CHAPTER 5 SLOPE AND DEFLECTION OF BEAMS Summary The following relationships exist between loading,shearing force(S.F.),bending moment (B.M.),slope and deflection of a beam: deflection =y (or 6) slope i or 0= dy dx bending moment M=EI d2y 2 shearing forceE dx3 loading =w=EI dxi In order that the above results should agree mathematically the sign convention illustrated in Fig.5.4 must be adopted. Using the above formulae the following standard values for maximum slopes and deftections of simply supported beams are obtained.(These assume that the beam is uniform,i.e.El is constant throughout the beam.) MAXIMUM SLOPE AND DEFLECTION OF SIMPLY SUPPORTED BEAMS Loading condition Maximum slope Deflection (y) Max.deflection (ymax)） Cantilever with concentrated WL2 W WL3 load W at end 2E1 i[2L-3Lx 3El Cantilever with u.d.l.across wL3 the complete span 6EI 24E3L4-4Lx+x的 wL4 8EI Simply supported beam with WL2 Wx WL> concentrated load W at the centre 16E1 48ei3L2-4] 48E1 Simply supported beam with wL3 WX 5wL4 u.d.I.across complete span 24E1 24E[L-2x+ 384E1 Simply supported beam with concentrated load W offset from centre (distance a from WL2 Wa one end b from the other) 0.062 El 92
CHAPTER 5 SLOPE AND DEFLECTION OF BEAMS Summary The following relationships exist between loading, shearing force (S.F.), bending moment (B.M.), slope and deflection of a beam: deflection = y (or 6) dY slope = i or 0 = - dx d2Y bending moment = M = EI ~ dx2 d3Y shearing force = Q = EIdx3 d4Y loading = w = EI- dx4 In order that the above results should agree mathematically the sign convention illustrated in Fig. 5.4 must be adopted. Using the above formulae the following standard values for maximum slopes and dejections of simply supported beams are obtained. (These assume that the beam is uniform, i.e. EI is constant throughout the beam.) MAXIMUM SLOPE AND DEFLECTION OF SIMPLY SUPPORTED BEAMS Loading condition Maximum slope Deflection ( y) Max. deflection (YId Cantilever with concentrated load Wat end WL2 2EI W 6E1 - ~2~3 - 3 ~2~ + x3~ WL3 3EI Cantilever with u.d.1. across the complete span wL3 6EI - W __ [3L4 - 4L3x + x4] 24EI wL4 8EI ~ Simply supported beam with concentrated load W at the centre WLZ 16EI wx 48EI __ [3L2 - 4x23 WL3 48EI ~ Simply supported beam with u.d.1. across complete span wL3 24EI ~ wx ~ [ L3 - 2Lx2 + x3] 24EI 5wL4 384EI ~ Simply supported beam with concentrated load W offset from centre (distance a from WLZ 0.062 __ El one end b from the other) 92
Slope and Defection of Beams 93 Here L is the length of span,EI is known as the flexural rigidity of the member and x for the cantilevers is measured from the free end. The determination of beam slopes and deflections by simple integration or Macaulay's methods requires a knowledge of certain conditions for various loading systems in order that the constants of integration can be evaluated.They are as follows: (1)Deflections at supports are assumed zero unless otherwise stated. (2)Slopes at built-in supports are assumed zero unless otherwise stated. (3)Slope at the centre of symmetrically loaded and supported beams is zero. (4)Bending moments at the free ends of a beam (i.e.those not built-in)are zero. Mohr's theorems for slope and deflection state that if A and B are two points on the deflection curve of a beam and B is a point of zero slope,then (1) slopetarc ofgrm betweend For a uniform beam,EI is constant,and the above equation reduces to 1 slope at AEarea of B.M.diagram between A and B N.B.-If B is not a point of zero slope the equation gives the change of slope between A and B M (2)Total deflection of A relative to B first moment of area of El diagram about A For a uniform beam total deflection of A relative to B= x first moment of area of B.M.diagram about A EI Again,if B is not a point of zero slope the equation only gives the deflection of A relative to the tangent drawn at B. Useful quantities for use with uniformly distributed loads are shown in Fig.5.1. anaaanaaaaanaaoananaaaoad A,9 元·碧 42 g碧 2 Fig.5.1
