If _x = l/2_ in equation (20), _S_ becomes equal to zero. In other words, there is no shear at the centre of the span of a beam uniformly loaded. Hence, if at each end of the span a vertical line _AK_ or _BL_ be laid off downward, and if straight lines _KD_ and _DL_ be drawn, any vertical intercept, as _GH_, between these lines and _AB_ will represent the shear at the corresponding point. Equation (20) also shows that the shear _S_ at any point is equal to the load resting on the beam between the centre _D_ and that point. Although this case of uniform loading is a special one it finds wide application in practical operations. =84. Greatest Shear for Uniform Moving Load.=—The preceding loads have been treated as if they were occupying fixed positions on the beams considered. This is not always the case. Many of the most important problems in connection with the loading of beams and bridges arise under the supposition that the load is movable, like that of a passing railroad train. One of the simplest of these problems, although of much importance, consists in finding the location of a uniform moving load, like that of a train of cars, which will produce the greatest shear at a given point of a simple beam, such as that represented in Fig. 15, in which a moving load is supposed to pass continuously over the span from the left-hand end _A_. It is required to determine what position of this uniform load will produce the greatest shear at the section _C_. [Illustration: FIG. 15.] Let the moving load extend from _A_ to any point _D_ to the right of _C_. The two reactions _R_ and _Rʹ_ may be found by the methods already indicated. Let _W_ represent the uniform load resting on the portion _CD_ of the span. The shear _S′_ existing at _C_ will be _Sʹ = Rʹ - W_. (21) Let _R‴_ be that part of _Rʹ_ which is due to _W_, and _Rʺ_ that part due to the load on _AC_. Evidently _R‴_ is less than _W_; then _Sʹ = Rʺ + R‴ - W_. (22) Since the negative quantity _W_ is greater than the positive quantity _R‴_, _S′_ will have its greatest value when both _W_ and _R‴_ are zero. Hence the greatest shear at the point _C_ will exist when _Sʹ = Rʺ_. (23) Obviously the loading must extend at least from _A_ to _C_ in order that _Rʺ_ may have its maximum value. Hence _the greatest shear at any section will exist when the uniform load extends from the end of the span to that section, whatever may be the density of the load_. If the segment of the span covered by the moving load is greater than one half the span, the maximum shear is called the _main shear_; but if that segment is less than one half the span, the maximum shear is called the _counter-shear_. The reason for these two names will be apparent later in the discussion of bridge-trusses.