Calculating stress and strain on Loading Systems.

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MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : LOADED BEAMS AND CYLINDERS

LESSON 1 : LOADED BEAMS

MP - 2 - 1

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Teesside University

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

This lesson extends the material covered in the previous lessons to include the

analyses of bent or deflected beams under load. From the study of Static

Engineering Systems in Engineering Science, you have already gained the

knowledge about simply supported beams. A quick revision may be necessary

to help you recall this and smooth the way for further study on loaded beams.

We are also going to see how we can manipulate the bending moment equation

to estimate the deflection and slope of beams subjected to simple loadings.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

After studying this lesson, you should be able to:

• recognise the relationship between bending moment, slope and

deflection for a loaded beam

• determine slope and deflection along loaded beams.

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________________________________________________________________________________________

REVISION ON BEAMS ________________________________________________________________________________________

A beam is a structural member that is designed to resist loads applied

transversely at various points along it, subjecting it to bending. Normally, we

consider beams as uniform, i.e. they have a constant cross-sectional area. In

this module, two types of loads are considered (shown in FIGURE 1):

(i) concentrated loads, acting on some points of the beam

(ii) uniformly distributed loads (UDLs) along all or part of the beam's length.

FIG. 1 Types of loads acting on beams: (a) concentrated loads;

(b) uniformly distributed loads (UDLs)

(a)

(b)

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Beams may be supported across a span in different ways, as shown in

FIGURE 2.

(a) A simply supported beam or a simple beam is one resting in a horizontal

position on two supports. It may be subjected to vertical forces but no

resisting moment.

(b) A cantilever is a beam built in or fixed at one end with the other end free

to move. When loads are applied to the cantilever a reaction and a

resisting moment occur at the fixed end.

resisting moments at both ends.

FIG. 2 Beams with different supports: (a) simply supported beam,

(b) cantilever beam, (c) built in beam

(a)

(b)

(c)

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When a beam is subjected to transverse loads it will bend and it will also tend

to shear. The bent beam experiences tension on one side and compression on

the other side, as shown in FIGURE 3. Shear force and bending moment are

examples of internal forces that are induced in a structure when loads are

applied to that structure. A sign convention is adopted to determine whether or

not a bending moment is positive, as shown in FIGURE 3.

(a) Positive bending moment (b) Negative bending moment

FIG. 3 The sign convention for bending moment

Engineers' theory of bending (refer to Lesson 1 - 4 in Engineering Science) is

used to describe relations between loadings, properties and deformation as

illustrated in:

where M = the bending moment at the section concerned (N m)

I = the second moment of area of section about neutral axis (m4)

σ = bending stress at any layer of material (N m–2 or Pa) y = distance from the neutral axis (m).

E = modulus of elasticity or Young's modulus for material of beam

(N m–2 or Pa)

R = radius of curvature of bent beam (m).

I y

R = =

σ ....................................... 1( )

"SAGGING" "HOGGING"

Compression

Tension

Top fibres in compression Bottom fibres in tension

Top fibres in tension Bottom fibres in compression

Compression

Tension

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________________________________________________________________________________________

INFLUENCE OF DEFLECTION ON BENDING MOMENT ________________________________________________________________________________________

We will now use an example to show how beam deflection affects bending

moment.

The two beams shown in FIGURE 4 carry a centrally concentrated load of

5 kN and are the same length. One beam is much more flexible than the other

and bends into a shape approximating to an arc of radius 15 m. Determine in

each case the maximum bending moment.

For both beams the end support reactions are 2.5 kN and the maximum

bending moment is at the centre (as you know from previous work). For the

'straight' beam, i.e. the one that has negligible deflection, the maximum

bending moment is:

But for the highly deflected beam, the maximum bending moment is slightly

less because the supported length is less.

2 5 3 7 5. . –× = kN m 1

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FIG. 4 (a) deflected beam (b) ‘straight’ beam

The supported length x is calculated as follows (see FIGURE 5).

FIG. 5

6 m

15 m Deflected beam

Reactions

5 kN

6 m

15 m

(a)

(b)

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and the maximum bending moment is now

The reduction of the bending moment is

In a linearly elastic system the deflection is directly proportional to the load

and, in the case of deflected beams, the bending moment at any section is also

directly proportional to the load. Here the bending moment is additionally

influenced by the deflection, and the greater the deflection, the more the

influence and the greater the degree of non-linearity.

