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Bending equation is considered to be a subsection of the bending theory. Bending theory states that a beam can undergo deformation when there is an external application of any form of force perpendicular to its longitudinal axis. Bending theory is also known as flexure theory. It basically signifies the effects of force on a beam. The bending equation is used to find the amount of stress applied on the beam.
Bending Equation = \(\frac{\sigma}{y} = \frac{M}{T} =\frac{E}{R}\) M = Bending Moment, y = Distance between the neutral axis and extreme fibres, σ = Stress of fibre at distance ‘y’ from neutral axis, E = Young’s modulus of the material of beam, R = Radius of curvature of the bent beam |
Also Read: Effects of force
Table of Content |
Key takeaways: Beam, Force, Stress, Strain, Bending, Pure Bending, Longitudinal axis, Deformation, cross-section, moment of inertia
Beam & its Bending
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Any structure can be a beam if one dimension of it is larger than the other two dimensions by a factor of 1/10 or less. When force is applied to this beam along its longitudinal axis, it tends to get deformed.
Bending equation is a mathematical equation that has been formed to calculate the amount of stress being applied to the beam. However, multiple assumptions have been made while forming it. Pure bending of beams is a hypothetical situation where the bending moment is solely applied on the beam without taking into account axial, shear or torsional forces.
Types of Beams
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Bending Equation
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Bending Equation is given by,
\(\frac{\sigma}{y} = \frac{M}{T} =\frac{E}{R}\)
Where,
- M = Bending Moment
- I = Moment of inertia on the axis of bending
- σ = Stress of fibre at distance ‘y’ from neutral axis
- E = Young’s modulus of the material of beam
- R = Radius of curvature of the bent beam
In case the distance y is replaced by the element c, then
\(\frac{M}{l} = \frac{\sigma max}{c}\)
Hence,
σmax = \(\frac{MC}{l} = \frac{M}{Z}\)
Where, Z = I / c
Z is called the section modulus of the beam.
Derivation of Bending Equation
[Click Here for Sample Questions]Consider an unstressed beam subjected to bending. It bends upto a radius R such that the bottom fibres expereince tension and top fibres experience compression as given in the figure below. The points with zero stress have a common locus. This locus is called as neutral axis.
Bending equation computes the bending stress that develops on the beam which experiences a force.
Strain in fibre AB = \(\frac{\text{ Change in length}}{\text{Original Length}}\)=\(\frac{A'B' - AB}{AB}\)
But, AB = CD and CD=C′D′
Thus, strain = \(\frac{A'B' - C'D'}{C'D'}\)
The stress on the neutral axis is found to be null because of the presence of CD and C′D′. Thus, the neutral axis does not experience any strain due to the applied force.
Bending of Beam
\(\frac{(R+y)\theta - R\theta}{R \theta}\)=\(\frac{R\theta +y \theta - R\theta}{R \theta}\) = \(\frac{y}{R}\)
We know that,
\(\frac{\text{Stress}}{\text{Strain}}\)= E (Young’s Modulus of Elasticity)
\(\frac{\sigma}{y} = \frac{y}{R} =\frac{E}{R}\) ……………(i)
Taking any arbitrary cross section of the beam, strain on its fibre is at a distance y from the neutral axis. The expression is:
σ = E / Ry
Considering a strip of area dA, Force on the strip
F = σδA = \(\frac{E}{R}\)yδA
Moment of the bending equation on the neutral axis will be
F y = \(\frac{E}{R}\)y2δA
Therefore, the total moment of bending equation for the entire cross-section will be
M = ∑ \(\frac{E}{R}\)y2δa = \(\frac{E}{R}\) ∑ y2δa
∑ \(\sigma \)/y2δa is taken to be the material property of the beam. It suggests the second moment of the area of cross section. ‘I’ is used to denote it.
Thus,
M = ERl …………...(ii)
Combining equations (i) and (ii), we get
\(\frac{\sigma}{y} = \frac{M}{I} =\frac{E}{R}\)
This is the derivation of the bending equation. It is also called as flexural formula.
