CHAPTER 7 SHEAR STRESS DISTRIBUTION Summary The shear stress in a beam at any transverse cross-section in its length,and at a point a vertical distance y from the neutral axis,resulting from bending is given by Ib or t-Tb where o is the applied vertical shear force at that section;A is the area of cross-section "above"y,i.e.the area between y and the outside of the section,which may be above or below the neutral axis (N.A.);yis the distance of the centroid of area A from the N.A.;I is the second moment of area of the complete cross-section;and b is the breadth of the section at position y. For rectangular sections, 62「d2 30 with Imax=2bd when y=0 For I-section beams the vertical shear in the web is given by -小+[] with a maximum value of 器+[ The maximum value of the horizontal shear in the flanges is d-4 2b For circular sections =-(] with a maximum value of 40 tmax=3πR2 The shear centre of a section is that point,in or outside the section,through which load must be applied to produce zero twist of the section.Should a section have two axes of symmetry, the point where they cross is automatically the shear centre. 154
CHAPTER 7 SHEAR STRESS DISTRIBUTION Summary The shear stress in a beam at any transverse cross-section in its length, and at a point a vertical distance y from the neutral axis, resulting from bending is given by QAj 7=- Ib where Q is the applied vertical shear force at that section; A is the area of cross-section “above” y, i.e. the area between y and the outside of the section, which may be above or below the neutral axis (N.A.); jj is the distance of the centroid of area A from the N.A.; I is the second moment of area of the complete cross-section; and b is the breadth of the section at position y. For rectangular sections, 7=- :$[: -- y2] with 7,,=- 3Q when y =O 2bd For I-section beams the vertical shear in the web is given by with a maximum value of The maximum value of the horizontal shear in the flanges is For circular sections with a maximum value of 4Q T,, = __ 3aR2 The shear centre of a section is that point, in or outside the section, through which load must be applied to produce zero twist of the section. Should a section have two axes of symmetry, the point where they cross is automatically the shear centre. 154
Shear Stress Distribution 155 The shear centre of a channel section is given by k2h2t e÷4l Introduction If a horizontal beam is subjected to vertical loads a shearing force (S.F.)diagram can be constructed as described in Chapter 3 to determine the value of the vertical S.F.at any section. This force tends to produce relative sliding between adjacent vertical sections of the beam, and it will be shown in Chapter 13,$13.2,that it is always accompanied by complementary shears which in this case will be horizontal.Since the concept of complementary shear is sometimes found difficult to appreciate,the following explanation is offered. Consider the case of two rectangular-sectioned beams lying one on top of the other and supported on simple supports as shown in Fig.7.1.If some form of vertical loading is applied the beams will bend as shown in Fig.7.2,i.e.if there is negligible friction between the mating surfaces of the beams each beam will bend independently of the other and as a result the lower surface of the top beam will slide relative to the upper surface of the lower beam. Fig.7.1.Two beams (unconnected)on simple supports prior to loading. Relotive sliding between beams Fig.7.2.Illustration of the presence of shear(relative sliding)between adjacent planes of a beam in bending. If,therefore,the two beams are replaced by a single solid bar of depth equal to the combined depths of the initial two beams,then there must be some internal system of forces, and hence stresses,set up within the beam to prevent the above-mentioned sliding at the central fibres as bending takes place.Since the normal bending theory indicates that direct stresses due to bending are zero at the centre of a rectangular section beam,the prevention of sliding can only be achieved by horizontal shear stresses set up by the bending action. Now on any element it will be shown in $13.2 that applied shears are always accompanied by complementary shears of equal value but opposite rotational sense on the perpendicular faces.Thus the horizontal shears due to bending are always associated with complementary vertical shears of equal value.For an element at either the top or bottom surface,however
