Consider a beam with a thin closed rectangular cross-section with variable thickness (Figure Q2a). Shear forces Sx = 1000N and Sy=2000N are applied through the shear centre. The material has Kc = 50 N/m/2 and shear modulus G = 1 GPa. b) q=10.8 N/cm q=10.8 N/cm >> D q=32 N/cm (500 (500 q=32 N/cm a) A D H 1cmA 0.5cm 1cmA q=10.8 N/cm q=10.8 N/cm 20cm G q=18.4 N/cm (46) =122.4 N/cm (1000) q=18.4 N/cm 0.5cm F 10cm 79=18.4 N/cm (a) Calculate the derivative of the shear flow 29/5 at E, F, G and H. s 1000 Figure Q2. a) Cross-section of the beam, which is thin-walled with variable thickness, with A,B,C,D as corners and E,F,G,H as the mid points of each straight section; b) the shear flow resulting from Sx = 1000 N; and c) the shear flow resulting from Sy = 2000N (the numbers inside circles coloured red are areas under the curve) q=18.4 N/cm [6] (b) The resulting shear flow for Sx= 1000N is shown in Figure Q2b, and that for Sy=2000N is shown in Figure Q2c. Calculate the location of the shear centre, detailing all the steps of the process. [3] (c) Calculate the maximum shear stress and its location for both cases (Sx= 1000N and Sy = 2000N). What is the angle of twist? [3] (d) If the maximum shear is 1 kPa, what is the maximum crack size it can tolerate? q=122.4 N/cm [2] (e) The location of the applied shear stress Sy is moved 10 cm to the right. Calculate and draw the new shear flow. Also calculate the angle of twist. [6]