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 = -__