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D.J.DUNN 1

Unit 2: Engineering Science

Unit code: L/601/1404

QCF Level: 4

Credit value: 15

OUTCOME 1 - TUTORIAL 1

STRESSES IN BEAMS DUE TO BENDING

1. Be able to determine the behavioural characteristics of elements of static engineering systems

Simply supported beams:determination of shear force; bending moment and stress due to bending;

radius of curvature in simply supported beams subjected to concentrated and uniformly distributed

loads; eccentric loading of columns; stress distribution; middle third rule

Beams and columns:elastic section modulus for beams; standard section tables for rolled steel beams;

selection of standard sections e.g. slenderness ratio for compression member, standard section and

allowable stress tables for rolled steel columns, selection of standard sectionsTorsion in circular shafts: theory of torsion and its assumptions e.g. determination of shear stress, shear

strain, shear modulus; distribution of shear stress and angle of twist in solid and hollow circular section

shafts

You should judge your progress by completing the self assessment exercises.

I t is assumed that students doing this tutorial are already fami li ar wi th the following work

at national l evel: -

The reaction forces in simply suppor ted beams.

Moments of f orce.

F ir st and second moments of area.

Basic stress and strain (di rect and shear)

I f you need to study this, you wil l f ind the mater ial elsewhere on the web site.

CONTENTS

1.

INTRODUCTION

2. THE BENDING FORMULA

2.1 Neutral Axis

2.2 Radius of Curvature

2.3 Relationship between Strain and Radius of Curvature

2.4 Relationship between Stress and Bending Moment

2.5 Standard Sections

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1 INTRODUCTION

A beam is a structure that is loaded laterally (sideways)

to its length. These loads produce bending and bending

is the most severe way of stressing a component.

Suppose you were given a simple rod or a ruler and

asked to break it. You would struggle to break it by

stretching it or twisting it but it would be easy to break

by bending it.

A beam may have point loads or a uniformly distributed load

(udl) such as might occur due to its weight or the weight of a

wall built along it.

We normally show the loads with simplified diagrams like this.

There are other ways of supporting beams as shown belowbut you dont need to study these.

The loads produce shear force and bending moments that vary with position along the length. You should

already know how to construct shear force and bending moment diagrams for simply supported beams.When a bending moment M is applied to a beam, one surface is compressed (negative stress) and the other

is stretched (tensile positive stress). The stress varies across the section from a maximum negative to a

maximum positive as shown. Somewhere in between there is a longitudinal layer that is not stressed

(neutral) and this layer lays on the NEUTRAL AXIS. The neutral axis is through the centroid for pure

bending and the centroid is at the middle of symmetrical sections. In addition the transverse forces produce

shearing on a given section as indicated.

The purpose of this tutorial is to enable you to calculate the stress due to the bending moment. We will

derive the following three part equation known as the bending equation.

R

E

yI

M

M is the bending moment at a given point along the length.

I is the second moment of area of the sectional area about the neutral axis. is the stress due to bending at a distance y from the neutral axis.

E is the modulus of elasticity.

R is the radius of curvature.

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WORKED EXAMPLE No. 3

A beam 4 m long rests on simple supports and carries a uniform load of 2.5 kN/m over the first 1.5 m as

shown. Calculate the reaction forces.

The total load is now 1.5 x 2.5 = 3.75 kN This acts at the middle of the length 0.75 m from the

end. Balancing moments about Rawe have

(Rbx 4) - (3.75 x 0.75) = 0 Rb= 2.8125/4 = 0.703 kN

Balancing vertical forces Rb+ Ra= 3.75 Ra= 3.75 - 0.703 = 3.05 Kn

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2. THE BENDING FORMULA

2.1 NEUTRAL AXIS

This is the axis along the length of the beam which remains unstressed, neither compressed nor stretched

when it is bent. Normally the neutral axis passes through the centroid of the cross sectional area. For simple

rectangular and circular sections, this is the axis along the centre line.

