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Beam Deflection Formula, in structural engineering terms, is the degree to which a part of a structural element like a beam is displaced by a considerable amount of load. It is 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.
Also Read: Mechanical Properties of Solids
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Key Terms: Beam Deflection, Cantilever Beam, Simply Supported Beams, Elasticity, Mechanical Properties of Solids, Young’s Modulus
Beam Deflection
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A beam is a long piece of a body that is capable of holding the load by resisting the bending. The deflection of the beam in a particular direction when force is applied to it is known as Beam Deflection. There are mainly four variables, which can determine the magnitude of the beam deflections:
- Loading on the structure
- The length of unsupported members
- The material (specifically the Young’s Modulus)
- The cross-section size (specifically the Moment of Inertia(I)
Beam Deflection Schematic
The deflection distance of a member under a load can be calculated by integrating the function that mathematically describes the slope of the member under that load. Generally, we calculate it by taking the double integral of the Bending Moment Equation, which means M(x) divided by the product of E (Young’s Modulus) and I (Moment of Inertia).
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Important Topics Related to Beam Deflection Formula | ||
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Important Thermal Properties of Matter | Mechanical Properties of Fluids | Shear Modulus |
Torque | Bulk Modulus | Moment of Inertia Vs Torque |
Rotational Inertia Formula | Law of Inertia | Effects of Force |
Balanced Forces | Types of Forces | Resultant Force Formula |
Beam Deflection Formula
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Beams can vary greatly in their geometry and composition. Some simple examples are shown below. The formulas expressed are approximations developed for long, slender, homogeneous, prismatic beams with small deflections and linear elastic properties.
Cantilever Beams
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, so that the slope and deflection at that end must be zero. Cantilever beams are further classified as:
- End-Loaded
- Uniformly Loaded
Cantilever Beam Schematic
End-Loaded Cantilever Beams
In this case of cantilever beams, the load is applied at a single point on the beam. Following is a schematic to understand this a little better.
End-Loaded Cantilever Schematic
To calculate the deflection of the cantilever beam with force at free end we can use the below equation:
\(\eth_ B = \frac{FL^{3}}{3EI}\)
Where,
δB is Beam deflection
F is Force at one end
L is Length of beam
E is Young’s Modulus
I is Moment of inertia
If the span doubles, the deflection increases eightfold. The deflection at any point x, along the span of an end loaded cantilever beam can be calculated using:
\(\eth_ x = \frac{Fx^{3}}{6EI} \)(3L-x)
When x = L (the end of the beam), δx is identical to δB in the equation above.
Also Read: Tensile Stress
Uniformly-Loaded Cantilever Beams
Uniformly-loaded cantilever beams have the force acting uniformly through out the length of the beam. The schematic is as follows:
Uniformly-Loaded Cantilever Schematic
The deflection, at the free end B, of a cantilever beam under a uniform load is given by:
\(\eth_ B = \frac{qL^{4}}{8EI}\)
Where,
q is Force at one end
L is Length of beam
E is Young’s Modulus
I is Moment of inertia
The deflection at any point x, along the span of a uniformly cantilevered beam can be calculated using:
\(\eth_ x= \frac{qx^{2}}{24EI}\)(6L2 -4Lx + x2)
Simply-Supported Beams
Simply-supported beams have support under their end which allows rotation, but not deflection. The simply supported beams can be further classified as:
Simply-Supported Beams
- Centre-Loaded
- Off-Centre
- Uniformly Loaded
Centre-Loaded Simple Beams
In centre-loaded beams, the force acts at the centre of the beam. The schematic is as follows:
Centre-Loaded Schematic
The special case of elastic deflection at the midpoint C of a beam, loaded at its centre, supported by two simple supports is given by:
\(\eth_ c= \frac{qL^{3}}{48EI}\)
The deflection at any point x, along the span of a centre loaded simply supported beam can be calculated using:
\(\eth_ x = \frac{Fx^{}}{48EI}\)I\((3L^2-4x^2)\)
Off-Centre Loaded Simple Beams
In this type of beam, the force acts at a point slightly away from the centre.
The maximum elastic deflection on a beam supported by two simple supports, loaded at a distance a from the closest support is given by:
\( \)
Where, a is distance from the load to the closest support
This maximum deflection occurs at a distance x1 from the closest support and it is given by:
x1 = \(\sqrt{\frac{L^2-a^2}{3}}\)
Also Read: Thermal Stress
Uniformly-Loaded Simple Beams
In this type of beams, the force acts uniformly along the length of the beam.
The elastic deflection at midpoint C on a beam supported by two simple supports, under a uniform load as shown in the picture is given by:
\(\eth_ c= \frac{5qL^{}}{384EI}\)
Where, q is uniform load on the beam (force per unit length)
Also Read: Strain Energy
Beam Deflection Formula Units
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The formulas stated above require the use of a consistent set of units. Most calculations are made in SI units.
