MD Rivet

RIVETED JOINTS

CLASSIFICATION OF FASTENINGSThe fastenings (i.e. joints) may be classified into

the following two groups :1. Permanent fastenings, and2. Temporary or detachable fasteningsThe permanent fastenings are those fastenings

which can not be disassembled without destroying the connecting components. The examples of permanent fastenings in order of strength are soldered, brazed, welded and riveted joints

The temporary or detachable fastenings are those fastenings which can be disassembled without destroying the connecting components. The examples of temporary fastenings are screws, keys, cotters, and pins

WHAT IS A RIVET?

A rivet is a permanent mechanical fastener. A rivet is a short cylindrical bar with a head integral to it

The cylindrical portion of the rivet is called shank or body and lower portion of shank is known as tail

RIVET Before the rivet is installed it consists of a

smooth cylindrical shaft with a head on one end. The end opposite the head is called the buck-

tail. On installation the rivet is placed in a punched

or pre-drilled hole. Then the tail is "upset" (i.e. deformed) so that it expands to about 1.5 times the original shaft diameter and holds the rivet in place.

Because there is effectively a head on each end of an installed rivet it can support tension loads (loads parallel to the axis of the shaft); however, it is much more capable of supporting shear loads (loads perpendicular to the axis of the shaft).

METHODS OF RIVETING The function of rivets in a joint is to make a

connection that has strength and tightness.

When two plates are to be fastened together by a rivet, the holes in the plates are punched and reamed or drilled

In structural and pressure vessel riveting, the diameter of the rivet hole is usually 1.5mm larger than the nominal diameter of the rivet

METHODS OF RIVETING The plates are drilled together and then separated

to remove any burrs or chips so as to have a tight flush joint between the plates

A cold rivet or a red hot rivet is introduced into the plates and the point (i.e. second head) is then formed

In hand riveting, the original rivet head is backed up by a hammer or heavy bar and then the die or set, is placed against the end to be headed and the blows are applied by a hammer

This causes the shank to expand thus filling the hole and the tail is converted into a point as shown

As the rivet cools, it tends to contract. The lateral contraction will be slight, but there will be a longitudinal tension introduced in the rivet which holds the plates firmly together

MATERIAL OF RIVETS

The material of the rivets must be tough and ductile.

They are usually made of steel (low carbon steel or nickel steel), brass, aluminum or copper

When strength and a fluid tight joint is the main consideration, then the steel rivets are used

TYPES OF RIVETED JOINTS

There are the two types of riveted joints, depending upon the way in which the plates are connected:

1.Lap Joint: A lap joint is that in which one plate overlaps the other and the two plates are then riveted together

2.Butt Joint: A butt joint is that in which the main plates are kept in butting each other and a cover plate (i.e. strap) is placed either on one side or on both sides of the alignment main plates. The cover plate is then riveted together with the main plates


Butt joints are of the following two types :

1. Single strap butt joint, and 2. Double strap butt joint.

Following are the types of riveted joints depending upon the number of rows of the rivets:

A single riveted joint is that in which there is a single row of rivets in a lap joint as shown

and there is a single row of rivets on each side in a butt joint as shown

TYPES OF RIVETED JOINTS A double riveted joint is that in which

there are two rows of rivets in a lap joint as shown


and there are two rows of rivets on each side in a butt joint as shown

NOTE REGARDING LAP AND BUTT JOINT

Since the plates overlap in lap joints, therefore the force P acting on the plates are not in the same straight line but they are at a distance equal to the thickness of the plate. These forces will form a couple which may bend the joint.

The forces P in a butt joint act in the same straight line, therefore, there will be no couple.

TERMS USED IN RIVETED JOINTS

Pitch. It is the distance from the centre of one rivet to the centre of the next rivet measured parallel to the seam as shown. It is usually denoted by p.

