Question

b) Two Grade 8.8 M20X bolts in standard holes are in a connection as shown Considering bolt strength only, what force. E, can be applied according to NZS3404 before it is considered that the following could occur: (0) slin (ii) bolt shear strength. (3) c) If the bearing check at a beam end support fails, what are two things that can be done to allow the same beam to be used satisfying NZS3404?

Answer 1

M2 degree so diameter of Bolt = 20mm according to NZ s3404 minimum thickness of plate is 10mm and width(B) = 90 mm Slip strength P_t = (p -d) t F_t Because [d = gross diameter = phi + 1.5 = 21.5 mm] = (90 - 21.50 times 10 times 150 = 102750 N = 102.75 KN Because [F_t = permissible of plate = 150 mpa] Shearing strength (according to question 2. way shear) P_s = 2 times rd^2/4 F_s Because[F_s = permissible shear of Bolt = 100MPa) = 2 times r/ 4 times (21.5)^2 times 100 = 72.61 KN P_b = dt F_b = 21.5 times 10 times 640 = 137.6 KN {F_b = permissible Bearing strength of BoH = 640 MPA in grade 8.8 to the bolt failure the slip strength & shearing strength of BOH are increase.