Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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- Pc=nxdxtxo =2x25x20x150 150000 N The strength of the riveted joint= Least of P, Ps , Pc =150 000 N %3D The lap join the edges of There are th %3D 000 050= 02I X 0Z X 001 ='91d=d 0000$ =D0.625 or 62.5 % %3D 0000 a-single tr Homework: 1-A single riveted lap joint is made in 1.5 cm thick plates with 2 cm diameter rivets Determine the strength of the joint, if the pitch of rivets is 6 cm? Take 6,=1200 kg/cm ; o,= 900 kg/cm² ; 6=1600 kg/cm 2-Two plates 16-mm thick are joined by á double riveted lap joint. The pitch of each row of rivets is 9 cm. The rivets are 2.5 cm in diameter. The permissible stresses are as aqnop -4 c-paraller [ans:2827kg] 2. G,=1400 kg/cm ; 6,=1100 kg/cm² ; 6.= 2400 kg/cm 2. Find the efficiency of the joint . 3-A double riveted double cover butt joint is made in 1.2 cm thich plates with 18 mm diameter rivets.find the efficiency of the joint for a pitch of 8 cm. 2. 2. ст Take :6,=1150 kg/cm ; 6, = 800 kg/cm² ; oc=1600 kg/cm 10 mm thick, the allowable stresses are : 6, =…arrow_forwardOnly mechanical engineering experts can do best. Subject deformablearrow_forward(2) h Q11 what the value of P Which are safer on the rivets as shown in figure below. given that : T=100.34 MPa, ob=370.3 MPa, 19mm diameter of .rivets 12.5m 19mm 12.5m 250mm P.arrow_forward
- Te - Effective throat thickness well in mm = 6 sin45° Consider the following drawing in which plate 1 is welded onto plate 2 as shown for tensile and shear loading. The plates are fillet welded with a weld thickness of 6 mm as shown for a length of 150 mm. The plates have an ultimate strength of 275 MPa and the welding was done with F70 welding electrodes, which possesses an ultimate strength of 483 MPa. Consider a partial factor of safety of 1.7 for the welded joint and determine the design strength of the weld joint, both in tension (Tdw) and in shear (Vdw) due to the forces acting on it. a) Determine the design strength of the following weld in tension ( ) due to vertical forces. b) Determine the design strength of the weld in shear ( ) due to horizontal forces. Attached pic is shear equation for calculation Tension equation for calculation is: Tdw = FyLwte /Ymwarrow_forwardplease help solve and explain and include FBD pleasearrow_forwardthe answers are pmax=195kN and the angle is 57.8. can you explain?arrow_forward
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