148 Mechanics of Materials 2 §5.6 0 Fig.5.6.Thin-walled closed section subjected to axial torque. At any point,then,the shear force on an element of length ds is =rtds =gds and the shear stress is g/t. Consider now,therefore,the element BC subjected to the shear force o=gds rt ds. The moment of this force about O =dT=Op where p is the perpendicular distance from O to the force O. dT=gds p Therefore the moment,or torque,for the whole section = apds-afpds But the area COB=base x height =ipds ie. dA=pds or 2dA pds torque T=2q dA T=2qA (5.14) where A is the area enclosed within the median line of the wall thickness. Now,since 9=t it follows that T=2tA T or t=2At (5.15) where t is the thickness at the point in question.148 Mechanics of Materials 2 $5.6 Fig. 5.6. Thin-walled closed section subjected to axial torque. At any point, then, the shear force Q on an element of length ds is Q = rt ds = q ds and Consider now, therefore, the element BC subjected to the shear force Q = qds = ttds. The moment of this force about 0 the shear stress is q/t. =dT=Qp where p is the perpendicular distance from 0 to the force Q. .. dT = qdsp Therefore the moment, or torque, for the whole section But the area COB = 4 base x height = ipds i.e. dA= ipds or 2dA=pds torque T = 2q dA .. s T = 2qA where A is the area enclosed within the median line of the wall thickness. Now, since q = rt it follows that or T = 2ttA - T t= - 2At (5.14) (5.15) where t is the thickness at the point in question