Also Read: Moment of force
Assumptions of Bending Equation
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Many assumptions were made for the derivation of the bending equation. They are:
- The cross-section of the beam is constant.
- The beam is straight.
- The cross-section is constant before and after the bending.
- The beam has a symmetrical longitudinal plane.
- The beam is made of a homogenous material.
- Elastic limit is the same for tension as well as compression.
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Things to remember
- The bending equation is \(\frac{\sigma}{y} = \frac{M}{I} =\frac{E}{R}\)
- Bending equation is used to find the stress applied on the beam.
- Deformation happens when an unstressed beam is subjected to stress acting perpendicular to its axis.
- When the direction of force applied is perpendicular to the longitudinal plane of the beam, the upper part of the beam undergoes compression while the lower part is experiences tension.
- Certain assumptions are made in the derivation of beam equation such as its straightness, cross section, material homogeneity and elastic limit.
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Sample Questions
Ques. Explain the bending theory and bending equation. [2 marks]
It is given by \(\frac{\sigma}{y} = \frac{M}{I} =\frac{E}{R}\)
Ques. What is flexural strength? [2 marks]
The unit of flexural strength is MPa.
Ques. What is ‘c’ in the derivation of the bending equation? [2 marks]
Ques. Explain the concept of pure bending. [3 marks]
However, such a situation cannot occur in real life. It is just an approximation that is made under multiple assumptions.
Ques. What is a bending moment? [3 marks]
It can be calculated by multiplying the magnitude of force with the distance of application of force from its support.
This is an extremely important concept in structural engineering for testing durability.
Ques. A beam of length 2m ruptures when a force of 2KN is applied to its free end. The cross-section of the beam is 40mm x 60mm. Calculate the value of stress at rupture. [4 marks]
= ( 40 x 60 x 60 x 60 ) / 12
= 7.2 x 105 mm4
Bending moment about the fixed end
M = WL
= 2 x 2
= 4 KN-m
Bending stress = M/I = \(\frac{\sigma}{y}\)
Here, y = depth / 2 = 60 / 2 = 30mm
=166.67 N/mm2
Ques. A load of 40KN/m is applied over the entire length of the beam. The beam has a rectangular section of depth 500 mm. If maximum stress is 120N/mm2 and the moment of inertia is 7 x 108 mm4, calculate the length of the beam. [3 marks]
= 40 x L2 / 8
= 5 x L2
Now,
M/I =\(\frac{\sigma}{y}\)
Substituting the values,
5 L2 / 7 x 108 = 120 / 250
Therefore, L = 8197.56mm
Ques. A cantilever beam of length 2m fails when a load of 2KN is applied at the free end. If the section is 40mmx60mm, find the stress at the failure. [4 marks]
Ans. Step 1: Data
Length of beam = 2m or 2000mm
load at failure = 2KN
Section dimensions = 40mm X 60mm
Step 2: Calculation of moment of inertia
I = bd3/12
= (40) (603)/12
= 7.2X105 mm4
Step 3: Calculation of bending moment about fixed end
M = WL
= (2)(2)
= 4KN-m
Step 4: Calculation of bending stress
M /I= σ / y
Substitute for above (where y = depth /2= 60/2 = 30mm)
There fore
σ = 166.67N/mm2
Ques. A beam is simply supported and carries a uniformly distributed load of 40KN/m run over the whole span. The section of the beam is rectangular having a depth of 500mm. If the maximum stress in the material of the beam is 120N/mm2 and the moment of inertia of the section is 7x108mm4, find the span of the beam. [5 marks]
Ans. Step 1: Data
Depth of beam = 500mm
maximum bending stress = σ = 120N/mm2.
moment of inertia =7x108mm4
Step 2: Calculation of bending moment for the above condition
M = wL2/8
= 40(L)2/8
= 5L2
Step 3: Calculation of length of beam
M /I= σ / y
Substitute for above (where y = depth /2= 500/2 =250mm)
5L2/7x108= 120 / 250
L=8197.56 mm
Ques. A rectangular beam 200mm deep and 300mm wide is simply supported over the span of 8m. What uniformly distributed load per metre the beam may carry, if the bending stress is not exceed 120N/mm2. [4 marks]
Ans. Length of beam = 8m or 8000mm
Section dimensions = 300mm X 200mm
Maximum bending stress = σ = 120N/mm2.
