chapter 3_design of mechanical joint

JJ513 Engineering Design Design of Mechanical Joint

THE BIG PICTURE

OBJECTIVES OF THIS CHAPTERAfter completing this chapter, you will be able to:

1. Explain the purpose of the key2. Sketch and name various types of key such as parallel keys, taper keys, gib-head keys, pin

keys, woodruff keys, etc.3. Apply mathematical analysis for rectangle and parallel key based on sheer stress and

compressed stress.4. List the advantages and disadvantages using pin and bolt and nut as the method of

joining.5. Analyze the factors of bolt and nut joint failures.6. Analyze bolt and nut joints by shearing load.7. Analyze bolt and nut joints shearing caused by eccentric load.8. Calculate the bolt size and maximum load.9. State the advantages and disadvantages of welded joints.10. Sketch the basic symbol for welded joint.11. Apply the mathematical analysis for welded joint12. Determine the weld throat and safe load.13. Determine the welded joint size caused by eccentric load.

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3 - 1 KEY DESIGN

3 - 2 DESIGN OF PIN JOINT, BOLT, AND NUT JOINT

3 - 3 WELDED JOINTS DESIGN


3-1 THE KEY DESIGNA key is a machinery component placed at the interface between a shaft and the hub of a power-transmitting element for the purpose of transmitting torque [figure 3-1]. Key is placed so that part of it lies in a groove cut on the shaft, called key seat, and part of it fits into a groove cut in hub, called key way. Therefore, after the assembly locked together by the key, the shaft and hub will rotate together.

Figure 3-1 key position applied to a shaft

Square and Rectangular Parallel KeysThe most common type of key for shafts up to 6 ½ inches in diameter is the square key, as illustrated in figure 3-2. The rectangular key is recommended for larger shafts and is used for smaller shafts where the shorter height can be tolerated.

Figure 3-2 Square key

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Figure 3-3 Illustration of keys. (a) dimensions of shaft with keyway in shaft and hub; (b) square parallel key; (c) rectangular parallel key

Taper Keys and Gib Head KeysTaper keys are designed to be inserted from the end of the shaft after the hub is in position rather than installing the key first and then sliding the hub over the key as with parallel keys. The gib head key has a tapered geometry inside the hub that is the same as that of the plain taper key. But the extended head provides the means of extracting the key from same end at which it was installed. This is very desirable if the opposite end is not accessible to drive the key out.

(a) (b)Figure 3-4 (a) Tapered key; (b) Gib Head key

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Pin KeysThe pin keys are a cylindrical pin placed in cylindrical groove in the shaft and hub. Lower stress concentration factors results from this design as compared with parallel or tapered keys. A close fit between the pin and the groove is required to ensure that the pin does not move and that the bearing is uniform along the length of the pin.

Figure 3-5 Pin keys

Woodruff KeysWhere light loading and relatively easy assembly and disassembly are desired, the woodruff key should be considered. The circular groove in the shaft holds the key in position while the mating part is slid over the key.

Figure 3-6 Woodruff keys

Stress Analysis

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There are two basic modes of potential failure for keys transmitting power: (1) shear across the shaft/hub interface(2) Compression failure due to the bearing action between the sides of the key and the shaft

or hub material.Figure 3-7 shows the idealized case in which the torque on the shaft creates a force on the left side of the key.

Figure 3-7 Forces on a key

Key design is successful if failure from shear or compression is prevented, both of which are considered here:

1. Failure due to shear:

P=2Td

(3.1)

The design shear stress is

τ design=PA s

= 2Tdwl (3.2)

Above equation (3.2) is independent of the height h. to avoid failure due to shear

τ design≤Ssyns

= yield stress∈shearsafety factor

(3.3)

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2. Failure due to compressive or bearing stress. The compression or bearing area of the key is

Ac=lh2

(3.4)

The compressive or bearing design stress is

σ design=PAc

= 2Tdlh/2

=4Tdlh (3.5)

Above equation (3.5) is independent of the width w. Failure due to compressive or bearing stress can be avoided if

σ design≤0.90S y

ns(3.6)

Noted: The allowable normal stress σalland the allowable shear stress τ allfor ferrous and nonferrous metals for various types of loading may be represented by,

Shear: τ all=0.40S y (3.7)

Compressive:σ all=0.90S y (3.8)

*Sy is a Yield Strength

Example problem 3-1Given: A 4-in-diameter shaft with a hub is made of high-carbon steel. A square key made of low-carbon steel has a width and height of 1 in. assume a torque value at 2-in radius.

