Joining Processes, Design & Economics

Joining is the preferred method of producing the desired shape when:

1. Product cannot be made as a single piece

2. Product is more economical as an assembly (use components for special purposes in design)

3. Products must be disassembled

4. Products are easier to transport disassembled

The main design advantage of assemblies is that the different components can have vastly different properties; a cooking vessel needs to have good conduction so water can boil, but the handle needs to be insulated.

Structures with lots of void space are easier to assemble than make as a single component; such as chairs, tables, and etc.

Complex structures are typically assemblies of components – cars, airplanes, computers, etc.

Joining Processes

1. Fasteners – Mechanical devices which join materials by clamping forces, pressure, or friction. They do not involve molecular bonding between the surfaces as the primary bonding mechanism.

2. Welding – The joining of two or more pieces of material by heat, pressure or both; with or without a filler metal to produce a localized union through fusion or recrystallization. The force of attraction is cohesion.

3. Adhesive Bonding – The joining of two or more pieces of material by the forces of attraction between the adhesive and the materials being joined. This includes processes such as brazing, soldering, gluing, and epoxies.

In many sources, adhesive bonding processes such as brazing and soldering are included in welding. Some processes involve two joining processes; for example a threaded stud is often welded to the base material and then used as a fastener.

IMSE 302

Joining Processes

1. Fasteners II. Adhesive Processes III. Cohesive Processes

A. Non-Permanent A. Metallic Adhesives A. Liquid Cohesion

(disassemble possible) Brazing Arc Welding

Threaded Fasteners Soldering Resistance Welding

Pins Braze Welding Electron Beam W.

Special Purpose

B. Permanent B. Organic Bonding B. Solid

(no disassemble) (gluing) Friction Welding

Rivets Glue Diffusion Bonding

Shrink Fits Epoxy Forge Welding

Seam(cans)

Major Fastener Categories(millions of fasteners)

1. Disassembly Permitted 2. Permanent Assembly

A. Threaded Fasteners A. Rivets(usually stronger

1. Nuts & Bolts & more economical than 2. 2. Screws fasteners)

B. Pins B. Metal Stitching

1. Machine Pins(straight, dowel,etc) C. Staples

2. Radial Locking Pins D. Seam Joints(cans)

C. Washers & Retaining Rings

1. Washers – distribute compressive stresses over

a wider area than the bolt fact.

2. Retaining Rings – provide a removable shoulder to

locate, retain or lock components.

D. Special Purpose Fasteners, Clips, etc.

1. Quick Opening-tend to operate against spring pressure

to snap into place.

2. Self-sealing –for leak joints

3. Spring Clips

Length of Engagement Design Considerations for Threaded Fasteners

Bolt Fastener failure can occur by:

A. Tensile Failure of the Bolt

B. Shear Failure of the Bolt(external threads)

C. Shear Failure of the Hole Component(internal threads)

Design so Tensile Load <=Shear Load

At St <= As Ss N L

L >= [1/N] [At / As ] [St /Ss]

L>= [P ] [At / As ] [St /Ss]

Where L = length of engagement

P = pitch of threads

N = number of threads per unit length = 1/P

At = tensile area of bolt St = UTS of bolt

As = shear area of bolt or hole Ss = USS of bolt or hold

Bolt Material Non-Bolt Material

Gr 5 UTS 120,000 psi Cast Iron UTS 27,000 psi

Gr 5 USS 92,000 psi Cast Iron USS 33,500 psi

Gr 1 UTS 60,000 psi PM Aluminum UTS 23,500 psi

Gr 1 USS 36,000 psi PM Aluminum USS 13,900 psi

SAE 903 Zinc UTS 41,000 psi

SAE 903 Zinc USS 31,000 psi

(see Table 15-3, page 308 for more materials)

Bolt Size A(tension) A(shear-bolt) A(shear-hole)

Dia-N (in 2 ) (in 2 /thread) (in 2 /thread)

?-13 0.1419 0.0599 0.0866

?-20 0.0318 0.01840 0.02699

L>= [P ] [At / As ] [St /Ss] = [P] [Pt (tension load)/Ps(shear load)]

For a ?-13 Grade 5 bolt in a cast iron block, the length values would be:

a) Shear of Bolt

L> = [P ] [At / As ] [St /Ss] = [1/13] [ 0.1419/0.0599] [ 120,000/92,000]

L> = 0.238 inches

b) Shear of Hole(cast iron block)

L>= [P ] [At / As ] [St /Ss] = [1/13] [ 0.1419/0.0866] [ 120,000/33,500]

L>= 0.451 inches

Therefore the length should be the largest of the two, or 0.451 inches. Both values must be calculated. The cast iron will shear before the bolt will shear, but the design is to make the bolt fail in tension before the shear failure will occur. A sheared component is difficult to repair whereas a tensile failure can be removed.

