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 beam158 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