tutorial 2 2.pdf

D.J.Dunn Material supplied from www.freestudy.co.uk 1

UNIT 62: STRENGTHS OF MATERIALS

Unit code: K/601/1409

QCF level: 5

Credit value: 15

OUTCOME 2 - TUTORIAL 2

STRUTS

2 Be able to determine the behavioural characteristics of loaded beams, columns and struts

Simply supported beams: use of Macaulays method to determine the support reactions, slope and deflection due to bending in cantilevers and simply supported beams with combined concentrated and

uniformly distributed loads

Reinforced concrete beams: theoretical assumptions; distribution of stress due to bending

Columns: stress due to asymmetrical bending; middle third rule for rectangular section columns and

walls; middle quarter rule for circular section columns

Struts: end fixings; effective length; least radius of gyration of section; slenderness ratio; Euler and

Rankine-Gordon formulae for determination of critical load

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

CONTENTS

1. CLASSIFICATION OF COMPRESSION MEMBERS

1.1 Introduction

1.2 Slenderness Ratio

1.3 Radius of Gyration k

2. STRUTS

2.1 Critical Load

2.2 Euler's Theory for Collapse

2.3 Validity Limit of Euler's Theory

3. DEFLECTION OF STRUTS

It is assumed that students doing this tutorial already familiar with the concepts of

second moments of area, bending stress, bending moments and deflection of

beams.


1. CLASSIFICATION OF COMPRESSION MEMBERS

1.1 INTRODUCTION

Compression members are loaded in the direction of their length and not transversely (beams).

They may be long relative to their cross section in which case they are STRUTS or short in which

case they are COLUMNS. There is obviously an in between case called intermediate members. In

this module you are required to study columns but you do need to appreciate the difference.

STRUTS fail by bending and buckling so they very limited as a structural element.

COLUMNS fail in compression. In civil engineering they are often made of brittle material which

is strong in compression such as cast iron, stone and concrete. These materials are weak in tension

so it is important to ensure that bending does not produce tensile stresses in them. If the

compressive stress is too big, they fail by crumbling and cracking. Structural steel is also used as

columns and the cross section properties of standard rolled steel columns (RSC) are found in

British Standard BS4 part 1.

Figure 1

1.2 SLENDERNESS RATIO

One way of deciding whether a compression member is long relative to its cross section is the use

of slenderness ratio. This is defined as:

k

LS.R.

L is the effective length and k is the radius of gyration for the cross sectional area. A strut is defined

as having a slenderness ratio is greater than 120 when made of steel and 80 when made of

aluminium.

1.3 RADIUS OF GYRATION k

The radius of gyration is defined as A

Ik

I is the 2nd moment of area and A is the cross sectional area.

These properties may be looked up in tables for standard RSC but must be calculated for other

sections.


WORKED EXAMPLE No.1

Derive formulae for the radius of gyration of a circle diameter D and a rectangle width B and

depth D.

Circle 4

D

D 64

D 4k

4

D A

64

DI

2

424

Rectangle 1212BD

BDk BDA

12

BDI

33 D

WORKED EXAMPLE No.2

Calculate the slenderness ratio of a strut made from a hollow tube 20 mm outside diameter and

16 mm inside diameter and 1.2 metres long.

For a hollow tube the second moment of area is

5.187mm 6.4

mm 1200

k

LS.R.

mm 6.4113.1

4637

A

Ik

mm 113.14

1620

4

dDA

4637mm64

1620

64

dDI

22222

44444

SELF ASSESSMENT EXERCISE No.1

1. Find the radius of gyration and the slenderness ratio of a strut made from 5 m length of hollow

tube 50 mm outer diameter and 40 mm inner diameter.

(Ans 16 mm and 312.3)


2. STRUTS

2.1 CRITICAL LOAD

The force applied to a strut is in an axial direction (the x direction) and not transversely (y

direction) as it is for beams. Consider a long thin strut resting against a solid surface at one end and

with a screw device at the other as shown. The distance from the end is x and the deflection is y.

When the screw is tightened, the strut is forced to deflect sideways. The more the screw is turned

the more the strut deflects. If the strut bends as shown, there must be a bending moment and a

bending stress in the material. The applied bending moment is Fy. The force applied by the screw

and the distance y will increase as the bending moment increases. At some point it will be found

that the screw can be turned with no further increase in the force. This can be explained because the

increase in deflection alone is sufficient to produce the required increase in the bending moment.

The strut will go on bending until it fails (usually by exceeding the yield stress in the material and

leaving it permanently bent). When this point is reached the strut has failed and the critical force Fc

has been reached.

Figure 2

Now consider a vertical strut with weights causing the compression.