Find: The critical length of the key while assuming a safety factor of 2 and considering both compression and shear.

Solution: from table A.1 (appendix) the yield strength for the shaft and hub is Sy = 55 ksi. From equation (3.7) and (3.8)

τ all=Ssy=0.40S y=22ksiσ design=Scy=0.90S y=49.5ksi

Thus, the design stresses are

τ design=Ssyns

=222

=11ksi

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σ design=Scyns

=49.52

=24.75ksi

For a shaft with circular cross section

J= πd4

32=π (4)4

32=25.13¿4

The maximum torque acting on the shaft at a radius of 2 in is

T max=τdes ignJ

r=11000 (25.13)

2=138,215lb .∈¿

The maximum force is

Pmax=T max

r=69,107.5lb

From table A.1 (Appendix) the yield strength for the key is Sy= 43 ksi. From equation (3.7) and (3.8)

τ all=Ssy=0.40S y=17.2ksiσ design=Scy=0.90S y=38.7 ksi

The design stresses are

τ design=Ssyns

=17,2002

=8600 psi

σ design=Scyns

=38,7002

=19,350 psi

For equation (3.2)

lcr=2T max

dw τdesign=2(138,215)4 (1 )(8600)

=8.036∈¿

Thus, to avoid key shear failure, the key should be at least 8.035 in long. From equation (3.5)

lcr=4Tmax

dhσdesign=4 (138,215)4 (1 )(19,350)

=7.144∈¿

To avoid key compressive or bearing failure, the key should be at least 7.144 in long. Therefore, failure will first occur from key shearing, and the key must be at least 8.036 in long.

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3-2 THE DESIGN OF BOLT AND NUT JOINT

Figure 3-8 Bolt and Nut

The bolt and nut shown in figure 3-8 can be thought of as a spring system, as shown in figure 3-9. The bolt is viewed as a spring in tension with stiffness kb. The joint, with a number of members being joined, is viewed as a compressive spring with stiffness kj.

Figure 3-9 bolt-and-nut assembly modeled as bolt-and-joint spring

Bolt stiffness A bolt with tread is considered as a stepped shaft. The root diameter is used for the threaded section of the bolt, and the crest diameter is used for the unthreaded section, called the shank.

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Figure 3-10 Bolt and nut. (a) Assembled; (b) stepped-shaft representation of shank and treaded section.

Figure 3-10(a) shows a bolt-and-nut assembly and figure 3-10(b), the stepped-shaft representation of the shank and the threaded section. In figure 3-10(b) the effective length of the shaft and the threaded section includes additional lengths extending into the bolt head and nut. Making use of equation:

1kb

= 4πE ( Ls+0.4dcdc

2 +Lt+0.4 dr

dr2 )

Wheredc = crest diameter, mdr = root diameter,m

Joint stiffnessDetermining joint stiffness is much more complicated than determining bolt stiffness. Thus, an approximation is that the stress included in the joint is uniform throughout a region surrounding

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a) b)


the bolt hole with zero stress outside this region. Figure 3-11 shows the conical frustum stress representation of the joint in a bolt-and-nut assembly.

Figure 3-11 Bolt-and-nut assembly with conical frustum stress representation of joint

In calculating stiffness, always use the smallest of frustum cone diameters. For the member closet to the bolt head or to the nut, di= dw= 1.5dc.

(a) Joint Stiffness for same material.

