# Introduction

In order to serve its intended purpose, restrictions are placed upon the amount of deflection suitable for any given beam. These restrictions are put in place so as to prevent physical deterioration of the beam, such as cracking in brittle materials like concrete and glass, and to ensure comfort standards are met for occupants, as dictated by serviceability limit states for structural design[1]. The Qualitative Deflected Shapes of a beam can be determined without any values however a method of numerical analysis is required to calculate the exact slope and deflections so as to ensure that the beams deflection restrictions are met. Following the procedure below, it is possible to use integration to directly solve for the deflection and slope of a beam at any given point.[2]

# Equation of the Elastic Curve

The analysis and design of a beam requires information on the deflection and the slope of the beam at different points. For sustainability of the structure, it is important to determine the location and value of the maximum deflection of the beam. Therefore, the relation obtained between the deflection $y$ measured a particular point, at a distance $x$ of that point from a fixed end is the equation of elastic curve. [1]

From elementary calculus, since the slope dy/dx is so small and its square is negligible in unity in the elastic curve of a beam, the curvature of a curve $y=f(x)$ becomes: [1]

$\frac{1}{\rho}=\frac{\frac{d^2 y}{dx^2}}{{[1+(\frac{dy}{dx})^2]^{(3/2)}}} \approx \frac{d^2 y}{dx^2}$

Since the curvature of a beam is given as: [3]

$\frac{1}{\rho}=\frac{M(x)}{EI}$

Thus, elastic curve equation is given as: [1]

$M(x)=EI\frac{d^2y}{dx^2}$

which is a second order linear differential equation. [1]

Substituting and integrating, we obtain beam slope and beam deflection (elastic curve) as follows: [4]

$EI\theta(x)\approx EI\frac{dy}{dx}=\int_{0}^{x} M(x)dx+C_1$
$EIy(x)=\int_{0}^{x} dx \int_{0}^{x} M(x)dx+C_1x+C_2$
 Constants being determined based on boundary conditions of the beam supports.

## Three Cases of Boundary Conditions for Statically Determinate Beams

1. Simply supported beam:[4]

 @ $x=0$; $y_A=0$ @ $x=L$; $y_B=0$

2. Overhanging beam: [4]

 @ $x=0$; $y_A=0$ @ $x=L$; $y_B=0$

3. Cantilever beam:[4]

 @ $x=0$; $y_A=0$ & $\theta_A=0$

# Equation of Elastic Curve for a Beam Subjected to Distributed Load

${M(x)=EI\frac{d^2v}{dx^2}$
$\Downarrow}$
${V(x)=\frac{dM(x)}{dx}=EI\frac{d^3v}{dx^3}}$
$\Downarrow}$
$-w(x)=\frac{dV(x)}{dx}=EI\frac{d^4v}{dx^4}$ [4]

After integrating four times: [4]

$EIy(x)=-\int_{0}^{x}dx \int_{0}^{x}dx \int_{0}^{x}dx \int_{0}^{x} w(x)dx+\frac{1}{6}C_1x^3+\frac{1}{2}C_2x^2+C_3x+C_4$
 Constants being determined based on boundary conditions of the beam supports.

## Two Cases of Boundary Conditions for Statically Determinate Beams

1. Simply supported beam:[4]

 @ $x=0$; $y_A =0$; $M_A=0$ @ $x=L; y_B=0; M_B=0$

2. Cantilever beam:[4]

 @ $x=0$; $y_A=0$; $\theta_A=0$ @ $x=L$; $V_A=0$; $M_B=0$

# Procedure for Deflection of Beams by Integration

A free-body diagram can be used to determine a single function of x for the bending moment M in the beam. The function $M(x)$ will then lead to an expression for the slope $(x)$ and for the deflection $y(x)$. However, when the beam consists of concentrated loads, reactions at supports, or any discontinuities in the distributed load, then it’s necessary to divide the beam into several portions, and to represent the bending moment by a different function of $M(x)$ in each of these portions of the beam. Also, since each expression for the slope and deflection contains two different constants of integration, a larger number of constants will have to be determined [1]