If the weights are less than the critical force Fc the strut is unlikely to

deflect very much as it is no longer forced to do so. However when

the critical value is reached, the strut collapses suddenly and fails as

there is nothing to stop it. This might be explained as follows.

If the load is critical, the strut will start to deflect. As the distance y

increases so will the bending moment. This in turn makes it deflects

even further. This is a run away or unstable condition and the strut

keeps on bending and fails. A strut is an unstable structure as

collapse is sudden and without warning.

Figure 3


2.2 EULER'S THEORY FOR COLLAPSE

No strut can be perfectly straight and a force applied as shown will make it bend slightly when an

axial compression load is applied. The direction in which it moves is random so lets sketch it so

that it bows upwards on the diagram. Note that the force is not a transverse force in the y direction

but an axial force applied in the x direction. Consider any distance x from the end. The strut has

deflected a distance y. The bending moment at this point is M = F y. This will be a maximum at the

point where the deflection is greatest so let this maximum value of y be denoted ym.

Figure 4

The applied bending moment is M = -Fy (minus because it hogs)

If the strut does not collapse, the internal bending moment must balance the applied moment and

this is given by bending theory as

EI

F expression therepresents c andn integratio ofconstant are B andA

Bsin(cx)Acos(cx)y

solution. standard aith equation w aldifferentiorder second a is This 0EI

Fy

dx

yd

RearrangeFy dx

ydEI

hogs)it because (minusFy dx

ydEIM

2

2

2

2

2

2

A and B are solved from boundary conditions. We know that for the case illustrated the deflection

is zero at the ends and a maximum at the middle.

x = 0 y = 0 and at x = L y = 0

First substitute y = 0 and x = 0 into the solution.

0 = Acos(0) + B(sin 0) = A(1) + B(0) hence A = 0

Next substitute y = 0 and x = L

0 = Acos(cL) + Bsin (cL) = 0 cos(cL) + B sin(cL) from which B sin (cL) = 0

If B is zero the solution is always zero and this clearly is not the case. It follows that sin (cL)= 0

and this occurs when cL = 0, 180o, 360o, and so on.

In radians this is cL= , 2, 3 and so on. In general cL = n where n is an integer.


We may state that

evaluated. bemay which Bsin(cL)y is deflection ingcorrespond theand L

EInF

nEI

FL

ncL

2

22

If the strut does not have a symmetrical cross section, it will buckle about the axis with least

resistance (smallest value of I). For a rectangular section B must always be the larger of the two

dimensions.

Figure 5

This is the formula usually given in exams and the above derivation should be practised prior to the

exam. n is called the mode and its meaning is very real. A node is any point where the strut does not

deflect. If the strut is restrained at any point (e.g. guy ropes on a mast) that point will be a node. The

diagram shows what happens when the restraints are placed at the middle (n = 2) and at equal

distances of 1/3 of the length (n = 3). n = 1 is the fundamental mode.

Figure 6

This derivation is due to Euler and the value of F is called Euler's critical load.

There are three other modes of importance which are governed by the way the ends are constrained.


Figure 7 Figure 8 Figure 9

Figure 7 shows the half mode with n = 0.5 occurs when one end is held rigidly and the other is

unrestrained.

Figure 8 shows another double mode with n = 2. This occurs when both ends are held rigidly.

Figure 9 shows an unusual mode with n = 1.43. This occurs when one end is held rigidly but the

other end is pinned (allowed to rotate) but not allowed to move sideways.

WORKED EXAMPLE No.3

A strut is 2 m long and has a rectangular cross section 30 mm x 20 mm. The bottom is built

into a ground socket and the top is completely unrestrained. Given E = 200 GPa calculate the

buckling load.

SOLUTION

F = n2

2EI/L

2

This case is as shown in fig. 7 with n = 0.5

I = BD3/12 = 2 x 10

-8 m

4

F = 0.52

2x200 x 10

9x 2 x 10

-8/2

2

F = 2470 N

WORKED EXAMPLE No.4

Repeat the previous problem but with the strut is pinned at the top and bottom and not allowed

to move sideways.

SOLUTION

F = n2

2EI/L

2

This case is as shown in fig. 6 with n = 1

I = BD3/12 = 2 x 10

-8 m

4

F = 12

2x200 x 10

9x 2 x 10

-8/2

2

F = 9870 N.


SELF ASSESSMENT EXERCISE No. 2

1. A steel strut is 0.15 m diameter and 12 m long. It is built in rigidly at the bottom but

completely unrestrained at the top.

Calculate the buckling load taking E = 205 GPa.

(Ans. 89.4 kN).

