Lecture Strength of Materials I: Chapter 7 - PhD. Tran Minh Tu

Lecture Strength of Materials I - Chapter 7: Bending. The following will be discussed in this chapter: Introduction, bending stress, shearing stress in bending, strength condition, sample problems, deflections of beam, statically indeterminate beams.

STRENGTH OF MATERIALS

CHAPTER

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BENDING

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7.1 Introduction

In previous charters, we considered the stresses in the bars caused

by axial loading and torsion Here we introduce the third fundamentalloading: bending When deriving the relationship between the bendingmoment and the stresses causes, we find it again necessary to makecertain simplifying assumptions

We use the same steps in the analysis of bending that we used fortorsion in chapter 6

7.1 Introduction

Classification of Beam Supports

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7.1 Introduction

 Limitation

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7.1 Introduction

Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane

7.2 Bending stress

 Simplifying assumptions

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7.2 Bending stress

The positive bending moment causes the

material within the bottom portion of the beam

to stretch and the material within the top portion

to compress Consequently, between these two

regions there must be a surface, called the

neutral surface, in which longitudinal fibers of

the material will not undergo a change in

length

Neutral axis

Due to bending moment Mx caused

by the applied loading, the

cross-section rotate relatively to each other

The Normal strain of the longitudinal

fiber cd that lies distance y below the

Consider a segment of the beam

bounded by two cross-sections that

are separated by the infinitesimal

distance dz

dA 

x

y

z K

MxBecause of the loads applied in the

plane yOz, thus: Nz=My=0 and Mx≠0

xyd A  I 



x – neutral axis (the neutral axispasses through the centroid C of thecross-section)

y - axis – the axis of symmetry ofthe cross-section

7.2 Bending stress

Mx>0: stretch top portion

Mx<0: compress top portion

y

z x

dA 

x

y

z K





EIx – stiffness of beam

Mx – internal bending moment

 – radius of neutral longitudinal fiber

x z

x

M

y I

 Flexure formula – section modulus

7.2 Bending stress

• Stress distribution

- Stresses vary linearly with

the distance y from neutral axis

• Maximum stresses at a cross-section

I W

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7.2 Bending stress

7.2 Bending stress

Properties of American Standard Shapes

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7.2 Bending stress

7.2 Bending stress

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7.2 Bending stress

7.3 Shearing stress in Bending

In general, a beam will support both shear and moment – non-uniformbending The shear force Qy is the results of a transverse shear-stressdistribution that acts over the beam’s cross-section

As a results of the shearing stress, the shear

strain will be developed and (these will) tend to

distort the cross-section This non-uniform

shear-strain will cause the cross-section warp

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7.3 Shearing stress in Bending

 The Shear formula

When a beam is subjected to loads that

produce both bending moment and shear force,

then both normal and shear stresses are

developed in the beam

Consider a beam of rectangular cross section

subjected to a positive shear force We assume

• The Shear stresses acting on the

cross-section are paralell to the shear force, that is,

parallel to the vertical sides of the cross-section

• The Shear stresses are uniformly distributed

across the width of the beam, although they

may vary along the height

7.3 Shearing stress in Bending

Let’s consider the horizontal force equilibrium of a portion (of theelement) taken from the beam in fig (a) A FBD of this element is shown infig (b) This distribution is caused by the bending moment M an M+dM

We have excluded the effect of V and V+dV and the transverse loading onthe FBD, because they are not involved in a horizontal force summation

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7.3 Shearing stress in Bending

Now consider the shaded top portion of the

element that has been sectioned at y’ from

the neutral axis This segment has the width

of bs, and the cross-sectional area of As The

Longitudinal shear  stress acts over the

bottom face of the segment (constant across

the width bs of the bottom face)

7.3 Shearing stress in Bending

7.3 Shearing stress in Bending

• Shearing stress distributions at

the rectangular cross-section:

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7.3 Shearing stress in Bending

• Shearing stress distributions at

the wide-flanged cross-section:

2

1 2

I d



2 1

7.3 Shearing stress in Bending

Consider a rectangular cross-section of the beam Normal and shearingstresses distribution on the cross-section are shown in figure.

