Chapter 16, Problem 1RP

To determine

Find the Equation of the elastic curve.

Answer to Problem 1RP

The Equation of the elastic curve is,

$\underset{\_}{v=\frac{1}{EI}\left[\begin{array}{l}-30x^3+46.25{\langle x-12\rangle }^3-11.7{\langle x-24\rangle }^3\\ +38703.24x-412,598.88\end{array}\right]\text{lb}\cdot {\text{in}}^3}$.

Explanation of Solution

Given information:

The length of the beam is 60 in..

EI is constant.

Calculation:

Sketch the Free Body Diagram of the beam as shown in Figure 1.

Refer to Figure 1.

Find the support reaction as shown below.

Apply the Equations of Equilibrium as shown below.

Take moment about A is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ B_y\times 48-70\times 12+180\times 12=0\\ 48B_y+1,320=0\\ B_y=-27.5\text{lb}\end{array}$

Summation of forces along vertical direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y-27.5-180-70=0\\ A_y-277.5=0\\ A_y=277.5\text{lb}\end{array}$

Consider the x distance from left support.

Take moment about the section as shown below.

$\begin{array}{c}M\left(x\right)=-180\langle x-0\rangle -\left(-277.5\right)\times \langle x-12\rangle -70\times \langle x-24\rangle \\ =-180x+277.5\langle x-12\rangle -70\langle x-24\rangle \end{array}$

Apply the slope and elastic curve as shown below.

$EI\frac{d^{2}v}{d{x}^2}=M\left(x\right)$

Here, EI is the flexural rigidity, $\frac{d^{2}v}{d{x}^2}$ is the second derivatives of the function v, and $M\left(x\right)$ is the moment at the section x.

Substitute $\left[-180x+277.5\langle x-12\rangle -70\langle x-24\rangle \right]$ for $M\left(x\right)$.

$EI\frac{d^{2}v}{d{x}^2}=-180x+277.5\langle x-12\rangle -70\langle x-24\rangle$

Integrate both sides of the Equation.

$\begin{array}{c}{\displaystyle \int EI\frac{d^{2}v}{d{x}^2}}={\displaystyle \int \left(-180x+277.5\langle x-12\rangle -70\langle x-24\rangle \right)}\\ EI\frac{dv}{dx}=-180\frac{x^2}{2}+277.5\frac{{\langle x-12\rangle }^2}{2}-70\frac{{\langle x-24\rangle }^2}{2}+C_1\\ =-90x^2+138.75{\langle x-12\rangle }^2-35{\langle x-24\rangle }^2+C_1\end{array}$

Integrate both sides of the Equation.

$\begin{array}{c}{\displaystyle \int EI\frac{dv}{dx}}={\displaystyle \int \left(-90x^2+138.75{\langle x-12\rangle }^2-35{\langle x-24\rangle }^2+C_1\right)}\\ EIv=-90\frac{x^3}{3}+138.75\frac{{\langle x-12\rangle }^3}{3}-35\frac{{\langle x-24\rangle }^3}{3}+C_{1}x+C_2\\ =-30x^3+46.25{\langle x-12\rangle }^3-11.67{\langle x-24\rangle }^3+C_{1}x+C_2\end{array}$ (1)

Apply the boundary conditions as shown below.

1. i. The value of $v=0$ at $x=12\text{in}\text{.}$.
2. ii. The value of $v=0$ at $x=60\text{in}\text{.}$.

Apply boundary condition (i) in Equation (1).

Substitute 0 for $v$ and 12 in. for $x$ in Equation (1).

$\begin{array}{l}EI\left(0\right)=-30{\left(12\right)}^3+46.25{\left(12-12\right)}^3+C_1\times 12+C_2\\ 12C_1+C_2-51,840=0\end{array}$ (2)

Apply boundary condition (ii) in Equation (1).

Substitute 0 for $v$ and 60 in. for $x$ in Equation (1).

$\begin{array}{l}EI\left(0\right)=-30{\left(60\right)}^3+46.25{\left(60-12\right)}^3-11.67{\left(60-24\right)}^3+C_1\left(60\right)+C_2\\ 60C_1+C_2-1,909,595.52=0\end{array}$ (3)

Solving Equations (2) and (3) we get,

$\begin{array}{l}{C}_1=38703.24\\ C_2=-412,598.88\end{array}$

Substitute $38703.24$ for $C_1$ and $-412,598.88$ for $C_2$ in Equation (1).

$\begin{array}{l}EIv=\left[\begin{array}{l}-30x^3+46.25{\langle x-12\rangle }^3-11.67{\langle x-24\rangle }^3\\ +38703.24x-412,598.88\end{array}\right]\\ v=\frac{1}{EI}\left[\begin{array}{l}-30x^3+46.25{\langle x-12\rangle }^3-11.67{\langle x-24\rangle }^3\\ +38703.24x-412,598.88\end{array}\right]\text{lb}\cdot {\text{in}}^3\end{array}$

Therefore, the Equation of the elastic curve is,

$\underset{\_}{v=\frac{1}{EI}\left[\begin{array}{l}-30x^3+46.25{\langle x-12\rangle }^3-11.67{\langle x-24\rangle }^3\\ +38703.24x-412,598.88\end{array}\right]\text{lb}\cdot {\text{in}}^3}$.