Shear Stress Distribution 155 The shear centre of a channel section is given by kZhZt e=- 41 Introduction If a horizontal beam is subjected to vertical loads a shearing force (S.F.) diagram can be constructed as described in Chapter 3 to determine the value of the vertical S.F. at any section. This force tends to produce relative sliding between adjacent vertical sections of the beam, and it will be shown in Chapter 13, 513.2, that it is always accompanied by complementary shears which in this case will be horizontal. Since the concept of complementary shear is sometimes found difficult to appreciate, the following explanation is offered. Consider the case of two rectangular-sectioned beams lying one on top of the other and supported on simple supports as shown in Fig. 7.1. If some form of vertical loading is applied the beams will bend as shown in Fig. 7.2, i.e. if there is neghgible friction between the mating surfaces of the beams each beam will bend independently of the other and as a result the lower surface of the top beam will slide relative to the upper surface of the lower beam. Fig. 7.1. Two beams (unconnected) on simple supports prior to loading. Relative sliding between beams Fig. 7.2. Illustration of the presence of shear (relative sliding) between adjacent planes of a beam in bending. If, therefore, the two beams are replaced by a single solid bar of depth equal to the combined depths of the initial two beams, then there must be some internal system of forces, and hence stresses, set up within the beam to prevent the above-mentioned sliding at the central fibres as bending takes place. Since the normal bending theory indicates that direct stresses due to bending are zero at the centre of a rectangular section beam, the prevention of sliding can only be achieved by horizontal shear stresses set up by the bending action. Now on any element it will be shown in 5 13.2 that applied shears are always accompanied by complementary shears of equal value but opposite rotational sense on the perpendicular faces. Thus the horizontal shears due to bending are always associated with complementary vertical shears of equal value. For an element at either the top or bottom surface, however
156 Mechanics of Materials 67.1 there can be no vertical shears if the surface is "free"or unloaded and hence the horizontal shear is also zero.It is evident,therefore,that,for beams in bending,shear stresses are set up both vertically and horizontally varying from some as yet undetermined value at the centre to zero at the top and bottom surfaces. The method of determination of the remainder of the shear stress distribution across beam sections is considered in detail below. 7.1.Distribution of shear stress due to bending Area A NA M+dM -dx (a) (b》 (M+dM)ybdy My bdy Force due to r dy (c) Fig.7.3. Consider the portion of a beam of length dx,as shown in Fig.7.3a,and an element AB distance y from the N.A.Under any loading system the B.M.across the beam will change from M at B to (M+dM)at A.Now as a result of bending, longitudinal stress a= My 1 longitudinal stress at=(M+dM)y 1 and longitudinal stress at B-My longitudinal force on the element atMM)by My and longitudinal force on the element at B= x bdy The force system on the element is therefore as shown in Fig.7.3c with a net out-of-balance force to the left
156 Mechanics of Materials $7.1 there can be no vertical shears if the surface is "free" or unloaded and hence the horizontal shear is also zero. It is evident, therefore, that, for beams in bending, shear stresses are set up both vertically and horizontally varying from some as yet undetermined value at the centre to zero at the top and bottom surfaces. The method of determination of the remainder of the shear stress distribution across beam sections is considered in detail below. 7.1. Distribution of shear stress due to bending C D // I I (Mt dM)ybd y I 1 I Fig. 1.3. Consider the portion of a beam of length dx, as shown in Fig. 7.3a, and an element AB distance y from the N.A. Under any loading system the B.M. across the beam will change from M at B to (A4 + dM) at A. Now as a result of bending, and .. longituLinr MY longitudinal stress o = - (M+dWY MY longitudinal stress at B = ~ I I longitudinal stress at A = I x ". (M+dWY I rce on the element at A = oA = Y MY and The force system on the element is therefore as shown in Fig. 7.3~ with a net out-of-balance force to the left longitudinal force on the element at B = - x bdy I