The expression for joint stiffness isk j=E idc A i e

Bidc/ L

Where Ai, Bi = numerical constants given in table 3-1 below:

Table 3-1 Constants used in joint stiffness formula

Material Modulus of elasticity,

E(GPa)

Numerical constantAi Bi

Steel 206.8 0.78715 0.62873Aluminum 71.0 0.79670 0.63816Copper 118.6 0.79568 0.63553Gray cast iron 100.0 0.77871 0.61616

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(b) Joint stiffness for different materials The equation is

k ji=1.813 E jdc

ln [ (1.15Li+d i−dc ) (d i+dc)(1.15Li+d i+dc ) (d i−dc) ]

Where;Li = axial length of the frustum conedi= frustum cone diameter; di = 1.5dc

The resulting joint stiffness is 1k j

= 1k j1

+ 1k j2

+ 1k j3

…

Once the joint and bolt stiffnesses are known, the dimensionless stiffness parameter can be calculated from equation

C k=kb

kb+k j

Strength The proof load of a bolt is the maximum load that a bolt can withstand without acquiring a permanent set. The proof strength is the limiting value of the stress determined by using the proof load and the tensile stress area. Although proof strength and yield strength have something in common, the yield strength is usually higher because it is based on a 0.2% permanent deformation.

The proof strength Sp, as defined by the Society of Automotive Engineers (SAE), the American Society For Testing and Materials (ASTM), and International Organization for Standardization (ISO) specifications, defines bolt grades or classes that specify material, heat treatment, and minimum proof strength for the bolt or screw. Table 3-1 gives the strength information for several metric grades. Metric grade numbers from 4.6 to 12.9, with higher numbers indicating greater strength.

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Table 3-2 strength of steel bolts for various sizes in millimeters

Table 3-2 gives the dimensions and the tensile stress areas for M coarse and fine threads.

Table 3-3 Dimensions and tensile stress areas for M coarse and fine threads.

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Bolt Preload-static loadingThe bolt failure safety factor

nsb=A tS p−Pi

Pmax, bC k

(3.9)

Where At = tensile stress area given by table 3-2Pi = preload, NPmax,b= maximum load applied to bolt, NCk = dimensionless stiffness parameter

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Figure 3-12 Separation

Equation 3.9 suggests that the safety factor is maximized by having zero preload on the bolt. For statiscally loaded bolts where separation is not a concern, this certainly true. However, a preload is often applied to bolted connections to make certain that members are tightly joined in order to minimize tolerances and provide tight fits. Further, many applications require members to avoid separation as shown in figure 3-12. Thus, the safety factor guarding against separation is

nsj=P i

Pmax , j (1−C k ) (3.10)

Where Pmax,j = maximum load applied to joint, N.

The amount of preload that is in practice applied to bolts under static conditions is therefore a compromise between bolt overloading (where zero preload is most beneficial) and separation (where a larger preload is desirable). The preload is given for reused and permanent connections as

Pi={0.75 Pp

0.90 Pp} (3.11)

Where Pp = proof load, SpAt. Example problem 3-2

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for reused connectionfor permanent connections


Figure 3-13. (a) Hexagonal bolt-and-nut assembly and dimensions. (b) Dimension of frustum cone. (All dimensions are in millimeters)

Given: A hexagonal bolt-and-nut assembly, shown in figure 3-12 is used to join two members. The bolt and the nut are made of steel, and the frustum cone angle is 30°. The thread crest diameter is 14 mm, and the root diameter is 12 mm.

Find:1. Find the bolt and joint stiffnesses as well as the dimensionless stiffness parameter. Use

the modulus of elasticity and Poisson’s ratio from table 3-1, and consider the cases where

a. Both members are made of steel. b. The 15 mm thick member is made of steel, and the 10 mm thick member made of

aluminum. 2. Determine the maximum load for bolt-and-nut joint failure while assuming reused

connections and a static safety factor of 2. Assume a 5.8 grade and coarse threads.

