Mr. Akash Assistant Professor

Department of Civil Engineering DCE Darbhanga

Module- 4

Flexural Stresses-Theory of simple bending

Introduction to Solid Mechanics

Course Code- 101205

Syllabus

Flexural Stresses-Theory of simple bending

• Assumptions – Derivation of bending equation: M/I = f/y = E/R - Neutral axis –

Determination of bending stresses – Section modulus of rectangular and circular sections

(Solid and Hollow), I,T, Angle and Channel sections – Design of simple beam sections.

Behaviour of Beam

• If we take any section of a beam, as in figure (a), and consider the free body on either side, we find that there must be

three internal stress resultants at the section in general to maintain the equilibrium of the free body.

• These internal stress resultants are equal and opposite to the external effects of loads on the section, i.e. the bending

moment, shear force, and axial force as in figure (b)

• Under the effect of the lateral loads, the beam bends and the longitudinal axis takes a curved shape. The beam produces

an internal resisting couple to balance the BM due to the applied loads. This couple is generated by the normal

compressive and tensile forces in the section, as shown in Figure (d). This couple is equal and opposite to the external

BM.

• To generate a force to balance the SF due to applied loads, the section has tangential stresses whose resultant is equal

and opposite to the SF (Figure (e)

Bending(Flexure) Stresses

• When a beam is subjected to a loading system or by a force couple acting on a plane passing through the axis, then the

beam deforms. In simple terms, this axial deformation is called as bending of a beam. Due to the shear force and

bending moment, the beam undergoes deformation. These normal stress due to bending are called flexure stresses.

Type of Bending Stress

1. Simple Bending Stress

• Bending will be called as simple bending when it occurs because of beam self-load and external load. This type of

bending is also known as ordinary bending and in this type of bending results both shear stress and normal stress in

the beam.

Pure Bending Stress

• Bending will be called as pure bending when it occurs solely because of coupling on its end.

• In that case there is no chance of shear stress in the beam. But, the stress that will propagate in the beam as a result

will be known as normal stress. Normal stress because it not causing any damages to beam. As shown below in the

picture, pure Bending stresses are those that results because of beam self load only.

Theory of Pure Bending: Bernoulli's Equation

• Consider a beam bent by pure couples so that the BM is constant and the SF is zero. The beam bends in such cases in

the form of an arc of constant radius. The beam is prismatic, i.e., of constant cross section and is symmetrical about a

vertical plane through its centroid.

• In deriving the theory of bending, the following assumptions are made

1. Sections which are plane before bending remain plane after bending. sections remain plane, there must be one layer

which is neither stretched nor shortened.

2. The plane of loading must contain a principal axis of the beam cross section. The plane containing the axis of

symmetry should also contain the loads and reactive forces.

3. The material is homogeneous and isotropic and obeys Hooke’s law.

4. The modulus of elasticity of the materials is the same in tension and compression. Stress diagram is linear and has

the same shape as the strain diagram.

5. The beam is initially straight and has constant cross section, i.e., a prismatic beam. Longitudinal axis of beam is

straight.

6. Each layer of the material is free to expand or contract under stress. This is necessary to ensure that we are able to

calculate the stress at any layer using the elastic properties. The layers are not affected by the presence of adjoining

layers, and are free to deform.

7. Beam is initially straight, self weight of beam is not considered and Poisson effect in beam is neglected.

In bending, one surface of the beam is subjected to tension and the opposite surface to compression there must be a

region within the beam cross-section at which the stress changes sign, i.e. where bending stress should be zero. The

layer of zero stress due to bending is called neutral layer and the trace of neutral layer in the cross-section is called

neutral axis.

• Due to pure bending, beams sag or hog depending upon the nature of bending moment. It can be easily observed

that when beams sag, fibres in the bottom side get stretched while fibres on the top side are compressed. In other

words, the material of the beam is subjected to tensile stresses in the bottom side and to compressive stresses in the

upper side. In case of hogging the nature of bending stress is exactly opposite, i.e., tension at top and compression

at bottom.

Neutral Axis

Bending Equation

Based upon these assumptions, the bending equation

can be derived as follows.

• Let us consider a beam initially unstressed as shown in figure. Now the beam is subjected to a constant bending

moment (i.e. ‘Zero Shearing Force') along its length as would be obtained by applying equal couples at each end.

The beam will bend to the radius R as shown in Fig below.

• The layers above n-n are compressed and those below n-n are stretched in the case shown of positive BM. If the BM

is negative, the upper layers will be in tension and the bottom layers will be in compression, with the beam bending

convex upwards. R is the radius to which the neutral layer bends.

Considering the deformation of a layer y above n-n, we have

dx= R d θ

where dθ is the angle subtended by the length dx at the centre of curvature,

and m-m = (R - y)dθ .

The length of the layer m-m before bending is dx = R dθ.

