Obviously these two values of the shear are equations of two parallel straight lines, that represented by equation (_f_) passing through _A_, and that represented by equation (_h_) passing through _B_, the constant vertical distance between them being _g_. Hence let _BF_ be laid off negatively downward and _AG_ positively upward, each being equal to _g_ by any convenient scale. The ordinates drawn from the various positions 1, 2, 3 ... 6 of _g_ on _AB_ to _AD_ and _BC_ will be the shears at _X_ produced by the load _g_ at any point of the span, and determined by equations (_f_) and (_h_). The influence line, therefore, for the section _X_ will be the broken line _ADCB_. When _g_ is at _X_ the sign of the shear changes, since the latter passes through a zero value. If a train of weights _W₁_, _W₂_, _W₃_, etc., passes across the span, the total shear at _X_ will be found by taking the sum of the vertical intercepts between _AB_ and _ADCB_, drawn at the positions occupied by the various single weights of the train. If those single weights are expressed in terms of the unit load _g_, the shear _S_ will have the value 1 _S_ = ---- ∑ _Wy_; _g_ _y_ being the general value of the intercept between _AB_ and the influence line. The latter shows that the greatest negative shear at _X_ will exist when the greatest possible amount of loading is placed on _AX_ only, while the greatest positive shear at the same section will exist when _BX_ only is loaded. If _BX_ is the smaller segment of span, the latter shear is called the “counter-shear,” and the former the “main shear.” If the loads are applied at panel-points of the span only, the treatment is the same in general character as that employed for bending moments. In Fig. 25_a_ let 4 and 5 be the panel-points between which the load _g_ is found, and let the panel length be _p_. Also, let _z′_ be the distance of the weight _g_ from panel-point 4. The reactions at _A_ and 4 will then be _l - z p - z′_ _R_ = ------_g_ and _R₄_ = ----------- _g_. _l p_ The shear at the section _X_ for any position of the weight _g_ will then be (_z′ z_) _S_ = _R - R₄_ = _g_(--- - ---). (_k_) (_p l_) As this is the equation of a straight line, with _S_ and _z_ or _z′_ for the coordinates, the influence line for the panel in which the section _X_ is located will be the straight line represented by _KL_ in Fig. 25_a_.