Solution: 1. The bolt stiffness

1kb

= 4πE ( Ls+0.4dcdc

2 +Lt+0.4 dr

dr2 )

¿ 4

π (206.8×109 ) ( 40×10−3+0.4 (16×10−3 )(16×10−3 )2

+(20×10−3 )+0.4 (13.83×10−3 )

(13.83×10−3 )2 )k b=0.7954GN /m

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d1

1st frusta

10

25

12.5

2nd frusta

3rd frusta

10

x

30°

21

d2


a) Steel joint. k j=E idc A i e

Bidc/ L

¿ (206.8×109 ) (14×10−3 ) (0.78715 ) e(0.62873) (14× 10−3) /(25×10−3 )

¿3.241 GN/m

The dimensionless stiffness parameter gives

C k=kb

kb+k j= 0.79540.7954+3.241

=0.197

b) Steel-aluminum jointAs shown in figure 3-12, the members have to consider as 3 frusta. The member stiffnesses are:

For the 1st frusta,We know that the material is steel; from table 3-1, we know E = 206.8 GPa and

d1=1.5dc=1.5 (10−3 )=21

k j1=1.813E jdc

ln [ (1.15Li+di−dc ) (d i+dc )(1.15Li+di+dc) (d i−dc ) ]

¿1.813 (206.8×109 ) (14×10−3 )

ln { [1.15 (12.5 )+21−14 ] (21+14 )[1.15 (12.5 )+21+14 ] (21−14 ) }

¿6.796GN /m

For the 2nd frusta,We need to obtain the d2 first as shown in figure below

tan30 °= x10

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x=5.77Thus, d2 = 21 + 2(5.77) = 32.54

k j2=1.813E jdc

ln [ (1.15Li+di−dc ) (d i+dc )(1.15Li+di+dc) (di−dc ) ]

¿1.813 (206.8×109 ) (14×10−3 )

ln { [1.15 (2.5 )+32.54−14 ] (32.54+14 )[1.15 (2.5 )+32.54+14 ] (32.54−14 ) }

¿6.232×1010N /m

For the 3rd frusta,We know that the material is aluminum; E= 71 GPa and the d3 = d1 = 21.

k j3=1.813E jdc

ln [ (1.15 Li+d i−dc ) (d i+dc )(1.15 Li+d i+dc) (di−dc ) ]

¿1.813 (71.0×109 ) (14×10−3)

ln { [1.15 (10 )+21−14 ] (21+14 )[1.15 (10 )+21+14 ] (21−14 ) }

¿ 2.620 GN/mThus,1k j

= 1k j1

+ 1k j2

+ 1k j3

¿ 1

6.796×109+ 1

6.232×1010+ 1

2.620×109

k j=1.835×109N /m

The dimensionless parameter for steel-aluminum joint is

C k=kb

kb+k j= 0.79540.7954+1.835

=0.3024

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2. From table 3-1 for 5.8 grade, SP = 380 MPa; and from the table 3-3 for crest diameter of 14 mm and coarse threads, At = (84.3 + 157)/2 = 120.7 mm2. For reused connections,

Pi=0.75 Pp=0.75 A t Sp=0.75 (120.7 ) (10−6 ) (380 ) (106 )=34,399.5N

The maximum load that the bolt can carry while assuming a safety factor of 2 is

Pmax ,b=At S p−Pi

nsbC k

=(120.7×10−6 ) (380×106 )−34,399.5

(2 ) (0.3024 )=18,519.35N

The maximum load before separation occurs is

Pmax , j=Pi

nsj (1−C k )= 34,399.52 (1−0.3024 )

=24,655.60 N

Thus, failure due to separation will occur before bolt failure.

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3-3 THE WELDED JOINTS DESIGN

The welding is a process of joining together two or more metal parts. It is done by heating the surfaces, to be connected, to a high temperature and then adding additional molten metal, which fuses with and combines the two surfaces. The molten or fused metal is deposited between the parent metal parts, which are also fused metal gets cooled; the parent metal parts are joined by this new metal.

Some advantages of welded joints over threaded fasteners are that they are inexpensive and there is no danger of joint loosening. The welding allows the arrangement of the structure components, in such a way, the joint provides maximum efficiency but it is possible in case of bolted joints. Additions and alterations can be easily made in the existing structures. In welded connections, the tension members are not weakened as in the case of bolted joint. A welded joint has a great strength. Sometimes, the members are of such a shape that they afford difficulty for riveting. But they can easily weld. The welding provides very rigid joints. It is possible to weld any part of structure at any point. But bolting requires enough clearance. The process of welding takes less time than the bolting. No need to drilling and measuring.