Back pitch. It is the perpendicular distance between the centre lines of the successive rows (pb)

Diagonal pitch. It is the distance between the centers of the rivets in adjacent rows of zig-zag riveted joint (pd)

Margin or marginal pitch. It is the distance between the centre of rivet hole to the nearest edge of the plate (m)

FAILURES OF A RIVETED JOINT

1. Tearing of the plate at an edge. A joint may fail due to tearing of the plate at an edge.

This can be avoided by keeping the margin, m = 1.5d, where d is the diameter of the rivet hole

FAILURES OF A RIVETED JOINT

2. Tearing of the plate across a row of rivets. Due to the tensile stresses in the main plates, the main plate or cover plates may tear off across a row of rivets as shown

The resistance offered by the plate against tearing is known as tearing resistance or tearing strength or tearing value of the plate

TEARING RESISTANCE

Let p = Pitch of the rivets,d = Diameter of the rivet hole,t = Thickness of the plate, andσt = Permissible tensile stress for the plate

materialTearing area per pitch length:

A t = (p – d ) t

Tearing resistance or pull required to tear off the plate per pitch length:

Pt = At.σt = (p – d) t.σt

When the tearing resistance (Pt) is greater than the applied load (P) per pitch length, then this type of failure will not occur.

FAILURES OF A RIVETED JOINT3. Shearing of the rivets. The plates which are

connected by the rivets exert tensile stress on the rivets, and if the rivets are unable to resist the stress, they are sheared off

The resistance offered by a rivet to be sheared off is known as shearing resistance or shearing strength or shearing value of the rivet

SHEARING RESISTANCE OF RIVETS

Let: d = Diameter of the rivet hole,τ = Safe permissible shear stress for the

rivet material, andn = Number of rivets per pitch length.

Shearing Area:As = (π/4)d2 (Single Shear)

= 2 (π/4)d2 (Double Shear)Shearing resistance or pull required to shear off the

rivet per pitch length:Ps = n(π/4)d2 τ (Single Shear)

= 2 × n (π/4)d2 τ (Double Shear) When the shearing resistance (Ps) is greater than

the applied load (P) per pitch length, then this type of failure will occur

FAILURES OF A RIVETED JOINT4. Crushing of the plate or rivets.

Due to crushing failure, the rivet hole becomes of an oval shape and hence the joint becomes loose. The failure of rivets in such a manner is also known as bearing failure.

The resistance offered by a rivet to be crushed is known as crushing resistance or crushing strength or bearing value of the rivet

CRUSHING RESISTANCE OF RIVETS

d = Diameter of the rivet hole,t = Thickness of the plate,σc = Safe permissible crushing stress for the rivet or

plate material, andn = Number of rivets per pitch length under crushingCrushing area per rivet (i.e. projected area per rivet):

Ac = d.t

Total crushing area = n.d.tCrushing resistance or pull required to crush the rivet

per pitch lengthPc = n.d.t.σc

When the crushing resistance (Pc) is greater than the applied load (P) per pitch length, then this type of failure will not occur

STRENGTH OF RIVETED JOINTS

The strength of a joint may be defined as the maximum force, which it can transmit, without causing it to fail

Pt, Ps and Pc are the pulls required to tear off the plate, shear off the rivet, and crush off the rivet.

If we go on increasing the pull on a riveted joint, it will fail when the least of these three pulls is reached,

EFFICIENCY OF RIVETED JOINTS The efficiency of a riveted joint is defined as

the ratio of the strength of riveted joint to the strength of the un-riveted or solid plate

Strength of the riveted joint = Least of Pt, Ps and Pc

Strength of the un-riveted or solid plate per pitch length, P = p × t × σt

∴ Efficiency of the riveted joint:η = (Least of Pt, Ps and Pc) / (p × t × σt)

p = Pitch of the rivets,t = Thickness of the plate, andσt = Permissible tensile stress of the plate

material

EXAMPLE 9.2.