Fortunately it is generally accepted that all typical engineering systems behave

linearly for stress calculation purposes.

7 5 7 45 7 5

100 0 667 . – .

. % . %× =

2 5 2 98 7 45. . . –× = kN m 1

Angle in radians 6

15 corresponding toα = = 0 4. ( 22.9 )

and sin

= × ⎛ ⎝⎜

⎞ ⎠⎟

= x

2 15

2 α

115 0 2

2 2 98

sin

therefore m

( )

= x

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________________________________________________________________________________________

SLOPE AND DEFLECTION FOR LOADED BEAMS ________________________________________________________________________________________

Often limits must be placed on the amount of deflection a beam may undergo

when it is subjected to a transverse load. Hence, we discuss various analytical

methods for determining the deflection and slope at special point on beams,

including the integration method, Macaulay's method and the principle of

superposition. These approaches enable us to establish mathematical

relationships that are important in beam design.

FIGURE 6 shows the side view of a deflected beam, which is subjected to a

transverse load. At a horizontal distance x along the x-axis and a vertical

distance y along the y-axis the slope of the deflected beam is .

FIG. 6

Deflected profile of beam

at point of maximum deflection

Slope is dy dx

d d

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From equation (1), we know that:

where EI is called the flexural stiffness, with a unit of N m2.

Since ds = R dθ, then, . In practice, the deflection of beams is very

small, which means dθ is very small. Thus, we can have dx = ds, and

. These give

and lead to

Equation (2) provides the fundamental relationship between the applied

bending moment M, the material property E, the shape of the beam cross-

section I, and the deflected profile of the beam .

How then is the relationship used?

The equation is of a type known as a differential equation and differential

equations are solved by integration.

d d

2 y

d d

2 y

EI2 = ....................................... 2( )

1 2 2R s

d y

x = =

⎛ ⎝⎜

⎞ ⎠⎟

= d d

d d

d d d

tan d d

θ θ= ≅ y

1 R s

= d d

=or 1

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Using the direct integration method, see if you can answer the following:

1. How many constants of integration will there be if our equation is integrated

through to y?

2. What would be the likely significance of being equal to zero?

3. How would the slope change with x if is a constant?

...................................................................................................................................................

________________________________________________________________________________________

d d

y x

d d

y x

d d

y x

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1. The expression

is a second order differential equation so it has to be integrated twice, once to derive

the slope, and again to derive the deflection, y.

Each integration process introduces one constant of integration. So there will be two.

2. Apart from the trivial conclusion that the beam is not loaded at all, the probability is

that the section where the slope is zero is at a maximum or a minimum deflection. In

other words, the profile of the beam is at the top of a 'mountain' or the bottom of a

'valley' where is zero.

3. If the rate of change of slope is constant then the slope is increasing or decreasing

uniformly. Put another way, if one plotted the slope against x it would be a straight

inclined line.

The task is now to integrate our expression for various beams under different

loading conditions. There are four standard cases of these:

1. Simply supported beam with a centrally concentrated load.

2. Simply supported beam with a uniformly distributed load.

3. Cantilever with a concentrated end load.

4. Cantilever with a uniformly distributed load.

We shall derive the deflection for cases 1 and 4 to show how it all works.

d d

x ,

x =

d d

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________________________________________________________________________________________

DEFLECTION BY INTEGRATION ________________________________________________________________________________________

SIMPLY SUPPORTED BEAM WITH CENTRALLY CONCENTRATED LOAD

FIG. 7

With reference to FIGURE 7, this is an outline sketch of case 1 above. The

first thing we notice is that the end support reactions must have equal values

such that:

The bending moment at x is:

Applying equation (2), the specific form of the differential equation is:

and it is ready for integration. The first integration becomes:

EI y

W x

d d

2 2 = ⎛

⎝⎜ ⎞ ⎠⎟

M P x W

x= × = ⎛ ⎝⎜

⎞ ⎠⎟

× 2

and it is positive.

P P P W

= = =1 2 2

x X

LP1 P2

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where A is the appropriate constant of integration.

Remember your maths? We now need to evaluate A. Well, we know that:

because the beam is symmetrically loaded and the deformed profile is

instantaneously horizontal at mid-span. We can use this fact by substituting

into our integrated form:

and the final slope equation becomes:

We are now ready to integrate again:

where B is the second constant of integration.

