How many 1/2 inch diameter hole that can be punch in one motion of a 1/8 inch thick plateusing a force of 50 tons. The ultimate shear stress is 52 ksi and factor of safety of 3. A. 7 B. 9 C. 8 D. 10Refer Machine Elements and Stresses Equations from the figure. Please solve the Problem elaborately. Your solution will be use as reference for my studies. Thank you so much your work will be appreciated much! V1Machine Elements1. Cylinders Rolling in opposite direction:Á. Tangential speedB. Relation of diameter and speedV, = V2 = T D, N, = r D2 N2DI N1 = D2 N2Speed of DriverC. Speed Ratio =Speed of the DrivenD. Center Distance = R, + R2 =D, + D22. Cylinders Rolling in the same directionVi = V2 = n D, N; = n D2 N2D, N1 = D2 N2A. Tangential speedB. Relation of diameter and speedSpeed of DriverSpeed of the DrivenD2 - D1C. Speed Ratio =D. Center Distance = R2 -R, =Stresses1. Stress (S) = a total resistance that a material offers to an applied load, Ib/in? , kg/cm² , KN/m?2. Ultimate stress (Su ) - is the stress that would cause failure3. Yield stress(Sy) - maximum stress without causing deformation4. Allowable stress(Sal) = Ultimate stress/Factor of SafetySy5. Design stress(Sa) - stress used in determining the size of a member. Sa =or Sa =FSFSwhere: FS = factor of safety1. Tensile Stress (S.)S =For solid circular cross-section: A =AD2#0,² -D²)For hollow circular cross-section: A =For rectangular cross-section: A = base x height = b x hFa2. Compressive Stress(Sc)S. =A3. Shearing Stress(S.)FA. For single bolt of rivet needed to join to plates together. S, =For single rivet: A = t/4 D²For double riveted joint: A = 2(TT/4 D²)where:DB. Shearing due to punching of hole.S =where A = 1 Dt (for punching a hole) A = 4 St (for square hole)Where: S = length of side of squareC. Pressure needed to punch a hole, F: F = d x t x 80, tonst = plate thicknesst = thickness, inwhere: A = DLWhere: d = hole diameter, in-Shear area4. Bearing Stress(Sb)Sp = Fb IAa. Based on yield strength FS = Sy / Sall5. Factor of safety(FS)b. Based on ultimate strength FS = S,/ SallD6. Torsional Shear Stress(Ss)Projected AreaD where: J = polar moment of inertia = T Dª /32 (for solid shaft)c = distance from neutral axis to the farthest fiberT = torquec = r (for circular cross section)7. Bending Stress(S:) Si =d = diameterFor rectangular beam S = 6 Mb h?Mcwhere: M = momentFc = distance of farthest fiber from neutral axis| = moment of inertia about the neutral axis| = bh/12 (for rectangular cross section)I Mc SfLZ = section modulusY8. Strain and Elongation Strain =StressFLStress =AEStrainFLY =AEAYwhere: y = elongation due to applied loadL = original lengthF = forceA = areaS = stress9. Thermal Elongation; Stressesy = kL (t2 - t1) S = E! = kE (t2 - t1)where: k = coeficient of thermal expansion, m/m-°CFor steel k = 6.5 x 10-6 in/in-F = 11.7 x 106 m/m-CE = 30 x 10° psiRelation between shearing and tensile stress based on theory of failure:Samax = Sty /2Smax = Sty10. Variable Stress1 Sm SaFS Sy S,Sy = yield pointwhere: FS = factor of safetySn = endurance limitSmax + SminS. = variable component stress =Smax - SminSm = mean stress =22Smax = maximum stress11. Poisson's Ratio(u) = is the ratio of lateral unit deformation to axial unit deformation.Smin = minimum stressEFE = strain =-AEwhere: G = shear modulus of elasticity2GW2 – W1Eyt2 - tyLateral StrainEy EzExL2 -L,ExL1EzLongitudin al StrainExt,

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How many 1/2 inch diameter hole that can be punch in one motion of a 1/8 inch thick plate using a force of 50 tons. The ultimate shear stress is 52 ksi and factor of safety of 3. A. 7 B. 9 C. 8 D. 10