§7.2 Shear Stress Distribution 157 (d +M)ndy ndy 1 dM 起了bdy Therefore total out-of-balance force from all sections above height y h [am For equilibrium,this force is resisted by a shear force set up on the section of length dx and breadth b,as shown in Fig.7.4. h aM ybdy Fig.7.4. Thus if the shear stress is t,then dM ybdy (7.10 h But ybdy first moment of area of shaded portion of Fig.7.3b about the N.A. =A5 where A is the area of shaded portion and y the distance of its centroid from the N.A. dM Also rate of change of the B.M. =S.F.O at the section t=24y Ib (7.2) or,alternatively, ydA where dA =bdy (7.3) 7.2.Application to rectangular sections Consider now the rectangular-sectioned beam of Fig.7.5 subjected at a given transverse cross-section to a S.F.O
$7.2 Shear Stress Distribution 157 Therefore total out-of-balance force from all sections above height y = jgybdy I Y For equilibrium, this force is resisted by a shear force set up on the section of length dx and breadth b, as shown in Fig. 7.4. h C D [ ybdy Y Fig. 7.4. Thus if the shear stress is T, then h But j ybdy = first moment of area of shaded iortion of Fig. 7.3b about the N.A. Y = Aj where A is the area of shaded portion and j the distance of its centroid from the N.A. dM ~ = rate of change of the B.M. dx Also = S.F. Q at the section .. QAY z=- lb or, alternatively, z = 2 ydA where dA = bdy lb Y 7.2. Application to rectangular sections Consider now the rectangular-sectioned beam of Fig. 7.5 subjected at a given transverse cross-section to a S.F. Q
158 Mechanics of Materials §7.3 NA Fig.7.5.Shear stress distribution due to bending of a rectangular section beam. Ib 8哈是到 -0 2×122 60「d2 -04 (i.e.a parabola) (7.40 Now 60、d230 Tmx=6dX4= 2bd when y=0 (7.5) and average t bd tmax=是X Taverage (7.6) 7.3.Application to I-section beams Consider the I-section beam shown in Fig.7.6. Para bolic dy Fig.7.6.Shear stress distribution due to bending of an I-section beam
158 Mechanics of Materials $7.3 Now and .. Fig. 7.5. Shear stress distribution due to bending of a rectangular section beam. = 6Q d2 [T - y2] (i.e. a parabola) 6Q d2 3Q bd3 4 2bd TmaX = - x - = - when y = 0 Q average z = - bd % max = t x %average 7.3. Application to I-section beams Consider the I-section beam shown in Fig. 7.6. ,Porobolic , Y Fig. 7.6. Shear stress distribution due to bending of an I-section beam
§7.3 Shear Stress Distribution 159 7.3.1.Vertical shear in the web The distribution of shear stress due to bending at any point in a given transverse cross- section is given,in general,by eqn.(7.3) d/2 0 T= ydA Ib In the case of the I-beam,however,the width of the section is not constant so that the quantity dA will be different in the web and the flange.Equation (7.3)must therefore be modified to h/2 d/2 t= 0 It tydy+ h/2 「h2 274 小 As for the rectangular section,the first term produces a parabolic stress distribution.The second term is a constant and equal to the value of the shear stress at the top and bottom of the web,where y =h/2, i.e. tA=T8= 0b「d2h2] 2lt44 (7.7) The maximum shear occurs at the N.A.,where y=0, Qh2,2b「d2h27 81+2m4-4 (7.8) 7.3.2.Vertical shear in the flanges (a)Along the central section YY The vertical shear in the flange where the width of the section is b is again given by eqn.(7.3) as d/2 T= d/2 (7.9) The maximum value is that at the bottom of the flange when yi=h/2, (7.10 this value being considerably smaller than that obtained at the top of the web
$7.3 Shear Stress Distribution 159 7.3.1. Vertical shear in the web The distribution of shear stress due to bending at any point in a given transverse crosssection is given, in general, by eqn. (7.3) dl2 .=-I Q ydA Ib Y In the case of the I-beam, however, the width of the section is not constant so that the quantity dA will be different in the web and the flange. Equation (7.3) must therefore be modified to hl2 di2 T =e It [ tydy+g I by,dy, Y hi2 As for the rectangular section, the first term produces a parabolic stress distribution. The second term is a constant and equal to the value of the shear stress at the top and bottom of the web, where y = hl2, i.e. The maximum shear occurs at the N.A., where y = 0, Qh2 Qb d2 h2 ‘Fmax=-+- --- 81 21t[4 41 (7.7) 7.3.2. Vertical shear in the flanges (a) Along the central section YY The vertical shear in the flange where the width of the section is b is again given by eqn. (7.3) as di2 Q Ib T = - y,dA Yl di2 = gj y,bdy, = lb Yl The maximum value is that at the bottom of the flange when y, = h/2, (7.9) (7.10) this value being considerably smaller than that obtained at the top of the web
