§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