5. Jan 2013•0 gefällt mir•6,931 views

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By Dr.S.G.A

hotman1991Folgen

- 2. Riveting applications Rivets Types of riveted joints Failure of riveted joints
- 3. Riveting applications: Pressure vessels, boilers tanks Bridges Hulls of ships Airplanes Cranes buildings Machinery in general
- 4. Rivets head A rivet is a round bar consisting of head shank. The rivet blank is heated to a red glow Inserted into the holes; Shank The head is held firmly against the Point plate The projecting end is formed into a second head, called the point,
- 5. Rivet material: •Tough and ductile low carbon steel •Nickel steel. •Brass •Aluminium
- 6. Types of riveted joints: Lap joints Rivet Strap butt joints
- 7. Riveted joint terminology. Gauge line Margin lap Pitch (p) distance ( m) Back pitch (pt)
- 8. Failure of riveted joints Bending of rivet or plate F F •In Lap connections the offset creates a moment • M=Ft/2 •Bending moment causes complex deformations and stresses •In most cases this offset moment is neglected •A suitable factor of safety is used.
- 9. b) Shearing of the rivets: Single shear FF F Double shear F/2 F F/2
- 10. Joint strength in shear 2 πd S s Fs (2 n 2 n1 ) 4 Where, n1: :number of rivets in single shear n2: number of rivets in double shear Ss: allowable shear stress d : diameter of rivet
- 11. c) Crushing of the rivets or the plates Crushing of margin F F Crushing of the rivet or plate occurs due to the pressure on the cylindrical surface of the rivet and the plate The resistance to crushing of rivets is, Fc (n 2 h 1 n 1 h 2 ) dS c where h1 h2: plate thickness Sc: allowable crushing stress
- 12. d) Rupture of plate by tension: F L F Rupture of plate occurs at the section between the rivets The resistance of rupture can be obtained from the expression: Where: Ft ( L nd ) hS t L: plate width h: plate thickness: St : allowable tensile stress n : number of rivet holes at the section
- 13. Undrilled Section F F L In the undrilled section: Ft LhS t Where L : width of plate
- 14. d) Tearing and shearing of the margin Tearing of the margin F F Margin Shearing of the margin F F For the riveted joint to resist tearing and shearing of the margin, the margin (m) = 1.5 d for double shear 2d for single shear.
- 15. Example: Fig.(3) shows a lap riveted joint, consists of two Rolled steel plates, SAE 1020, of 0.5 in thickness. The plates are riveted together with four rivets 0.375 inch in diameter of low carbon steel, SAE 1010. Estimate the maximum value of the force F that the joint can stand while considering a factor of safety equals 2 and the rivets are driven by hand hammer. 0.5 in F F 4 in. 2 1 1
- 16. Solution Shearing of the rivets: d = 0.375 fs = 2 π 2 n1 = 4 F (2 n2 n1) d Ss s n2 =0 4 fs π 0 .375 Rivet-driving 2 Load carrying member Type of stress 10000 Rivets acting in Rivets acting in Fs 4 power single shear double shear Rolled steel, SAE 1020 Tension 4 2 ….. 18000 18000 Shear Power 13500 13500 Rivets, SAE1010 FS = Shear lb 2209 Hand 10000 10000 Crushing Power 24000 30000 Crushing Hand 16000 20000
- 17. Crushing of the rivets: (n 2 h1 n 1 h 2 ) dS c Fc fs Load carrying member Type of stress 4 0 . 375 Rivet-driving Rivets acting in 0 . 5 16000 shear Rivets acting in Fc power single double shear Rolled steel, SAE 1020 Tension Shear 2 ….. Power 18000 13500 18000 13500 Shear Hand 10000 10000 Rivets, SAE1010 Crushing Power 24000 30000 F = 6000 lb C Crushing Hand 16000 20000
- 18. Rupture of plate by tension (L nd ) hS t Ft fs Load carrying member Type of stress Rivet-driving Rivets acting in Rivets acting in (4 2 0 . 375 ) 0 . 5 power 18000 single shear double shear Rolled steel, SAE 1020 Ft Tension ….. 18000 18000 Shear 2 Power 13500 13500 Shear Hand 10000 10000 Rivets, SAE1010 Ft = 14625 lb Crushing Crushing Power Hand 24000 16000 30000 20000
- 19. Crushing of the plates (n 2 h1 n 1 h 2 ) dS c Fc fs Load carrying member Type of stress Rivet-driving Rivets acting in Rivets acting in 4 0 .5 0 . 375 power 36000 single shear double shear Rolled steel, SAE 1020 Fc Tension ….. 18000 18000 Shear 2 Power 13500 13500 Shear Hand 10000 10000 Rivets, SAE1010 FC = 13500 lb Crushing Crushing Power Hand 24000 16000 30000 20000 Maximum value of F = 2209 lb
- 20. Design procedure for structural joints The following sequence of steps applies to the calculations for structural joints: • The load on each member is determined analytically or graphically • The shape and size of each member is determined based on the load • The diameter of the rivets is determined by the thickness of the structural shapes apply the equation: 1 d 2h 16
- 21. •The number of rivets required is based upon the shearing or crushing stress which ever determine the cause of failure. •The rivets in the joint are spaced in order to utilize the material economically •avoiding eccentric loading as far as possible •The margin of the edge parallel to the load m1 1 .5 d •The margin of the edge normal to the load m2 2d •Pitch limit: 16 h p 3d where h is the thickness of the thinnest plate used in the joint
- 22. Thank You