2. Cut your beam at an appropriate section and write the $M(x)$ equation for the selected cut-section

3. Obtain the differential equation for the elastic curve using $M(x)$

4. Integrate the differential equation obtained in step 3 twice

5. Identify your boundary conditions for $x=0$ and $x=L$

6. Use your boundary conditions to find $C_1$ and $C_2$ and if more than one differential equations exist, determine all constants $C_3$, $C_4$, etc

7. Substitute $C_1$ and $C_2$ to find the elastic curve equation

8. If more constants such as $C_3$, $C_4$, etc exist, substitute all the variables to the appropriate differential equation to find the elastic curve equation

9. Identify your point of zero slope ($\theta=0$) or point of maximum deflection ($y_{max}$) (determining $X_{max}$)

10. Determine your maximum deflection within the cut-section of the beam specified in step 2 (determining $y_{max}$)

11. Determine you maximum slope ($\theta_{max}$) of the beam if necessary [4].

# Sample Problems

The following sample problems provide step by step examples of the two most common cases for the use of direct integration for calculating beam slopes and deflections.

## Example 1

Determine the slope and deflection at point B when E = 200Gpa and I = 160x106mm4

Example 1

Step 1 First find support reactions

$\rightarrow + \Sigma F_x = 0;$ $A_x = 0$

\begin{align} \uparrow + \Sigma F_y = 0; \qquad A_y - [5Kn\cdot \frac{5m}{2}] &= 0 \\ A_y &= 12.5KN \end{align}

\begin{align} \Sigma M_A = 0; \qquad M_A - \left [12.5KN\cdot \left (5m\cdot\frac{2}{3} \right ) \right ]&=0 \\ M_A&=41.7Kn\cdot m \end{align}

Step 2

Cut the beam at point(x) and determine moment(x)

Beam cut at (x)

EXAMPLE IS INCORRECT FROM THIS POINT FORWARD

\begin{align} \Sigma M_x =0; \qquad M_x + 41.7KN\cdot m -12.5x + \frac{1}{2} (x)(x) \left ( \frac{x}{3} \right ) &=0 \\ M_x &= 12.5x - \frac{x^6}{6} -41.7 \end{align}

Step 3

Integrate the function of moment (x) in terms of x

\begin{align} EI \frac{d^2v}{dx^2} &= M_x \\ EI \frac{d^2v}{dx^2} &= 12.5x - ( \frac{x^3}{6} ) - 41.7 \\ EI \frac{dv}{dx} &= \int_0^x 12.5x - ( \frac{x^6}{6} ) -41.7 dx \\ \end{align}

\begin{align} EI \frac{dv}{dx} &= \frac{12.5x^2}{2} - \frac{x^4}{24} -41.7x + C_1 \\ EIv &= \frac{12.5x^3}{6} - \frac{x^5}{120} - \frac{41.7x^2}{2} +C_1x +C_2 \\ \end{align}

Step 4

Using boundary conditions, substitute values for x in order to determine unknown C values, \begin{align} @x =0 & \frac{dv}{dx} = 0 \end{align}

We know this because the support at x=0 is fixed and not permitted to rotate, causing the slope to be zero.

$0= \frac{1}{EI} \left [ 6.25(0)^2 - \frac{o^4}{24} - 41.7(0) + C_1 \right ] \\ \therefore C_1 = 0 \\$

$Also, @ x= 0, v=0 \\ 0= \frac{1}{EI} \left [ 2.08(x)^3 - \frac{(0)^5}{120} - 20.85(0)^2 + C_1(0) + C_2 \right ] \\ \therefore C_2 = 0$

\begin{align} \theta &=\frac{1}{EI} \left [ 6.25x^2 - \frac{x^4}{24} - 41.7x \right ] \\ V &= \frac{1}{EI} \left [ 2.08x^3 - \frac{x^5}{120} - 20.85x^2 \right ] \end{begin}

Step 5

Substitute values into final equations

\begin{align} \theta &= \frac{1}{(20000x10^9)(160x10^-6)} \cdot \left [ 6.25(5)^2 - \frac{(5)^4}{24} - 41.7(5) \right ] \cdot10^6 \\ \theta &= -0.0024 rads (negative slope) \\ \end{begin}

\begin{align} V &= 31.24x10^(-12)\cdot \left [ 2.08(5)^3 - \frac{(5)^5}{120} - 20.86(5)^3 \right ]\cdot10^6 \\ V &= -0.00897m \\ &= -8.97mm \end{begin}

## Example 2

Determine the maximum deflection of beam AB and the slope at A.

