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# Deflection of Beams: Deflection Formula, Deflection Unit

By BYJU'S Exam Prep

Updated on: September 25th, 2023

**Deflection of Beams **is the representation of deflection of the structure from its original unloaded position. Beams are widely used in different types of structures. Therefore, it is very important to know how to predict the deflection of beams when different types of load and their load combinations are applied.

It is very important to limit the deflection of beams as too much deflection might cause damage to other parts of the structures. If too much deflection of the beam is allowed, it will fail the serviceability criteria; even though the stresses acting in the beam are low and safe, the structure will fail to serve the purpose of its construction. Let us know more about the deflection of beams discussed in the upcoming sections.

Table of content

**What is the Deflection of Beams?**

The deflection is measured from the beam’s original neutral surface to the deformed beam’s neutral surface. The configuration assumed by the deformed neutral surface is known as the elastic curve of the beam.

### Deflection of Beams Definition

Deflection of beams is the transverse deformation that occurs due to the shear force and bending moment. The deflection of beams due to the shear force is insignificant compared to the deflection of beams due to the bending moment. Therefore, the deflection of beams due to shear force is ignored.

Let us now see the slope and deflection of beams in brief.

**The slope of a Beam**: The slope of a beam is the angle between deflected beam to the actual beam at the same point.**Deflection of Beam:**Deflection is defined as the vertical displacement of a point on a loaded beam. Many methods determine the slope and deflection at a section in a loaded beam.

The maximum deflection occurs where the slope is zero. The maximum deflection position is found by equating the slope equation to zero. Then the value of x is substituted in the deflection equation to calculate the maximum deflection.Deflection of Beams is an essential topic in the Strength of Materials subject.

**Deflection of Beams Formula**

The deformed shape of the bean is known as an elastic curve. The slope and deflection of beams formula for some standard cases are listed below. These formulae can be directly for solving many problems of deflection of beams.

y is the vertical deflection of the beam. The objective of the beam is defined as the shape given by:

**θ= dy/dx**

The moment Curvature equation is used to calculate the deflection of beams.

**1/R=M/EI**

Where

- M = Bending Moment
- R = Radius of Curvature of deformed shape
- EI = Flexural Rigidity of the beam
- 1/R = Curvature of beam

**Deflection Differential Equation**

d^{2}y/dx^{2} = M/EI

(a)

Θ_{B}= ML/EI

∆_{B}= ML^{2}/2EI

(b)

Θ_{B}= PL^{2}/2EI

∆_{B}= PL^{3}/3EI

(c)

Θ_{B}= WL^{3}/6EI

∆B= WL^{4}/8EI

### DIFFERENTIAL EQUATION OF THE DEFLECTION CURVE OF BEAM

**Deflection of Beams Formula for Cantilever Beams:**

**Deflection of Beams Formula for Simply supported Beams:**

**Methods of Determining Beam Deflections**

There are various methods through which deflections of beams and structures can be found depending upon the type of load and nature of the beam. Some important methods to determine the deflection of beams are:

- Double Integration Method
- Macaulay’s Method
- Moment Area Method
- Conjugate Beam Method

**Double Integration Method**

The double integration method is suitable for prismatic beams having constant EI. This method is suitable when the bending moment equation remains constant throughout the length of the beam. According to the double integration method,

EId^{2}y/dx^{2} =Mx

Where

EI = Flexural Rigidity

Mx = Bending Moment at Section

Integrating the above equation,

EIdy/dx= ∫Mx+ c_{1}

This gives the equation of slopes where dy/dxis slope or rotation.

Integrating the above equation,

EI.y= ∬Mx+ ∫c_{1}+ c_{2}

Where y is deflection.

**Macaulay’s Method**

Macaulay’s Method is an improvement over the double integration method. It becomes difficult to solve deflections of the beam that need to be split into many sections. Therefore, this method is used where there is no need to split the beam. This method is suitable for prismatic bars having different bending moments at different cross-sections.

