Mechanics of Materials

 6.3. Bending deformation of a straight member6.4. The flexural formula6.6. Composite beams   

Bending

Readings

Mechanics of Materials

Bending

Mechanics of Materials

Bending Deformation: Straight Member

Cross sections of the beam remain plane and perpendicular to longitudinal axis

Deformation of the cross section will be neglected

Deformation of a Beam: Bending Moment

Neutral axis

Do not change lengthForms a curvaturePasses through the Neutral Surface

Axis of symmetry

Longitudinal axis

Neutral surface

Bottom fibers : Elongates (Tension)Top fibers : Contracts (Compression)

Mechanics of Materials

Normal strain

Strain along cd,

Longitudinal normal strain,

Before deformation cd = pqAfter deformation pq = ρ dθ = cd c’d’ = (ρ-y) dθ

ρ is the radius of curvaturedθ angle between the cross sectional sides

How the BM strains the material

Varies linearly with y from neutral axis

Mechanics of Materials

Normal strain distribution

Maximum normal strain,                                               

Sign Convention

Fibers located above neutral axis (+y): contracts (-ε) Fibers located below neutral axis (-y): elongates (+ε)

Where c is the distance from the neutral axis to the farthest fiber

For positive bending moment:

max

maxc

y

maxc

y

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These strain components cause the cross sectional dimensions to become smaller below the neutral axis and larger above the neutral axis.

Normal strain

Strain in other directions

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Normal Stress distributionFor a linear elastic Material

Distribution of Normal Strain

Distribution of Normal Stress

E

Hooke’s law Normal Strain

y

y

E

Normal Stress

Sign Convention

For positive bending moment:

Fibers located above neutral axis (+y): contracts (-ε) Fibers located below neutral axis (-y): elongates (+ε)

max

c

y

The resultant of the normal stresses consists of two stress resultants1. The force acting in the x direction2. Bending couple acting about z axis

Stress resultants

Mechanics of Materials

Neutral Axis

Condition 1: The resultant force in the x direction is equal to zero

For an elemental area dA (at distance y from the neutral axis)The force, dF = σdA

AA

dAdF 0Total force

0max A

ydAc

A

dAc

y0max

0A

ydA

The first moment of the area of the cross section about the neutral axis must be zero. This condition can only be satisfied if the neutral axis is also the horizontal centroidal axisConsequently once the centroid of the cross sectional area is determined, that is the location of the neutral axis.

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Condition 2: The resultant moment of normal stresses acting over the cross section is equal to the internal resultant moment.

Flexural Formula

The moment of dF about the neutral axis

For an elemental area, dA located at a distance y from the neutral axis subjected to an elemental force, dF =σdA.

dAyydFdM

A

dAyM

A

dAc

yyM max

A

dAyc

M 2max

is the moment of inertia of the cross sectional area.A

dAy2

Ic

M max I

Mcmax

σmax : The maximum normal stress in the top/bottom most fiber of the beamM : Resultant internal momentI : Moment of inertia of the cross sectional area with respect to the neutral axisc : Perpendicular distance from the neutral axis where σmax acts

I

My

yc

max

Normal Stress Distributions

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If the cross sectional area is not symmetric after deformation, Flexural formula is not valid Flexural formula is valid only for prismatic membersMaximum normal stress/strain occurs at outermost fiber and linearly varies from the neutral axisFlexural formula is valid only for homogeneous material For linear elastic-material the neutral axis passes through the centroid of the cross sectional area which is based on the fact that the resultant normal force acting on the cross section must be zero

Flexural FormulaPoints to Ponder

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Example Problems

A member having the dimensions shown is to be used to resist an internal bending moment of M = 2kN·m. Determine the maximum stress in the member if the moment is applied (a) About the z axis(b) About the y axis. Sketch the stress distribution for each case.

Example Problem 1 (6.43)

Mechanics of Materials

Example Problems

Example Problem 1 (6.54 & 6.55)

The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment M = 8 kN·m,

(i). Determine the bending stress acting at points A and B, and show the results acting on volume elements located at these points.(ii). Determine the maximum bending stress in the beam, and sketch a three-dimensional view of the stress distribution acting over the entire cross-sectional area

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Composite Beam

Beams constructed of two or more different materials are referred as composite beams Example:

Concrete beams reinforced with steel rods Wooden Blocks reinforced with steel plates

Assumptions•The total cross sectional are remain plan after deformation which is valid for pure bending regardless of the nature of material

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For a linearly elastic material

Composite Beam

For material 1 : σ1 = E1 εFor material 2 : σ2 = E2 ε

Normal strain,

Normal Strains•vary linearly, from zero at neutral axis to maximum at farthest fiber

Consider a composite beam made of two materials 1 and 2 of height h and width b. Bending moment, M is applied with respect to z axis. Assume, material 1 is stiffer than material 2, hence E1 > E2.

The normal stress acting on the cross section

EHooke’s law

y

Mechanics of Materials

How to transform

Composite Beam

Since E1 > E2When transforming material 1 into less stiffer material 2, the area of the transformed portion beam has to be with bigger area with keeping the height h constant to maintain the linear strain distribution

The material 1 (top portion) is be transformed into an equivalent material 2, thus the entire beam will be made of material 2.

What is Transformed-section Method

The method consists of Transforming the cross section of a composite beam into an equivalent single material

beam (transformed section) with keeping h constant The flexural formula is then applied to the transformed section of beam Finally, the stresses in the transformed section are converted into the actual beam

How to apply Flexural Formula to composite beamFlexural formula is valid only for homogeneous material. Composite beam has to be transformed into a single material beam

Mechanics of Materials

Composite Beam

For the transformed section: To increase the area, width, b is increased to nb1 keeping the height, h constant

To produce same moment, the force acting on both the beam must be same.

Hence the area of the transformed beam: Force acting on area dA:

For the composite beam:Consider an area at distance y from the neutral axis Force acting on area, dA : dydzEdAdF )( 11

dyndzEdAdF )( 22

dydzdA

)(ndzdydA

dyndzEdydzE )()( 21 2

1

E

En

The n is called the transformation factor

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The stress found on the transformed section has to be multiplied by the transformation factor n to get the stress value for the original material.

Composite Beam

Once the beam has been transformed into a single material then the normal stress distribution over the transformed cross section will be linear. Consequently the neutral axis, moment Inertia for the transformed area can be determined and flexure formula can be applied in the usual manner to determine the stress at each point on the transformed beam.

dAdAdF 21

dzndydydz 21

21 n

Since the area of the transformed material (dy n dz) is n times the area of the actual material (dy dz). That is

Why?

Mechanics of Materials

Example Problem 1 (6.120) 

The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). If the height h = 40 mm, Determine, the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is, σ al = 128 MPa and for the brass, σ br = 35 MPa?

Example Problems

Mechanics of Materials

A wood beam is reinforced with steel straps at its top and bottom as shown. Determine the maximum bending stress developed in the wood and steel if the beam is subjected to a bending moment of M = 5kN·m. Sketch the stress distribution acting over the cross section. Take Ew = 11 GPa, Est = 200 GPa.

Problem No: 6-121