Slope and Depection of Beams 93 Here Lis the length of span, El is known as the flexural rigidity of the member and x for the cantilevers is measured from the free end. The determination of beam slopes and deflections by simple integration or Macaulay's methods requires a knowledge of certain conditions for various loading systems in order that the constants of integration can be evaluated. They are as follows: (1) Deflections at supports are assumed zero unless otherwise stated. (2) Slopes at built-in supports are assumed zero unless otherwise stated. (3) Slope at the centre of symmetrically loaded and supported beams is zero. (4) Bending moments at the free ends of a beam (i.e. those not built-in) are zero. Mohr's theorems for slope and deflection state that if A and B are two points on the deflection curve of a beam and B is a point of zero slope, then M. El slope at A = area of - diagram between A and B (1) For a uniform beam, El is constant, and the above equation reduces to 1 El slope at A = - x area of B.M. diagram between A and B N.B.-If B is not a point of zero slope the equation gives the change of slope between A and B. M. El (2) Total deflection of A relative to B = first moment of area of - diagram about A For a uniform beam total deflection of A relative to B = - x first moment of area of B.M. diagram about A Again, if B is not a point of zero slope the equation only gives the deflection of A relative to Useful quantities for use with uniformly distributed loads are shown in Fig. 5.1. 1 EI the tangent drawn at B. I I Fig. 5.1
94 Mechanics of Materials §5.1 Both the straightforward integration method and Macaulay's method are based on the d2y relationship M=EI dx2(sce§5.2andS5.3 Clapeyron's equations ofthree moments for continuous beams in its simplest form states that for any portion of a beam on three supports 1,2 and 3,with spans between of L and L2,the bending moments at the supports are related by -M1L1-2M2(L1+L2)-M3L2= 6A+A22 LLL2】 where A,is the area of the B.M.diagram,assuming span L simply supported,and is the distance of the centroid of this area from the left-hand support.Similarly,A2 refers to span L2,with 2 the centroid distance from the right-hand support(see Examples 5.6 and 5.7).The following standard resultsare usefu for 6Ax (a)Concentrated load W,distance a from the nearest outside support 6Ax Wa LT(L2-d2) (b)Uniformly distributed load w 6AX wL3 (see Example 5.6) L 4 Introduction In practically all engineering applications limitations are placed upon the performance and behaviour of components and normally they are expected to operate within certain set limits of,for example,stress or deflection.The stress limits are normally set so that the component does not yield or fail under the most severe load conditions which it is likely to meet in service. In certain structural or machine linkage designs,however,maximum stress levels may not be the most severe condition for the component in question.In such cases it is the limitation in the maximum deflection which places the most severe restriction on the operation or design of the component.It is evident,therefore,that methods are required to accurately predict the deflection of members under lateral loads since it is this form of loading which will generally produce the greatest deflections of beams,struts and other structural types of members. 5.1.Relationship between loading,S.F.,B.M.,slope and deflection Consider a beam AB which is initially horizontal when unloaded.If this deflects to a new position A'B'under load,the slope at any point C is dy i= dx
