JOINTS
One of the most vexing problems confronting a structural designer is the design of
joints. Among the common type of fastenings are bolts, rivets, welds, pins,
threaded couplings, and glued joints.
Types of Bolted or Riveted or Pinned Joints
1. Butt joint
2. Single lap joint
3. Double lap joint
Types of Welded Joints
1. Groove or Butt Weld
2. Fillet Weld
3. Slot Weld
4. Plug Weld
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2
Glued Joint
Types of Bolted or Riveted Joint Failures
1. Shear failure – Single or Double
2. Bearing failure
3. Tear-out failure (of plates)
4. Tension failure (of plates)
Types of Welded Joint Failures
1. Tensile or Compressive failure
2. Shear failure
Type of Glued Joint Failure
1. Shear failure
Bolted or Riveted Joint
Shear Failure – Single Shear
3
V F
τ ave
A A
where A is the cross-sectional area of the bolt or rivet ൌ ߨ݀ଶ /4, d being the
diameter of the bolt or rivet.
Shear Failure – Double Shear
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V F
τ ave
A 2A
where A is the cross-sectional area of the bolt or rivet ൌ ߨ݀ଶ /4, d being the
diameter of the bolt or rivet.
Example
Two Steel plates are connected together by two bolts as shown. The bolts have
diameters of 25 mm. the connection transmits 150 kN of load. Determine the
average shear stress in the bolts.
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Bearing Failure
F F
ave σb
A b dt
where d is the diameter of the bolt or rivet and t is the minimum thickness of the
two plates that the bolt or rivet connects
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Tear-out Failure
F
Shearing of these two planes
e
F
F F
ave τ up
A s 2et
where e is the distance of the center of the bolt or rivet from the edge of the plate
and t is the minimum thickness of the two plates that the bolt or rivet connects
Tension Failure
t
F
F F
t tu
A net (b - d)t
where b is the width of the plate, t is the minimum thickness of the two plates that
the bolt or rivet connects and d is the diameter of the bolt or rivet
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Joint Efficiency for Single-Lap Joints using Rivets
Shear failure
A r u
N
Joint Efficiency = = A
sheet σ tu
Ar = cross-sectional area of one rivet
u=ultimate shearing stress of rivets
N=number of rivets in the joint
Asheet = cross-sectional area of each plate or sheet
tu =ultimate stress of sheets in tension
Basic Assumption: The cross-shearing stress, , in the rivets is uniformly
distributed over all shear areas. All rivets take equal loads.
Bearing failure
Contact Area σ bu N
Joint Efficiency = A sheet σ tu
Assumption: Ultimate bearing stress, bu, between the rivets and the plates is
assumed to be uniformly distributed over the projection of the contact area.
Contact Area is the plate thickness, t, multiplied by the rivet diameter, d=2r.
Tear-out failure
2 t e up
η
Joint Efficiency = to
Ne
Asheetσ tu
Ne=Number of rivets in the row closest to the edge of either plate
t = thickness of the plate
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e = distance of rivet center from the edge
up = ultimate shear stress of the plate
Tension failure
P
P
Joint Efficiency = =
Pult on net area at row A net σ tu A net
Pult fraction at row N-n N-n
A sheet σ tu A sheet
N N
n = total number of rivets in all previous rows
P
t1=?
t2=?
t3=?
P
Assumption: Tensile stress in the plate is assumed to be uniformly distributed over
the net area of the plate at each row of rivets.
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Some rules of thumb for designing with rivets
P
d
e
1. Distance to the edge.
P e / d 2 .0
2. Inter rivet spacing
3. Distance to the side
4. Rivet length
1.5 D
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Welded Joint
Tension of Compression failure
Groove weld can see failure in tension or compression. The average stress (tensile
or compressive) in the weld is
F F
A weld tl
Shear failure
Fillet welds come in 1/8 in. to ½ in. sizes in increments of 1/16 in. In fillet welds,
area A is in tension and area B is in shear so these planes need to be analyzed
separately for the critical plane. However, the shear stress is taken on a plane
between A and B where the weld metal is thinnest. Because shear strength is
normally lower than tensile strength, the fillet weld strength is based on the
allowable shear stress.
F F
A weld 0.707tl
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