EIy W x

WL x

B= × ⎛ ⎝⎜

⎞ ⎠⎟

+ 4 3 16

3 2–

EI y

W x

WLd d

= ⎛ ⎝⎜

⎞ ⎠⎟4 16

2 2

–

0 2

2 2

= ⎛ ⎝⎜

⎞ ⎠⎟

⎛ ⎝⎜

⎞ ⎠⎟

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

+ W

i.e. AA WL

= – 2

at , d d

x L y

x = =

2 0

EI y

W x A

d d

= ⎛ ⎝⎜

⎞ ⎠⎟

× + 2 2

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We can now use another fact: if x is zero, then y is also zero because it is at

support height. Substituting into our deflection equation gives B = 0. Our

final form for the deflection equation is therefore:

and this equation is valid as long as the original bending moment equation is

valid. The expression would not be true once x is greater than L/2 because

then the bending moment equation would have to be modified by taking into

account the negative bending effect of the central load. There will be more of

this later.

We are also interested in the deflection at the centre under the load. We can

find this by substituting L/2 for x thus:

Note the negative sign indicating downwards deflection in accordance with

FIGURE 6.

In order to know how to deal with a uniformly distributed load, the formula for

the deflection, y, in case (4) above will now be derived.

EIy W

L WL L

= ⎛ ⎝⎜

⎞ ⎠⎟

⎛ ⎝⎜

⎞ ⎠⎟ ( )

× ⎛ ⎝⎜

⎞ ⎠⎟

4 2 3 16 2

–

– LL

y WL

48 or = –

EIy W x

WL x

= ⎛ ⎝⎜

⎞ ⎠⎟

× 4 3 16

3 2–

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CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD

FIG. 8

Refer to FIGURE 8. What we require is the end deflection under the uniform

load w/unit length. The method is similar to the previous case. We need to

write down the bending moment expression at XX and then integrate it twice.

We have:

Note the negative sign for bending convex upwards. Now we substitute for M

thus:

and this is our equation to integrate. The first integration gives:

EI y

wx A

d d

= +– 3

EI y

wxd d

2 = –

M w x x wx

= ×( ) × =– – 2 2

x X

w/unit length

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Find A by using the end condition: when x = L, dy/dx = 0 because there can

be no slope at the 'built in' end.

From this

and the specific form of the equation for slope becomes:

Integrate again for y:

End condition again: when x = L, y = 0, so,

and our final form for the deflection equation is:

From which we can find the end deflection because this is at x = 0.

y wL

EI = –

EIy wx wL x wL

= +– – 4 3 4

24 6 8

B wL

= – 4

EIy wx

wL x

B= + +– 4

24 6

EI y

wx wLd d

= +– 3 3

6 6

A wL

= 3

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Repeat the analysis for the other two standard cases. The final results are given to

assist you.

Case (3) end deflection,

Case (2) central deflection,

...................................................................................................................................................

________________________________________________________________________________________

y wL EI

= – 5

384

4⎛ ⎝⎜

⎞ ⎠⎟

y WL

EI = –

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As a summary, some standard cases for slope and deflection of beams are

listed in TABLE 1.

TABLE 1 Some standard cases for slope and deflection of beams

( )WEI x 3 L3x L3

6 2 3 – +–

( )wEI x 4 L3x L4

24 6 8 – +–( )

( )

Case

x2 L2

4 16 –( )

Slope Deflection (y) dy

dx max

y max

x3 L2x

12 16 –

WL2

16EI

at A & B

WL3

48EI

at C

–

Lx2 x3 L3

4 6 24 –

wL3

24EI

at A & B

5wL4

384EI

at C

–

L2 x2

2 2 –

WL2

2EI

at A

WL3

3EI

at A

–

L3 x3

6 6 –

wL3

6EI

at A

wL4

8EI

at A

–

A B

– ( )wEI Lx 3 x4 L3x

12 24 24 – –

CL/2

L/2

dx( )

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PRINCIPLE OF SUPERPOSITION

The principle of superposition states that at a given location the total deflection

is the algebraic sum of deflections due to various loadings. For example, the

cantilever shown in FIGURE 9 is subject to both a concentrated load at its free

end and a uniformly distributed load over its whole length. According to this

principle and referring to TABLE 1, the maximum deflection is at the free end

with the magnitude of ymax, given by:

FIG. 9 Cantilever beam subject to loads causing deflection

Example 1

A simply supported beam is 5 m long with a uniformly distributed load of

5 kN m–1. The second moment of area for the beam is 700 × 106 mm4. When the limitation of deflection at the middle is 1 mm, calculate the minimum

Young's modulus for the material of the beam and the slope at the supporting

point.