Find the efficiency of the following riveted joints :1. Single riveted lap joint of 6mm plates with 20mm

diameter rivets having a pitch of 50mm.2. Double riveted lap joint of 6mm plates with

20mm diameter rivets having a pitch of 65mm.Assume: Permissible tensile stress in plate = 120MPa Permissible shearing stress in rivets = 90MPa Permissible crushing stress in rivets = 180MPa

SOLUTION – EXAMPLE 9.2

Given: t = 6mm; d = 20mm; σt = 120MPa = 120N/mm2; τ = 90MPa = 90N/mm2; σc = 180MPa = 180N/mm2

Solution:1.Efficiency of the first joint:

p = 50mm(i) Tearing resistance of the plate

Pt = (p – d)t × σt = (50 – 20) 6 × 120 = 21600N

(ii) Shearing resistance of the rivetPs = π/4 × d2 × τ = π/4 (20)2 90 = 28278N

(iii) Crushing resistance of the rivetPc = d × t × σc = 20 × 6 × 180 = 21600N

SOLUTION – EXAMPLE 9.2∴ Strength of the joint = Least of Pt, Ps and Pc = 21600 N

Strength of the unriveted or solid plate:P = p × t × σt = 50 × 6 × 120 = 36000N

∴ Efficiency of the joint:η = (Least of Pt , Ps and Pc) / P = 21600/36000 = 60%

2. Efficiency of the second jointPitch, p = 65mm

i) Tearing resistance of the platePt = (p – d)t × σt = (65 – 20) 6 × 120 = 32400N

(ii) Shearing resistance of the rivetPs = n × π/4 × d2 × τ = 2 × π/4 (20)2 90 = 56556N

(iii) Crushing resistance of the rivetPc = n × d × t × σc = 2 × 20 × 6 × 180 = 43200N

SOLUTION – EXAMPLE 9.2

The strength of the unriveted or solid plate:P = p × t × σt = 65 × 6 × 120 = 46800N

∴ Efficiency of the joint:η = (Least of Pt , Ps and Pc) / P = 32400/46800

= 69.2%

Examples 9.1 and 9.3 -> Do yourself

RIVETED JOINT FOR STRUCTURAL USE –JOINTS OF UNIFORM STRENGTH

A riveted joint known as Lozenge joint used for roof, bridge work or girders etc. is shown (triple riveted double strap butt joint.)

In such a joint, diamond riveting is employed so that the joint is made of uniform strength

1. Diameter of Rivetd = 6(t)1/2

The sizes of rivets for general purposes are given in the table 9.7

RIVETED JOINT FOR STRUCTURAL USE –JOINTS OF UNIFORM STRENGTH2. Number of RivetsThe number of rivets required for the joint may be

obtained by the shearing or crushing resistance of the rivets

Let Pt = Maximum pull acting on the joint. This is the tearing resistance of the plate at the outer row which has only one rivet

= ( b – d ) t × σt

n = Number of rivetsSince the joint is double strap butt joint, therefore the

rivets are in double shear. It is assumed that resistance of a rivet in double shear is 1.75 times than in single shear in order to allow for possible eccentricity of load and defective workmanship.

RIVETED JOINT FOR STRUCTURAL USE –JOINTS OF UNIFORM STRENGTH∴ Shearing resistance of one rivet,Ps = 1.75 × (π/4) × d2 × τand crushing resistance of one rivet,Pc = d × t × σc

Number of rivets required for the joint,n = Pt / (Least of Ps or Pc)

3. From the number of rivets, the number of rows and the number of rivets in each row is decided.

4. Thickness of the butt strapsThe thickness of the butt strap,t1 = 1.25t, for single cover strap

= 0.75t, for double cover strap


At section 1-1, there is only one rivet hole

∴ Resistance of the joint in tearing along 1-1,

Pt1 = (b – d ) t × σt

At section 2-2, there are two rivet holes.


Pt2 = (b – 2d ) t × σt + Strength of one rivet in front of section 2-2


Similarly at section 3-3 there are three rivet holes.


