Derivation of Flexure formula for pure bending

                                     Fig.1-Before bending

Consider any two normal sections AB and CD of a beam for a small length ‘dx’ apart as shown in fig. After the action of the transverse loading on the beam, the beam will be bends A’B’C’D’ as shown in fig. Below

Fig 2 -After bending

Let A’ B’ and C’ D’ intersect each other at O.

Let,  R = Radius of curvature of Neutral layer FF.

 Θ = Subtended angle at o

Consider a layer GH at a distance y from Neutral axis

Strain in layer GH due to bending

Let GH=EF

From fig-2

Length of arc = Subtended angle × Radius

Original length

Length of layer

Change in length

By definition, modulus of elasticity

From Equation (1) and (2)

Since E and R are constant, the stress is directly proportional to the distance from Neutral axis.

    Now consider cross-section of the beam.

Consider an elemented strip at ‘y’ from NA and Area of strip ‘dA’.

Force on the strip = Stress on the strip × Area on the strip

Moment of this force about N.A

Total moment of whole section about N.A                         

But                                         Fig. 3

From (3) and (4)

This is known as Bending formula or Flexure formula.

Key Takeaways-

This is known as Bending formula or Flexure formula

• The general formula for bending or normal stress on the section is given by:

=      Where,

M = Internal bending moment about the section’s neutral axis

y = Perpendicular distance from the neutral axis to a point on the section

I = Moment of Inertia of the section area about the neutral axis

• It is obvious to see that the bending stress will be maximised by the distance from the neutral axis (y). Thus, the maximum bending stress will occur either at the TOP or the BOTTOM of the beam section depending on which distance is larger.

=

Where,

c = Perpendicular distance from the neutral axis to the farthest point on the section

Key Takeaways-

• The general formula for bending or normal stress on the section is given by:

=     

To calculate the section modulus, the following formula applies:

Z =

Where, 

I = moment of inertia,

y = distance from centroid to top or bottom edge of the rectangle

For symmetrical sections the value of Z is the same above or below the centroid.

For asymmetrical sections, two values are found: Z max and Z min.

Key Takeaways-

• To calculate the section modulus, the following formula applies:

Z =

For a simple floor beam, I-sections are used.

=

M = σ

  = Z (section modulus)

Therefore, M = z σ

When beams are loaded, bending stresses are developed at all sections.

The bending stresses developed in beams can be determined by the equation theory of simple bending.

For laterally supported beams, the permissible bending stress in tension as well as in compression should not exceed σbc or σbt = 0.66fy

For laterally unsupported beams, the permissible stress in bending compression is calculated by using tables from the IS code book (IS:800).

Load carrying capacity of the Beam

From structural steel tables for the given beam, the section modulus () is obtained.

Depending upon whether the beam is laterally restrained or unrestrained; the value of permissible stress in bending compression (σbc) is calculated.

The moment of resistance of the beam is found out.

MR = σ bc

Equating the moment of resistance to the maximum bending moment equation, the total load (w) the beam can carry is calculated.

Key Takeaways-

• The moment of resistance of the beam is found out.

MR = σ bc

A)                Torsion equation

Torsion equation or torsion constant is defined as the geometrical property of a bar’s cross-section that is involved in the axis of the bar that has a relationship between the angle of twist and applied torque.

The SI unit of Torsion equation is m4.

The torsion equation is given as follows:

Where,

T - torsional moment, N-mm

J - polar moment of inertia, mm4

 - shear stress in the element, N/mm2

r- distance of element from centre of shaft, mm

G - modulus of rigidity, N/mm2

- angle of twist, rad

L- length of shaft, mm

B)                Derivation

Consider a solid circular shaft with radius R that is subjected to a torque T at one end and the other end under the same torque.

Angle in radius =

Arc AB = RӨ = Lγ

Where,

A and B: two fixed points on the circular shaft

γ: angle subtended by AB

   (Modulus of rigidity)

Where,

𝞃: shear stress

γ: shear strain

Consider a small strip of radius with thickness dr that is subjected to shear stress.

Where,

r: radius of small strip

dr: thickness of the strip

γ: shear stress

(torque at the centre of the shaft)

  (substituting for 𝛕’)

 (after integrating and substituting for R)

Therefore, this equation is Torsion equation.

C)               Assumptions

Following are the assumptions made for the derivation of torsion equation:

• The material is homogeneous (elastic property throughout)
• The material should follow Hook’s law
• The material should have shear stress proportional to shear strain
• The cross-sectional area should be plane
• The circular section should be circular
• Every diameter of the material should rotate through the same angle
• The stress of the material should not exceed the elastic limit

Key Takeaways-

              Therefore, this equation is Torsion equation

• Following are the assumptions made for the derivation of torsion equation:

The expressions for the polar moments of circular areas are

A)                Solid Circular Shaft:

= =                              

                                                                              Fig.4 Solid shaft

B)                Hollow Circular Shaft:   

                                                                         Fig.4 Hollow shaft

= =                  

Key Takeaways-

• For Solid Circular Shaft = =                              
• For Hollow Circular Shaft:    = =

From the torsion equation,

Angle of twist,

=

Where,

T - Torsional moment, N-mm

J - Polar moment of inertia, mm4

G - Modulus of rigidity, N/mm2 (sometimes denoted by C)

 - angle of twist, rad

For a given specimen, the shaft properties like length L, polar modulus J and material properties like rigidity modulus G are constants and hence the angle of twist is directly proportional to the twisting moment or torque producing the twist.

