CHAPTER 6 BUILT-IN BEAMS Summary The maximum bending moments and maximum deflections for built-in beams with standard loading cases are as follows: MAXIMUM B.M.AND DEFLECTION FOR BUILT-IN BEAMS Loading case Maximum B.M. Maximum deflection Central concentrated WL WL3 load w 8 192EI Uniformly distributed wL2 WL WL4 WL load w/metre (total load W) 12=12 384Ei=384E Concentrated load W not at mid-span Wab2 Wa'b 2Wab2 2aL d or D 3E1(L+2a2 at x= (L+2a) where a< L Wab3 3EIL under load Distributed load w' varying in intensity MA=- w儿-对k between x =x1 and L X =X2 X2 w'(亿-x)x2 M8=- dx 140
CHAPTER 6 BUILT-IN BEAMS Summary The maximum bending moments and maximum deflections for built-in beams with standard loading cases are as follows: MAXIMUM B.M. AND DEFLECTION FOR BUILT-IN BEAMS Loading case Central concentrated load W Uniformly distributed load w/metre (total load W) Concentrated load W not at mid-span Distributed load w’ varying in intensity between x = x, and x = x2 Maximum B.M. WL 8 - wL2 WL 12 12 _-- Wab2 Wa2b ~ or - L2 L2 w‘(L -x)Z MA= - dx Maximum deflection WL3 192EI __ wL4 WL3 38481 384EI __=- 2 Wa3b2 2aL at x=- 3EI(L + 2a)2 (L + 2a) where a < - Wa3b3 3EIL3 =- under load 140
§6.1 Built-in Beams 141 Effect of movement of supports If one end B of an initially horizontal built-in beam AB moves through a distance o relative to end A,end moments are set up of value 6E18 MA=-MB=Li and the reactions at each support are 12EI6 RA=-RB=L3 Thus,in most practical situations where loaded beams sink at the supports the above values represent changes in fixing moment and reaction values,their directions being indicated in Fig.6.6. Introduction When both ends of a beam are rigidly fixed the beam is said to be built-in,encastred or encastre.Such beams are normally treated by a modified form of Mohr's area-moment method or by Macaulay's method. Built-in beams are assumed to have zero slope at each end,so that the total change of slope along the span is zero.Thus,from Mohr's first theorem, area of M Ediagram across the span-0 or,if the beam is uniform,EI 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 Mohr's second theorem: 、M first moment of area ofdiagram about one end0 and,if EI 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 built-in beam into two parts: (a)that resulting from the loading,assuming simply supported ends,and known as the free-moment diagram; (b)that resulting from the end moments or fixing moments which must be applied at the ends to keep the slopes zero and termed the fixing-moment 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-moment diagram and A that of the fixing-moment diagram
$6.1 Built-in Beams 141 Efect of movement of supports If one end B of an initially horizontal built-in beam AB moves through a distance 6 relative to 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 values represent changes in fixing moment and reaction values, their directions being indicated in Fig. 6.6. Introduction When both ends of a beam are rigidly fixed the beam is said to be built-in, encastred or encastri. Such beams are normally treated by a modified form of Mohr’s area-moment method or by Macaulay’s method. Built-in beams are assumed to have zero slope at each end, so that the total change of slope along the span is zero. Thus, from Mohr’s first theorem, M. El area of - diagram 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 Mohr’s second theorem: M EI first moment of area of - diagram about one end = 0 and, 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-moment 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-moment diagram and A, that of the fixing-moment diagram
142 Mechanics of Materials $6.2 W Free'moment diagram Fixing moment diagrom M、 ⊙ Total B.M.diagram WL Γ Fig.6.1. By symmetry the fixing moments are equal at both ends.Now from eqn.(6.1) Aa+Ap=0 WL 支×L×年=-ML WL M= 8 (6.3) The B.M.diagram is therefore as shown in Fig.6.1,the maximum B.M.occurring at both the ends and the centre. Applying Mohr's second theorem for the deflection at mid-span, first moment of area of B.M.diagram between centre and 1 one end about the centre E =[经✉竖xL)*)+(竖x)] 贤+]紧既] WL =-192E1 (i.e.downward deflection) (6.4) 6.2.Built-in beam carrying uniformly distributed load across the span Consider now the uniformly loaded beam of Fig.6.2