Consider the above beam carrying two point loads L_{1} and L_{2} at a distance of ‘a’ and ‘b’ from the left support. Consider a section x-x at a distance of x from left support. Finding moment about x-x:

Mx= R_{A}x- L_{1}(x-a)- L_{2}(x-b)

EId^{2}y/dx^{2}=R_{A}x- L_{1}(x-a)- L_{2}(x-b)

Integrating the above equation,

EI y= R_{A}x^{3}/6+ c_{1}x+ c_{2}-[L_{1}(x-a)^{3}/6]- [L_{2}(x-b)^{3}/6]

**Moment Area Method**

The moment area method is used when we need to analyze a beam in which we are interested in calculating the slope and deflection of the beam at a single location. This method is suitable for prismatic and non-prismatic members. This method is based on the area under the bending moment diagram.

**Theorem 1**

Change in slope from section A to section B B-A will equal the area of the M/EI diagram between Θ_{A} and Θ_{B}.

Θ_{B}-Θ_{A}=A

**Theorem 2**

The deflection of section B concerning the tangent at section A is the area of the M/EI diagram between A and B about point B.

**Conjugate Beam Method**

A conjugate beam is an imaginary beam whose loading diagram is the M/EI diagram of the real beam. The conjugate beam method is suitable for both prismatic and non-prismatic members; the only requirement is the M/EI diagram should not be completed; it must be easy to determine the area and centroid of the M/EI diagram. If the M/EI diagram is positive, then loading in the conjugate beam will be upward. If the M/EI diagram is negative, then loading in the conjugate beam will be downward.

**Theorem 1**

The slope at any section on the real beam becomes a shear force at the corresponding section on the conjugate beam. The shear force diagram of the conjugate beam represents the slope diagram of the real beam.

**Theorem 2**

The deflection at any section on the real beam becomes a bending moment at the corresponding section in a conjugate beam. The bending moment diagram of the conjugate beam represents the deflection diagram of the real beam.

**Method of Superposition:** The method of superposition, in which the applied loading is represented as a series of simple loads for which deflection formulas are available. Then the desired deflection is computed by adding the contributions of the component loads(principle of superposition).

The most direct formula is used in questions. Hence it is advised to look for the beam deflection formula, which is directly asked from this topic, rather than going for long derivations.