Pt3 = (b – 3d ) t × σt + Strength of 3 rivets in front of section 3-3

The strength of unriveted plate,P = b × t × σt

∴ Efficiency of the joint,η = (Least of Pt1, Pt2 , Pt3, Ps or

Pc)/P


6. The pitch of the rivets is obtained by equating the strength of the joint in tension to the strength of the rivets in shear. See table 9.8 for details

7. The marginal pitch (m) should not be less than 1.5 d.

8. The distance between the rows of rivets is 2.5d to 3d

EXAMPLE 9.11. Two lengths of mild steel tie rod having width 200

mm and thickness 12.5mm are to be connected by means of a butt joint with double cover plates. Design the joint if the permissible stresses are 80MPa in tension, 65MPa in shear and 160MPa in crushing. Make a sketch of the joint.

Given: b = 200mm; t = 12.5mm; σt = 80MPa = 80N/mm2 ; τ = 65MPa = 65N/mm2; σc = 160MPa = 160N/mm2

Solution: 1. Diameter of rivet

d = 6 (t)1/2 = 6 (12.5) 1/2 = 21.2mm.From table 9.7, we take diameter of hole as

21.5mm and diameter of rivet as 20mm

EXAMPLE 9.11.

2. Number of rivetsMaximum pull acting on the joint,Pt = (b – d ) t × σt = (200 – 21.5) 12.5 × 80 =

178500NSince the joint is a butt joint with double cover

plates, therefore, the rivets are in double shear. Assume that the resistance of the rivet in double shear is 1.75 times than in single shear

∴ Shearing resistance of one rivet:Ps = 1.75 × (π/4) × d2 × τ = 1.75 × (π/4) × (21.5)2

65 = 41300NAnd crushing resistance of one rivet:

Pc = d × t × σc = 21.5 × 12.5 × 160 = 43000N

EXAMPLE 9.11.

Since the shearing resistance is less than the crushing resistance, therefore number of rivets required for the joint:

N = Pt / Ps = 178500/41300 = 4.32 say 5

3. The arrangement of the rivets is shown

EXAMPLE 9.11.

4. Thickness of butt strapst1 = 0.75 t = 0.75 × 12.5 = 9.375 say 9.4mm

5. Efficiency of the jointFind the resistances along the sections 1-1, 2-2

and 3-3.At section 1-1, there is only one rivet hole.

EXAMPLE 9.11.

∴ Resistance of the joint in tearing along section 1-1:Pt1 = (b – d)t × σt = (200 – 21.5) 12.5 × 80 = 178500N

At section 2-2, there are two rivet holes. In this case, the tearing of the plate will only take place if the rivet at section 1-1 (in front of section 2-2) shears

∴ Resistance of the joint in tearing along section 2-2:Pt2 = (b – 2d)t × σt + Shearing resistance of one rivet

= (200 – 2 × 21.5 ) 12.5 × 80 + 41300 = 198300NAt section 3-3, there are two rivet holes. The tearing of

the plate will only take place if one rivet at section 1-1 and two rivets at section 2-2 shears

EXAMPLE 9.11.

∴ Resistance of the joint in tearing along section 3-3:Pt3 = (b – 2d) t × σt + Shearing resistance of 3 rivets

= (200 – 2 × 21.5) 12.5 × 80 + 3 × 41300 = 280900N

Shearing resistance of all the 5 rivets:Ps = 5 × 41300 = 206500N

and crushing resistance of all the 5 rivets,Pc = 5 × 43000 = 215000N

Since the strength of the joint is the least value of Pt1, Pt2 , Pt3, Ps or Pc, therefore, strength of the joint:

= 178500N along section 1-1Strength of the un-riveted plate:= b × t × σt = 20 × 12.5 × 80 = 200000N

EXAMPLE 9.11.∴ Efficiency of the joint:

