18.2 Assumption in the Theory of Simple Bending
The following assumption is made in the theory of simple bending:
18.3 Theory of Simple Bending
Consider a small length dx of a simply supported beam subjected to a bending moment M. Now consider two section AB and CD, which are normal to the axis of the beam RS. Due to the action of the bending moment, the beam as a whole will bend. Since we are considering a small length of dx of the beam, therefore, the curvature of the beam, in this length, is taken to be circular. A little consideration will show, that all the layer of the beam, which were originally of the same length doing not remain of the same length any more. The top layer of the beam AC has suffered compression, and reduced to A1C1. As we proceed towards the lower layer of the beam, we find that the layers have, no doubt, suffered compression, but to a lesser degree; until we come across the layer RS, which has suffered no change in its length, through bent into R1S1. If we further proceed towards the lower layers are stretched; the amount of extension increases as we proceed lower, until we come across the lowermost layer BD which has been stretched to B1D1.
Fig. 18.1 simple bending
Now we see that the layers above RS have been compressed and that bellow have stretched. The amount, by which a layer is compressed or stretched, depends upon the position of the layer with reference to RS. This layer RS, which is neither compressed nor stretched, is known as neutral plane or neutral layer. The theory of bending is called theory of simple bending.
18.4 Bending Stress
Fig. 18.2 Bending stress
Consider a small length dx of a beam subjected to a bending moment. As a result of this moment, let this small length of beam bend into an arc of a circle with O a center.
Let M= Moment acting at the beam,
θ = angle subtended at the center by the arc, and
R= Radius of curvature of the beam
Now consider a layer PQ at a distance y from RS the neutral axis of the beam. Let this layer be compressed to P1Q1 after bending.
Decrease in length of this layer
Strain e = (δl / Original length) = (PQ - P1Q1) / (PQ)
Now from the geometry of the curved beam, we find that the two section OP1Q1 and OR1S1 are similar
(P1Q1) / (R1S1) = (R - y) / R
1 - (P1Q1) / (R1S1) = 1 - (R - y) / R
(R1S1) - (P1Q1) / (PQ) = ( y / R)
(PQ) - (P1Q1) / (PQ) = ( y / R) ( PQ = R1S1 = RS = Neutral axis)
e = (y / R) ( e =(PQ) - (P1Q1) / (PQ) )
It is thus obvious, that the stain of a layer is proportional to its distance from the neutral axis.
We also know that the stress,
f = Strain X Elasticity = e.E
= (y / R) x E = y x (E / R) ( e = y / R)
Since E and R are constants in this expression, therefore the stress at any point is directly proportional to y, i.e., the distance of the point from the neutral axis.
The above expression may also be written as:
f / y = E / R ----------(1)
18.5 Position if Neutral Axis
The line of intersection of the neutral layer, with any normal cross-section of a beam, is known as neutral axis of the section. We have that on one side of the neutral axis there are compressive stresses, whereas on the other there are tensile stresses. At the neutral axis, there is no stress of any kind.
Fig. 18.3 Position of neutral axis
18.6 Moment of Resistance
Fig. 18.4 Moment of resistance
We have already known that on one side of the neutral axis there are compressive stresses, and on the other there are tensile stresses. These stresses from a couple, whose moment must be equal to the external moment M. The moment of this couple, which resist the external bending moment, is known as moment of resistance. Consider a section of the beam.
Let NA be the neutral axis of the section. Now consider a small layer PQ of the beam section at a distance y from the neutral axis.
18.7 Section Modulus
M / I=б/y=E / R
M / I= б / y
б or ƒs =M / I*y
б max =M /I /ymax where Z = I/ymax
= M / Z Z=section modulus.
18.8 Examples of Section Modulus for Different Sections
Fig. 18.5 Examples of section modulus for different sections
I = Moment of inertia, m^4 ,m4 Z = Section modulus, m^3 , m3
y = Centroidal distance, m
I = b * h^3 / 12 = 1/12 b.h3
Z = b * h^2 / 6 = 1/6 b.h2
y = h / 2
I = b * h^3 / 36 = b.h3/36
Z = b * h^2 / 12 = b.h2/12
y = h / 3
I = ∏d4 / 64
y = d / 2