deflection in the beam which

7/29/2019 Deflection in the Beam Which

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When a beam is subjected to transverse loads itcauses deflection in the beam which producesbending. When the loads producing bending lie in

the centroidal plane such that bending is notaccompanied by torsion then the bending is saidto be simple bending. When the beam issubjected to such a system of loads so that theshear force in the beam iszero then the beam issaid to be subjected to pure bending. Bending isquite a common

phenomenon. Beams and plates used in civilengineering construction, transmission shafting,tall structures like chimneys and distillationcolumns subjected to wind load, etc.are some ofthe common examples of bending. We shallstudy the tensile, compressive andshear stresses

produced in beams due to simple bending. Someof the terms associated with simple bending ofbeams are explained below.

Consider a beam subjected to bending as shownin Fig. 4.1.

4.1 DEFINITIONS


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symmetry and plane of bending coincide. GHKL isthe plane of of bending.

Neutral Surface. It is a surface on which thebending stress is zero ar d the beam fibres donot undergo any tension or compression. ABDC isthe neutral surface.

Neutral Axis. It is a line where the neutralsurface intersects with the area of cross- sectionof the beam. AC and BD are the neutral axes.

Centroid : The centroid of a plane area is thatpoint at which the whole of the area of the planemay be assumed to be concentrated. If a planearea is subdivided into small areas

A1, A2 ,and the perpendicular distances of the

centroids of these areas from two chosen

coordinate axes and y are y1, Y2 , and x1, x2,respectively, then the coordinates of

the centroid (, ) of the whole area are given by:

Moment of Inertia. The moment of inertia or thesecond moment of area of a plane figure about a

given axis is the sum of the product of theelementary areas into which plane figure may be


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divided and the square of the distance of thecentroids of elementary areas from the chosenaxis. Thus

Parallel Axis Theorem. This theorem states that

the moment of inertia of a plane figure about anyaxis in its plane, is equal to the total sum of themoment of inertia of the figure about a parallelaxis through the centroid of the figure plus theproduct of the total area of the figure and thesquare of the distance between the parallelcentroidal axis and

the axis of reference.

For example,

4.2 THEORY OF SIMPLE BENDING

The theory of simple bending was developed byBernoulli. The following assumptions

are made in this theory:


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1. The material of the beam is assumed to behomogeneous, perfectly elastic and isotropic.Homogeneous material means that the

composition is uniform throughout. Perfectlyelastic means, Hookes law is obeyed and nopermanent set is formed on unloading. Isotropicmeans that the elastic constants like E, G, K aresame in all directions in the beam.

2. All transverse sections of the beam, which are

planes before bending, remain plane afterbending. -

3. The radius of curvature of the beam beforebending is very large in comparison to itstransverse dimensions. This implies that thebeam is initially straight.

4. The resultant pull or push across anytransverse section of the beam is zero. Thisimplies that total tensile force is equal to thecompressive force in the beam cross- section.

5. The Youngs modulus of elasticity is same intension and compression.

6. The stresses are within the proportional limit.

Consider the simple bending of a beam as shownin Fig. 4.2 (a). In the unloaded state, let GH bethe fibres at a distance y from the centroidal axisKL, its length being determined by two

transverse parallel planes AD and BC. Fig. 4.2(b) shows the beam after bending, when loaded,


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and the planes AD and BC assume the positionsA1D1 and B1C1 respectively.The planes A1D1and B1C1 intersect at point 0, the centre of

curvature, and are inclined at an angle 0. Let Rbe the radius of curvature of the centroidal fibresat E1F1. Then


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Therefore, the bending stress at any point in thecross-section is proportional to its distance fromthe centroidal axis.

Now consider an elementary area dA at adistance y from the centroidal axis, as shown inFig. 4.2 (c). Force acting on dA is,


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Further, since there is no resultant force acrossany transverse cross-section therefore,

y4A = 0, i.e. the first moment of area about thecentroidal axis is zero, which is possible

only if the centroidal axis coincides with theneutral axis.

Now moment of the force acting on cIA about theneutral axis is,

The total moment of all the forces acting onvarious elements composing the cross-

section forms a couple which is equal to thebending moment M. This total moment is

called the moment of resistance.


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Eq. (4.1) represents the bending formula. Theimplications of the bending formula are:

The variation of bending stress (tensile orcompressive) through the beam cross

section is linear.

The bending stress is directly proportional tothe distance of the beam fibres fro:

the neutral axis. Therefore, maximum stressoccurs on the outermost fibres oft!

beam.

The neutral axis coincides with the centroidalaxis.

4.3 SECTION MODULUS


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Those sections which have higher value of

section modulus will develop lower val

of bending stress for the same value of bendingmoment applied and same area of cros

mefit

section.

4.4 CALCULATION OF BENDING STRESSES

4.4.1 Circular Section (Fig. 4.3 (a)


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4.4.5 Channel Section (Fig. 4.3 (e))


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Example 4.1 The cross-section of a T-

section beam is shown in Fig. 4.4 (a). It issubjected to a bending moment of 15 kNm.Plot the distribution of stress due tobending in the beam.


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The stress distribution is shown in Fig. 4.4 (b).

Example 4.2 The cross-section of an I-beamis shown in Fig. 4.5. The bending moment atthe section is 20 kN.m. Plot the distribution

of bending stress in the beam.


