mechanics of materials, volume 1

8/6/2019 Mechanics of Materials, Volume 1

1/14

CHAPTER 6

BUILT-IN BEAMS

Summary

The maximum bending moments and maximum deflections for built-in beams withstandard loading cases are a s follows:

MAXIMUM B.M. AN D DEFLECT ION FO R BUILT-IN BEAMS

Loading case

Central concentratedload W

Uniformly distributedload w/metre(total load W )

Concentrated load Wnot at mid-span

Distributed load wvarying in intensitybetween x = x, andx = x2

Maximum B.M.

WL8

-

wL2 WL12 12

_ - -

Wab2 Wa2b~ or

-2 L 2

w(L -x)ZM A = - dx

Maximum deflection

WL3192EI_ _

wL4 WL338481 384EI_ _ = -

2 Wa3b2 2aLa t x=-

(L +2a)EI(L+2a)2

where a < -Wa3b33EIL3

- under load

140


2/14

$6.1 Built-in Beam s 141

Efec t of movement of supports

If one end B of an initially horizontal built-in beam A Bmoves through a distance 6 relativeto end A , end moments are set up of value

and the reactions at each support are

Thus, in most practical situations where loaded beams sink at the supports the above valuesrepresent changes in fixing moment and reaction values, their directions being indicated inFig. 6.6.

Introduction

When both ends of a beam are rigidly fixed the beam is said to be built-in, encastred orencastri. Such beams are normally treated by a modified form of Mohrs area-momentmethod or by Macaulays method.

Built-in beams are assumed to have zero slope at each end, so that the total change of slopealong the span is zero. Thus, from Mohrs first theorem,

M .

Elarea of - iagram across the span = 0

or, if the beam is uniform, El is constant, and

area of B.M. diagram = 0 (6.1)

Similarly, if both ends are level the deflection of one end relative to the other is zero.Therefore, from Mohrs second theorem:

MEI

first moment of area of - iagram about one end = 0and, if EZ is constant,

first moment of area of B.M. diagram about one end= 0 (6.2)To make use of these equations it is convenient to break down the B.M. diagram for the

(a) that resulting from the loading, assuming simply supported ends, and known as the

(b) that resulting from the end moments or fixing moments which must be applied at the

built-in beam into two parts:

free-moment diagram;

ends to keep the slopes zero and termed the fixing-mom ent diagram.

6.1. Built-in beam carrying central concentrated load

Consider the centrally loaded built-in beam of Fig. 6.1. A , is the area of the free-momentdiagram and A, that of the fixing-moment diagram.


3/14

142 Mechanics of Materials 46.2

iA- re e' m e n t diagramFixing moment diagmm %I1 = - %

Fig. 6.1.

By symmetry the fixing mom ents are equal at b oth ends. Now from eqn.(6.1)

A , + & = 0

. .W L3 x L X = - M L4

The B.M. diagram is therefore a s shown in Fig.6.1, he maximum B.M. occurring at both theends an d the centre.

Applying Mohr's second theorem fo r the deflection a t mid-span,

first momentof area of B.M. diagram between centre andone end about the centre

= [

1 L M L L

1 [W L ~M L ~ 1 W L ~ W L ~EZ 96 8 E l 96

(i.e. dow nward deflection)L J

192EZ- --

6.2. Built-in beam carrying uniformly distributed load across the span

Consider now the uniformly loaded beam of Fig.6.2.


4/14

$6.3 Built-in Beams 143

'Free' moment diagram

I IFixing moment diagram A b 1 Z - d2

I I

Fig. 6.2.

Again, for zero change of slope along the span,

& + A , = 0

. .2 W L 2- x - x L = - M L3 8

The deflection at the centre is again given by Mohr's second theorem as the moment of one-half of the B.M. diagram about the centre.

. . 6 = [3x $ ;)(; x ;)+ ( y $)]AI

wL4384 E I

The negative sign again indicates a downwards deflection.

- --

6.3. Built-in beam carrying concentrated load offset from the centre

Consider the loaded beam of Fig. 6.3.Since the slope at both ends is zero the change of slope across the span is zero, i.e. the total

area between A and B of the B.M. diagram is zero (Mohr's theorem).


5/14

144 Mechanics of Materials 46.3

. .

L

Fig. 6.3.

Also the deflection ofA relative to B s zero; therefore the m oment of the B.M.diagrambetween A and B about A is zero.

... [ ~ x ~ x a ] ~ + [ j x ta b x b I a + - :) ( M A L x -)+ ( M B L x -t )O

W abM A + 2 M ~ = - - [ 2 a 2 + 3 a b + b 2 ]

L3Subtracting (l),

W abL3

- - [ 2 a 2 +3ab+ b2- L 2 ]-but L = a + b ,

W abL 3

[2a2+3ab +b Z- 2 - ab - 2 ]-.

