Me1302-nol

Failure Theories ✔ Stress in machine components should be

accurately computed. ✔ Designer must understand material limits to

ensure a safe design.

Design Factor ✔ Factor of Safety (N)

Expected Stress N= Stress at Component Failure ✔ Suitable values depend on inherent danger,

certainty of calculations, certainty of material properties, etc.

Static Stresses - Brittle Materials  Percent elongation < 5% • •

N= N=

sut

σ suc

σ

1 σ1 σ 2 = + • N sut suc

for parts in tension for parts in compression for parts with general stress

Example The Gray Cast Iron (Grade 40) cylinder carries an axial compressive load of 75,000 lbs and a torque of 20,000 in lbs. Compute the resulting design factor. R0.25” R0.25”

Ø4.00” Ø5.00”

Static Stresses - Ductile Materials  Percent elongation > 5%  Distortion Energy Theory Define von Mises Stress σ ' = σ 12 + σ 22 − σ 1σ 2 sy

• For nominal stress

N=

• For localized stress

su N= σ'

σ'

Static Stresses - Ductile Materials  Percent elongation > 5%  Maximum Shear Stress Theory s ys

• For nominal stress

N=

• For localized stress

sus N= τ max

τ max

=

sy 2τ max

Example Specify a diameter for the middle portion of the rod, if it is to be made from AISI 1040-hot rolled steel. 450

5000 lbs

Example For the seat support shown, specify a standard structural tube to resist static loads shown. The tube has properties similar to AISI 1020 hot-rolled steel. Use a design factor of 3. 400 lb

14”

200 lb

20”

Stress

Repeated Loads

σalt

σmean Time

Example The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Determine the mean and alternating nominal stresses.

1.25”

0.1” R 1”

.75”

Fatigue Strength ✔ R.R. Moore Test

Motor

Alternating Stress, σa

Endurance Strength, sn 103

104

105

106

107

108

Cycles to Failure, N (log)

Endurance Strength ✔ sn = Endurance strength • Listed in tables • If no information is available, use (Steel) • sn ≈ 0.5 su • sn ≈ 0.4 su

(Aluminum)

Adjusted Endurance Strength ✔The data from the standard R.R. Moore

test is adjusted for a particular application. ✔sn’ = Adjusted endurance strength = (Cs) (Cm) (Cst) (CR) (sn)

Size and Stress Type Factors – Cs = Size Factor • D< 0.4 in

Cs = 1.0

• 0.4 < D 2.0 in

Cs = (D/0.3)-0.068

• 2.0 < D 10.0 in

Cs = D-0.19

For rectangular sections, D=.808(h b)1/2 – Cst = Stress Type Factor • = 1.0 for bending • = 0.80 for axial tension • = 0.50 for torsion

Material and Reliability Factor – Cm = Material Factor • = 1.0 for wrought steel • = 0.80 for cast steel • = 0.70 for cast iron – CR = Reliability Factor • 50%

CR = 1.0

• 90%

CR = 0.90

• 99%

CR = 0.81

• 99.9%

CR = 0.75

Example The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Determine the endurance limit of the material.

1.25”

0.1” R 1”

.75”

Repeated Stresses - Ductile Materials  Distortion Energy Theory Define repeated von Mises Stress σ 'm = σ 12m + σ 22m − σ 1mσ 2 m σ 'a = σ 12a + σ 22a − σ 1aσ 2 a

• Solderberg criterion 1 σ 'm K t σ 'a = + N sy s 'n

Repeated Stresses - Ductile Materials  Maximum Shear Stress Theory 1 (τ m ) max K t (τ a ) max = + N ssy s 'sn

• ssy = 0.5 sy • s’sn = 0.5 sn

Example The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Comment on the robustness of the design.

1.25”

0.1” R 1”

.75”

Example Comment on the robustness of a 1-1/4” round bar made from AISI 1213 C-D steel. It carries a constant tensile load of 1500 lbs, a bending load that varies from 0 to 800 lbs at the senter of the 48” length and a constant torque of 1200 in lbs.

48”

Shafts ✔ Connect power transmission components.

✔ Inherently subjected to transverse loads and

torsion.