W w/unit length

y WL

EImax = – –

3 4

3 8

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Solution

This is the same loading situation as case (2) listed in TABLE 1. Thus, the

maximum deflection is at the middle of the beam with the magnitude of

From the question, it is known that ymax ≤ 1mm. When ymax = 1 mm, we have:

Therefore, minimum Young's modulus for material of the beam is 58.13 GPa.

According to previous study, the slope at the supporting point is the maximum

slope, which is obtained by

d d

max

⎛ ⎝⎜

⎞ ⎠⎟

= × ×

× × ×

3 3

5 10 5 24 58 13 10. 7700 10 10

6 4 10

6 12

× ×

= ×

–

–.

E wL

y I =

= × × ×

× × × ×

5 384

5 5 10 5 384 1 10 700 10

3 4

3 6

max

– ××

= ×

58 13 10

–

. Pa

58.13 GPa

y wL

E wL

max

which gives

5 384

mmax I

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Example 2

A cantilever beam of length 2.5 m has a concentrated load of 10 kN at the free

end and a uniformly distributed load of 500 N m–1 along its entire length as

shown in FIGURE 9. The material of the beam and the shape of cross-section

area are the same as those in Example 1. Calculate the slope and deflection at

the free end.

Solution

According to the principle of superposition, at the free end both the slope and

deflection are the algebraic sum of those when the beam is subject to the

concentrated load and the uniformly distributed load separately.

Using TABLE 1, it can be found that at the free end the total deflection is

and the total slope is

Substituting W = 10 kN = 10 000 N, E = 58.13 GPa = 58.13 × 109 Pa, L = 2.5 m, w = 500 N m–1, I = 700 × 106 mm4 = 700 × 106 × 10–12 m4

to the above equations, we have

= × =

= ×

1 34 10

8 0 10

–

m 1.34 mm

d d

EI = +

2 3

2 6

y WL

EI = +

3 4

3 8

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________________________________________________________________________________________

DEFLECTION BY MACAULAY'S METHOD ________________________________________________________________________________________

Macaulay's method for beam deflection is a very clever development of the

technique of direct integration. By using an artifice of integration, Macaulay

gets over the mathematical intractability of dealing with deflection on beams

carrying anything other than simple symmetrical load systems, which have

been dealt with in the previous section.

CONCENTRATED LOADS

Refer to FIGURE 10. Here we have a beam under three concentrated loads

W1, W2 and W3, supported by reactions at P and D.

FIG. 10

W1 W2

X P

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Rules of the Method

1. Measure x from one end and write down an expression for the bending

moment in the last section of the beam, enclosing all distances less than x

in special brackets like these ⟨.......⟩. Apply equation (2),

In the case shown in FIGURE 10, the bending moment for XX section is

2. Subject to the proviso that all terms for which the quantity in the pointy

brackets is negative are omitted (i.e. given a value of zero) . . . . . .

[For example, if x is less than c then the last term in Equation (3) is

omitted; likewise if x is less than b both the last two terms are omitted.]

3. . . . . . The pointy bracket terms are to be integrated as a whole – as an

entity.

[There is nothing wrong with this from the mathematical viewpoint, as can be

demonstrated by the following.

Suppose we had an expression, EI = W (x – s) and we want to integrate

it. If we multiply out the bracket first, we would obtain:

EI y

Wx Wsx A

d d

= + 2

2 –

d d

EI y

x M W x P x a W x b W x c

d d

2 1 2 3 3= = + (– – – – – – ......... ))

d d

or d d

2 2y

EI EI

x M

2 2 = =

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However, integrating it the way required by the Macaulay method we get:

Multiplying out the bracket term gives:

Comparing the two ways, B = A – Ws2/2 but, since we are going to evaluate

B from the end conditions anyway, it does not matter that it contains terms of

A, W and s as they are all constants anyway.]

Getting back to the Rule, let us integrate Equation (3) in the manner of

Macaulay.

This is the slope equation.

And integrating again,

This is the deflection equation.

EIy W x P

x a W

x b

W x c Ax B

= +

+ +

– – – –

– – .....

1 3

3 2 3

3 3

6 6 6

6 .................. 5( )

EI y

W x P x a

W x b

W x c A

d d

= +

– – – –

– – .....