Torque producing twist in a shaft is similar to the bending moment producing bend or deflection in a beam. Similar to the flexural rigidity in beams expressed by EI, torsional rigidity is expressed as GJ which can be defined as the torque required to produce a twist of unit radian per unit length of the shaft.

Key Takeaways-

• Angle of twist,

                                   =

• In most practical transmission situations shafts which carry torque are also subjected to bending, if only by virtue of the self-weight of the gears they carry. Many other practical applications occur where bending and torsion arise simultaneously so that this type of loading represents one of the major sources of complex stress situations.
• In the case of shafts, bending gives rise to tensile stress on one surface and compressive stress on the opposite surface while torsion gives rise to pure shear throughout the shaft.
• For shafts subjected to the simultaneous application of a bending moment M and torque T the principal stresses set up in the shaft can be shown to be equal to those produced by an equivalent bending moment, of a certain value Me acting alone.
• Figure

Fig 5. Combine Torsion and Bending

• Maximum direct stress () & Shear stress () in element A

= +

=

• Principal normal stresses () & Maximum shearing stress ()

=

= =

• Maximum Principal Stress () & Maximum shear stress ()

= (M + )

• Location of Principal plane (θ)

=

• Equivalent bending moment () & Equivalent torsion (Te)

= ()

=

Important Note,

• Uses of the formulas are limited to cases in which both M & T are known.
• Under any other condition Mohr’s circle is used.
• Safe diameter of shaft (d) on the basis of an allowable working stress.

• in tension, d =

• in shear, d =

Key Takeaways-

• Maximum direct stress () & Shear stress () in element A

= +

=

• Principal normal stresses () & Maximum shearing stress ()

=

= =

A)                Principal Stress –

There exists an angle qp where the shear stress tx'y' becomes zero. That angle is found by setting tx'y' to zero in the above shear transformation equation and solving for q (set equal to qp). The result is,

=

The angle qp defines the principal directions where the only stresses are normal stresses. These stresses are called principal stresses and are found from the original stresses (expressed in the x, y, z directions) via,

=

The transformation to the principal directions can be illustrated as:

Fig 6. Principal Stress

B)                Maximum Shear Stress –

Another important angle, s, is where the maximum shear stress occurs. This is found by finding the maximum of the shear stress transformation equation, and solving for . The result is,

=

=

The maximum shear stress is equal to one-half the difference between the two principal stresses,

=   = 

The transformation to the maximum shear stress direction can be illustrated as:

Fig 7. Maximum Shear Stress

Part C) Shear Stresses –

• The bending stress is due to bending moment at the section.
• The bending stress acting longitudinally and its intensity is directly proportional to its distance from neutral axis.
• A typical beam section is subjected to shear force in addition to bending moment.
• The variation of shearing stress, which is due to the presence of shear force is studied.

Key Takeaways-

• Principal stresses are found from the original stresses (expressed in the x, y, z directions) via,

=

• The maximum shear stress is equal to one-half the difference between the two principal stresses,

=   = 

Consider, an elemental length of a beam between the section AA and BB separated by a distance , as shown in the following figure.

Let the moments acting at AA and BB be M and M+dm respectively.

Let CD be a fibre fo thickness dy at a distance y from the neutral axis.

Then bending stress of left side of the fibre

Force on the left side of the layer

Therefore, unbalanced force, towards right on the layer CD is  

There are a number of such elements  above the section CD

Hence the unbalanced horizontal force above the section,

This horizontal force is resisted by the resisting force provided by shearing stresses acting horizontal on the plane at CD.

Let the intensity of shear stress be Equating the resisting force provided by the shearing stress to the unbalanced horizontal force, we have

Where, is area of the element.

Where the term

But the term the shear force

Substituting in the expression we obtain

Where

F = Shear force at a section in a beam

a = Area above or below a fibre

= Distance from N . A to the centroid of the shaded area

I = M.I  of the entire section about the N.A

b= breadth of the fibre

Fig.9 Rectangular Section

Consider a rectangular section of width b and depth d subjected to shearing force F.

Let AA be a fibre at a distance y from the NA

Where

And , substituting

Thus the shear stress variation is parabolic when

(i) at

(ii) at

(iii) is max and its value is

,  For a rectangular section and this occurs at the neutral axis, where

B) Circular Section

Consider a circular section of diagram d as shown in fig. Let AA be a fibre at a vertical distance y and angle from NA on which shear stress is to be determined. To find moment of area above the fibre AA about the N.A.

Consider an element of thickness dz at a vertical distance z from the N.A

Let the angular distance of the element be θ

Area of element A=b.dx

Width of the element

                                          Fig.10 Circular Section

Area of the element

Moment of area of the element about the N.A=A×Z

Therefore moment of the entire area, above the fibre AA about the N.A = a

If and is integration

Moment of inertia of the section

Substituting in the expression for shear stress

Where b=width of the fibre AA =

Hence shear stress varies parabolically over the depth. Its value is zero at the extreme fibres where and its value is maximum when at the N.A and is given by

Thus in circular sections shear stress is maximum at the centre and is equal to 4/3 times the average stress

C) Triangular Section

                                                     Fig.11 Triangular Section

Let AA be a fibre distance y from the top.

Shear stress in general

i)  at

ii) at

At the centroid   substituting in (1)

Or substituting in the expression for ,

Thus maximum shear stress occurs at the half the depth and its value is 1.5 times the average shear stress in the case of an isosceles triangle.

Key Takeaways-

• In case of rectangular section Thus the shear stress variation is parabolic when

(i) at

(ii) at

(iii) is max and its value is