142 Mechanics of Materials 46.2 iA- Free' ment diagram Fixing moment diagmm %I1 M=-% 8 Fig. 6.1. By symmetry the fixing moments are equal at both ends. Now from eqn. (6.1) A,+& = 0 .. WL 3 x LX -= -ML 4 The B.M. diagram is therefore as shown in Fig. 6.1, the maximum B.M. occurring at both the ends and the centre. Applying Mohr's second theorem for the deflection at mid-span, first moment of area of B.M. diagram between centre and one end about the centre 6=[ 1L ML L 1 [ WL~ ML~ 1 W L ~ W L ~ EZ 96 8 El 96 (i.e. downward deflection) WLJ 192EZ - -- 6.2. Built-in beam carrying uniformly distributed load across the span Consider now the uniformly loaded beam of Fig. 6.2
§6.3 Built-in Beams 143 w/metre A 'Free'moment diagram Ap Fixing moment diagram Total B.M.diagram 12 12 Fig.6.2. Again,for zero change of slope along the span, A。+Ab=0 2.wL2 x8×L=-ML 3 M= L2 12 (6.5) 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. -[(眉ד答)层*)+（些）]日 +]-站] wL+ 384E1 (6.6) 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)
$6.3 Built-in Beams 143 'Free' moment diagram I I Fixing moment diagram Ab1Z-d 12 I I Fig. 6.2. Again, for zero change of slope along the span, &+A, = 0 .. 2 WL2 -x- xL=-ML 38 The deflection at the centre is again given by Mohr's second theorem as the moment of onehalf of the B.M. diagram about the centre. .. 6 = [ (3 x $ x ;)(; x ;)+ (y x $)]A EI wL4 384 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)
144 Mechanics of Materials §6.3 Wab Free moment diagram Ma王 Fixing moment diagram t Total B.M.diagram -Wo2 b Fig.6.3. (径x"2xL）+w+aL=0 Wab M+M8=- L (1) Also the deflection of A relative to B is zero;therefore the moment of the B.M.diagram between A and B about A is zero. [且x"空xa]g+"兜x]e+)+(wL×号)+Mu×号)=o w+2w+"2+"2(+)-0 Wab. M4+2Ma=-L8[2a2+3ab+b2] (2) Subtracting (1), wab」 Ma=-"p[2a2+3ab+b2-L2] but L=a+b, Wab Ma=-[2a+3ab+b2-a2-2ab-b] Wab Wa'bL 3Ca>+ab]=- Wa2b =- L2 (6.7)
144 Mechanics of Materials 46.3 .. L Fig. 6.3. Also the deflection of A relative to B is zero; therefore the moment of the B.M. diagram between A and B about A is zero. ... [~x~xa]~+[jxt wabxb I( a+- :) + ( +MALx- 4) + ( +MBLx- 't) =O Wab MA+2M~= - ---[2a2+3ab+b2] L3 Subtracting (l), Wab L3 M 8- - - -[2a2 + 3ab+ b2 - L2] but L=a+b, Wab L3 .. M B- [2a2 + 3ab + bZ - a2 - 2ab - b2] Wab Wa'bL [a +ab] = -___ L3 L3 = -__
§6.4 Built-in Beams 145 Substituting in (1), Wab Wa?b MA=- L+ L2 Wab(a+b)Wa?b 宁一 L2 +2 Wab2 三一 D (6.8) 6.4.Built-in beam carrying a non-uniform distributed load Let w'be the distributed load varying in intensity along the beam as shown in Fig.6.4.On a short length dx at a distance x from A there is a load of w'dx.Contribution of this load to M Wab2 一 L2 (where W=w'dx) w'dx×x(L-x)2 w'x(L-x)2dx total M=- L2 (6.9) w'/metre A dx L-X- Fig.6.4.Built-in (encastre)beam carrying non-uniform distributed load. Similarly, MB=一 w(L-x)x2 dx (6.10) If the distributed load is across only part of the span the limits of integration must be changed to take account of this:i.e.for a distributed load w applied between x=x and x =x2 and varying in intensity, M4=- 背wxL-x好d (6.11) w(L-x)x2 MB=- L dx (6.12)
56.4 Built-in Beams 145 Substituting in (l), Wab Wa2b L L2 M ---+- AWab(a + b) Wa2b =- +- Wab’ LZ L2 L2 = -__ 6.4. Built-in beam carrying a non-uniform distributed load Let w’ be the distributed load varying in intensity along the beam as shown in Fig. 6.4. On a short length dx at a distance x from A there is a load of w’dx. Contribution of this load to MA Wab2 = -~ L2 (where W = w’dx) w’dx x x(L - x)’ L2 1 w’x( 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 be changed to take account of this: i.e. for a distributed load w’ applied between x = xl and x = x2 and varying in intensity, (6.1 1) (6.12)