1 2

2 2 2

3 2

2 2 2

2 ...................... 4( )

EI y

W x xs s B

Wx Wxs

Ws B

d d

= +( )

= + +

2 2

–

EI y

W x s B

d d

= ( )

+ –

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We can put numbers in from the real problem and find the values of A and B in

the usual way. The dodge is not to pursue the algebra: use numbers as soon as

possible.

We shall do a numerical example to show what is meant. In FIGURE 10, let

a = 2, b = 3, c = 4, L = 5, W1 = 50, W2 = 70, W3 = 60. Loads in kN,

distances in m. Find the slope at P and the deflection at W2.

By statics P = 150 (check this), and substituting into Equation (3) we have:

Integrate

and again:

At x = 2, y = 0, therefore, ignoring negative bracket terms,

0 8 33 8 2 66 64 2

2 66 64

= ×( ) + + = + +

+ =

– . – .

A B A B

A Bi.e. orr 2 66 64A B– . –=

EIy x x x x Ax B= + + +– . – . – – – –8 33 11 67 3 10 4 25 23 3 3 3

EI y

x x x x x A

d d

= + +– – – – – –25 35 3 30 4 75 22 2 2 2

EI y

x x x x x

d d

2 50 70 3 60 4 150 2= +– – – – – –

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Also, at x = 5, y = 0,

or, rearranging,

Combining the two equations for –B gives

hence:

Slope equation:

At P, x = 2,

EI y

d d

other terms neglected

d d

= + ( )– .100 134 1

xx EI =

34 1.

EI y

x x x x x

d d

= + +– – – – – – .25 35 3 30 4 75 2 134 12 2 2 2

5 469 2 66 64

134 1 201 6

A A

A B

– – .

. – .

= = and

– –B A= 5 469

0 1041 11 67 8 10 25 27 5= ×( ) + ×( ) + +– – . – A B

or 00 1041 93 36 10 675 5

5 469

= + + +

+ =

– – . – A B

A B

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Deflection equation;

At W2, x = 3,

Example 3

FIGURE 11 shows a beam subjected to two loads supported by point reactions

at its ends. Write the bending moment equation in the Macaulay way and

show that, subject to his rules, the equation gives the correct bending moment

in the sections X1X1, X2X2 and X3X3. Find the constants A and B.

FIG. 11

P1 P2

EIy = + + ×( )– . . – .224 9 25 134 1 3 201 6 other terms negglected( )

=y EI

0 8.

EIy x x x x

= +

– . – . – – – –

. –

8 33 11 67 3 10 4 25 2

134 1

3 3 3 3

2201 6.

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Solution

The last section of the beam is X3X3. The bending moment equation by Rule 1

is:

If we are concerned with section X1X1, then by Rule 2, are

both negative and are omitted. We are left with:

If we are concerned with X2X2, then is omitted and the BM is:

Integrating Equation (a), remembering Rule 3, we get,

and integrating again

EIy P x W x a W x b

Ax B= + + ( )1 3

1 3

2 3

6 6 6 –

– –

– ...... e

EI y

P x W x a W x b A

d d

= +1 2

1 2

2 2

2 2 2 –

– –

– ............. dd( )

EI y

x P x W x a

d d

2 1 1 = – – ................................................ c( )

x b–

EI y

x P x

d d

2 1 = ...................................................................... b( )

x a x b– –and

EI y

x P x W x a W x b

d d

2 1 1 2 = – – – – ........................... a( )

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We have two constants of integration and two end conditions. So, at P1,

x = 0 and y = 0, then

The second and third terms are omitted so therefore B = 0.

Also, at P2, x = L and y = 0 giving:

so,

and this would be placed in Equations (d) and (e) for slope and deflection.

You can see why it is wise to put in numbers early: it cuts down the algebra.

As I said earlier about the technique, it is not economical to pursue the algebra

very far if we have a numerical problem.

Note also how simple this procedure is compared with the direct integration

method.

A P L W

L a W

L b L= ⎧ ⎨ ⎩

⎫ ⎬ ⎭

– – – – – /1 3

1 3 2 3

6 6 6

EI P L W

L a W

L b AL0 6 6 6

1 3

1 3 2 3( ) = +– – – – ....................... g( )

EI P W

a W

b A B0 0

6 6 0

6 0 01

3 1 3 2 3( ) = ( ) + ( ) +– – – – ............ f( )

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UNIFORMLY DISTRIBUTED LOADS

FIG. 12

We run up against a slight difficulty with distributed loads. Have a look at

FIGURE 12. We have a uniformly distributed load running from AA to BB,

the last section of the beam is XX and we want the BM due to w. We can write

it down all right, the centre of gravity of the uniform load is half way along its

length, i.e. (a + (b – a)/2) from the left-hand end, and the BM at XX is:

What happens if XX is slightly to the left of the centre of gravity of the load?