η = (Strength of the joint) / (Strength of the unriveted plate)

= 178500/200000 = 89.25%6. Pitch of rivetsp = 3 d + 5 mm = (3 × 21.5) + 5 = 69.5 say

70mm7. Marginal pitch:m = 1.5d = 1.5 × 21.5 = 33.25 say 35mm8. Distance between the rows of rivets= 2.5 d = 2.5 × 21.5 = 53.75 say 55mm

Examples 9.12 & 9.13 (Do yourself)

HOMEWORK

Chapter No. 9 (Khurmi Book)Problems 1 – 7; 13 – 15

PROBLEM # 5

A double riveted lap joint with chain riveting is to be made for joining two plates 10mm thick. The allowable stresses are : σt = 60MPa ; τ = 50MPa and σc = 80MPa. Find the rivet diameter, pitch of rivets and distance between rows of rivets. Also find the efficiency of the joint

Given: t = 10mm; σt = 60MPa; τ = 50MPa; σc =

80MPa; d = ?; p = ?; pd = ?; η = ?

Solution:d = 6 (t)1/2 = 6 (10) 1/2 = 18.97mmWe take d = 20mm as directed in table 9.3

PROBLEM # 5

Ps = n × π/4 × d2 × τ = 2 × π/4 (20)2 × 50 = 31415.9N

Pc = n × d × t × σc = 2 × 20 × 10 × 80 = 32000N

Since Pc > Ps, we take Pmax = Pc = 32000N

Pmax = Pt = (p – d)t × σt

32000 = (p – 20) × 10 × 60 p = 73mmpb = 0.33p + 0.67d = 0.33(73) + 0.67(20) = 37.49 or

38mmStrength of the unriveted or solid plate:P = p × t × σt = 73 × 10 × 60 = 43800N

η = (Least of Pt , Ps and Pc) / P = 31415.9/43800 = 71.7%

PROBLEM # 13

Two mild steel tie bars for a bridge structure are to be joined by a double cover butt joint. The thickness of the tie bar is 20mm and carries a tensile load of 400kN. Design the joint if the allowable stresses are : σt = 90MPa ; τ = 75MPa and σc = 150Mpa. Assume the strength of rivet in double shear to be 1.75 times that of in single shear

Given:t = 20mm; Pt = 400000N; σt = 90MPa ; τ =

75MPa; σc = 150Mpa; d = ?; b = ?; n = ?; η = ?

Solution:Diameter of rivet: d = 6 (t)1/2 = 6 (20) 1/2 = 26.82

or 27mm

PROBLEM # 13Width of plate: Pt = (b – d)t × σt

400000 = (b – 27) × 20 × 90 b = 249mm or 250mmNumber of rivets per plate:Ps = 1.75 × π/4 × d2 × τ = 1.75 × π/4 (27)2 × 75

= 75148Nn = Pt/Ps = 400000/75148 = 5.32 or 6

Efficiency of the joint:Strength of unriveted plate = P = btσt =

249.20.90P = 448200NPc = d × t × σc = 27 × 20 × 150 = 81000N

PROBLEM # 13

The sketch of the joint is as shown.

Resistance of the joint in tearing along 1-1:

Pt1 = (b – d)t × σt = 400000N


Pt2 = (b – 2d)t × σt + Shearing resistance of 1 rivet

= (250 – 2.27).20.90 + 75148 = 427947.8N


Pt2 = (b – 3d)t × σt + Shearing resistance of 3 rivets

= (250 – 3.27)20.90 + 3.75148 = 529643N

PROBLEM # 13

Shearing resistance of all 6 rivets:= 6Ps = 6.75148 = 450886.8N

Crushing resistance of all 6 rivets:= 6Pc = 6.81000 = 486000N

Since the strength of the joint is the least value of Pt1, Pt2 , Pt3, Ps or Pc, therefore, strength of the joint:

= 400000Nη = 400,000/P = 400000/448200 = 89.2%

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