160 Mechanics of Materials §7.3 At the outside of the flanges,where y=d/2,the vertical shear(and the complementary horizontal shear)are zero.At intermediate points the distribution is again parabolic producing the total stress distribution indicated in Fig.7.6.As a close approximation, however,the distribution across the flanges is often taken to be linear since its effect is minimal compared with the values in the web. (b)Along any other section SS,removed from the web At the general section SS in the flange the shear stress at both the upper and lower edges must be zero.The distribution across the thickness of the flange is then the same as that for a rectangular section of the same dimensions. The discrepancy between the values of shear across the free surfaces CA and ED and those at the web-flange junction indicate that the distribution of shear at the junction of the web and flange follows a more complicated relationship which cannot be investigated by the elementary analysis used here.Advanced elasticity theory must be applied to obtain a correct solution,but the values obtained above are normally perfectly adequate for general design work particularly in view of the following comments. As stated above,the vertical shear stress in the flanges is very small in comparison with that in the web and is often neglected.Thus,in girder design,it is normally assumed that the web carries all the vertical shear.Additionally,the thickness of the web t is often very small in comparison with b such that eqns.(7.7)and (7.8)are nearly equal.The distribution of shear across the web in such cases is then taken to be uniform and equal to the total shear force O divided by the cross-sectional area (th)of the web alone. 7.3.3.Horizontal shear in the flanges The proof of $7.1 considered the equilibrium of an element in a vertical section of a component similar to element A of Fig.7.9.Consider now a similar element B in the horizontal flange of the channel section (or I section)shown in Fig.7.7. The element has dimensions dz,t and dx comparable directly to the element previously treated of dy,b and dx.The proof of $7.1 can be applied in precisely the same way to this flange element giving an out-of-balance force on the element,from Fig.7.9(b), (M+dM)y.tdMy.td 1 dM Iy.tdz with a total out-of-balance force for the sections between z and L y.tdz
160 Mechanics of Materials $7.3 At the outside of the flanges, where y, = d/2, the vertical shear (and the complementary horizontal shear) are zero. At intermediate points the distribution is again parabolic producing the total stress distribution indicated in Fig. 7.6. As a close approximation, however, the distribution across the flanges is often taken to be linear since its effect is minimal compared with the values in the web. (b) Along any other section SS, removed from the web At the general section SS in the flange the shear stress at both the upper and lower edges must be zero. The distribution across the thickness of the flange is then the same as that for a rectangular section of the same dimensions. The discrepancy between the values of shear across the free surfaces CA and ED and those at the web-flange junction indicate that the distribution of shear at the junction of the web and flange follows a more complicated relationship which cannot be investigated by the elementary analysis used here. Advanced elasticity theory must be applied to obtain a correct solution, but the values obtained above are normally perfectly adequate for general design work particularly in view of the following comments. As stated above, the vertical shear stress in the flanges is very small in comparison with that in the web and is often neglected. Thus, in girder design, it is normally assumed that the web carries all the vertical shear. Additionally, the thickness of the web t is often very small in comparison with b such that eqns. (7.7) and (7.8) are nearly equal. The distribution of shear across the web in such cases is then taken to be uniform and equal to the total shear force Q divided by the cross-sectional area (th) of the web alone. 