Wab Wa'bL[ a + a b ] = - _ _ _

L3 L3= - _ _


6/14

56.4 Built-in Beam s 145

Substituting in (l),Wab Wa2b

L L2M - - - + -

A -

Wab(a+b ) Wa 2 b= - +-

WabLZ

L2 L 2

= - _ _

6.4. Built-in beam carrying a non-uniform distributed load

Let wbe the distributed load varying in intensity along the beam as shown in Fig. 6.4. On ashort length dx at a distance x from A there is a load of wdx.Contribution of this load to M A

Wab2= -~ L2 (where W = wdx)

wdx x x ( L- )L 2

1x(L;x)dx- -

total MA = -0

w/me tre\

Fig. 6.4. Built-in (encostre) beam carrying non-uniform distributed load

Similarly,

(6.9)

(6.10)

0

If the distributed load is across only part of the span the limits of integration must bechanged to take account of this: i.e. for a distributed load w pplied between x = x l andx = x 2 and varying in intensity,

(6.1 1)

(6.12)


7/14

146 Mechanics of Materials $6.5

6.5. Advantages and disadvantagesof built-in beams

Provided that perfect end fixing canbe achieved, built-in beams carry smaller maximumB.M.s (and hence are subjected to smaller maximum stresses) an d have smaller deflectionsthan the corresponding simply suppo rted beams w ith the same loads applied; in other words

built-in beams ar e stronger an d stiffer. Although this would seem to imply that built-in beamsshould be used whenever possible, in fact this is not the case in practice. The principal reasonsare as follows:

(1) The need fo r high accuracy in aligning the supports and fixing the ends during erection

( 2 ) Small subsidence of either support can set up large stresses.(3) Changes of temperature can also set up large stresses.(4) The end fixings are normally sensitive to vibrations and fluctuations in B.M.s, as in

These disadvantages can be reduced, however, if hinged joints are used at points o n thebeam where the B.M. is zero, i.e. atpoints of inflexionor contraflexure. The beam is theneffectively a central beam supported on two end cantilevers, and for this reason theconstruction is sometimes termed thedouble-cantileverconstruction. The beamis then free toadjust to changes in level of the supports and changes in temperature (Fig.6.5).

increases the cost.

applications introducing rolling loads (e.g. bridges, etc.).

oints of inflexion

Fig. 6.5. Built-in beam using d oub leantileve r construction.

6.6. Effect of movement of supports

Consider a beamAB initially unloaded w ith its ends a t the sam e level. If the slope is toremain horizontal at each end whenB moves through a distance6 relative to endA, th emom ents must be as shown in Fig.6.6. Taking moments abou t B

RAx L = MA+ MB

M A = M g = Mnd, by symmetry,

2ML. R A = -

Similarly,2M

LR B = -

in the direction shown.


8/14

$6.6 Built-in Beams 147

-M

Fig. 6.6. Effect of support movement on B.M.s.

Now from Mohrs second theorem the deflection of A relative to B is equal to the firstmoment of area of the B.M. diagram about A x l /E I .

. .

12EISand R A = R e = -

E1SM=---L 2 L 3

(6.14)

in the directions shown in Fig. 6.6.

beam under load when one end sinks relative to the other.These values will also represent the changes in the fixing moments and end reactions for a

Examples

Example 6.1

An encastre beam has a span of 3 m and carries the loading system shown in Fig. 6.7.Drawthe B.M. diagram for the beam and hence determine the maximum bending stress set up. Thebeam can be assumed to be uniform, with I = 42 x m4 and with an overall depth of200 mm.

Solution

Using the principle ofsupe rpositionthe loading system can be reduced to the three cases forwhich the B.M. diagrams have been drawn, together with the fixing moment diagram, inFig. 6.7.


9/14

148 Mechanics of Materials

4 0 k N

A 0

2 0 kN I I

1

M= -3 4 kN rnM,=-25.4 k N rnI

---_-----

Bending moment diagrams

( a ) u .d .1 .

( b ) 4 0 k N load

( c ) 2 0 kN load

( d ) F ix ing-moment d iagram

Tota l bending- momentdiagram on base offixing moment line

Total bending-momentdiagram re-drawn onconventiona I horizontabase

-3 4

Fig. 6.7. Illustration of the application of the principle of superposition to Mohrs area-moment methodof

solution.

