"Let ABCNML be a beam built into a vertical wall at the section ABC, and supposed to consist of a number of parallel fibres perpendicular to the wall... and equal to AN in length. Let H' be a point on the 'base of fracture,' and H'E [which is perpendicular to AC] = y, AE= x. Then if a weight Q be attached by means of a pulley to the extremity of the beam, and be supposed to produce a uniform horizontal force over the whole section NML, \; Q = r \cdot \int ydx where r is the resistance of a fibre of unit sectional area and the integration is to extend over the whole base of fracture. Q is by later writers termed the absolute resistance and is given by the above formula. Now suppose the beam to be acted upon at its extremity by a vertical force P instead of the horizontal force Q. All the fibres in a horizontal line through H' will have equal resistance, this may be measured by a line HK drawn through H in any fixed direction where H is the point of intersection of the horizontal line through H and the central vertical BD of the base. As H moves from B to D, K will trace out a curve GK which gives the resistance of the corresponding fibres. Take moments for the equilibrium of the beam about ACP \cdot l = \iint uydxdywhere l = length of the beam DT and u = HK."