Chapter 9, Problem 9.7.8P

The tapered cantilever beam AB shown in the figure has a solid circular cross section. The diameters at the ends A and B are d_Aand d_B= 2d_A, respectively. Thus, the diameter d and moment of inertia / at distance v from the free end are, respectively,

   in which I_Ais the moment of inertia at end A of the beam.

Determine the equation of the deflection curve and the deflection S_Aat the free end of the beam due to the load P.

  

To determine

The equation for the deflection curve and the deflection ${\delta }_A$ should be determined at free end of the tapered cantilever beam AB due to load P.

Answer to Problem 9.7.8P

The equation for the deflection curve and the deflection ${\delta }_A$ are $v=\frac{P{L}^4}{6E{I}_A}\left[\frac{L}{{\left(L+x\right)}^2}-\frac{3}{\left(L+x\right)}\right]-\frac{P{L}^3}{12E{I}_A}x+\frac{7P{L}^3}{24E{I}_A}$ and ${\delta }_A=\frac{P{L}^3}{24E{I}_A}$ at free end of the beam due to load P.

Explanation of Solution

Given Information:

We have,

Length of the tapered cantilever beam AB, L

Load at point A as P

Diameter at point A, $d_A$

Diameter at point B, $d_B=2d_A$

Moment of inertia, I

The bending moment is $M=-Px$

So, the 2^nd degree diffrential equation would be,

  $\begin{array}{l}EI\frac{d^{2}v}{d{x}^2}=M\\ EI\frac{d^{2}v}{d{x}^2}=-Px\mathrm{.....}(1)\\ \text{}\text{But, we have I =}\frac{I_A}{L^4}{\left(L+x\right)}^4\\ E\frac{I_A}{L^4}{\left(L+x\right)}^4\frac{d^{2}v}{d{x}^2}=-Px\\ \frac{d^{2}v}{d{x}^2}=-\frac{P{L}^4}{E{I}_A}\left(\frac{x}{{\left(L+x\right)}^4}\right)\end{array}$

In the above equation, taking integration on both sides :

  $\begin{array}{l}\frac{d^{2}v}{d{x}^2}=-\frac{P{L}^4}{E{I}_A}\left(\frac{x}{{\left(L+x\right)}^4}\right)\\ \frac{dv}{dx}=-\frac{P{L}^4}{E{I}_A}{\displaystyle \int \frac{\left(L+x\right)-L}{{\left(L+x\right)}^4}dx}\\ \frac{dv}{dx}=-\frac{P{L}^4}{E{I}_A}\left[{\displaystyle \int \frac{1}{{\left(L+x\right)}^3}dx-L{\displaystyle \int \frac{L}{{\left(L+x\right)}^4}dx}}\right]+C_1\\ \frac{dv}{dx}=-\frac{P{L}^4}{E{I}_A}\left[\frac{-1}{2{\left(L+x\right)}^2}+\frac{L}{3{\left(L+x\right)}^3}\right]+C_1\\ \frac{dv}{dx}=-\frac{P{L}^4}{6E{I}_A}\left[\frac{-3L-3x+2L}{{\left(L+x\right)}^3}\right]+C_1\\ \frac{dv}{dx}=-\frac{P{L}^4}{6E{I}_A}\left[\frac{L+3x}{{\left(L+x\right)}^3}\right]+C_1\end{array}$

According to boundary condition at x=L,

  $\begin{array}{l}\frac{dv}{dx}=0\\ 0=\frac{P{L}^4}{6E{I}_{A}8L^3}+C_1\\ 0=\frac{-P{L}^3}{12E{I}_A}+C_1\\ C_1=\frac{-P{L}^3}{12E{I}_A}\end{array}$

  $\frac{dv}{dx}=-\frac{P{L}^4}{6E{I}_A}\left[\frac{L+3x}{{\left(L+x\right)}^3}\right]-\frac{P{L}^3}{12E{I}_A}$

In above equation taking integration again,

  $\begin{array}{l}v=\frac{P{L}^4}{6E{I}_A}{\displaystyle \int \left[\frac{L}{{\left(L+x\right)}^3}+\frac{3\left(L+x\right)}{{\left(L+x\right)}^3}-\frac{3L}{{\left(L+x\right)}^3}\right]}dx-\frac{P{L}^3}{12E{I}_A}x+C_2\\ v=\frac{P{L}^4}{6E{I}_A}\left[{\displaystyle \int \frac{-2L}{{\left(L+x\right)}^3}+\frac{3}{{\left(L+x\right)}^2}}\right]dx-\frac{P{L}^3}{12E{I}_A}x+C_2\\ v=\frac{P{L}^4}{6E{I}_A}\left[\frac{2L}{2{\left(L+x\right)}^2}-\frac{3}{\left(L+x\right)}\right]-\frac{P{L}^3}{12E{I}_A}x+C_2\\ v=\frac{P{L}^4}{6E{I}_A}\left[\frac{L}{{\left(L+x\right)}^2}-\frac{3}{\left(L+x\right)}\right]-\frac{P{L}^3}{12E{I}_A}x+C_2\end{array}$

And at 2^nd Boundary conditions, x = L and v = 0.

  $\begin{array}{l}0=\frac{P{L}^4}{6E{I}_A}\left[\frac{L}{{\left(4L\right)}^2}-\frac{3}{2L}\right]-\frac{P{L}^3}{12E{I}_A}+C_2\\ 0=\frac{P{L}^4}{6E{I}_A}\left[\frac{1-6}{4L}\right]-\frac{P{L}^3}{12E{I}_A}+C_2\\ 0=\frac{P{L}^4}{6E{I}_A}\left[\frac{1-6}{4L}\right]-\frac{P{L}^3}{12E{I}_A}+C_2\\ 0=\frac{-5P{L}^3}{24E{I}_A}-\frac{P{L}^3}{12E{I}_A}+C_2\\ 0=\frac{-7P{L}^3}{24E{I}_A}+C_2\\ C_2=\frac{7P{L}^3}{24E{I}_A}\\ v=\frac{P{L}^4}{6E{I}_A}\left[\frac{L}{{\left(L+x\right)}^2}-\frac{3}{\left(L+x\right)}\right]-\frac{P{L}^3}{12E{I}_A}x+C_2\\ v=\frac{P{L}^4}{6E{I}_A}\left[\frac{L}{{\left(L+x\right)}^2}-\frac{3}{\left(L+x\right)}\right]-\frac{P{L}^3}{12E{I}_A}x+\frac{7P{L}^3}{24E{I}_A}\end{array}$

At x = 0, then

  $\begin{array}{l}v=\frac{P{L}^4}{6E{I}_A}\left[\frac{L}{{\left(L+x\right)}^2}-\frac{3}{\left(L+x\right)}\right]-\frac{P{L}^3}{12E{I}_A}x+\frac{7P{L}^3}{24E{I}_A}\\ v=\frac{P{L}^4}{6E{I}_A}\left[\frac{L}{{\left(L\right)}^2}-\frac{3}{\left(L\right)}\right]+\frac{7P{L}^3}{24E{I}_A}\\ v=\frac{-2P{L}^3}{6E{I}_A}+\frac{7P{L}^3}{24E{I}_A}\\ v=\frac{-P{L}^3}{24E{I}_A}\end{array}$

At point A, the deflection would be below.

  $\begin{array}{l}{\delta }_A=-v_{x=0}\\ {\delta }_A=-\left(\frac{-P{L}^3}{24E{I}_A}\right)\\ {\delta }_A=\frac{P{L}^3}{24E{I}_A}\end{array}$

Conclusion:

The answers are calculated according to deflection and deflection curve.