Consider that the beam is bent into an arc of a circle

through angle radians. AB is on the neutral axis

and is the same length before and after bending. The

radius of the neutral axis is R.

Remember the length of an arc i s radius x angle in

radians

2.2 RADIUS OF CURVATURE

Normally the beam does not bend into a circular arc. However, what ever shape the beam takes under thesideways loads; it will basically form a curve on an x y graph. In maths, the radius of curvature at any

point on a graph is the radius of a circle that just touches the graph and has the same tangent at that point.

2.3 RELATIONSHIP BETWEEN STRAIN AND RADIUS OF CURVATURE

The length of AB AB = R

Consider a layer of material distance y from the neutral axis as shown. This layer is stretched because it

must become longer and the material has stress and strain in it in a lengthwise direction as a result. (If y

was to the inside of the neutral axis it would be compressed and become shorter).

The radius of the layer is R + y.

The length of this layer is the line DC. DC = (R + y)

This layer is strained and strain () is defined as = change in length/original length

Substitute AB = R and DC = (R + y)

R

y

R

RyR

AB

ABDC

The modulus of Elasticity (E) relates direct stress () and direct strain () for an elastic material and the

definition is as follows.

yR

E

y

REand

R

ySubstitute

strain

stressE

It follows that stress and strain vary along the length of the beam depending on the radius of curvature.

We will now go on to show that the radius of curvature depends upon the bending moment M acting at anygiven point along the length of the beam.

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2.4 RELATIONSHIP BETWEEN STRESS AND BENDING MOMENT

Consider a beam with a consistent shape along its length. An arbitrary oval shape is shown here. Think of

the beam as being made of many thin layers of material running the length of the beam and held together by

molecular forces.

Consider one such elementary layer at a given point along the length at a distance y from the neutral axis.

When the cross section is viewed end on it appears as an elementary strip width b and thickness y.

The cross sectional area is A.

The elementary strip is a small part of the total cross sectional Area and is denoted in calculus form as A.The strip may be regarded as a thin rectangle width b and height y so A = b y

The stress on the strip is = Ey/R

If the layer shown is stretched, then there is a small force F pulling normal to the section trying to slide the

layer out of the material in a lengthwise direction. This force must be the product of the stress and the area

and is a small part of the total force acting on the section F.

AR

EyFand

R

EySubstituteAF

Consider that the whole beam s made up of many such layers. Some are being stretched and pull normal tothe section and some are compressed and push. The total force acting on the section is the sum of all these

small forces.

AR

EyFF

In the limit as y tends to zero, the number of strips to be summed tends to infinity. The small quantities y

and A become the differential coefficient dy and dA. The total force is given by the integration

dAyR

EdA

R

EyF

top

bot tom

top

bot tom

The expression dAy

top

bottom is by definition the first moment of area of the shape about the centroid.

Evaluating this expression would give zero since any first moment of area is zero about the centroid.

The centroid in this case is on the neutral axis. The areas

above and below the neutral axis are equal. Half the force

is a compressive force pushing into the diagram, and half is

tensile pulling out. They are equal and opposite so it

follows that F = 0 which is sensible since cross sections

along the length of a beam obviously are held I

equilibrium.

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2.5 STANDARD SECTIONS

At this stage, don't have to worry about how M is found, it is covered later. For simple sections the value of

I may be determined by mathematics. The good news is that for standard engineering sections, they may be

looked up in tables.

Steel and other products used in structural engineering are manufactured with standard cross sections and

sizes and in the UK they are made to British Standard BS4. In this you will find Universal Beams, Universal

Columns and much more with the sizes and sectional properties listed. You will find I sections, T

sections, U sections and more. You will find names like RSJ (Rolled Steel Joists) and RSC (Rolled Steel

Columns) which refer to the method of manufacture. A sample of the table is attached for I section beams.