Do we make the bracket term then zero? Well, no. What about all the rest of

the uniform load between the C of G and AA? Do we suddenly neglect its

bending effect? No, we don't. There is a rule, which we shall call Rule 4, to

deal with this.

EI y

x Px w b a x a

b ad d

2 2 = ( ) + ( )⎛

⎝⎜ ⎞ ⎠⎟

⎛

⎝⎜ ⎞

⎠⎟ – – –

– ................. 6( )

a A

A b

Centre of gravity of load

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FIG. 13

We have re-drawn FIGURE 12 and called it FIGURE 13. Can you see the

difference between the two figures? In order to obtain an expression for the

bending moment at a distance x from the end which will apply to all values of

x, it is necessary to continue the loading right up to the section XX, but

compensate it with an equal negative load from BB to XX. This is Rule 4.

The charming thing about this is that it makes the BM easier to write:

EI y

x Px w x a

x a w x b

x b

EI y

x Px

w x

d d

2 2 = +

– – –

– –

– –– –

....................... a w x b

2 2

2 2 7+ ( )

a A

A b

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APPLYING MACAULAY'S TECHNIQUE TO A NUMERICAL EXAMPLE

Let us play with some numbers. Go back to FIGURE 11 and assign some

dimensions:

a = 2 m, b = 6 m, L = 8 m, W1 = 50 kN, W2 = 95 kN.

This is shown in FIGURE 14.

FIG. 14

Now, we are not just going to substitute numbers into the algebra of the

previous question, we are going to use numbers from the start. We want the

maximum deflection in terms of E and I. First we need the reaction at the left

end, P1.

To find P1, take moments about the right end,

therefore P1 = 61. 25 kN.

The bending moment equation for the last section is:

EI y

x x x x

d d

2 61 25 50 2 95 6= . – – – – ............................ 8( )

8 6 50 2 95 01P – –×( ) ×( ) =

6 m

2 m

50 kN

95 kN

8 m

P1 P2

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Integrate,

Integrate again,

To evaluate the constants:

at x = 0, y = 0, so, in Equation (10), B = 0

at x = 8, y = 0, so, Equation (10) becomes:

So the complete slope equation is:

EI y

x x x x

d d

= 30 63 25 2 47 5 6 412 72 2 2. – – – . – – . ............. 11( )

0 10 21 8 8 33 8 2 15 83 8 6 8

412 7

3 3 3= × +

. – . – – . –

– .

EIy x x x Ax B= + +10 21 8 33 2 15 83 6 103 3 3. – . – – . – ...... (( )

EI y

x x x x A

d d

= +30 63 25 2 47 5 62 2 2. – – – . – .................... 9( )

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Now, we will make a guess that the slope is zero (and therefore that y is a

maximum) somewhere near the centre, i.e. to the left of the 95 kN load in

section X2X2. So, the slope equation (11) becomes:

This is a quadratic equation and x evaluates to:

So substituting this value into equation (12) gives:

Now, that was a bit of a fiddle and, thinking about the problem, you should

have spotted it. The method above has guesswork in it, and guesswork always

has to be treated with caution. How did I know that the maximum deflection

would be in section X2X2? Well, I didn't really, but experience told me that it

was very likely. What does a learner with little experience do about this? He

or she might assume that the maximum y is in section X1X1, or X3X3. Does

the method break down because the user is not all that experienced? It would

not be much of a method if it did – and it doesn't!

EIymax = ×( ) ×( ) ×10 21 4 15 8 33 2 15 412 7 4 13 3. . – . . – . . 55

729 74 82 79 1712 71

1065 7

( )

. – . – .

– .

y EImax

x = 4 15. m

0 30 63 25 2 412 7

0 30 63 25 4 4 4

2 2

= +

. – – – .

. – – –

x x

x x x 112 7

0 5 63 100 100 412 7

0 5 63 100 5

. – – .

. –

= +

x x

x x 112 7.

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Let us make an assumption that the maximum deflection occurs in section

X1X1. The slope equation becomes,

that is,

which cannot be right because it is greater than 2 m, in fact in section X2X2.