94 Mechanics of Materials #5.1 Both the straightforward integration method and Macaulay’s method are based on the relationship M = El, d2Y (see 5 5.2 and 0 5.3). dx Clapeyron’s equations of three moments for continuous beams in its simplest form states that for any portion of a beam on three supports 1,2 and 3, with spans between of L, and L,, the bending moments at the supports are related by where A, is the area of the B.M. diagram, assuming span L, simply supported, and X, is the distance of the centroid of this area from the left-hand support. Similarly, A, refers to span L,, with f2 the centroid distance from the right-hand support (see Examples 5.6 and 5.7). The following standard results are useful for -: 6Af L (a) Concentrated load W, distance a from the nearest outside support 6Af Wa LL -- (L2 - a2) -~ (b) Uniformly distributed load w 6Af wL3 L 4 -- -- (see Example 5.6) Introduction In practically all engineering applications limitations are placed upon the performance and behaviour of components and normally they are expected to operate within certain set limits of, for example, stress or deflection. The stress limits are normally set so that the component does not yield or fail under the most severe load conditions which it is likely to meet in service. In certain structural or machine linkage designs, however, maximum stress levels may not be the most severe condition for the component in question. In such cases it is the limitation in the maximum deflection which places the most severe restriction on the operation or design of the component. It is evident, therefore, that methods are required to accurately predict the deflection of members under lateral loads since it is this form of loading which will generally produce the greatest deflections of beams, struts and other structural types of members. 5.1. Relationship between loading, S.F., B.M., slope and deflection Consider a beam AB which is initially horizontal when unloaded. If this deflects to a new position A‘B under load, the slope at any point C is dx
§5.1 Slope and Deflection of Beams 95 A O Fig.5.2.Unloaded beam AB deflected to A'B'under load. This is usually very small in practice,and for small curvatures ds =dx Rdi (Fig.5.2) di 1 dy But i= dx d2y I (5.1) dx2R Now from the simple bending theory 1 M R= El Therefore substituting in eqn.(5.1) M=EI d2y dx? (52) This is the basic differential equation for the deflection of beams. If the beam is now assumed to carry a distributed loading which varies in intensity over the length of the beam,then a small element of the beam of length dx will be subjected to the loading condition shown in Fig.5.3.The parts of the beam on either side of the element EFGH carry the externally applied forces,while reactions to these forces are shown on the element itself. Thus for vertical equilibrium of EFGH, o-wax =o-do do=wdx
$5.1 Slope and Defection of Beams 95 Fig. 5.2. Unloaded beam AB deflected to A’B’ under load. This is usually very small in practice, and for small curvatures ds = dx = Rdi (Fig. 5.2) di 1 dx R -- -- But .. . dY I=- dx d2y 1 dx2 R -=- Now from the simple bending theory ME IR -- -- Therefore substituting in eqn. (5.1) M=EI- d2Y dx2 This is the basic differential equation for the deflection of beams. If the beam is now assumed to carry a distributed loading which varies in intensity over the length of the beam, then a small element of the beam of length dx will be subjected to the loading condition shown in Fig. 5.3. The parts of the beam on either side of the element EFGH carry the externally applied forces, while reactions to these forces are shown on the element itself. Thus for vertical equilibrium of EFGH, .. Q-wdx = Q-dQ dQ = wdx
96 Mechanics of Materials s5.1 Non-uniform loading w/unit length 一d 4Q-do Applied w Seeeeeec Reaction w/unit length M+dM M+dM Q-do Fig.5.3.Small element of beam subjected to non-uniform loading (effectively uniform over small length dx). and integrating, 2=wdx (5.3) Also,for equilibrium,moments about any point must be zero. Therefore taking moments about F, M+dM0+wdx?=M+2d成 Therefore neglecting the square of small quantities, dM =Odx and integrating, M=Qdx The results can then be summarised as follows: deflection =y dy slope= dx bending moment =EI dy dx shear force =EI d3y dx load distribution =EI dx In order that the above results should agree algebraically,i.e.that positive slopes shall have the normal mathematical interpretation of the positive sign and that B.M.and S.F.conventions are consistent with those introduced earlier,it is imperative that the sign convention illustrated in Fig.5.4 be adopted