2. A steel strut has a solid circular cross section and is 8 m long. It is pinned at the top and bottom

but unable to move laterally at the ends. The strut collapses under a load of 200 kN. Taking E =

205 GPa calculate the diameter of the strut.

(Ans 106.5 mm).

3. A shaft is made from alloy tubing 50 mm outer diameter and 30 mm inner diameter. The shaft

is placed between bearings 3 m apart so that the ends are constrained to remain horizontal. The

shaft also has to take a horizontal axial load. Taking E = 120 GPa determine the maximum

axial load before buckling occurs.

(Ans. 140 kN).

4. A strut is 0.2 m diameter and 15 m long. It is pinned at both ends. Calculate Euler's critical

load.

(Ans. 706.25 kN)


2.3 VALIDITY LIMIT OF EULER'S THEORY

Euler's theory is inaccurate when the slenderness ratio is small. If the strut is very thin, then the

material will simply crush under the axial compression. The slenderness ratio limit depends upon

the material but generally if the ratio is less than 120 for steel or less than 80 for aluminium and its

alloys, the crushing becomes important and failure will occur at loads smaller than those predicted

by Euler.

3. DEFLECTION OF STRUTS

It is of interest to know the deflection of a strut at loads less than the buckling value. This can be

very complicated work but we do not have to go into full details.

When the strut buckles, it fails because it reaches the elastic limit of the material in compression.

The strut is put into compression by the load and the direct compressive stress is

D= F/A

Bending stress is also induced in the strut which tends to be tensile on one side and compressive on

the other. This is given by :

B= M/I

Figure 10

is the distance from the neutral axis to the extreme edge in compression (this is denoted y in beam

stress problems but y is unsuited to this case). M is the bending moment Fy. The total compressive

stress is hence

= F/A + M/I

At the point of collapse this is the elastic limit in compression c and the deflection is ym.

F

I

A

Fy

I

Fy

A

F

cm

mc


WORKED EXAMPLE No.5

A strut is made is made from 16 mm diameter steel bar. The buckling load is 2400 N. The

elastic limit in compression is 320 MPa. Calculate the deflection just prior to collapse.

SOLUTION

I = D4/64 = 3.217 x 10

-9 m4.

A =D2/4 = 201 x 10

-6 m

2.

c =320 x 106 N/m

2.

= D/2 = 0.008 m

mm 51.6or m 0.0516 0.008 x 2400

10 x 3.217 x

10 x 201

240010 x 320y

-9

6-

6

m


SELF ASSESSMENT EXERCISE No. 3

1.a A uniform slender elastic column of length L is pin jointed at each end and subjected to an

axial compression load P.

Show that the Euler crippling load occurs when P = 2EI/L2 where I is the relevant second

moment of area of the column and E is the modulus of elasticity of the material. State any

assumptions made.

b A straight steel rod 0.5 m long and 0.01 m diameters loaded axially until it buckles. Assuming

that the ends of the rod are pin jointed, determine the Euler crippling load. Assume E = 206

GPa.

(Ans. 3.992 kN)

2.a The Euler buckling load P for a slender strut of length L and second moment of area I, pin

jointed at each end , is given by P = 2EI/L2

E is the modulus of elasticity of the material. Using this expression without proof, obtain the

formula for the Euler buckling load when the strut is

i. fixed (built in) at each end.

ii. fixed at one end and pin jointed at the other.

b) Fig. 11 shows a vertical pole 6 m long, pinned at the lower end and supported by a wire at the

upper end. The pole consists of a tube 50 mm outside diameter and 40 mm inside diameter and

the wire has an effective diameter of 6 mm. What is the maximum load P that this system can

withstand before failure occurs?

For steel assume that the modulus of elasticity E is 206 GPa and for the wire assume that the

ultimate stress is 480 MPa.

Figure 11

(Answer, the rope breaks before the pole buckles so the maximum value of P is 9.59 kN)


3.a A uniform slender strut of length L which is clamped at one end and free at the other is

subjected to an axial compression load P as shown in fig.12. Show that according to EULER'S

theory, the strut will buckle when P= (/2L)2EI where I is the minimum second moment of

area of the strut and E is the modulus of elasticity for the material.

b A straight steel rod 9 mm diameter is rigidly built into a foundation, the free end protruding 0.5

m normal to the foundation. An axial load is applied to the free end of the rod which deflects as

shown in fig.12.

Determine the following.

i. Eulers buckling load. (636 N)

ii. The deflection of the free end of the rod when the total compressive stress reaches the

elastic limit. (32.6 mm)

For steel assume E = 200GPa and the stress at the elastic limit is 300MPa.

Figure 12

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