The clamping force load which can be supported by the bolt would be

P = At UTSt = (0.1419 in2 ) (120,000 psi) =17,028 pounds

However, only 5-25 percent of the force is for clamping – the other 75-95 percent of the force is used to overcome thread friction and face friction. This is one reason why torque wrenches are a poor indicator of the actual clamping force. Fasteners with good lubrication would have more clamping force than those with poor lubrication. At Ford, they used the rate of torque increase to indicate whether the fastener was engaging properly rather than the amount of torque.

Adhesive Bonding(Gluing)

The general advantages of adhesives over fusion(welding) bonding are:

1. Dissimilar metal/materials can be joined relatively easy.

2. Joining is at lower temperatures

3. Easy to bond thin gage materials

4. Used as a sealant

5. Used as an insulator

6. Easy to apply

7. Cost is generally low

The general disadvantages of the use of adhesives are:

1. Low joint strength(3-6,000 psi for most organic adhesives and 20-60,000 psi for most metallic adhesives)

2. Low peel strength

3. Generally restricted to room temperature applications

4. Generally organics are poor conductors

5. Curing time and temperature may be needed(lowers production rates)

A. Organic(Epoxy and Gluing) Adhesives

I. Types of Bonding that may occur are:

1. Secondary Bonding(electrostatic or hydrogen)

2. Mechanical Bonding(interlocking; especially in porous materials)

3. Diffusion Bonding(across interface)

4. Chemical Reaction(primary bonding – ionic or covalent)

II. Application Procedure for Organic Adhesives

1. Surface Preparation

a. Surface must be clean from dirt, oxides, and etc.

b. Surface must be reactive and not oxidized

c. A large surface area is needed as adhesive strengths are low

2. Adhesive Preparation

a. Adhesive have short shelf life and must be prepared immediately prior to use.

3. Apply Adhesives

a. Adhesives are generally applied with a brush or spray

b. Factors in adhesive application are:

i. Number of coats

ii. Time between coats

iii. Rate of spread of adhesive

4. Apply Pressure

a. Pressure is generally applied to help adhesive make good contact with the surface of the adherends.

5. Cure

a. Primary factors in curing of the adhesive are the time and temperature. High temperatures and long times hinder application in high production applications.

II. Application Procedure for Metallic Adhesives using Capillary Attraction

1. Good Fit and Proper Clearance

The joint clearance is typically 0.001 to 0.010 inches(0.025 to 0.250mm) depending upon the base metal, flux, and filler metal.

2. Clean Surfaces

The surfaces must be clean from dirt, grease, or oxides.

3. Flux

The purposes of the flux are:

i. Decompose oxides

ii. Protect surface during heating

iii.Reduce surface tension between filler and base metal so flow is better.

4. Assemble and Support

The pieces must be supported during the heating stage and the filler is either preplaced or applied during the heating.

5. Heat & Flow of Alloy

Must have controlled heating to melt the filler metal and have capillary flow into the joints.

6. Cooling and Cleaning of Assembly

Must avoid residual stressed from heating and cooling and must remove fluxes or they may cause corrosion.

III Adhesive Design Considerations(Usually Lap Joints in shear)

t = P/A where

t = shear stress of adhesive

A = total shear area

P = load to be supported

This can be rewritten and solved for the area by:

A = P/t

The expression for the length of the lap(L) for flat parts would be:

L = A/W

where W = width of the Piece

and thus L = P/tW

The length of lap for circular parts, such as pipes, would be

L = A/pD

Where D = outside diameter of smaller pipe

Factor of Safety

Design Load Actual Area Used

FS = Factor of Safety = Actual Load = Area for Actual Load

Material UTS

= Design UTS

If one designs a part to handle a 2,000 pound load, but the actual load expected is only 1,000 pounds, the FS is 2.

The area required to support the load is 2 square inches, but the area used to support the load is 4 square inches, FS = 4/2 = 2

A material has a tensile strength of 40,000 psi, but the 2,000 pound load uses an area of 0.1 square inches, the actual stress is 20,000 psi and the factor of safety is 40,000/20,000 = 2

Example Problem A lap joint is made of an aluminum strip with a tensile strength of 30,000 psi and a stainless steel strip with a tensile strength of 50,000 psi. The strips are 2 inches wide, the aluminum strip is 0.25 inches thick and the stainless steel is 0.20 inches thick. The brazing filler metal has a shear strength of 12,000 psi. Use a factor of safety of 2 and determine the length for the lap.

What is the load the strips can support

P(aluminum) = 30,000 psi (2 inch) (0.25 inch) = 15,000 lb

P(stainless steel) = 50,000 psi (2 inch) ( 0.20inch) = 20,000 lb

Weakest is aluminum at 15,000lb

Design load is FS (15,000) = 30,000 lb(joint will hold 30,000lb)

A(joint) = 30,000lb/12,000psi = 2.5 inches = LxW = 2L

L = 2.5/2 = 1.25 inches in length of lap

(the aluminum strip will fail first at a load of 15,000 lb)

IMSE 302