Shaft Forces ✔ Gears

As before Wt Wr T

Shaft Forces ✔ Chains Ftight

Ftight

2T = D D T

Fslack = 0

Shaft Forces ✔ V-belts

Ftight

Ftight 2.5T = D D

Fslack

T = 2D

Fslack

T

Shaft Forces ✔ Flat belts Ftight 3T Ftight = D D

Fslack

T = D

Fslack

T

Material Properties ✔ For steady load (torsion)

sys=.5sy

✔ For fatique load ( bending)

sn’=cs cR sn cT = 1 (bending) cm = 1 (wrought steel)

Stress Concentrations ✔ Keyseats – Sled Runner Kt = 1.6

– Profile

Kt = 2.0

– Woodruff

Kt = 1.5

Stress Concentrations ✔ Shoulders – Sharp, Bearing (r/d ≈.03) Kt = 2.5 – Round, Gear Bore (r/d ≈.17) Kt = 1.5

✔ Grooves – Retaining Rings Kt = 1.5

Try not to let Kt’s overlap. Leave .10 - .15” between

Strength Analysis K M c K M t t ✔ Bending stress σ = = I S For round sections

✔ Torsion stress

π D3 S= 32

Tr T τ= = J 2S

For round sections

J I =2 r c

Strength Analysis ✔ Mohr’s circle and Solderberg 3 2 ( K t M / s n ') + ( T / s y ) 1 4 = N S 2

Suggested Design Factors: • N=2 smooth operation • N=3 typical industrial operation • N=4 shock or impact loading

Minimum Acceptable Diameter ✔ The designer must size the shaft. – Solve for appropriate diameters  32 N D=  π

3 2 ( K t M / s n ') + ( T / s y )  4  2

Example Determine a suitable diameter for a shaft made from AISI 1144 OQT 1000. It is subjected to a reversing bending moment of 3000 ft lbs and a steady torque of 1800 ft lbs. The shaft has a profile keyway.

Example The shaft shown is part of a grain drying system ✔ At A, a 34 lb. propeller-type fan requires 12 hp when rotating at 475 rpm. ✔ A flat belt pulley at D delivers 3.5 hp to a screw conveyor handling the grain. ✔ All power comes to the shaft through the v-belt at C. Using AISI 1144 cold drawn steel, determine the minimum acceptable diameter at C.

Example

B

A 12”

Sheave C

D

C 10”

10”

E 4”

150 Sheave D

Shafts Accessories ✔ Components used to securely mount power

transmitting elements on a shaft. ✔Axial ✔Rotational

Keys ✔ Allow torque to be transferred from a shaft

to a power transmitting element (gear, sprocket, sheave, etc.)

Key Design ✔ Use a soft, low strength material H L

(ie, low carbon steel)

W

✔ Standard size H=W=1/4 D ✔ Design length

based on strength

Standard Key Sizes Shaft Dia. (in)

W (in)

W T

H S

D − H + D2 −W 2 S= 2 D + H + D2 −W 2 T= + .005 in. 2

Key Design T 2T F= = D/2 D

✔ Key Shear

F 2T τ= = A DLW

✔ Failure Theory N = ✔ Length

4TN L= DWs y

sy 2τ

=

s y LW 4TD

Example Specify a key for a gear (grade 40, gray cast iron) to be mounted on a shaft (AISI 1144, hot rolled) with a 2.00 in. diameter. The gear transmits 21000 lb-in of torque and has a hub length of 4 in.

Retaining Rings ✔ Also known as snap rings ✔ Provides a removable shoulder to lock

components on shafts or in bores. ✔ Made of spring steel, with a high shear

strength. ✔ Stamped, bent-wire, and spiral-wound.

Retaining Ring Selection ✔ Based on shaft diameter & thrust force

Set Screws ✔ Setscrews are fasteners that hold collars,

pulleys, or gears on shafts. ✔ They are categorized by drive type and

point style.

Standard Set Screw Sizes

Set Screw Holding

Pins ✔ A pin is placed in double shear ✔ Holds torsion and axial loads

8T N d= π D sy

D d

✔ Hole is made slightly smaller than the pin

(FN1 fit)

Example Specify a pin for a gear (grade 40, gray cast iron) to be mounted on a shaft (AISI 1144, hot rolled) with a 2.00 in. diameter. The gear transmits 21000 lb-in of torque and has a hub length of 4 in.

