|---------------------------------------------| Both shearing stresses may act at the same time. Thus the weight carried by a beam tends to shear it off at right angles to the axis; this stress is equal to the resultant force acting perpendicularly at any point, and in a beam uniformly loaded and supported at either end is maximum at the points of support and zero at the centre. In addition there is a shearing force tending to move the fibres of the beam past each other in a longitudinal direction. (See Fig. 12.) This longitudinal shear is maximum at the neutral plane and decreases toward the upper and lower surfaces. [Illustration: FIG. 12.--Horizontal shear in a beam.] Shearing across the grain is so closely related to compression at right angles to the grain and to hardness that there is little to be gained by making separate tests upon it. Knowledge of shear parallel to the grain is important, since wood frequently fails in that way. The value of shearing stress parallel to the grain is found by dividing the maximum load in pounds (P) by the area of the cross section in inches (A). { P } { Shear = --- } { A } Oblique shearing stresses are developed in a bar when it is subjected to direct tension or compression. The maximum shearing stress occurs along a plane when it makes an angle of 45 degrees P with the axis of the specimen. In this case, shear = -----. When 2 A the value of the angle [Greek: theta] is less than 45 degrees, P the shear along the plane = --- sin [Greek: theta] cos [Greek: A theta]. (See Fig. 13.) The effect of oblique shear is often visible in the failures of short columns. (See Fig. 14.) [Illustration: FIG. 13.--Oblique shear in a short column.] [Illustration: FIG. 14.--Failure of short column by oblique shear.]