sigma_{1}=\frac{M_{x}}{S_{x}}=\frac{32 P x}{\pi\left[d_{A}+\left(d_{B}-d_{A}\right)(x / L)\right]^{3}}                  (5-32)

We can see by inspection of the beam that the stress \sigma_{1} is tensile at the top of the beam and compressive at the bottom.
Note that Eqs. (5-30), (5-31), and (5-32) are valid for any values of d_{A} and d_{B}, provided the angle of taper is small. In the following discussion, we consider only the case where d_{B} = 2d_{A}.
Maximum stress at the fixed support. The maximum stress at the section of largest bending moment (end B of the beam) can be found from Eq. (5-32) by substituting x = L and d_{B} = 2d_{A}; the result is

\sigma_{B}=\frac{4 P L}{\pi d_{A}^{3}}                        (a)

Maximum stress in the beam. The maximum stress at a cross section at distance x from the end (Eq. 5-32) for the case where d_{B} = 2d_{A} is

\sigma_{1}=\frac{32 P x}{\pi d_{A}^{3}(1+x / L)^{3}}                              (b)

To determine the location of the cross section having the largest bending stress in the beam, we need to find the value of x that makes \sigma_{1} a maximum. Taking the derivative d\sigma_{1}/dx and equating it to zero, we can solve for the value of x that makes \sigma_{1} a maximum; the result is

x=\frac{L}{2}                          (c)

The corresponding maximum stress, obtained by substituting x = L/2 into Eq. (b), is

\sigma_{\max }=\frac{128 P L}{27 \pi d_{A}^{3}}=\frac{4.741 P L}{\pi d_{A}^{3}}                                  (d)

In this particular example, the maximum stress occurs at the midpoint of the beam and is 19% greater than the stress \sigma_{B} at the built-in end.
Note: If the taper of the beam is reduced, the cross section of maximum normal stress moves from the midpoint toward the fixed support. For small angles of taper, the maximum stress occurs at end B.