Now from eqn. (6.1)A1 +A2 +A4 = A3

( ~ X 3 3 . 7 5 X 1 0 3 x 3 ) + ( ~2 8 . 8 ~ 1 0 ~ X 3 ) + [ $ ( M A + M ~ ) 3 ] = ( ~ ~ 1 4 . 4 ~ 1

67.5 x lo3 +43.2 x lo3 + 1.5(MA+MB) = 21.6 x l o 3M A +M B = - 59.4 x 103 (1)

Also, from eqn. (6.2), taking moments of area about A ,

A121 +A222+A424 = A323


10/14

Built-in Beams 149

and, dividing areas A , and A , into the convenient triangles shown,

2 x 1.83

(67.5 x IO3 x 1.5)+ (3x 28.8 x lo3 x 1.8)- + (5 x 28.8 x lo3 x 1.2)(1.8+ $ X 1.2)+ (3MA 3 x $ x 3)+ ( f MB x 3 x 5 x 3) = (4 x 14.4 x lo3 x 12)3 x 1.2

+ ( f x 14.4 x lo3 x 1.8) 1.2 +-Y)(101.25 +31.1 +38.0)103 +1.5MA+ 3MB = (6.92 +23.3)103

1.5 MA + 3MB = - 140 X lo3

MA +2MB = - 93.4 x lo 3 (2)

M B = - 3 4 x 1 0 3 N m = - 3 4 kN mand from (l),

The fixing moments are therefore negative and not positive as assumed in Fig. 6.7. The totalB.M.diagram is then found by combining all the separate loading diagrams and the fixingmoment diagram to produce the result shown in Fig. 6.7. It will be seen that the maximumB.M. occurs at the built-in end B and has a value of 34kNm. This will therefore be theposition of the maximum bending stress also, the value being determined from the simplebending theory

MA= - 2 5 . 4 ~ 03 Nm = -25.4kNm

MY

34x

103x

io0x

10-3omax=-I 42 x= 81 x lo6 = 81 MN/mZExample 6.2

A built-in beam, 4 m long, carries combined uniformly distributed and concentrated loadsas shown in Fig. 6.8. Determine the end reactions, the fixing moments at the built-in supportsand the magnitude of the deflection under the 40 kN load. Take E l = 14 MN m2.

30 kN/m40 kN

I X

Fig. 6.8.

Solution

Using Macaulay's method (see page 106)


11/14


N ot e that the unitof load of kilonewton is conveniently accountedfor b y dividingEZ by lo3. Itcan then be assum ed in further calculation that RAis in kN and MA in kN m .

Integrating,

_ -I d y = M , x + R A - - - [ ( ~ - 1 . 6 ) ~ ] - - [ ( ~ - 1 . 6 ) ~ ]2 40 30 + Alo 3 dx 2 2 6

andEI x2 x3 40 30

2 6 6 24y MA- +RA- - -[ (x- 1.6y] - -[(x- 1.6)4]+ A X+ B

Now, when x = 0 , y = O ... B = O

dYand w h e n x = O , - O .. A = Odx

When x = 4 , y = O

42 43 40 302 6 6 24

= MA X - RAx - - 2.4)3- - 2.4)40 = 8MA + 10.67 RA -92.16-41.47

133.6= 8MA+ 10.67 RA N lq

dYWhen x = 4, - Odx42 40 302 2 6

= MA+ -RA- -(2.4)- -(2.4)3

Multiply (2)x 2,

(3)- I),

Now

. .

368.64 = 8MA+16RA

= 44.1 kN35.045.33

A=-

R A + R B= 40-t 2.4 X 30) = 112kN

RB = 112 -44.1 = 67.9 kN

Substituting in (2),4MA+ 352.8= 184.32

. . MA = (184.32- 52.8) = - 2.12 kN m

i.e. MA is in the opposite direction to that assumed in F ig. 6.8.


12/14

Built-in Beams 151

Taking moments about A,

MB+4RB-(40X 1 . 6 ) - ( 3 0 ~ 2 . 4 ~ 2 . 8 ) - ( - 4 2 . 1 2 ) = 0

. .

i.e. again in the opposite direction to that assumed in Fig. 6.8.(Alternatively, and more conveniently, this value could have been obtained by substitutioninto the original Macaulay expression with x = 4, which is, in effect, taking moments about B.The need to take additional moments about A is then overcome.)

M B = - 67.9 x 4) +64 +201.6 - 2.12 = - 48.12 kN m

Substituting into the Macaulay deflection expression,

xz 4 4 . 1 2 2 0

GYy 2 6 3= -42.1- + --

X - .613- [x - .614l

Thus, under the 40 kN load, where x = 1.6 (and neglecting negative Macaulay terms),

- 42.1 x 2.56) (44.1 x 4.1)y = E [Z 2 + 6 - 0 - 0 1

23.75 x l o 3

14 x l o 6- = - 1 . 7 ~ 0 + m

= - 1.7mm

The negative sign as usual indicates a deflection downwards.

