Deflection, in structural engineering terms, means the movement of a beam or node from its original position. It happens due to the forces and loads being applied to the body. Deflection also referred to as displacement, which can occur from externally applied loads or from the weight of the body structure itself. It can occur in beams, trusses, frames and basically any other body structure. In this article, we will discuss the beam deflection formula with examples. Let us learn it!
What is Beam Deflection?
Deflection is the degree to which a particular structural element can be displaced with the help of a considerable amount of load. It can also be referred to as the angle or distance. The distance of deflection of a member under a load is directly related to the slope of the deflected shape of the body under that load. It can be computed by integrating the function which is used to describe the slope of the member under that load.
The Beam is a long piece of a body that is capable to hold the load by resisting the bending. The deflection of the beam towards in a particular direction when force is applied to it is known as Beam deflection.
The beam can be bent or moved away from its original position. This distance at each point along the member is the representation of the deflection. There are mainly four variables, which can determine the magnitude of the beam deflections. These include:
- How much loading is on the structure?
- The length of the unsupported member
- The material, specifically the Young’s Modulus
- The Cross-Section Size, specifically the Moment of Inertia [I]
There is a variety of range of beam deflection equations that can be used to calculate a basic value for deflection in different types of beams. Generally, we calculate deflection by taking the double integral of the Bending Moment Equation means M[x] divided by the product of E and I [i.e. Young’s Modulus and Moment of Inertia].
The unit of deflection, or displacement, will be a length unit and normally we measure it in a millimetre. This number defines the distance in which the beam can be deflected from its original position.
The formula for Beam Deflection:
Cantilever beams are the special types of beams that are constrained by only one given support. These types of objects would naturally deflect more due to having support at one end only. To calculate the deflection of the cantilever beam we can use the below equation:
D= \[ \frac{WL^3}{3EI} \]
Where,
D Beam deflection W Force at one end L Length of beam E Young’s Modulus I Moment of Inertia
Solved Examples
Q: Calculate the deflection of a cantilever beam of length 2 meter which has support at one end only. Young’s modulus of the metal is \[ 200\times 10^9\] and the moment of inertia is 50 Kg m². At the end force applied is 300 N.
The conjugate beam method, developed by Heinrich Muller-Breslau in 1865, is one of the methods used to determine the slope and deflection of a beam. The method is based on the principle of statics.
A conjugate beam is defined as a fictitious beam whose length is the same as that of the actual beam, but with a loading equal to the bending moment of the actual beam divided by its flexural rigidity, \[EI\].
The conjugate beam method takes advantage of the similarity of the relationship among load, shear force, and bending moment, as well as among curvature, slope, and deflection derived in previous chapters and presented in Table 7.2.
\[Table 7.2\]. Relationship between load-shear-bending moment and curvature-slope-deflection.
Supports for Conjugate Beams
The supports for conjugate beams are shown in Table 7.3 and the examples of real and conjugate beams are shown in Figure 7.4.
\[Table 7.3\]. Supports for conjugate beams.
\[Table 7.4\] Real beams and their conjugate.
Sign Convention
For a positive curvature diagram, where there is a positive ordinate of the \[\frac{M}{E I}\] diagram, the load in the conjugate should point in the positive \[y\] direction [upward] and vice versa [see Figure 7.14].
\[Fig. 7.14\]. Positive curvature diagram.
If the convention stated for positive curvature diagrams is followed, then a positive shear force in the conjugate beam equals the positive slope in the real beam, and a positive moment in the conjugate beam equals a positive deflection [upward movement] of the real beam. This is shown in Figure 7.15.
\[Fig. 7.15\]. Shear and slope in beam.
Procedure for Analysis by Conjugate Beam Method
•Draw the curvature diagram for the real beam.
•Draw the conjugate beam for the real beam. The conjugate beam has the same length as the real beam. A rotation at any point in the real beam corresponds to a shear force at the same point in the conjugate beam, and a displacement at any point in the real beam corresponds to a moment in the conjugate beam.
•Apply the curvature diagram of the real beam as a distributed load on the conjugate beam.
•Using the equations of static equilibrium, determine the reactions at the supports of the conjugate beam.
•Determine the shear force and moment at the sections of interest in the conjugate beam. These shear forces and moments are equal to the slope and deflection, respectively, in the real beam. Positive shear in the conjugate beam implies a counterclockwise slope in the real beam, while a positive moment denotes an upward deflection in the real beam.
Example 7.11
Using the conjugate beam method, determine the slope and the deflection at point \[A\] of the cantilever beam shown in the Figure 7.16a. \[E = 29,000 \mathrm{ksi}\] and \[I = 280 \mathrm{in.}^{4}\]
\[Fig. 7.16\]. Conjugate beam.
Solution
[\[M/EI\]] diagram. First, draw the bending moment diagram for the beam and divide it by the flexural rigidity, \[EI\], to obtain the \[\frac{M}{E I}\] diagram shown in Figure 7.16b.
