Question 6.10: A cantilever beam of circular solid cross-section is fixed a...
A cantilever beam of circular solid cross-section is fixed at one end and carries a concentrated load P at the free end. The diameter of the section is 200 mm at free end and increases uniformly to 400 mm at the fixed end over a length of 2 m. At what distance from the free end will the bending stresses in the cantilever be maximum? Also calculate the value of the maximum bending stress if the concentrated load, P = 30 kN.
In Figure 6.25, the beam is shown. Let the diameter of the section at a distance x from the free end be d. Clearly,
\frac{d-200}{400-200}=\frac{x}{2}
or d = (200+100 x) = 100(x+2) = 0.1(x+2) mm
Now, bending moment at a distance x from the free end is
M_x=P x
Therefore, maximum bending stress at the extreme fibre is
\sigma_{\max }=\frac{\left(M_x\right) d / 2}{\pi d^4 / 64}=\frac{32 M_x}{\pi d^3}=\left\lgroup \frac{32 P}{\pi\left(0.1^3\right)} \right\rgroup \frac{x}{(x+2)^3} N / m ^2
where P is in N.
Thus for given P, maximum bending stress occurs when x/(x + 2)³ is maximum or in other words, (x + 2)³/x is minimum. Therefore, let
f(x)=\frac{(x+2)^3}{x}=x^2+6 x+12+\frac{8}{x}
For minimum f (x),
f^{\prime}(x)=0 \Rightarrow 2 x+6-\frac{8}{x^2}=0
Therefore,
\begin{aligned} & x^3+3 x^2-4=0 \\ \Rightarrow & x^3+4 x^2-x^2-4=0 \\ \Rightarrow & x^2(x-1)+4(x+1)(x-1)=0 \\ \Rightarrow & (x-1)\left(x^2+4 x+4\right)=0 \Rightarrow(x-1)(x+2)^2=0 \\ \Rightarrow & x=1,-2 \end{aligned}
Clearly, x = -2 is an inadmissible root. Now,
f^{\prime \prime}(x)=2+\frac{4}{x^3}
so f”(x) > 0 for x = 1. Hence, f (x) is minimum at x = 1. Thus, \sigma_{\max } is maximum at a distance x = 1 m from the free end of the beam.
And \sigma_{\max } \text { for } P=30 kN =30 \times 10^3 N is
\begin{aligned} \sigma_{\max } & =\frac{32 \times 30 \times 10^3}{\pi(0.1)^3} \times \frac{1}{27} N / m ^2 \\ & =11.318 \times 10^6 N / m ^2 \approx 11.32 MPa \end{aligned}
Hence, maximum bending stress is 11.32 MPa.