146 Mechanics of Materials §6.5 6.5.Advantages and disadvantages of built-in beams Provided that perfect end fixing can be achieved,built-in beams carry smaller maximum B.M.s (and hence are subjected to smaller maximum stresses)and have smaller deflections than the corresponding simply supported beams with the same loads applied;in other words built-in beams are stronger and stiffer.Although this would seem to imply that built-in beams should be used whenever possible,in fact this is not the case in practice.The principal reasons are as follows: (1)The need for high accuracy in aligning the supports and fixing the ends during erection increases the cost. (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 applications introducing rolling loads (e.g.bridges,etc.). These disadvantages can be reduced,however,if hinged joints are used at points on the beam where the B.M.is zero,i.e.at points of inflexion or contraflexure.The beam is then effectively a central beam supported on two end cantilevers,and for this reason the construction is sometimes termed the double-cantilever construction.The beam is then free to adjust to changes in level of the supports and changes in temperature (Fig.6.5). Pin joints Points of inflexion Fig.6.5.Built-in beam using"double-cantilever"construction. 6.6.Effect of movement of supports Consider a beam AB initially unloaded with its ends at the same level.If the slope is to remain horizontal at each end when B moves through a distance 6 relative to end A,the moments must be as shown in Fig.6.6.Taking moments about B RAXL=MA+MB and,by symmetry, M=Ma=M 2M RA二L 2M Similarly, RB in the direction shown
146 Mechanics of Materials $6.5 6.5. Advantages and disadvantages of built-in beams Provided that perfect end fixing can be achieved, built-in beams carry smaller maximum B.M.s (and hence are subjected to smaller maximum stresses) and have smaller deflections than the corresponding simply supported beams with the same loads applied; in other words built-in beams are stronger and stiffer. Although this would seem to imply that built-in beams should be used whenever possible, in fact this is not the case in practice. The principal reasons are as follows: (1) The need for 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 on the beam where the B.M. is zero, i.e. at points of inflexion or contraflexure. The beam is then effectively a central beam supported on two end cantilevers, and for this reason the construction is sometimes termed the double-cantilever construction. The beam is then free to adjust 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 “doubleantilever” construction. 6.6. Effect of movement of supports Consider a beam AB initially unloaded with its ends at the same level. If the slope is to remain horizontal at each end when B moves through a distance 6 relative to end A, the moments must be as shown in Fig. 6.6. Taking moments about B RA x L = MA+ MB and, by symmetry, MA= Mg= M 2M L .. RA=- Similarly, 2M L RB=- in the direction shown
§6.6 Built-in Beams 147 Ma=M R B.M.diagram -M Fig.6.6.Effect of support movement on B.M.s. Now from Mohr's second theorem the deflection of A relative to B is equal to the first moment of area of the B.M.diagram about A x 1/EI. [ww] ML2 ML2 24E1 -1+5)= (6.13) 6EI M- 6EIS and】 12EI6 A=Rn=L3 (6.14) in the directions shown in Fig.6.6. These values will also represent the changes in the fixing moments and end reactions for a beam under load when one end sinks relative to the other. Examples Example 6.1 An encastre beam has a span of 3 m and carries the loading system shown in Fig.6.7.Draw the B.M.diagram for the beam and hence determine the maximum bending stress set up.The beam can be assumed to be uniform,with I =42 x 106 m*and with an overall depth of 200mm. Solution Using the principle ofsuperposition the loading system can be reduced to the three cases for which the B.M.diagrams have been drawn,together with the fixing moment diagram,in Fig.6.7
$6.6 Built-in Beams 147 -M Fig. 6.6. Effect of support movement on B.M.s. Now from Mohr’s second theorem the deflection of A relative to B is equal to the first moment of area of the B.M. diagram about A x l/EI. .. 12EIS and RA=Re=- 6E1S M=--- L2 L3 (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. Draw the B.M. diagram for the beam and hence determine the maximum bending stress set up. The beam can be assumed to be uniform, with I = 42 x m4 and with an overall depth of 200 mm. Solution Using the principle ofsuperposition the loading system can be reduced to the three cases for which the B.M. diagrams have been drawn, together with the fixing moment diagram, in Fig. 6.7