Roll Pins ✔ Easier disassembly

Collars ✔ Creates a shoulder on shaft without

increasing stock size. ✔ Held with either set screw or friction (clamped)

Mechanical Couplings ✔ Couplings are used to join two shafts ✔ Rigid couplings are simple and low cost.

But they demand almost perfect alignment of the mating shafts. ✔ Misalignment causes undue forces and

accelerated wear on the shafts, coupling, shaft bearings, or machine housing.

Mechanical Couplings ✔ In connecting two shafts, misalignment is

the rule rather than the exception. It comes from such sources as bearing wear, structural deflection, thermal expansion, or settling machine foundations. ✔ When misalignment is expected, a flexible

coupling must be used.

Mechanical Couplings ✔ Selection factors include: • Amount of torque (or power & speed) • Shaft Size • Misalignment tolerance

Fasteners, Powers Screws, Connections Helical thread screw was an important invention. Power Screw, transmit angular motion to liner motion Transmit large or produce large axial force It is always desired to reduce number of screws

Definition of important Terminologies

Major diameter d, Minor diameter dr Mean dia or pitch diameter dp Lead l, distance the nut moves for one turn rotation

Single and Double threaded screws

Double threaded screws are stronger and moves faster

Screw Designations ✔ United National

Standard UNS ✔ International Standard Organization

Roots and crest can be either flat or round

Coarse thread Designated by UNC ✔ Fine Thread UNF, is more resistance to

loosening, because of its small helix angle. ✔ They are used when Vibration is present ✔ Class of screw, defines its fit, Class 1 fits have widest tolerances, Class 2 is the most commonly used ✔ Class three for very precision application ✔ Example:1in-12 UNRF-2A-LH, A for Ext. Thread and B for Internal, R root radius ✔ Metric M10x1.5 10 diameter mm major diameter,1.5 pitch

Some important Data for UNC, UNF and M threads ✔ Lets Look at the Table 8-1 on Page 398

Square and Acme Threads are used for the power screw

Preferred pitch for Acme Thread d, in p,in

1/4 1/16

5/16 1/14

3/8 1/12

1/2 1/10

5/8 1/8

3/4 1/6

7/8 1/6

1 1/5

1 1/4 1/5

Mechanics of Power Screws

Used in design to change the angular motion to linear motion, Could you recall recent failure of power screw leading to significant causalities

What is the relationship between the applied torque on power screw and lifting force F

Torque for single flat thread Fd m l + πfd m TR = ( ) 2 πd m − fl Fd m πfd m − l TL = ( ) 2 πd m + fl If the thread as an angle α, the torque will be

Fd m l + πfd m sec α TR = ( ) 2 πd m − fl sec α Wedging action, it increases friction

Stresses in the power Screw

Shear stress in the base of the screw Bearing stress Bending stress at the root of the screw Shear stress in the thread nt number of engaged thread

τ=

16T πd 3

σB = −

F πd m nt p / 2

6F σb = πd r nt p

τ=

3V 3F = 2 A πd r nt p

Loading to the fasteners and their Failure considerations

Bolts are used to clamp two or more parts It causes pre tension in the bolt Grip length is the total thickness of parts and washers l

d

t

l

l

ld

h

t2

lt=L’- ld

L’ effective grip= h+t2 if t2
Failure of bolted or riveted joints

Type of Joints ✔ Lap Joint (single Joint)

But Joint

Example 1

Example 2

Example 2

Example 3

Weld

Weld under Bending

Springs Flexible machine elements Used to: ✔ Exert force ✔ Store energy

Spring Rate ✔ Effective springs have a linear deflection

curve. ✔ Slope of the spring deflection curve is the rate Force k 1 Deflection

∆F k= ∆L

Example A compression spring with a rate of 20 lb/in is loaded with 6 lbs and has a length of 1.5 in. Determine the unloaded spring length (free length)

Geometry Di

Dw

Dw

L

Do

✔ Wire diameter, Dw (Standard gages) ✔ Mean Diameter, Dm

Dm = Do - Dw

Spring Parameters ✔ Spring index

Dm C= Dw

C > 5 (manufacturing limits) ✔ Active coils, Na = N for plain ends = N-1 for ground ends = N-2 for closed ends

Deflection ✔ Deflection for helical springs

8 FDm3 N a 8 FC 3 N a δ= = 4 GDw GDw G = Shear modulus

✔ Spring rate for helical springs

GDw k= 3 8C N a

Example A helical compression spring is formed from 35 gage music wire with 10-1/4 turns and an O.D. if 0.850 in. It’s ends are squared. The free length is 2 inches. Determine the force to press the spring solid.