K, N – uniaxial stressC- pure shearing stressB- special plane stress

7.4 Strenght condition

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7.4 Strenght condition

• Bending stress criterion

Cross-section need to control: cross-section have maximum bending

moment (ductile: absolute magnitude, brittle : maximum positive annegative bending momet)

• Shear stress criterion

Cross-section need to control : cross-section have maximum shear

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Problem 7.5.1:

For the timber beam and loading shown

• Draw the shear and bend-moment diagrams

• Determine the maximum normal stress due to bending

• Check strength condition of the beam, known [  ]=1,5 kN/cm2

7.5 Sample Problem

• Maximum bending moment

0.080 m 0.250 m833.33 10 m

50 10 N m833.33 10 m

x

x

W b h

M W

7.5 Sample Problem

A simply supported steel beam is

to carry the distributed and

concentrated loads shown

Knowing that the allowable normal

stress for the grade of steel to be

used is 160 MPa, select the

• Determine the minimumacceptable beam section modulus.Choose the best standard sectionwhich meets this criteria

Problem 7.5.2:

  5 m 60 kN 1.5 m  50 kN 4 m 0 58.0 kN

58.0 kN 60 kN 50 kN=0 52.0 kN

5866

Mmax

MkNm

7.5 Sample Problem

Support reactions:

• Bending stress criterion:

 max

67.6 kN m 16kN/cm

M W

44

W250

549 7

38

W310

474 9

32

W360

637 38.8

9 32

360 

W

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7.6 Deflection of Beam

Because the design of beams is frequently governed by rigidity ratherthan strength, the computation of deflection is an integral component ofbeam analysis For example, building codes specify limits on deflections

as well as stresses

7.4.1 Differential equation of elastic curve

• The deformed axis of the beam is called

elastic curve

• Consider a cantilever beam with concentrated

load acting downward at the free end Chosen

coordinates are shown in fig

7.6 Deflection of Beam

Assumption

- Displacements and slopes are very small

- The stresses are below the elastic limit

• The deflection y is the displacement in the y

direction of any point on the axis of the beam

Deflection is positive when downward

• The angle of slope  (angle of rotation) is

angle between cross-section before and after

deformation

• From geometry can obtain

 (z) = tg  = y’(z)

7.6 Deflection of Beam

• From equilibrium of bending beam – the curvature of neutrallongitudinal fiber (relationship between bending moment andcurvature for pure bending)

y z

y z y

• Second integration give the deflection’s function

• C an D are constants of integration to be determined from theprescribed constrains (boundary conditions)

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7.6 Deflection of Beam

Problem 7.4.1: The cantilever beam shown

in figure is subjected to a vertical load P at

its end Determine the deflection and angle

of slope at free end EI = const

SOLUTION

M   F L  z

B F

L-zL

EIxz

Substitution to diff equation of elastic curve:

• For a beam subjected to a distributed load,

• Integrating four times yields

• Constants are determined from boundary

 Direct Determination of the Elastic Curve From the Load Distribution7.6 Deflection of Beam

• Consider a beam subjected non-uniform bending consist nsegments, is numbered 1,2,…,i, i+1, , n in order from left to right.Bending rigidities of each segments are: E 1 I 1 , E 2 I 2 ,…, E n I n. Considertwo adjacent segments (i) and (i+1) Between them there is aspecial connection that deflection and slope have “jump” In thecross-section between two segments there are concentrated loadingand moment, and distributed loading also have “jump”

7.7 Initial Parameters Method

• After mathematical manipulation (Fourier expansion deflection’sfunction at z=a), using relations among bending moment, shear forceand transverse distributed load, we obtained recurring formula ofdeflection’s function (deflection (i+1)-th segment is calculated throughdeflection of i-th segment)

 Positive sign of couples, concentrated load, and distributed

load is shown in figure

 If connection between (i)-th and (i+1)-th segment is pinned,

Problem 7.8:

Use initial parameters method

to determine deflection at point

C and slope at point D of the

beam subjected by loading as

M=qa P=4qa

a a

qa 6EI

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7.7 Statically Indeterminate Beams

• Consider beam with fixed support at A and roller support at B.

• From free-body diagram, note that there are fourunknown reaction components

• Conditions for static equilibrium yield

• Also have the beam deflection equation,

which introduces two unknowns but providesthree additional equations from the boundaryconditions:

At z  0,  0 y  0 At z  L y,  0

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