Condition: uniformly distributed load for simply supported beam
Calculate the bending moment for above condition:
M = wL2/8
= w (8)2/8
= 8wX106
Moment of Intertia or I = bd3/12
= (300) (2003) /12
= 2X108 mm4
M /I= σ / y
Substitute for above (where y = depth /2= 200/2 =100mm)
8wX106 /2X108= 120 / 100
w =3X104 N/m or 30 N/mm
Ques. Calculate the maximum stress induced in a cast iron pipe with an external diameter of 40mm, of the internal diameter of 20mm, and a length of 4m when the pipe is supported at its ends and carries a point load of 80N at its center. [4 marks]
Ans. Length of beam = 4m or 4000mm
Internal diameter = 20mm
External diameter = 40mm
Now, M= W L /4
M = 80 X 4000 /4
M = 80 KN-m
I = π (D 4 –d 4)/64
I = π (40 4 –20 4)/64
I = 117809.7mm4
M /I= σ / y
Substitute for above (where y = depth /2= 40/2 =20mm)
80X1000/117809.7 = σ / 20
σ = 13.58 N/mm2
Ques. A square beam 20mmx20mm in section and 2m long is supported at the ends. The beam fails when a point load of 400N is applied at the centre of the beam. What uniformly distributed load per meter length will break a cantilever of the same material 40 mm wide, 60 mm deep, and 3 m long? [5 marks]
Ans. Length of beam = 2m or 2000mm
Cross section of the beam = 20mm x 20mm
M= W L /4
M= (400) (2000) /4
M= 200x103
I = bd3/12
= (20) (203)/12
= 13333.33mm4
M /I= σ / y
2 X105 /13333.33= σ / 10
σ = 150N/mm2
Length of beam = 3m or 3000mm
Width of beam = 40mm
Depth of beam = 60mm
M= W L2 /2 = W (3000)2 /2
M /I= σ / y
W (3000)2 /2 /72x104= 150 / 30
Ques. When a beam is subjected to simple bending, the Modulus of elasticity is the same in both tension and compression for the material. Explain. [2 marks]
Ans. Modulus of elasticity is one of the most important assumptions made in the theory of simple bending that is the modulus of elasticity that is Young’s modulus [E] is the same in both tension and compression for the material and the stress in a beam does not exceed the elastic limit.
Ques. What are the units of flexural rigidity? [2 marks]
Ans. The product of young’s modulus (E) of the material and the moment of inertia (I) of the beam section about its neutral axis is called flexural rigidity.
Units for E are N/m2
Units for I are m4
Their product is Nm2
Ques. Why is a positive bending moment known as hogging? [2 marks]
Ans. The positive bending moment in a section is considered because it causes convexity downwards. Such bending moment is called a sagging bending moment or positive bending moment.
Ques. Maximum bending moment in a cantilever beam subjected to udl (w) over the entire span (l). [2 marks]
Ans. In a cantilever beam, the maximum bending moment occurs at the fixed end. The moment at the free end is 0 and the maximum at the fixed end. The maximum shear force is w×l.
Ques. Which stress comes when there is an eccentric load applied? [2 marks]
Ans. When there is an eccentric load it means that the load is at some distance from the axis. This causes compression in one side and tension on the other. This causes bending stress.
Ques. Consider a 250mmx15mmx10mm steel bar which is free to expand is heated from 15C to 40C. what will be developed? [2 marks]
Ans. If we resist to expand then only stress will develop. Here the bar is free to expand so there will be no stress.
Ques. The safe stress for a hollow steel column that carries an axial load of 2100 kN is 125 MN/m2. if the external diameter of the column is 30cm, what will be the internal diameter? [3 marks]
Ans. Area of the cross section of column = π/4 (0.302 – d2) m2
Area = load / stress.
So, π/4 ( 0.302 – d2) m2 = 21 / 125
d = 26.19cm
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