**Deflection for Common Loadings**

**1. Concentrated load at the free end of the cantilever beam (origin at A):**

Maximum Moment, **M=−PL**

The slope at the end:θ=PL^{2}/2EI

Maximum deflection:δ=PL^{3}/3EI

Deflection Equation(y is positive downward): EIy = (Px^{2})(3L−x)/6

**2. Concentrated load at any point on the span of a cantilever beam**

Maximum Moment:**M = -wa**

The slope at the end:**θ = wa ^{2}/2EI**

Maximum deflection:>**δ = wa ^{3}(3L−a)/6EI**

Deflection Equation(y is positive downward),

EIy = Px^{2}(3a−x)/6for0^{2}(3x−a)/6fora

**3. Uniformly distributed load over the entire length of a cantilever beam**

Maximum Moment: **M = −wL ^{2}/2**

Slope at end: **θ = wL ^{3}/6EI**

Maximum deflection:**δ = wL ^{4}/8EI**

Deflection Equation (y is positive downward),

EIy=wx^{2}(6L^{2}−4Lx+x^{2})/120L

**4. Triangular load, full at the fixed end and zero at the free end**

Maximum Moment: M=−wL^{2}/6

Slope at the end: **θ = wL ^{3}/24EI**

Maximum deflection, **δ = wL ^{4}/30EI**

Deflection Equation: (y is positive downward),

EIy = wx^{2}(10L^{3}−10L^{2}x+5Lx^{2}−x^{3})/120L

**5. Moment load at the free end of a cantilever beam**

Maximum Moment: **M = −M**

The slope at end:**θ=ML/EI**

Maximum deflection:**δ=****ML ^{2}/**

**2EI**

Deflection Equation (y is positive downward),

EIy=Mx^{2}/2

**6. Concentrated load at the midspan of a simple beam**

Maximum Moment: **M=PL/4**

Slope at end:**θ _{A}=θ_{B}= WL^{2}/16EI**

Maximum deflection:**δ=PL ^{3}/48EI**

Deflection Equation (y is positive downward),

EIy=Px{(3/4)L^{2}−x^{2})}/12 for 0

**7. Uniformly distributed load over the entire span of a simple beam**

Maximum Moment:**M = wL ^{2}/8**

The slope at the end:**θ _{L}=θ_{R}=wL^{3}/24EI**

Maximum deflection:**δ=5wL ^{4}/384EI**

Deflection Equation (EIy=wx(L^{3}−2Lx^{2}+x^{3})/24

9. **Triangle load with zero at one support and full at the other support of the simple beam**

Maximum Moment:**M = w _{o}L^{2}/9√3**

Slope at the end,

**θ _{L}=7wL^{3}/360EI**

**θ _{R}=8wL^{3}/360EI**

Maximum deflection:**δ=2.5wL ^{4}/384EI at x = 0.519L**

Deflection Equation (y is positive downward),

EIy = wx(7L^{4}−10L^{2}x+3x)/360L

10. **Triangular load with zero at each support and full at the midspan of a simple beam**

Maximum Moment:** M = wL ^{2}/12**

The slope at the end, **θ _{L }= θ_{R }= 5wL^{3}/192EI**

Maximum deflection: **δ = wL ^{4}/120EI**

Deflection Equation(y is positive downward),**EIy=w _{o}x(25L^{4}−40L^{2}x^{2}+16x^{4})/960L** for 0

**Deflection of Beams Questions**

**Problem 1:** What is the slope and direction of a cantilever beam at the free endpoint load acting at the free end?

**Solution-**

The differential equation for deflection

M= -Px

EId^{2}y/dx^{2}=-Px

∫EId^{2}y/dx^{2}= ∫-Px

EIdy/dx= -Px^{2}/2+ c_{1}

Boundary Conditions at x=L, dy/dx = 0,

c_{1}= PL^{2}/2

EIdy/dx= -Px^{2}/2+ PL^{2}/2

Integrating this, we get,

EIy= -Px^{3}/6+ PL^{2}x/2+ c_{2}

Boundary condition x=L, y=0

c_{2}= -PL^{3}/3

EIy= -Px^{3}/6+ PL^{2}x/2- PL^{3}/3

Slope and deflection at the free end

Slope dy/dx= PL^{2}/2EI

Deflection y= PL^{3}/3EI

**Problem 2:** Simply supported beam of span 6m is loaded as shown in the figure. I = 78 x 10^{6} mm^{4} E = 2.1 x 10^{5} N/mm^{2}. Find the central deflections.

**Solution-**

We know that

y_{c}= WL^{3}/48EI= 60 × 10^{3} ×6000^{3}/(48 ×2.1 × 10^{5} ×78 × 10^{6})= 16.48mm

**Problem 3:** A simply supported beam of a span of 5m carrying a load of 10N/mm^{2}. Given E= 10^{4} N/mm2. The maximum permissible bending stress is 8N/mm^{2,} and the maximum deflection of the beam is 1 cm. Determine the width and depth of the beam.

**Solution-**

(σ_{b})_{max}= M_{max}/z= 8

10 × 5000^{2} ×6/(8 ×BD^{2})=23437500 mm^{3}

Maximum Deflection= 5WL^{4}/384EI

5WL^{4}/384EI=10mm

5WL^{4}/384 ×10^{4} ×B × D^{3}=10mm

BD^{3}=9765625000 mm^{4}

D= BD^{3}/BD^{2}= 9765625000/23437500= 416.67 mm

BD^{2}=23437500

B = 134.99mm

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