Area under shear is established by the shear rake angle, the material thickness and the material ductility. : =, where: τ = the shear stress; F = the force applied; A = the cross-sectional area of material with area parallel to the applied force vector. The width of cut and uncut thickness are 2 mm and 0.2 mm, respectively. The shear must produce more force than that needed to cut the thickest material. The shear strength is 350 MN/m 2.. Should you be shearing Aluminum, a shear sized for 1/2 the amount of For example: if cutting the Q235 steel plate with 10mm thickness and 6000mm length, the shear force is 10 × 6000 × 23.5 = 1410000(N) = 141 (Tons). Solution 115 If your application is Stainless Steel, you should select a shear that is rated for approximately 1.5-2x the material thickness you are using. Problem 115 What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? Other forms Pure. Question: Mild steel is being machined at a cutting speed of 200 m/min with a tool rake angle of 10°. The length of the edge 15 7.50 in. Sheet Rb 40,000 276 .80 25% ASTM A-36 BHN 119-159 58-80,000 1.20 20-25% 45-50 Carbon HR Sheet BHN 200 80,000 552 1.60 25% Spring Steel 1074, 1095 Hardened to Spring Temper If you're interested in the force required to break the bolt, simply substitute the Ultimate Shear Strength for Yield Shear Strength. Shearing force is roughly the product of material shear strength and the area under shear. If your application is Stainless Steel, you should select a shear that is rated for approximately 1.5-2x the material thickness you are using. Compute the force required to shear a straight edge of a sheet of SAE 1040 cold-drawn steel having a thickness of 0.105 in. When using SI units, it’s best to use units of [math][N][/math] and [math][mm][/math] when solving strength of material questions like this. is the cross sectional area of the bolt in the shear plane. General shear stress. Steel ruptures when a shear of 3.5*10^8 N/m^2 is applied.The force needed to punch a 1cm diameter hole in a steel sheet 0.3cm thick is nearly asked Jan 21, 2020 in Physics by Vannsh Jani ( … Shear Strength N/mM 2: material multiplier: Recommended Die Clearance In % Of Thickness Low Carbon HR Steel Rb 70 50,000 345 1.00 20% Low Carbon C.R. If it’s 304 steel, then the shear force is 141 × 2 = 282 tons. where P d = power required to drive the chipper disc, F = centripetal force acting on the periphery of the disc = mV 2 /R, m = Total mass of the chipper unit, R = Radius of disc., V = Disc velocity = p Dn, n is given by the ratio of n between the pulley and the disc.. Power required in driving the chipper disc is 14.15kW. All shears are rated on their capability to cut steel and should be selected according to the capacity required as related to steel. Since the shear plane is across the threaded portion of the bolt, use the minor diameter for an external M8x1 thread, 6.596mm, to calculate the area. The formula to calculate average shear stress is force per unit area. If the average value of co‐efficient of friction between the tool and the chip is 0.5 and the shear stress of the work material is 400 N/mm 2, then determine: [ESE 2005] (i) The shear angle by the shearing force required. If it’s Q345 steel plate, then the shear force should be 141× 1.4 = 197.4 tons. All shears are rated on their capability to cut steel and should be selected according to the capacity required as related to steel. 4-11.