96 Mechanics of Materials §5.1 Fig. 5.3. Small element of beam subjected to non-uniform loading (effectively uniform over small length dx). (5.3) and integrating, Q = f wdx Also, for equilibrium, moments about any point must be zero. Therefore taking moments about F, dx (M+dM)+wdxT = M+Qdx Therefore neglecting the square of small quantities, dM = Qdx and integrating, M = f Qdx The results can then be summarised as follows: deflection = y d2y bending moment = El ~ d3 shear force = El -JJ d4 In~ti tii~trihlltinn = 1':1 ~ .~-- -.~...~-..~.. --dx4 In order that the above results should agree algebraically, i.e. that positive slopes shall have the normal mathematical interpretation of the positive sign and that B.M. and S.F. conventions are consistent with those introduced earlier, it is imperative that the sign convention illustrated in Fig. 5.4 be adopted
§5.2 Slope and Deflection of Beams 97 (a)Deflection y=8 positive upwards Positive dy (b)Slope slope d d" Postve dx (c)B.M. B.M (d)S.F (e)Loading Upward loading positive Fig.5.4.Sign conventions for load,S.F..B.M.,slope and deflection. N in' 5.2.Direct integration method If the value of the B.M.at any point on a beam is known in terms of x,the distance along the beam,and provided that the equation applies along the complete beam,then integration of eqn.(5.4a)will yield slopes and deflections at any point, ie. dx+A or y=∬(0x++8 where A and B are constants of integration evaluated from known conditions of slope and deflection for particular values of x. (a)Cantilever with concentrated load at the end (Fig.5.5) Fig.5.5
45.2 Slope and Deflection of Beams 97 (a) Deflection y=8 positive upwards (e) Loading Upward loading positive +a .,:.i XEI , Fig. 5.4. Sign conventions for load, S.F., B.M., slope and deflection. Nlq' 5.2. Direct integration method If the value ofthe B.M. at any point on a beam is known in terms of x, the distance along the beam, and provided that the equation applies along the complete beam, then integration of eqn. (5.4a) will yield slopes and deflections at any point, i.e. or dxy s" El y = Is( Zdx) dx +Ax + B d2Y M = EI, and -= --dx+A dx where A and B are constants of integration evaluated from known conditions of slope and deflection for particular values of x. (a) Cantilever with concentrated load at the end (Fig. 5.5) w Fig. 5.5
98 Mechanics of Materials $5.2 Mx=E d2y dx2=-Wx EI dy Wx2 =-2+A dx assuming El is constant. E=、 W 6+A+B Now when x=bk y=0.A=) and when x=L,y=0.B= WL3 WL2.WL3 6- -L= 2 3 Wx3 WL2x WL3 E-6+2-3 (5.5) This gives the deflection at all values of x and produces a maximum value at the tip of the cantilever when x =0, WL3 i.e. Maximum deflection ymax=- (5.6) 3EI The negative sign indicates that deflection is in the negative y direction,i.e.downwards. Wx2 WL2 Similarly (5.7) and produces a maximum value again when x=0. dy WL2 Maximum slope (5.8) dx 2EI (positive) max (b)Cantilever with uniformly distributed load (Fig.5.6) w/metre Fig.5.6. Mxx=EI d2y wx2 2 2 Eldy wx3 dx 6+A Ely =wxt 24 +Ax+B
98 Mechanics of Materials $5.2 .. d2Y M,, = EIy = - WX dx dy Wx2 dx 2 EI-= --+A assuming EI is constant. wx3 EIy= --+Ax+B 6 Now when x=L, -- dy - 0 :. dx w12 2 A=---- and when .. WL3 WLZ w13 6 2 3 x=L,y=Q .’. B=---L= -- --+--- EI 6 2 (5.5) This gives the deflection at all values of x and produces a maximum value at the tip of the cantilever when x = 0, i.e. w13 Maximum deflection = y,= - - 3e1 The negative sign indicates that deflection is in the negative y direction, i.e. downwards. Similarly dY 1 wx2 WL2 dx EI and produces a maximum value again when x = 0. Maximum slope = (2) , =- w12 (positive) 2EI (b) Cantilever with uniformly distributed load (Fig. 5.6) Fig. 5.6. d2y wx2 dx2 2 M =EI-=---- xx dy wx3 dx 6 wx4 EIy= --+Ax+B 24 EI-= --+A (5.7)
