In shear cutting operation, a sheet of 5 mm thickness is cut along a length of 200 mm. The cutting blade is 400 mm long (see figure) and zero-shear (S = 0) is provided on the edge. The ultimate shear strength of the sheet is 100 MPa and penetration to thickness ratio is 0.2. Neglect friction. 400 S I Assuming force vs displacement curve to be rectangular, the work done (in J) is (a) 100 (c) 250 (b) 200 (d) 300

a) Calculate the reaction forces in terms of F andtheta.b) Calculate the internal shear force and moment as a function of x using cuts and summingforces and moments in terms of F, L, andtheta.c) Calculate the shear force and moments by integration in terms of x, L, F andtheta.d) Assuming a rectangular cross section compute the average normal stress for the case F =10kNat 30° where b =0.04 m, h = 0.08 m, at the point x = L/4, L = 1 m at the centroid.e) With the cross section given above compute the average normal stress for the case F =10kN at30° at the point x = L/4, L = 1 m at each surface of the beam.f) Calculate the shear stress at the top and bottom surface, at the neutral axis and at x = 0.25 mand y = 0.02 m above the neutral axis.

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Q2(c) The steel alloy shaft having a diameter of 38 mm is used to transmit torque, T as shown in Figure Q2(c). If the allowable shear stress of the shaft is 80kPa, i)Determine the maximum torque T that can be transmitted by the shaft. ii)What would be the maximum torque T' if a 25 mm diameter hole is bored through the shaft? Figure Q2(c.) |Q2 c) (i) T c The torsion formula :7 = J Т (0.019) T (0.019)^4 ; Therefore, the max torque T = x10 Nm 80 2 Q2 c) (ii) For point lying on the outside suurface of the shaft: x10 ="(0.019) * (0.019*-0.0125 ) ; Therefore the max torque T' Nm 80 2 Q2 c) (iii) The shear stress at point located at inner surface of shaft, T'c T = -= J ) (0.0125) = 7 (0.019*-0.01254) T'c kPa 4. 2 2.