Question

6. Calculate the principal and maximum shearing stresses at points A and B on the bar with rectangular cross-section. 0.7m :0.5mm 30mm 10mm 5 kN 1 kN 20mm

Answer 1

The various forces acting on the section containing point A & B are F_X = 5 KN (Normal force) F_y = 1 KN (shear force) F_Z = 2 KN (shear force) rightarrow the vertion Bending moments acting on the sector containing point A & B are M_y = 2 KN (0.7 m) = 1.4 KN - m M_z = 1 KN (0.7 m + 0.5 m) = 1.2 KN - m \r\n section properties: - cross - sectional are A = (20 mm times 40 mm) = 800 mm^2 Area moment as generator I_y = 40 (20)^3/12 = 26666.67 mm^4 I_z = 20 (40)^3/12 = 106666.67 mm^4 \r\n point A Stresses: - F_x = 5 KN = 5000 N produces normal strees: J, J_1x = F_x/A = 5000 N/800 mm^2 = 6.25 N/mm^2 (Tensite) F_y = 1 KN produces no stress at A F_Z = 2 KN produces streer stress at A: T z = z_max = 3/2 (z_ary)