148 Mechanics of Materials 40KN 30 kN/m cocoococccccxxeoccooccocxecco 出B 2m+06m+-2m 20 kN Bending moment diagrams 晋3a75w A 8 (o)u.d.. (b)40kN load ab.40x8x1.2:288kNm A2 A3wO2.延：44kNm (c)20 kN load Me--34 kN m Ma=-25.4 kN m Aa (d)Fixing-moment diagram 51.6 28.84 Total bending-moment 25.4 272 3056 diagram on bose of fixing moment line 2104 -1.64 Total bending-moment diagram re-drawn on conventional horizontal -25.4 base Fig.6.7.Illustration of the application of the"principle of superposition"to Mohr's area-moment method of solution. Now from eqn.(6.1) A1+A2+A4=A3 (号×33.75×103×3)+(3×28.8×103×3)+[(M4+M)3]=(生×14.4×103×3) 67.5×103+43.2×103+1.5(M4+M)=21.6×103 MA+MB=-59.4×103 (1) Also,from egn.(6.2),taking moments of area about A, A1X1+A2X2+A44=A3x3
148 Mechanics of Materials 40 kN A 0 20 kN I I 1 I M,= -34 kN rn M,=-25.4 kN rn I ---_ -- -- --- -- Bending moment diagrams (a) u.d.1. (b) 40kN load (c) 20 kN load (d) Fixing-moment diagram Total bending- moment diagram on base of fixing moment line Total bending-moment diagram re-drawn on conventiona I horizonta I base -34 Fig. 6.7. Illustration of the application of the “principle of superposition” to Mohr’s area-moment method of solution. Now from eqn. (6.1) A1 + A2 + A4 = A3 (~X33.75X103x3)+(~ ~28.8~10~X3)+[$(MA+M~)3]=(~~14.4~10~~3) 67.5 x lo3 +43.2 x lo3 + 1.5(MA + MB) = 21.6 x lo3 MA + MB = - 59.4 x 103 (1) Also, from eqn. (6.2), taking moments of area about A, A121 + A222 + A424 = A323
Built-in Beams 149 and,dividing areas A2 and A4 into the convenient triangles shown, 67.5×103×15)+(生×28.8×103×1.82×318+（生×28.8×103×121.8+青×1.2 +(2MA×3×号×3)+（全MB×3×3×3)=(位×14.4×103×1.2）号×1.2 +×144x0x1(2+》 (101.25+31.1+38.0)103+1.5M4+3MB=(6.92+23.3)103 1.5M4+3MB=-140×103 MA+2Mg=-93.4×103 (2) (2)-(1, MB=-34×103Nm=-34kNm and from (1), M4=-25.4×103Nm=-25.4kNm The fixing moments are therefore negative and not positive as assumed in Fig.6.7.The total B.M.diagram is then found by combining all the separate loading diagrams and the fixing moment diagram to produce the result shown in Fig.6.7.It will be seen that the maximum B.M.occurs at the built-in end B and has a value of 34kNm.This will therefore be the position of the maximum bending stress also,the value being determined from the simple bending theory 0n=M_34×103×100×10-3 42×10-6 =81×106=81MN/m2 Example 6.2 A built-in beam,4 m long,carries combined uniformly distributed and concentrated loads as shown in Fig.6.8.Determine the end reactions,the fixing moments at the built-in supports and the magnitude of the deflection under the 40kN load.Take EI =14 MNm2. 30 kN/m 40kN Fig.6.8. Solution Using Macaulay's method (see page 106) EI d2y M最-M+Rx-0x-1-0[-6]
Built-in Beams 149 and, dividing areas A, and A, into the convenient triangles shown, 2 x 1.8 3 (67.5 x IO3 x 1.5)+ (3 x 28.8 x lo3 x 1.8)- + (5 x 28.8 x lo3 x 1.2)(1.8+ $ X 1.2) + (3 MA x 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 lo3 (2) M B = - 3 4 x 1 0 3 N m = -34kNm and from (l), The fixing moments are therefore negative and not positive as assumed in Fig. 6.7. The total B.M. diagram is then found by combining all the separate loading diagrams and the fixing moment diagram to produce the result shown in Fig. 6.7. It will be seen that the maximum B.M. occurs at the built-in end B and has a value of 34kNm. This will therefore be the position of the maximum bending stress also, the value being determined from the simple bending theory MA= -25.4~ 103Nm = -25.4kNm MY 34 x 103 x io0 x 10-3 omax= - - I 42 x = 81 x lo6 = 81 MN/mZ Example 6.2 A built-in beam, 4 m long, carries combined uniformly distributed and concentrated loads as shown in Fig. 6.8. Determine the end reactions, the fixing moments at the built-in supports and the magnitude of the deflection under the 40 kN load. Take El = 14 MN m2. 30 kN/m 40 kN I X Fig. 6.8. Solution Using Macaulay's method (see page 106)