s5.2 Slope and Deflection of Beams 99 Again,when =0andA=wL x=L dx x=L y=0 and B=wL4wLi wL4 24 6、 8 y= wx4 wL3x wL4 24+ (5.9) EI 68 wL4 dy wL3 Atx=0, and (5.10) 8EI dx m 6EI (c)Simply-supported beam with uniformly distributed load(Fig.5.7) w/metre xbecccoxceecc wL Fig.5.7. M=E1=wLx、w dr= 2 2 El dywLx2 wx dx 4 6+A Ely =WLx3 wxt 1224+Ax+B At x=0,y=0 ∴.B=0 At x=L,y=0 0=WL4 wL4 1224+AL A=、wL3 24 1「wLx3wx4wL3x7 y=El12-24-24 (5.11) In this case the maximum deflection will occur at the centre of the beam where x=L/2. ()-()()】 5wL+ (5.12) 384EI Similarly dy WL3 dx 4E at the ends of the beam. (5.13)
$5.2 Again, when Slope and Deflection of Beams dY w13 x=L, -=0 and A=- dx 6 .. At x = 0, wL4 w13 y,= -__ and (2) =- 8El rmx 6El (c) Simply-supported beam with uniformly distributed load (Fig. 5.7) I' w/metre - WL - WL 2 2 Fig. 5.1. d2y wLx wx2 dx2 2 2. M =El-=--- xx dy wLx2 wx3 dx 4 6 wLx3 wx4 12 24 EI- = __ +A - ~ Ely = ~ - __ +Ax+B At x=O, y=O .'. B=O At 99 (5.9) (5.10) (5.11) In this case the maximum deflection will occur at the centre of the beam where x = L/2. .. - 5wL4 - -__ 384El WL3 , 24EI Similarly (2) =*- at the ends of the beam. (5.12) (5.13)
100 Mechanics of Materials §5.2 (d)Simply supported beam with central concentrated load (Fig.5.8) Fig.5.8. In order to obtain a single expression for B.M.which will apply across the complete beam in this case it is convenient to take the origin for x at the centre,then: =最-作--以坠 E少=WL、Wx2 4+A Ely -WLx2 Wx 812+Ax+B At y=0.A=0 +=⊙、x L x=2y=0 ∴.0= WL3 WL3 3296 +B WL3 B=- 48 (5.14) WL3 ymax= at the centre 48EI (5.15) and WL2 at the ends (5.16) /max =士16E1 In some cases it is not convenient to commence the integration procedure with the B.M. equation since this may be difficult to obtain.In such cases it is often more convenient to commence with the equation for the loading at the general point XX on the beam.A typical example follows:
100 Mechanics of Materials $5.2 (d) Simply supported beam with central concentrated load (Fig. 5.8) W Fig. 5.8. In order to obtain a single expression for B.M. which will apply across the complete beam in this case it is convenient to take the origin for x at the centre, then: WLX2 wx3 8 12 Ely = ~-__ +Ax+B At dY x=o, -=o :. dx L 2’ x=- y=o WL3 WL3 O=- -__ +B 32 96 (5.14) 12 48 Y=- = -___ .. wL3 at the centre ymax 48EI and at the ends WLZ (5.15) (5.16) In some cases it is not convenient to commence the integration procedure with the B.M. equation since this may be difficult to obtain. In such cases it is often more convenient to commence with the equation for the loading at the general point XX on the beam. A typical example follows:
§5.2 Slope and Deflection of Beams 101 (e)Cantilever subjected to non-uniform distributed load (Fig.5.9) + w 3w Fig.5.9. The loading at section XX is -票-+6w-引-(+) Integrating, 装-(+)+A (1) Erdzy=wt3 (2+3+Ax+B (2) -（+品 /x3,x4 Ax +Bx+C 3 /x4 E1y=-w24+ Ax3 Bx2 + 60L 6* 2+Cx+D (4) Thus,before the slope or deflection can be evaluated,four constants have to be determined; therefore four conditions are required.They are: At x=0,S.F.is zero .from (1)A=0 At x =0,B.M.is zero from (2) B=0 At x =L,slope dy/dx =0(slope normally assumed zero at a built-in support) from (3) 0=-w +) 6 +C C= 4 At x=L,y=0 /L4 from (4) 0=-w ,L4 WL4 24 60 +4+D D=、 23wL4 120
$5.2 Slope and DeJIection of Beams 101 (e) Cantilever subjected to non-uniform distributed load (Fig. 5.9) Fig. 5.9. The loading at section XX is w‘ = El- d4Y = - [ w + (3w - w)’] = - w (1 + %) dx4 1 Integrating, E~-=-w d2y (; -+- ;I) +A x+B dx2 (;: 6.6,) Ax3 Bx2 (4) Ely= -W -+- +-+--++x+D 62 (3) Thus, before the slope or deflection can be evaluated, four constants have to be determined; therefore four conditions are required. They are: At x = 0, S.F. is zero .‘. from (1) A=O At x = 0, B.M. is zero .’. from (2) B=O At x = L, slope dyldx = 0 (slope normally assumed zero at a built-in support) .’. from (3) At x=L, y=O ... from (4) o=-w -+- +C (: ti) O= -w($+$)+F+D .. 23wL4 120 D= -~