E = 200 GPa, I=39.9x10-6mm4

Step 1 Find the support reactions

\begin{align} \Sigma M_A = 0; \qquad -\frac{1}{2}(60KN/m)(6m)(3m) + B_y(6m) &= 0 \\ B_y &=90KN \end{align}

\begin{align} \uparrow + \Sigma F_Y = 0; \qquad 90Kn -180KN +A_y &= 0 \\ A_y &= 90KN \end{align}

$\rightarrow + \Sigma F_x = 0;$ $A_x = 0$

Resulting Free Body Diagram

Step 2

Cut beam at point (x) and determine moment (x)

Beam cut at (x)

\begin{align} \Sigma M_A = 0 \qquad M(x) + \frac{1}{2} (30x)(x)(\frac{x}{3}) - 90x &=0 \\ M(x) &= 90x -5x^3 \end{align}

Step 3

Integrate the function of moment (x) in terms of x

\begin{align} EI \frac{d^2v}{dx^2} &=M(x) \\ EI \frac{d^2v}{dx^2} &= 90x -5x^3 \\ EI \frac{dv}{dx} &= \int_0^x (90x - 5x^3) dx \\ \end{align}

\begin{align} EI \frac{dv}{dx} &= \left ( 45x^2 - \frac{5x^4}{4} + C_1 \right ) \\ EI v &= \left ( \frac{-x^5}{4} + 15x^3 + C_1 + C_2 \right ) \\ \end{align}

Step 4 Using boundary conditions, substitute values for x in order to determine unkown C values.

Due to symettry $\frac{dv}{dx} = 0 @ x=3m$

\begin{align} EI(0) &= 45(3^2) - \frac{5(3)^4}{4} + C_1 \\ C_1 &= -303.75KNm^2 \end{align}

Also at x=0, V=0

\begin{align} EI(0) &= \left ( \frac{-(0)^5}{4} + 15(0)^3 +C_1(0) + C_2 \right )\\ C_2 &= 0 Knm^2 \end{align}

Substituting constants C1 and C2 into $\frac{dv}{dx}$ and v

$\frac{dv}{dx} = \frac{1}{EI} \left ( 45x^2 - \frac{5x^4}{4} - 303.75 \right )$

$v= \frac{1}{EI} \left ( \frac{-x^5}{4} +15x^3 - 303.75x \right )KNm^2$

Step 5 Substitute values into final equations

At A, x=0;

\begin{align} \theta_A &= \frac{1}{EI}\left ( 45(0)^2 - \frac{5(0)^4}{4} - 303.75 \right )\\ &= \frac{-303.75 KNm^2}{EI} \\ \end{align}

\begin{align} \hspace{20mm}&= \frac{-303.75KNm^2}{(200\cdot10^6 KN/m^2)(39.9\cdot10^-6m^4)}\\ \hspace{20mm}&= 0.381 rad \downarrow \end{align}

Due to the symmetry of the beam Vmax occurs at mid span or x=3m;

\begin{align} v_m_a_x &= \frac{1}{EI} \left [ \frac{-(3)^5}{4} +15(3)^3 - 303.75(3) \right ] \\ &= \frac{-567Knm^3}{(200\cdot10^6 KN/m^2)(39.9\cdot10^-6m^4)}\\ \end{align}

\begin{align} \hspace{20mm}&= 0.0748m \downarrow \\ \hspace{20mm}&= 74.8 mm \downarrow \end{align}

# References

1. Hibbeler, R.C. (2011). Mechanics of Materials (8th ed.). Upper Saddle River, NJ: Pearson Prentice Hall.
2. Kassimali, A (2011). Structural Analysis: SI edition (4th ed.) Stanford, CT: cengage Learning.
3. Verterra, Romel. Double Integration Method | Beam Deflections. Retrieved from http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/double-integration-method-beam-deflections.
4. Fathifazl, Reza (2013). Lecture 11: Deflection of Beams (Part I: Integration Method). CIVE 3202, Mechanics of Materials. Carleton University, Ottawa.