The areas and second moments of area are listed in the standards and since the distance y to the edge is also

known they list a property called the ELASTIC MODULUS and this is defined asy

Iz . The stress at the

edge of the beam is then found from the equation:

Z

M

I

My .

For standard shapes the second moment of area can be calculated with the formulae shown. This is covered

in the pre-requisite tutorial on moments of area. The following formulae apply to standard shapes.

For more complex shapes such as TEE and U sections, you will need to study the pre-requisite level tutorial

in order to solve the second moment of area.

There are also many computer programmes for solving beam problems that contain the standard

information or calculate the second moment of area when the dimensions are supplied.

The Archon Engineering web site has many such programmes.

You will currently (2005) find details of steel sections at www.roymech.co.uk

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WORKED EXAMPLE No.4

A beam has a rectangular cross section 80 mm wide and 120 mm deep. It is subjected to a bending

moment of 15 kNm at a certain point along its length. It is made from metal with a modulus of

elasticity of 180 GPa. Calculate the maximum stress on the section.

SOLUTION

B = 80 mm, D = 100 mm. It follows that the value of y that gives the maximum stress is 50 mm.

Remember all quantities must be changed to metres in the final calculation.

26

6

3

464633

N/m10x112.510x6.667

0.05x10x15

I

My

yI

M

m10x6.667mm10x6.66712

100x80

12

BDI

WORKED EXAMPLE No.5

A beam has a hollow circular cross section 40 mm outer diameter and 30 mm inner diameter. It is made

from metal with a modulus of elasticity of 205 GPa. The maximum tensile stress in the beam must not

exceed 350 MPa.

Calculate the following.

(i) the maximum allowable bending moment.

(ii) the radius of curvature.

SOLUTION

D = 40 mm, d = 30 mm

I = (404- 304)/64 = 85.9 x 103mm4 or 85.9 x 10-9m4.

The maximum value of y is D/2 so y = 20 mm or 0.02 m

m11.7110x350

0.02x10x205

EyR

R

E

y

MNm1.503orNm15030.02

10x85.9x10x350

y

IM

y

I

M

6

9

9-6

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WORKED EXAMPLE No.6

A beam is made from a universal column with an I section to BS4. The size of the beam is 356 x 127

x 39. The modulus of elasticity of 205 GPa. The maximum tensile stress in the beam must not exceed

350 MPa.

Calculate the following.

(i) The maximum allowable bending moment.

(ii) The radius of curvature.

SOLUTION

It is normal to arrange the I section so that it bends about the x-x axis.

From the table (at the end of the tutorial) the elastic modulus z is 576 cm3

(576 x 10-9

m3)

The depth of the section is 353.4 mm so y = 176.7 mm

m103.410x350

0.1767x10x205

EyR

R

E

y

Nm20110x576x10x350zM

z

M

6

9

9-6

SELF ASSESSMENT EXERCISE No.1

1. A beam has a bending moment (M) of 3 kNm applied to a section with a second moment of area (I) of

5 x 10-3m4.

The modulus of elasticity for the material (E) is 200 x 109N/m2.

Calculate the radius of curvature. (Answer 333.3 km).

2. The beam is Q1 has a distance from the neutral axis to the edge in tension of 60 mm. Calculate the

stress on the edge. (Answer 36 kPa).

3. A beam under test has a measured radius of curvature of 300 m. The bending moment applied to it is 8

Nm. The second moment of area is 8000 mm4. Calculate the modulus of elasticity for the material.

(Answer 300 GPa).

4. An I sectionuniversal beam made to BS4 had dimensions 610 x 305 x 238. Assuming the modulus of

elasticity is 200 GPa, calculate the stress and radius of curvature when a bending moment of 500 kNm

is applied about the x axis. (Answer 75.9 MPa and 838 m)

5. A beam must withstand a bending moment of 360 Nm. If the maximum stress must not exceed 250

MPa, determine the elastic modulus z and select an appropriate I section from the table for BS3

Universal Beams. (Answer z = 1440 cm3 so select 457 x 152 x 82)

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