MET 304 Revited joints
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Transcript of MET 304 Revited joints
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Riveted Joints
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Riveting applications RivetsTypes of riveted joints Failure of riveted joints
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Riveting applications:
Pressure vessels, boilerstanksBridgesHulls of shipsAirplanesCranesbuildingsMachinery in general
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RivetsA rivet is a round bar consisting
of head shank.The rivet blank is heated to a
red glow Inserted into the holes;The head is held firmly against
the plateThe projecting end is formed
into a second head, called the point,
head
PointShank
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Rivet material:
•Tough and ductile low carbon steel
•Nickel steel. •Brass•Aluminium
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Types of riveted joints:
Lap joints
butt joints
Strap Rivet
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Riveted joint terminology.
(pPitch )
Gauge line
tBack pitch (p )
( )
Margin lapdistance m
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Failure of riveted joints
Bending of rivet or plate FF
•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.
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b) Shearing of the rivets:
Single shear
Double shear
FF/2
F/2
F
F
F
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Joint strength in shear
Where, n1: :number of rivets in single shear n2: number of rivets in double shear Ss: allowable shear stress d : diameter of rivet
4Sπd)nn(2F s
212s
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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,
c2112c dS)hnh(nF whereh1 و h2: plate thicknessSc: allowable crushing stress
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d) Rupture of plate by tension:
Rupture of plate occurs at the section between the rivets
The resistance of rupture can be obtained from the expression:
Where: L: plate width h: plate thickness: St : allowable tensile stress n : number of rivet holes at the section
tt hS)ndL(F
FF
L
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In the undrilled section:
Where L : width of plate
tt LhSF
FF
Undrilled Section
L
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d) Tearing and shearing of the margin
For the riveted joint to resist tearing and shearing of the margin, the margin (m) =
1.5 d for double shear2d for single shear.
Shearing of the margin
Tearing of the margin
F F
F F
Margin
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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.
FF
0.5 in
4 in. 2
1 1
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Solution
s
s2
12s f4Sdπ)nn(2F
2410000.3750π4F
2
s
Shearing of the rivets:d = 0.375fs = 2n1 = 4n2 =0
FS = 2209 lb
Load carrying member Type of stress Rivet-driving power
Rivets acting in single shear
Rivets acting in double shear
Rolled steel, SAE 1020 Tension ….. 18000 18000
Rivets, SAE1010
Shear Power 13500 13500Shear Hand 10000 10000
Crushing Power 24000 30000Crushing Hand 16000 20000
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Crushing of the rivets:
s
c2112c f
dS)hnh(nF
2160005.0375.04Fc
FC = 6000 lb
Load carrying member Type of stress Rivet-driving power
Rivets acting in single shear
Rivets acting in double shear
Rolled steel, SAE 1020 Tension ….. 18000 18000
Rivets, SAE1010
Shear Power 13500 13500Shear Hand 10000 10000
Crushing Power 24000 30000Crushing Hand 16000 20000
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Rupture of plate by tension
s
tt f
hS)ndL(F
2180005.0)375.024(Ft
Ft = 14625 lb
Load carrying member Type of stress Rivet-driving power
Rivets acting in single shear
Rivets acting in double shear
Rolled steel, SAE 1020 Tension ….. 18000 18000
Rivets, SAE1010
Shear Power 13500 13500Shear Hand 10000 10000
Crushing Power 24000 30000Crushing Hand 16000 20000
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Crushing of the plates
s
c2112c f
dS)hnh(nF
236000375.05.04Fc
FC = 13500 lb
Maximum value of F = 2209 lb
Load carrying member Type of stress Rivet-driving power
Rivets acting in single shear
Rivets acting in double shear
Rolled steel, SAE 1020 Tension ….. 18000 18000
Rivets, SAE1010
Shear Power 13500 13500Shear Hand 10000 10000
Crushing Power 24000 30000Crushing Hand 16000 20000
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Design procedure for structural joints
161h2d
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:
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• The margin of the edge parallel to the load
d3ph16
d5.11m
d22m • The margin of the edge normal to the load
•Pitch limit:
where h is the thickness of the thinnest plate used in the joint
•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
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Thank You