Change in length = R dθ - (R -y)dθ =y dθ

and

Since R is a constant, strain at any layer is proportional to its distance y from the neutral layer. Since E is constant

and is the same in tension and compression, stress σ at layer m-m = strain x E.

or, …………………(1)

The stress diagram is thus linear following the variation in strain. This diagram is obtained by multiplying the

strain diagram by the constant E.

• Considering an elementary strip dy of width b of the cross section, the force acting on the elementary area is given

by

Or, from equation (1)

This can be summed up for the entire cross section and since C = T, such summation is equal to zero. If y1 and y2 , are the extreme fibre depths above and below the NA, then

Moment of dF about the neutral axis would be , dM = dF y

• Total Moment for the whole cross-section is therefore equal to

is the property of the material and is called as a second moment of area of the cross-section

and is denoted by a symbol “I”

Therefore,

or,

Since,

Or,

This equation is known as the Bending Equation or flexural formulae. The above proof has involved the assumption of

pure bending without any shear force being present. Therefore this termed as the pure bending equation. This equation gives

distribution of stresses which are normal to cross-section i.e. in x-direction.

Locating Neutral Axis

Consider an elemental area δA at a distance y from neutral axis

• If ‘σ ’ is the stress on it, force on it F= σ δA But σ=Ey/R, from eqn..(1)

∴ Force on the element,

Hence total horizontal force on the beam

Since there is no other horizontal force, equilibrium condition of horizontal forces gives

As is not zero . …………………..(2)

If A is total area of cross-section, from eqn. (2), we get

is moment of area about neutral axis

is distance of centroid of the area from the neutral axis.

Hence means the neutral axis coincides with the centroid of the cross-section

Section Modulus

• Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members.

• Section Modulus is moment of inertia of the section divided by the distance to the extreme fibre, either to the top or

to the bottom, from the neutral axis.

• Section modulus is usually denoted by the symbol Z . The unit of section modulus Z is

• From Bending equation i.e.

Hence bending stress is maximum, when y is maximum. It means maximum stress occurs in the extreme fibres.

If is distance of extreme fibres from neutral fibre, then

In a design is restricted to the permissible stress in the material. If is permissible stress , then from above

equation

1

2

3

4

5

6. I – Section

Example : 1. A timber beam has to carry a load of 2 kN/m over a span of 3 m. The permissible stresses are 12 MPa in

compression and 8 MPa in tension. Design the section if the width is half of the depth.

Solution :

Example : 2. A steel beam has a span of 16 m and carries a load of 14 kN/m. Design a suitable I-section if the

permissible stress is 150 MPa.

Solution :

Example : 3. Shows the cross-section of a cantilever beam of 2.5 m span. Material used is steel for which maximum

permissible stress is 150 . What is the maximum uniformly distributed load this beam can carry?

Solution : Since it is a symmetric section, centroid is at mid depth.

I = MI of 3 rectangles about centroid

Load Carrying Capacity • Given a beam section and the span on which it is required to support loads, we can find the load that the section can

carry. This is generally done for UD load.

• More appropriate is the moment carrying capacity called moment of resistance of a section. This is independent of the

span and the loads.

• Given a section and the maximum permissible stresses in tension and compression, the moment carrying capacity can

be calculated. This can be compared with the maximum moment in a span and the load condition.

• The examples to illustrate both the load carrying capacity and moment of resistance are given later

Question: A rectangular section, 200 mm x 400 mm, spans a distance of 2 m. Find what UD load can the beam section

carry on this span. Permissible stresses are 7 in tension and 15 in compression.

Solution:

Example: 1 A circular log of wood is used as a beam. If the diameter of the log is 200 mm, find the moment of

resistance of the section. Permissible stresses are 10 in tension and 18 in compression.

Solution:

Example: 2. Design a rectangular section to carry a load of 32 kN/m over its whole span of 3 m, including the self-

weight. Permissible stresses are 8 in tension and 15 in compression.

Solution:

Example: 3. The section shown in Fig. 5.34 is a reinforced concrete beam section. Find the moment capacity of the

section if the permissible stress in concrete is 5 in compression and 140 in steel in

tension. The modular ratio is 18.

Solution:

Example: 4. Find the moment resistance of the section shown in Figure. Permissible stresses are 100 MPa in tension and

compression. What UD load can it carry over a span of 8 m?

Solution :

Example: 5. An I-section girder, 200 mm wide by 300 mm deep, with flange and web of thickness 20 mm is used as a

simply supported beam over a span of 7 m. The girder carries a distributed load of 5 kN/m and a

concentrated load of 20 kN at mid-span. Determine: (a) the second moment of area of the cross-section of

the girder, (b) the maximum stress set-up.

Solution :

Example: 6. A uniform T-section beam is 100 mm wide and 150 mm deep with a flange thickness of 25 mm and a web

thickness of 12 mm. If the limiting bending stresses for the material of the beam are 80 in compression

and 160 in tension, find the maximum u.d.l. that the beam can carry over a simply supported span of 5 m.

Solution :

Thanks