Conjugate beam. The conjugate beam loaded with the \[\frac{M}{E I}\] diagram is shown in Figure 7.16c. Notice that the free end in the real beam becomes fixed in the conjugate beam, while the fixed end in the real beam becomes free in the conjugate beam. The \[\frac{M}{E I}\] diagram is applied as a downward load in the conjugate beam because it is negative in Figure 7.16b.
Slope at \[A\]. The slope at \[A\] in the real beam is the shear at \[A\] in the conjugate beam. The shear at \[A\] in the conjugate is as follows:
\[V_{A}=\left[\frac{1}{2}\right][12]\left[\frac{36}{E I}\right]=\frac{216 \mathrm{k}-\mathrm{ft}^{2}}{E I}\]
The same sign convention for shear force used in Chapter 4 is used here.
Thus, the slope in the real beam at point \[A\] is as follows:
\[\theta_{\mathrm{A}}=\frac{216 \mathrm{k}-\mathrm{ft}{2}}{E I}=\frac{216[12]{2}}{[29,000][280]}=0.0038 \mathrm{rad}\]
Deflection at \[A\]. The deflection at \[A\] in the real beam equals the moment at \[A\] of the conjugate beam. The moment at \[A\] of the conjugate beam is as follows:
\[M_{A}=-\left[\frac{1}{2}\right][12]\left[\frac{36}{E I}\right]\left[\frac{2}{3} \times 12\right]=-\frac{1728 \mathrm{k}-\mathrm{ft}^{3}}{E I}\]
The same sign convention for bending moment used in Chapter 4 is used here.
Thus, the deflection in the real beam at point \[A\] is as follows:
\[\Delta_{\mathrm{A}}=-\frac{1728[12]^{3}}{[29,000][280]}=-0.37 \mathrm{in} \quad \Delta_{A}-0.37 \mathrm{in} \downarrow\]
Example 7.12
Using the conjugate beam method, determine the slope at support \[A\] and the deflection under the concentrated load of the simply supported beam at \[B\] shown in Figure 7.17a.
\[E = 29,000 \mathrm{ksi}\] and \[I = 800 \mathrm{in.}^{4}\]
\[Fig. 7.17\]. Simply supported beam.
Solution
[\[M/EI\]] diagram. First, draw the bending moment diagram for the beam and divide it by the flexural rigidity, \[EI\], to obtain the moment curvature [\[\frac{M}{E I}\]] diagram shown in Figure 7.17b.
Conjugate beam. The conjugate beam loaded with the \[\frac{M}{E I}\] diagram is shown in Figure 7.17c. Notice that \[A\] and \[C\], which are simple supports in the real beam, remain the same in the conjugate beam. The \[\frac{M}{E I}\] diagram is applied as an upward load in the conjugate beam because it is positive in Figure 7.17b.
Reactions for conjugate beam. The reaction at supports of the conjugate beam can be determined as follows:
\[A_{y}=B_{y}=-\frac{1}{E I}\left[\frac{1}{2}\right][30][180][0.5]=-\frac{1350 \mathrm{k} . \mathrm{ft}^{2}}{E I} \text { due to symmetry in loading }\]
Slope at \[A\]. The slope at \[A\] in the real beam is the shear force at \[A\] in the conjugate beam. The shear at \[A\] in the conjugate beam is as follows:
\[V_{A}=-\frac{1350 \mathrm{k} \cdot \mathrm{ft}^{2}}{E I}\]
Thus, the slope at support \[A\] of the real beam is as follows:
\[\theta_{A}=-\frac{1350 \mathrm{k} \cdot \mathrm{ft}{2}}{E I}=-\frac{1350[12]{2}}{[29,000][800]}=-0.0084 \mathrm{rad}\]
Deflection at \[B\]. The deflection at \[B\] in the real beam equals the moment at \[B\] of the conjugate beam. The moment at \[B\] of the conjugate beam is as follows:
\[M_{B}=\frac{1}{E I}\left[-[1350][15]+\left[\frac{1}{2}\right][15][180]\left[\frac{15}{3}\right]\right]=-\frac{13500 \mathrm{k} \cdot \mathrm{ft}^{3}}{E I}\]
The deflection at \[B\] of the real beam is as follows:
\[\Delta_{B}=-\frac{33750 \mathrm{k} \cdot \mathrm{ft}{3}}{E I}=-\frac{13500[12]{3}}{[29,000][800]}=-1.01 \text { in. } \quad \Delta_{B}=1.01 \text { in. } \downarrow\]
Chapter Summary
Deflection of beams through geometric methods: The geometric methods considered in this chapter includes the double integration method, singularity function method, moment-area method, and conjugate-beam method. Prior to discussion of these methods, the following equation of the elastic curve of a beam was derived:
Method of double integration: This method involves integrating the equation of elastic curve twice. The first integration yields the slope, and the second integration gives the deflection. The constants of integration are determined considering the boundary conditions.
Method of singularity function: This method involves using a singularity or half-range function to describe the equation of the elastic curve for the entire beam. A half-range function can be written in the general form as follows:
\[\langle x-a\rangle^{n}=\left\{\begin{array}{c} 0 \text { for }[x-a]