Example 6.3

Determine the fixing moment a t the left-hand end of the beam shown in Fig. 6.9 when theload varies linearly from 30 kN/m to 60 kN/m along the span of 4 m.

30 N h I kN/rn

A

Fig. 6.9.

Solution

From $6.4

Now w' = (30 +F) o 3 = (30+7.5x)103N/m


13/14


. . M A = - /30 +7.5x)103 4 - ) ~x dx42

0

4

103- - - (30 +7 . 5 ~ ) ( 1 6 x +x2)x dx16 04

103= -

- 4 8 0 ~ 4 0 2 +30x3+ 1 2 0 2 - 0x3+7 . 5 ~ ~ )x16

0

4

103-- - - 4 8 0 ~ 2 0 x 2- 0x3+7.5x4)dx16 0

103= - - [ 2 4 0 ~ 1 6 - 4 0 ~ 6 4 - 3 0 ~ 6 4 + 2 . 5 ~ 1 0 2 4 ]

16

= - 1 2 0 x 1 0 3 N m

The required moment atA is thus 120 kN m in the opposite direction to that shown inFig. 6.8.

Problems

6.1 (A/B). straight beam ABCD is rigidly built-in at A and D and carries point loads of 5 kN at B and C.

A B = B C = C D = 1.8m

If the second moment of area of the section is 7 x 10-6m4 and Youngs modulus is 210GN/mZ, calculate:

(a) the end moments;(b ) the central deflection of the beam. [U.Birm.l[-6kNm; 4.13mm.l

6.2 (A/B). beam of uniform section with rigidly fixed ends which are at the same level has an effective spanof10 m. It carries loads of 30 kN and 50 kN at 3 m and 6 m respectively from the left-hand end. Find the verticalreactions and the fixing moments at each end of the beam. Determine the bending moments at the two points ofloading and sketch, approximately to scale, the B.M. iagram for the beam.

c41.12, 38.88kN; -92, -90.9, 31.26, 64.62kNm.l

6.3 (A/B). beam of uniform section and of 7 m span is fixed horizontally at the same level at each end. I tcarries a concentrated load of 100 kN at 4m from the left-hand end. Neglecting the weight of the beam and workingfrom first principles, find the position and magnitude of the maximum deflection if E = 210GN/m2 andI = 1% x m4. C3.73 from 1.h. end; 4.28mm.l

6.4 (A/B). uniform beam, built-in at each end, is divided into four equal parts and has equal point loads, eachW ,placed at the centre of each portion. Find the deflection at the centre of this beam and prove that it equals thedeflection at the centre of the same beam when carrying an equal total load uniformly distributed along the entirelength.

[U.C.L.I.] [--.IL 9 6 ~ 1


14/14

Built-in Beams 153

6.5 (A/B). A horizontal beam of I-section, rigidly built-in at the ends and 7~1 long, came sa total uniformlydistributed load of90 kN as wellas a concentrated central loadof 30 kN. If the bending stress is limited to90 MN/m2and the deflection must not exceed 2.5 mm , find the dep thof section required. Prove the deflection formulae ifused,or work from first principles.E = 210GN/m2. [U.L.C.I.] [583 mm.]

6.6 (A/B). A beam of uniform sectionis built-in a t each endso as to havea clearspan of 7 m. Itcame a uniformlydistributed load of 20 kN /mon he left-hand half of the beam, together w ith a 120kN load at5 m from the left-hand

end. Find the reactions an d the fixing mom ents at the ends an d drawa B.M. diagram fo r the beam, inserting theprincipal values. [U.L.][-lO5.4, -14 8kN ; 80.7, 109.3kN m.l6.7 (A/B). steel beam of 10m span is built-in at both ends andcames two point loads, each of 90kN,

at points 2.6m from the ends of thebeam. The middle 4.8m has a section for which the second momentof area is 300x m4 and the 2.6m lengths at either end have a section for which the second mom ent of area is400 x m4. Find the fixing mom ents at the ends and calculate the deflection at m id-span. TakeE = 210 GN /mzand neglect the weight of the beam. [U.L.][ M a = M B= 173.2 kN m; 8.1 mm.]

m4.As aresult of subsidence one end m oves vertically through 12 m m . Determine the changes in the fixing mom ents andreactions. For the beam materialE = 210GN/m2. C21.26 kN m ; 5.32 kN.]

6.8 (B.) oaded horizontal beam has its ends securely built-in; the clear span is8 m andI = 90 x

mechanics of materials, volume 1

Documents