Might it be in the last section, X3X3? The slope equation would become:

This reduces to the quadratic equation:

from which x = 11.3 m or 4.7 m.

It cannot be either of those because 11.3 m is longer than the beam and 4.7 m

is in section X2X2, not X3X3.

So it is not in section X1X1 or X3X3, which means that the maximum

deflection must occur in section X2X2.

0 16 53 082= +x x– .

0 30 63 25 2 47 5 6 412 72 2 2= . – – – . – – .x x x

EI y

x x

d d

= =

0 30 63 412 7

3 67

2. – .

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For comparison we can calculate the deflections at the two load points, as

follows.

By Equation (12) at x = 2,

By Equation (12) at x = 6,

EIy

= ×( ) ×( ) ×( )

10 21 216 8 33 64 412 7 6

804 0

. – . – .

– .

EEI

EIy

= ×( ) ×( )

10 21 8 412 7 2

81 68 825 4

743 7

. – .

– .

EEI

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Example 4

For the beam shown in FIGURE 15 determine the deflection of the beam at its

centre, if EI = 100 MN m2.

FIG. 15

Solution

We first find the reactions at the supports by taking moments about RB.

Since 30 + 50 + 20 × 4 = RA + RB, then

RB kN= 84

10 30 7 20 4 5 50 3

10 760

A kN

= ×( ) + × ×( ) + ×( )

4 m

R A

20 kN m–1 3 m 3 m

30 kN 50 kN

X R

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Taking the origin as RA we apply the Macaulay method shown previously. So

by considering section XX to the extreme right of the beam and taking

moments of the forces on the left of XX we get:

and since

then

therefore

and

Now the boundary conditions yield the following values. When x = 0, y = 0,

so that B = 0, since all the Macaulay terms are negative and so equate to zero.

When x = 10, y = 0, now all the Macaulay terms are positive and so we find

A = –879.3.

d d

x EI

x x x=

⎡ ⎣⎢

10 76 2

30 2

3 50 2

7 3 2 2 2

– – – –

– – –

–

20 6

3 20 6

10 76 6

30 6

3 3

x x A

y EI

+ + ⎤ ⎦⎥

= ⎡ ⎣⎣⎢

x x

– – –

– –

3 50 6

20 24

3 20 24

3 3

4 –– 7

4 + + ⎤ ⎦⎥

Ax B

d d

x EI x x x x

310 76 30 3 50 7

20 2

= – – – – – – 33

20 2

⎡ ⎣⎢

+ ⎤ ⎦⎥

x –

M x x x x= ⎡ ⎣⎢

10 76 30 3 50 7 20 2

33 2

– – – – – –

++ ⎤ ⎦⎥

20 2

7 2

x –

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At the centre of the beam x = 5, then:

y EI

= ( ) ( ) ( ) ( )

⎡

⎣ ⎢

10 76 5

6 30 6

2 20 24

2 4396 7 3 3

3 4 – – – .

⎢⎢

⎤

⎦ ⎥ ⎥

= ( ) ( )

10 2866 7

100 10

– .

giving –– . –0 0287 m or 28.7 mm

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. A cantilever of length 2 m is made from a steel with E = 210 GPa. A

concentrated load of 40 kN exerts at the free end. If the maximum

deflection is 1.5 mm, determine:

(i) the second moment of area for the beam

(ii) the maximum slope.

2. A simply supported beam is 10 m long, with a load of 500 kN at the

middle and a uniformly distributed load of 500 N m–1 along the whole

beam. If E and I are the same as those in Question 1, calculate the

maximum deflection and the slope at the end.

3. FIGURE 16 (opposite) shows a beam under a variety of loading.

Write the bending moment expression after Macaulay, integrate it for

slope and deflection, find the deflection at the left end and the slope at the

left-hand support. You may use the following values for E and I.

= ×

200 10

5 10

N m

–

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FIG. 16

80 kN20 kN

P2P12 m

4 m

7 m

9 m

25 kN m–1

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. (i) The second moment of area I is easily calculated using TABLE 1.

From the question, W = 40 kN = 40 × 103 N, L = 2 m, E = 210 GPa = 210 × 109 Pa and ymax = 1.5 × 10–3 m

(ii) Again using TABLE 1, the maximum slope is:

d d

max

⎛ ⎝⎜

⎞ ⎠⎟

= × ×

× × ×

3 2

40 10 2 2 210 10 3 3. 886 10

1 125 10

= ×

–

–.

