Question

Two metallic plates of thickness e=5 mm are joint together using joint covers of the same thickness t=2,5 mm and width b=5 cm. Other dimensions of the joint are given by: s=1 cm and r=1 mm. The joint is realized with four screws without pretensioning of diameter 8 mm. The screw holes have a diameter that is one millimeter larger than the diameter of the screws. The admissible shear stress in screws or plates is Tadm=110 MPa, the admissible compressive stress in the screws is O admc=235 MPa, and the admissible tensile stress for the plates is O admt=155 MPa. Calculate the maximum external load P (in kN) that the joint can withstand without failure. EEEEEE ܂ s | s ܟ݂ ;]3]

Answer 1

solo: A Joint can fail in either ways given below. Shear failure of screw & crushing failure of screw ③ Tearing of plate To ensure safety of a joint it should be a safe in all the fallure criteria. 1. Shear failure of screw :- there are 4 screws with diameter d = 8 mon Total Sheared area As = 2x4x I (8) ² 4 Sheared area in a screw.

As = 402.123 mm admission shear stress giver in screw 2 = 110MPa Therefore maximum permissible load & will be P - Asx T = 402. 123x110 = 44233. 624 P = 44.233 KN

© crushing failure of screw Crushing area Ac = axe = 8x5 = 40mm2 Toral crushing area in 4 screw A=4&A - Ibomma given permissible compressive Stress 6= 235 MPa Now maximum permessible load P. - Ax6= 160x235 = 37.6 KN. 1ooo

loo moott A Tearing plane 3. Tearing of plates :- Tearing for Plates Teaning area Az = (b-24)xe At = (50-2x8) x 5 = 170mm² given permissible tensile stress in the plate 6 = 155 MPa Therefore maximum permissible load P P = At * r = 170x155 = 26350 = 26-35 KN Tearing for Cover : Tearing area At = (b-2d) xt = (50-2x8) x 2.5 At = 85 mm2 Maximum permissible load p=85x155 I 1000 13. 175 KNV

To ensure the safety of entire joint we will minimum Value of load P which is equal to 13.175 KN This maximum external load that can joint can withstand without failure is P = 13.175 KN