Some disadvantages of welded joints over threaded fasteners are that they produce residual stresses; they distort the shape of the member, metallurgical changes occur, and disassembly is a problem. As there is an uneven heating and cooling, during the fabrication, therefore the members may get distorted or additional stresses may develop. It requires a highly skilled labor and supervision. As the welded joints are rigid joints, therefore they have to be treated as such in their design. The inspection of welding work is more difficult than bolting work.

There are many configurations of welds and manufacturing process for producing them. Figure 3-12 shows the standard symbols for welds, indicating that weld joints offer considerable design flexibility. Although welded joints are available in a wide variety of forms, only the fillet weld is considered here.

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Figure 3-14 Basic weld symbolsA fillet weld, shown in figure 3-13 , is made with equal legs. The thinnest section is then at the throatof the weld, at 45° from the legs. The governing stress in fillet welds is shear on the throat of the weld, as show in figure 3-13. Observe from this figure that in the fillet weld aligned parallel to the load, the shear stress occurs along the throat of the fillet parallel to the load. In a fillet weld aligned transverse to the load; the shear stress occurs at 45° to the load, acting transverse to the axis of the fillet.

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Figure 3-15 Fillet weld. (a) Cross section of weld showing throat and legs; (b) shear planes.

Static Axial and Direct shear loadingTotal weld throat area is

A=0.707he lw

Where he = weld leg length, mlw = weld length, m

Welding-distortion or deformation or warping of weldments during welding is a natural outcome of intrinsic non uniform heating and cooling of the joint. The distortion energy theory says that failure occurs due to distortion of a part. Distortion energy theory is

SSy=0.58S y

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A B

C D

l =50 mm


Where Sy= Yield Strength of the material part.

Eccentricstatic load, Fis

F=Ssy A

ηsfWhere ηsf= safety factor

Example problem 3-2

Weld joints subjected to static axial and direct shear loading. Estimate the static strength of a parallel loaded fillet weld. The plates in figure below are 12 mm thick and made of steel having yield strength, SY 350 MPa. They are welded together by convex fillet welds along sides AB and CD, each of which is 50 mm long. With a safety factor of 3, what static load, F can be carried using a 6 mm weld leg?

AnalysisTotal weld throat area is

A=0.707he lw¿0.707 (6 ) (100 )¿424.2mm2

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Distortion energy theory isSSy=0.58S y

¿0.58 (350 )¿203MPa

Thus, the eccentric static load is

F=Ssy A

ηsf

F=(203×106 ) (424.2×10−6 )

3¿28.7kN

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APPENDIX

TABLE A.1: Properties of ferrous metals

TABLE A.2: Properties of non-ferrous metals

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REFERENCES

Anon. (2010, februari 7). Parallel Keys – Flat Keys – Machine Keys – Shaft Keys Sizes and Prices. Retrieved may 30, 2012, from CTS Enterprises Ltd (Machine Key Manufacturer) web site: http://www.machinekey.org/parallel-keys-machine-keys/

Bernard J.Hamrock, S. R. (2005). fundamentals of machine elements 2nd edition. kuala lumpur: The McGraw Hill Companies.

KHURMI, R. S. (2007). Strength of Materials. new delhi: S. Chand & Co.

Krok, P. (2006, march). wikepedia foundation inc. Retrieved june 18, 2012, from wikepedia wed site: http://eo.wikipedia.org/wiki/Dosiero:Bolt-with-nut.jpg

Mott, r. L. (2004). Machine Elements in Mechanical Design 4th Edition. New Jersey: Pearson prentice hall.

Robert C. Juvinall, K. M. (2000). Fundamentals of Machine Component Design 3rd Edition. new york: John Wiley & Sons, Inc.

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chapter 3_design of mechanical joint

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