∴ = × ×

× × × ×

= ×

I 40 10 2

3 210 10 1 5 10

3 386

3 3

9 3.

–

110 4– m 4

y WL

I WL

max

∴ =

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2. From the question, the loading situation is as shown in FIGURE 17.

FIG. 17

Applying the principle of superposition to the beam and using TABLE 1,

we have

and

The maximum deflection ymax is at the middle, while the maximum slope

is at the end.

Since W = 500 kN = 500 × 103 N, L = 10 m, w = 500 N m–1, E = 210 GPa = 210 × 109 Pa and I = 3.386 × 10–4 m4

then y

max

m 147 mm

d d

= =

⎛ ⎝⎜

⎞ ⎠⎟

0 147

0 0442

y WL

max

d d

= +

⎛ ⎝⎜

⎞ ⎠⎟

3 4

48 5 384

16 ++

10 m

W = 500 kN

w = 500 N m–1

5 m

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3. Refer to FIGURE 18.

FIG. 18

First find the reaction, P1 by taking moments about the right-hand end,

Now the bending moment equation for section XX.

EI y

x x x x x

d d

2 3 210 20 120 4 2 80 4 12 5 4= +⎡⎣ – . – – – – . –

+ ⎤⎦12 5 7 2

. – ....................x .................................. 1( )

7 20 9 80 5 75 3 5

120 4

= ×( ) + ×( ) + ×( )

. kN

80 kN20 kN

P2P12 m

4 m

7 m

9 m

25 kN m–1

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Integrate for the slope equation:

Integrate again for the deflection equation

At x = 2, y = 0, so Equation (3) becomes:

At x = 9, y = 0, so Equation (3) becomes:

By Equations (5) and (4),

2153 4 9 26 64 2

7 2180

311 4

26 64 2

. .

–

– .

. –

+ = +

A A

B AA = + =26 64 622 8 649 4. . .

0 2427 6 6880 6 1666 2 650 16 64 9

0 215

= + + + +

– . . – . – . A B

33 4 9. ..................................+ +A B ........................ 5( )

0 26 64 2= + +– . ..............................A B ........................... 4( )

EIy x x x x= +10 3 33 20 06 2 13 33 4 1 04 43 3 3 3 4– . . – – . – – . –⎡⎡⎣ + + + ⎤⎦ 1 04 7

4 . – ................x Ax B .............................. 3( )

EI y

x x x x x

d d

= +⎡⎣10 10 60 2 2 40 4 4 17 4 3 2 2 2 3– . – – – – . –

+ + ⎤⎦4 17 7 3

. – ......................x A ........................... 2( )

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Specific deflection equation,

and slope equation

For deflection at the left end, substitute x = 0 into equation (6)

At left-hand support x = 2.

For slope at left-hand support, substitute x = 2 into equation (7)

EI y

d d

(neglecting negative= [ ]10 40 311 43 – – . and zero terms)

351.4 103= ×–

EIy = ×649 4 103. (neglecting negative and zeroo terms)

y = ×

× × ×

= ×

649 4 10 200 10 5 10

6 49 10

9 4

–

–33 m

Deflection 6.49 mm=

EI y

x x x x x

d d

= +⎡⎣10 10 60 2 2 40 4 4 17 4 3 2 2 2 3– . – – – – . –

+ ⎤⎦4 17 7 311 4 3

. – – . .................x .............................. 7( )

EIy x x x x= +10 3 33 20 06 2 13 33 4 1 04 43 3 3 3 4– . . – – . – – . –⎡⎡⎣ + + ⎤⎦ 1 04 7 311 4 649 4

4 . – – . . .......x x ........................... 6( )

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Putting in values for E and I we obtain:

d d

slope

= ×

× × ×

= =

– .

–

351 4 10 200 10 5 10

9 4

551 10 3× –

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

In this lesson, we have studied two different methods of finding the deflection

of simply supported beams and cantilevers. By integration, we can easily

determine the slope and deflection for some standard cases; while Macaulay’s

method can be applied to any loading situation.

But not all beams, for example, are supported on only two props. Those with

three or more are called 'continuous beams' and they are better tackled in other

ways. It is like the old saying 'horses for courses'; choose the best technique

for the job in hand.

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/ITA 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<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> >> >> setdistillerparams << /HWResolution [2400 2400] /PageSize [612.000 792.000] >> setpagedevice