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The stress distribution is shown in Fig. 4.5(b).

Example 4.3 A cast iron channel carries

water as shown in Fig. 4.6. It is supportedat two points 12m apart. Density of water is


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1000 kg/m3 and that of cast iron 7000kg/m3. Calculate the depth of water in thechannel if the tensile and compressive

stresses in bending are not to exceed 20MPa and 50 MPa respectively.

Solution.


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Example 4.4 A cantilever beam with

triangular cross-section as shown in Fig. 4.7carries a uniformly distributed load ofintensity 2 kN/m over its entire length. Ifthe depth h = 2b, determine the minimumdimensions allowing tensile or compressivestress to be 25 MPa.


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4.5 COMPOSITE OR FLITCHED BEAMS

A composite or flitched beam is made of two ormore different materials and rigidly connectedtogether. Such beams are used when onematerial, if used alone, requires a large cross-sectional area.


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Consider a beam made of timber and steel. Let tand s be the subscripts for timber and steelrespectively. At the common surface,


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Example 4.5 A timber beam 20 cm wide and30 cm deep is reinforced by a steel plate20cm wide, 2 cm thick and bolted to itsbottom surface. Calculate the moment ofresistance if safe stresses in timber and

steel are 10 MPa and 150 MPa respectively.Take E5 20 E.

Solution. The composite beam is shown in Fig.4.8 (a)


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Example 4.6 A flitched beam is made of twotimber joists each 10 cm wide by 30 cm

deep with a steel plate 2 cm thick and 25

cm deep placed symmetrically and firmly


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fixed to them. Calculate the moment ofresistance of beam when the maximumstress in timber reaches 9 MPa. Also find

the maximum uniformly distributed loadwhich the beam can carry on a simplySupported span of 6m. E = 200 GPa and E= 10 GPa.

Solution. Stress rn timber at 12.5 cm from NA(Fig. 4.9)


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4.6 HORIZONTAL SHEAR STRESS IN BEAMS

The vertical shear force in a beam subjected tobending gives rise to horizontal shear force dueto its complementary nature. Consider a beam of

uniform cross-section su1ectedto bendingmoment M at the section AC and M + dM at BD,the two sections being dx apart, as shown in Fig.4.10 (a) Let a be the bending stress at E due toM on an elementary area of width b andthickness dy, as shown in Fig. 4.10 (b), and a1at F on the corresponding area of the cross-section. Then


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cross-shaded area, about the neutral axis.


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If A = cross-shaded area above the sectionconsidered.

and = distance of centroid of area A from theneutral axis.

1. Rectangular cross-section

Consider a beam of rectangular cross-section ofwidth b and depth h as shown in Fig. 4.11 (a).Let F be the shear force of the sectionconsidered. Consider an elementary strip of thebeam at a distance y from the neutral axis and of

depth dy.


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This represents the equation of a parabola. Thusthe horizontal shear stress distribution in a beamof rectangular cross-section is parabolic in

nature, as shown in Fig. 4.11 (b).


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3. Triangular cross-section

Consider a triangular cross-section beam of based and height Ii. Consider an

elementary strip of width b and depth dy at adistance y from the apex 0 of the triangle as

shown in Fig. 4.13(a). Then


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5. Rectangular section with a coaxialHole Refer to Fig. 4.15.


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Example 4.7. A 150mm x 60mm I-beam issubjected to a shearing force of 15 kN.

Determine the distribution of horizontalshear stress in the beam. Find what

percentage of the total shear force iscarried by the web. Web thickness = 4mm,flange thickness = 6mm.

Solution. Refer to Fig. 4.17.

Consider (Fig. 4.18


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(Fig. 4.18).

Example 4.8 Draw the horizontal shear

stress distribution in a T beam shown in


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Fig. 4.19 and subjected to a shear force of10 kN.

Solution:


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Area Moment of Inertia, also known as Second Moment of Inertia -I, is a property ofshape that is used to predict deflection and stress in beams.

Imperial units

inches4

Metric units

mm4 cm4

m4

Converting between units

1 cm4 = 10-8 m = 104 mm 1 in4 = 4.16x105 mm4 = 41.6 cm4
http://www.engineeringtoolbox.com/beam-stress-deflection-d_1312.htmlhttp://www.engineeringtoolbox.com/beam-stress-deflection-d_1312.html


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1 cm3 = 10-6m = 103 mm

Definition of Second Moment of Inertia

Ix = y2 dA (1)

where

Ix = second moment of inertia

y = the perpendicular distance from axis x to the element dA

dA = an elemental area

Area Moment of Inertia for some common Cross Sections

Solid Square Cross Section

Ix = b4 / 12 (2)

where

b = side

Iy = b4 / 12 (2b)

Solid Rectangular Cross Section


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Ix = b h3

/ 12 (3)where

b = width

h = height

Iy = b3 h / 12 (3b)

Solid Circular Cross Section

Ix = r4 / 4

= d4 / 64 (4)

where


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r = radius

d = diameter

Iy = r4 / 4

= d4 / 64 (4b)

Hollow Cylindrical Cross Section

Ix = (do4 - di4) / 64 (5)

where

do = cylinder outside diameter

di = cylinder inside diameter

Iy = (do4 - di4) / 64 (5b)

deflection in the beam which

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