- Force: Newtons (N)
- Length: Metres (m)
- Modulus of elasticity: N/m2
- Moment of Inertia: m4
Other units may be used as well, as long as they are self-consistent. For example, sometimes the kilogram-force (kgf) unit is used to measure loads. In such cases, the modulus of elasticity must be converted to kgf/m2.
Also Read: Stress Vs Pressure
Solved Example
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Ques: A cantilever beam of length 2 metres is supported at one end only. The Young’s Modulus of a metal is 200 x 109 Nm-2. The moment of inertia is 50 kg m2. If a force of 300 N is applied at one end, calculate the deflection in the metal beam.
Solution: From the question,
E= 200 x 109 Nm-2
I= 50 kgm2
L= 2m
F= 300N
Using the formula for calculating deflection of cantilever beam , \(\eth_ B = \frac{FL^{3}}{3EI}\)
= \(\frac{300*2*2*2}{3*200*10^9*50}\)
=0.08m
Things to Remember
- Deflection Consideration: The maximum deflection of a loaded beam should be within a certain limit so that the strength and efficiency of the beam should not be affected.
- Deflection in beams can be broadly classified as: Cantilevers and Simply Supported Beams.
- Based on the location of application of force, cantilever and simply supported bemas can be further classified.
- The units of beam deflection formula are expressed in terms of force, length, moment of inertia or elasticity modulus.
- Young’s modulus of elasticity of the beam is equal to the ratio of stress and strain within the limit of proportion.
- Young’s Modulus is the property of the material which does not depend on the external factors like force, moment etc.
- Three elastic modules which are young’s modulus, shear modulus and bulk modulus are used to describe the elastic behaviour of objects as they respond to deforming forces that act on them.
- The young’s modulus and shear modulus are relevant only for solids.
- The bulk modulus is relevant for solids, liquids and gases.
Also read:
Important Topics From Chapter: Mechanical Properties of Solids | ||
---|---|---|
Stress | Stress Formula | Causes of Stress |
Stress-Strain Curve | Strain | Solid Deformation |
Poisson’s Ratio | Elastic Moduli | Elastic Limit |
Hooke’s Law | Yield Strength | Elastic Constants Relation |
Previous Year Questions on Beam Deflection Formula
- Choose the correct relation between the parameters of elesticity. [JEE Main 2021]
- Ratio of weights of steel and brass wires upon suspension. [NEET 2015]
- Which graph is a straight line. [NEET 2014]
- Ratio of increase in length of steel and copper wires. [NEET 2013]
- Calculate the force required to stretch the wire. [NEET 2018]
- Calculate the elastic potential energy in the elongated wire. [NEET 2019]
- Calculate the stress relation between two extended wires. [KCET 2018]
- What is Young’s Modulus. [KCET 2017]
- Which substance has highest elasticity? [KCET 2010]
- Ratio of hydraulic stress to strain is? [KEAM]
- Determine the ratio of e=increase in length of two wires of same material. [KEAM]
- Determine the length of the cord when Young’s Modulus is given. [KEAM]
- Calculate the energy stored in the wire. [KEAM]
- Calculate the displacement when force is applied on a cube. [JEE Main 2018]
- Calculate the force needed to push the cork into the bottle. [JEE Main 2016]
- Determine the Young’s Modulus. [JEE Main 2019]
- Choose the wrong statement. [KEAM]
- Determine the stress change factor. [JEE Main 2017]
- Calculate the tensile stress in the wire. [JEE Main 2019]
- Calculate the fractional decrease in wire radius. [JEE Main 2013]
Sample Questions
Ques. Calculate the deflection of a cantilever beam of length 2 metre which has support at one end only. Young’s modulus of the metal is 200 × 109 and the moment of inertia is 50 Kg m2. At the end force applied is 300 N. ( 4 Marks)
Ans. Given parameters are, E = 200 × 109 Nm-2
I = 50 kgm2
L = 2 m
W = 300 N
Now, using the beam deflection formula,
D = WL3/3EI
D = 300 × 23/ 3 × 200 × 50
D = 2400/30000
D= 0.08 m
Hence, the value of beam deflection will be 0.08 m or 8 cm.
Ques. In case of bending of a beam, how the depression δ depends on Young modulus of elasticity Y? ( 4 Marks)
Ans. Considering any beam, but for simplicity let us consider a simply supported beam subjected to a point load at its centre.
we know that the deflection of a simply supported beam of length L which is subjected to a point load at its midpoint is given by:
D = FL3/ 48YI
Here Y is Young's modulus for beam material and L is the moment of inertia of the beam.
