Flexural stresses

FMD-3141907
Content
• Theory of simple bending,
• Assumptions
• derivation of equation of bending, neutral axis,
• determination of bending stresses,
• section modulus of rectangular & circular (solid & hollow), I,T, Angle, channel
sections
Flexural stresses

• When some external load acts on a beam the shear force and bending moments
are set up at all sections of beam.
• due to this shear force and bending moment the beam undergoes certain
deformation.
• If length of a beam is subjected to a constant bending moment and no shear
force then the stresses will be set up in that length of the beam due to bending
moment only and that length of the beam is said to be in pure bending or simple
bending.
• Stresses set up in that length of the beam are known as bending stresses or
Flexural stresses
Flexural stresses
Due to pure bending,
• beams sag or hog depending upon the nature of bending
moment as shown
• 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.
• Thus bending stress varies from compression at one edge to
tension at the other edge.
• Hence somewhere in between the two edges the 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
Assumptions

• The material is homogeneous and isotropic.

• Modulus of elasticity is the same in tension and in compression.
• Stresses are within the elastic limit.
• Plane section remains plane even after deformations.
• The beam is initially straight and every layer of it is free to expand or contract.
• The radius of curvature of bent beam is very large compared to depth of the
beam.
BENDING EQUATION

The Equation can be derived by

1. Relationship between bending stresses and radius of curvature.
2. Relationship between applied bending moment and radius of curvature.

1.Relationship between bending stresses and

radius of curvature:
Consider an elemental length AB of the beam
as shown in Fig.
Let EF be the neutral layer and CD the
bottommost layer. If GH is a layer at distance y
from neutral layer EF, initially AB = EF = GH =
CD.
BENDING EQUATION
Let after bending A, B, C, D, E, F, G and H take positions
A′, B′, C′, D′, E′, F′, G′ and H′ respectively as shown in Fig.
(b).
Let R be the radius of curvature and φ be the angle
subtended by C′A′ and D′B′ at centre of radius of
curvature. Then,
EF = E′F′, since EF is neutral axis
= Rφ ……………………………….(1)

Strain in GH =(Final length – Initial length)/Initial length

BENDING EQUATION

(2)
Since strain in GH is due to tensile forces, strain in GH = f/E ...(3)
where f is tensile stress and E is modulus of elasticity.
From eq 2 and 3

A
BENDING EQUATION

2. Relationshipbetween bending moment and

radius of curvature:
Consider an elemental area δa at distance y
from neutral axis as shown in Fig.
From eqn. of stress on this element is
BENDING EQUATION

B
 BENDING EQUATION

From eqns. and we get

Where
M = bending moment at the section
I = moment of inertia about centroid axis
f = bending stress
y = distance of the fiber from neutral axis
E = modulus of elasticity and
R = radius of curvature of bent section.

Eq. C is known as bending equation

Condition of Simple Bending

• Bending formula is applicable to member with pure bending only.

• Beam should be free from shear force
• In actual practice a member is subjected to both shear and bending moment at every section
but shear force is zero at the section where the B.M is maximum.
• Stresses produced by maximum B.M by flexure formula give satisfactory result to design beams
and structure.
LOCATING NEUTRAL AXIS

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

If ‘f ’ is the stress on it, force on it = f δa
But f =(E/R)y, from eqn. A

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

2
Section modulus
 Section modulus

The eqn. gives permissible maximum moment on the section and is known as moment
carrying capacity of the section. Since there is definite relation between bending
moment and the loading given for a beam it is possible to find the load carrying capacity
of the beam by equating maximum moment in the beam to moment carrying capacity of
the section. Thus
Mmax = fper Z
If permissible stresses in tension and compressions are different for a material, moment
carrying capacity in tension and compression should be found separately and equated
to maximum values of moment creating tension and compression separately to find the
load carrying capacity. The lower of the two values obtained should be reported as the
load carrying capacity.
section modulus of standard section
section modulus of standard section
section modulus of standard section
section modulus of standard section
section modulus of standard section
Example 1
A simply supported beam of span 3.0 m has a cross-section 120 mm × 180 mm. If
the permissible stress in the material of the beam is 10 N/mm2,
determine
(i) maximum udl it can carry
(ii) maximum concentrated load at a point 1 m from support it can carry.
Neglect moment due to self weight.
Ex 2. A circular steel pipe of external diameter 60 mm and thickness 8 mm is used
as a simply supported beam over an effective span of 2 m. If permissible stress in
steel is 150 N/mm2, determine the maximum concentrated load that can be
carried by it at mid span.
Given D = 60 mm
Thickness = 8 mm
Ex.3 Figure shows the cross-section of a cantilever beam of 2.5 m span. Material
used is steel for which maximum permissible stress is 150 N/mm2. What is the
maximum uniformly distributed load this beam can carry?

symmetric section, so centroid is at mid depth.

I = MI of 3 rectangles about centroid