§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