To achieve the highest degree of shearing accuracy, the machine components such as the bed, knife bar and end frames must be of sufficient size and strength, and properly ribbed and braced to offset deflection where maximum load occurs. A reliable design should incorporate “controlled deflection,” meaning that the expected deflection of the knife bar and bed will be within controlled limits according to the thickness and type of material being sheared. Depending on the speed of the machine a full range of material thicknesses with one blade setting. This is particularly true with mechanical shears as compared to the slower acting hydraulic shears. The load required to shear metal will vary according to the square of the thickness, the adjusted shear stress of the material, and inversely as to the rake of the upper blade. The term “rake” is used to designate the angel of the upper blade and is usually expressed in inches per foot. A 3/8-inch rake means that the upper blade slopes at a rate of 3/8 inches per foot of length. Because of this slope, there is only a portion of the metal that comes into contact with the blades at any one time. On a full width sheet, the shear force remains constant as the blade moves through the downward portion of the shear cycle. In a normal shearing cycle, a portion of the material is actually sheared and the balance is fractured through due to the shearing action. The depth that the blade penetrates into the material is determined by the hardness and shear strength of the material and is reflected in the factor “elongation” shown in steel charts. For example a material with an elongation factor of 37% (mild steel) would indicate that the blade would penetrate 37% into the material before fracture occurs. Elongation, or depth of penetration is an important factor in determining shear load. The formula for shear load is: SL = ST²P / (R/12)

SL = Shear Load  S = Shear Strength of Material T = Metal Thickness P = Penetration or Factor of “Elongation” R12 = Rake of Top Blade in Inches Per Foot (1' =12")

Example: 1/4" x 10' shear used to shear M1020 mild steel.  Using a standard steel chart it is determined this grade of steel has a tensile strength of 63,000 psi and the shear strength is factored at 50,000 psi, or 80% of the tensile strength (rule of thumb is that shear strength is 75-80% of the tensile strength of low carbon steels). The steel chart also tells us for this grade of steel the elongation factor is 38% (how far the shear blade must penetrate the metal before fracturing occurs).  By applying these values to the formula, we find that the shear load (tonnage) required to shear 1/4" M1020 grade steel is 38,000 lbs. or 19 tons.

SL = 50,000 (0.250")² (0.38) / 0.375 divided by 12 = (1187.5/0.0312) = 38,061 lbs or 19.3 tons