Deflection of Beams: Deflection Formula, Deflection Unit

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,

EId²y/dx² =Mx

Where

EI = Flexural Rigidity

Mx = Bending Moment at Section

Integrating the above equation,

EIdy/dx= ∫Mx+ c₁

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

Integrating the above equation,

EI.y= ∬Mx+ ∫c₁+ c₂

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 L₁ and L₂ 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_Ax- L₁(x-a)- L₂(x-b)

EId²y/dx²=R_Ax- L₁(x-a)- L₂(x-b)

Integrating the above equation,

EI y= R_Ax³/6+ c₁x+ c₂-[L₁(x-a)³/6]- [L₂(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=PL²/2EI

Maximum deflection:δ=PL³/3EI

Deflection Equation(y is positive downward): EIy = (Px²)(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 = wa²/2EI

Maximum deflection:>δ = wa³(3L−a)/6EI

Deflection Equation(y is positive downward),
EIy = Px²(3a−x)/6for0EIy = Pa²(3x−a)/6fora

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

Maximum Moment: M = −wL²/2

Slope at end: θ = wL³/6EI

Maximum deflection:δ = wL⁴/8EI

Deflection Equation (y is positive downward),
EIy=wx²(6L²−4Lx+x²)/120L

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

Maximum Moment: M=−wL²/6

Slope at the end: θ = wL³/24EI

Maximum deflection, δ = wL⁴/30EI

Deflection Equation: (y is positive downward),
EIy = wx²(10L³−10L²x+5Lx²−x³)/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=ML²/2EI

Deflection Equation (y is positive downward),
EIy=Mx²/2

6. Concentrated load at the midspan of a simple beam

Maximum Moment: M=PL/4

Slope at end:θ_A=θ_B= WL²/16EI

Maximum deflection:δ=PL³/48EI

Deflection Equation (y is positive downward),
EIy=Px{(3/4)L²−x²)}/12 for 0

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

Maximum Moment:M = wL²/8

The slope at the end:θ_L=θ_R=wL³/24EI

Maximum deflection:δ=5wL⁴/384EI

Deflection Equation (EIy=wx(L³−2Lx²+x³)/24

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

Maximum Moment:M = w_oL²/9√3

Slope at the end,

θ_L=7wL³/360EI

θ_R=8wL³/360EI

Maximum deflection:δ=2.5wL⁴/384EI at x = 0.519L

Deflection Equation (y is positive downward),
EIy = wx(7L⁴−10L²x+3x)/360L

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

Maximum Moment: M = wL²/12

The slope at the end, θ_L = θ_R = 5wL³/192EI

Maximum deflection: δ = wL⁴/120EI

Deflection Equation(y is positive downward),
EIy=w_ox(25L⁴−40L²x²+16x⁴)/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

EId²y/dx²=-Px

∫EId²y/dx²= ∫-Px

EIdy/dx= -Px²/2+ c₁

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

c₁= PL²/2

EIdy/dx= -Px²/2+ PL²/2

Integrating this, we get,

EIy= -Px³/6+ PL²x/2+ c₂

Boundary condition x=L, y=0

c₂= -PL³/3

EIy= -Px³/6+ PL²x/2- PL³/3

Slope and deflection at the free end

Slope dy/dx= PL²/2EI

Deflection y= PL³/3EI

Problem 2: Simply supported beam of span 6m is loaded as shown in the figure. I = 78 x 10⁶ mm⁴ E = 2.1 x 10⁵ N/mm². Find the central deflections.

Solution-

We know that

y_c= WL³/48EI= 60 × 10³ ×6000³/(48 ×2.1 × 10⁵ ×78 × 10⁶)= 16.48mm

Problem 3: A simply supported beam of a span of 5m carrying a load of 10N/mm². Given E= 10⁴ 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 × 5000² ×6/(8 ×BD²)=23437500 mm³

Maximum Deflection= 5WL⁴/384EI

5WL⁴/384EI=10mm

5WL⁴/384 ×10⁴ ×B × D³=10mm

BD³=9765625000 mm⁴

D= BD³/BD²= 9765625000/23437500= 416.67 mm

BD²=23437500

B = 134.99mm

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