7.3.3. Horizontal shear in the flanges The proof of $7.1 considered the equilibrium of an element in a vertical section of a component similar to element A of Fig. 7.9. Consider now a similar element E in the horizontal flange of the channel section (or I section) shown in Fig. 7.7. The element has dimensions dz, t and dx comparable directly to the element previously treated of dy, b and dx. The proof of $7.1 can be applied in precisely the same way to this flange element giving an out-of-balance force on the element, from Fig. 7.9(b), My. tdz y.tdz---- - (M + dM) - I I dM I = -y.tdz with a total out-of-balance force for the sections between z and L
§7.3 Shear Stress Distribution 161 dz Element B dx dy Element A o =(M +dM)y I (a) (b】 (c} Fig.7.7.Horizontal shear in flanges. This force being reacted by the shear on the element shown in Fig.7.9(c), ttdx L [dm rtdx ·ytdz L dM 1 and r= dx'It tdzy dM But tdz.y=Ay and =0. dx QAy t= (7.110 It Thus the same form of expression is obtained to that of eqn(7.2)but with the breadth b of the web replaced by thickness t of the flange:I and y still refer to the N.A.and A is the area of the flange 'beyond'the point being considered. Thus the horizontal shear stress distribution in the flanges of the I section of Fig.7.8 can
$7.3 Shear Stress Distribution 161 (a) (C) Fig. 7.7. Horizontal shear in flanges. This force being reacted by the shear on the element shown in Fig. 7.9(c), = Ttdx z and But .. T=- dM - 1tdz.y dx 'It z dM dx tdz.y = Aj and ~ = Q. (7.11) Thus the same form of expression is obtained to that of eqn (7.2) but with the breadth b of the web replaced by thickness t of the flange: 1 and y still refer to the N.A. and A is the area of the flange 'beyond the point being considered. Thus the horizontal shear stress distribution in the flanges of the I section of Fig. 7.8 can
162 Mechanics of Materials §7.4 b/2 22 NA Fig.7.8. now be obtained from egn.(7.11): QAy It with _d_=d-t) 9=22 A=t dz t=t b12 Thus 1- e d-k,dk=号d-ag The distribution is therefore linear from zero at the free ends of the flange to a maximum value of Qb (dt)at the centre (7.12) 7.4.Application to circular sections In this case it is convenient to use the alternative form of eqn.(7.2),namely (7.1), dM tbdx bydy bydy Consider now the element of thickness dz and breadth b shown in Fig.7.9
162 Mechanics of Materials 97.4 N.A. -- I Fig. 7.8. now be obtained from eqn. (7.11): with Thus A = t,dz t = t, bl2 +(d - tl)tldz = - Q (d - tl) [z]:” 21 0 The distribution is therefore linear from zero at the free ends of the flange to a maximum value of (7.12) Qb rmax = - (d - tt) at the centre 41 7.4. Application to circular sections In this case it is convenient to use the alternative form of eqn. (7.2), namely (7.1), Consider now the element of thickness dz and breadth b shown in Fig. 7.9
§7.4 Shear Stress Distribution 163 Strip element .b- NA Fig.7.9. Now b =2R cos a,y=z=R sina and dz=R cos ada and,at section distance y from the N.A.,b 2R cos1, r/2 Q t= -I×2Rcos1J 2R cos a R sin a R cos a da 2×4 2 R cos1×πR4 2Rcosa sima da since 40 c0s3x1π/2 元R2cos￥1 3 1 4Q a 4Q cos2a1 3πR2cosx1 cos3 a1 3元R2 40 40 3R[1-sin2a]=3R2 -(食)] (7.13) i.e.a parabola with its maximum value at y =0. 40 Thus 3πR2 Now 0 mean stress πR2 40 maximum shear stress 3rR24 (7.14) mean shear stress 03 元R2 Alternative procedure Using eqn.(7.2),namely t= QAy Th,and referring to Fig.7.9, b =(R2-22)112-R coso and sina=
$7.4 Shear Stress Distribution 163 m Fig. 1.9. I .aw b = 2R cos a, y = z = R sin a and dz = R cos ada anL, at section distance y from the N.A., b = 2R cosal, nlZ 2R cosa R sin a R cosada I x 2R cos a, .. z= nR4 2R3 cos a sin a du since I = __ 2R cosal x nR4 4 - - - 4Q [cos3al] = 4Qcos2a1 - 4Q 3nR2 cos a1 n/Z 3ZRZ [I - sinZ all = - 3nRZ 3nR2 -- i.e. a parabola with its maximum value at y = 0. Thus Now 4Q z msx = - 3nR2 Q mean stress = nR 4Q maximum shear stress 3aR2 4 .. =-=- mean shear stress Q3 ZR2 Alternative procedure Using eqn. (7.2), namely 7 = w, and referring to Fig. 7.9, Ib (7.13) (7.14) b Z - = (Rz -z2)'/' = R cosa and sin a = - 2 R