From the above expression, we can say that deflection is inversely proportional to Young’s modulus of elasticity.
D ∝ 1 / Y
This deflection is the depression of the beam so that we can write:
δ ∝ 1/ Y
δ ∝ Y-1
Therefore, in case of bending of a beam depression δ is inversely proportional to the young modulus of elasticity of the beam material.
Ques. A simply supported beam of length 6m is subjected to a uniformly distributed load of 75 kN/m over the whole span. Determine the deflection of the beam when both ends are fixed when the value of EI is 5.06 × 1010 Nmm2. ( 4 Marks)
Ans. Given parameters are, l = 6 m = 6000 mm
Load, W = 75 kN/m = 75 × 103 N/m
= 75 mm
Value of EI = 5.06 × 1010 Nmm2
Simply supported beam subjected to uniformly distributed load, deflection is given by
δ = 5qL4/384EI
δ = 5 × 75 × (6000)4 / 384 × 5.06 × 1010
δ = 25 mm
Therefore, the value of beam deflection is 25 mm.
Ques. In case of bending of a beam, how the depression δ depends on Young modulus of elasticity Y? ( 4 Marks)
Ans. We can find the answer to this question by considering any beam, but for simplicity let us consider a simply supported beam subjected to a point load at its centre.
we know that the deflection of a simply supported beam of length L which is subjected to a point load at its midpoint is given by:
D = FL3/ 48YI
Here Y is Young's modulus for beam material and L is the moment of inertia of the beam.
From the above expression, we can say that deflection is inversely proportional to Young’s modulus of elasticity.
D ∝ 1 / Y
This deflection is the depression of the beam so that we can write:
δ ∝ 1/ Y
δ ∝ Y-1
Therefore, in case of bending of a beam depression δ is inversely proportional to the young modulus of elasticity of the beam material.
Ques. A simply supported beam of length 6m is subjected to a uniformly distributed load of 75 kN/m over the whole span. Determine the deflection of the beam when both ends are fixed when the value of EI is 5.06 × 1010 Nmm2. ( 4 Marks)
Ans.
Given parameters are, l = 6 m = 6000 mm
Load, W = 75 kN/m = 75 × 103 N/m
= 75 mm
Value of EI = 5.06 × 1010 Nmm2
Simply supported beam subjected to uniformly distributed load, deflection is given by
δ = 5qL4/384EI
δ = 5 × 75 × (6000)4 / 384 × 5.06 × 1010
δ = 25 mm
Therefore, the value of beam deflection is 25 mm.
Ques. What will be the ratio of the deflections of the free end of a cantilever beam due to an isolated load at 1/3rd and 2/3rd of the span? ( 3 Marks)
Ans. 1st condition: for a cantilever beam subjected to load W at distance of L/3 from free end, the deflection is given by:
Yc1 = W/3EI × (2L/3)3 + W/2EI(2L/3)2 × L/3
Yc1 = 28WL3/162EI
2nd condition: for a cantilever beam subjected to load W at distance of 2L/3 from free end:
Yc2 = W/3EI × (L/3)3 + W/2EI(L/3)2 × 2L/3
Yc1 = 8WL3/162EI
Ratio (r) = Yc1/ Yc2 = 28/8 = 7/2
∴ Yc1/ Yc2 = 7/2
Ques. A simply supported beam carries a uniformly distributed load of 20 kN/m over the length of 5m. If flexural rigidity is 30000 kN.m2, what is the maximum deflection in the beam? ( 3 Marks)
Ans. Given parameters are load W = 20 kN/m
Length of the beam = 5 m
Flexural rigidity (EI) = 30000 kN.m2
Maximum deflection in the beam δmax = (5/384)(WL4/EI)
δmax = (5/384)(20 × 54 / 30000)
δmax = 5.42 × 10-3 m = 5.42 mm
Ques. A simply supported beam of length 5 m is loaded at the centre with a concentrated load of 1200 N and deflects 12 mm at the centre. Determine the flexural rigidity (EI). ( 3 Marks)
Ans. Given parameters are length L = 5 m = 5000 mm
Load W = 1200 N
Deflection δ = 12 mm
Simply supported beam subjected concentrated load at centre, deflection is given by,
δ = WL3/ 48EI
EI = WL3/ 48δ
EI = 1200(5000)3/48×12
EI = 2.6 × 1011 N/mm2
Ques. If a simply supported beam of span 4 m is subjected to a terminal couple of 4 kN-m at both the ends, then what will be the magnitude of the central deflection? ( 3 Marks)
Ans. Given parameters are L= 4m
M = 4 kN-m
When a simply supported beam subjected to terminal couples at the both ends then,
The maximum central deflection is given by: ML2/8EI
δmax = 4(4)2/8EI
δmax = 8/EI (upwards if couples are hogging type)
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