Stress Analysis ✔ Spring wire is in torsion

V

T r 8K F C τ= = 2 J π Dw ✔ Wahl factor, K Accounts for the curvature of the wire 4C − 1 .615 K= + 4C − 4 C

T

F

Example A helical compression spring is formed from 35 gage music wire with 10-1/4 turns and an O.D. if 0.850 in. It’s made from A228 and the ends are squared. The free length is 2 inches. If the spring is repeatedly compressed to 1.3 in, do you expect problems?

Design Procedure ✔ Select a material ✔ Compute required spring rate ✔ Estimate Dm based on size constraints ✔ Determine required Dw (use K=1.2) ✔ Select standard wire ✔ Verify actual stress is satisfactory. ✔ Compute number of coils required.

Example Design a helical compression spring to exert a force of 22 lbs when compressed to a length of 1.75 in. When its length is 3.0 in, it must exert 5 lb. The spring will be cycled rapidly. Use ASTM A401 steel wire.

Rolling Element Bearings ✔ Provides support for machine elements,

while allowing smooth motion.

µ=0.001 - 0.005

Types Single-row Radial Ball

Angular Contact Ball

Radial Roller

Angular Roller

Types Spherical Roller

Tapered Roller

Needle Thrust

Ball Bearings

Stress Analysis ✔ Contact Stress

σc=300,000 is not unusual ✔ Balls, rollers and races are made from

extremely high strength steel ex. AISI 52100 sy = 260,000 psi su=322,000 psi

Bearing Load/Life ✔ Test (fatigue) data Radial Load (lbs)

Empirical relationship:

L2  P1  =   L1  P2 

k

•k=3.0 (ball) •k=3.33 (roller)

L10 Life (cycles)

Example A bearing is mounted on a shaft rotating at 1200 rpm. The bearing has been tested to have a L10 life of 300 hrs, when loaded with 500 lbs. Determine the expected L10 life, if the load is increased to 700 lbs.

Manufacturer’s Data ✔ Vendors publish the

Basic Dynamic Load rating (C) of a bearing at an L10 life of 1 million cycles.

Bearing Selection ✔ Determine the design life (in cycles) ✔ Determine the design load Pd = V R

•V=1 for inner race rotation •V=1.2 for outer race rotation

✔ Calculate the required basic dynamic load 1  Ld  k Creq 'd = Pd  6   10  ✔ Select a bearing with (C > Creq’d) and a bore

that closely matches the shaft diameter.

Example Specify suitable bearings for a shaft used in an grain dryer. The shaft rotates at 1700 rpm. The required supporting loads at the bearing are R =589 lb Bx

RBy=164 lb

and the minimum acceptable diameter is 2.16”.

Mounting of Bearings ✔ Shaft/bearing bore has a light interference

fit. ✔ Housing/outer race has a slight clearance fit. Check manufacturers catalog ✔ Match maximum permissible fillet radius. ✔ Shaft or housing shoulders not to exceed

20% of diameter.

Mounted Bearings ✔ Pillow block

✔ Bearing is inserted into a cast housing, with

base or flange slots, which can be readily attached to a machine base.

Bearings with Varying Loads ✔ Compute a weighted average load based on

duty cycle.

[

 ∑ ( Fi ) p N i Fm =   ∑ Ni 

]   

1 p

Fm=equivalent load Fi= load level for condition i Ni= cycles for condition i p = exponent for load/life

Example Bearing 6211 is carrying the following load cycle, while rotating at 1700 rpm. Stage Load (lbs) Time (min) 1 600 480 2 200 115 3 100 45 Compute the bearing L10 life in minutes.

Radial & Thrust Loads ✔ Calculate an equivalent load

P=VXR +YT T=thrust load X factors depending = Y on bearing

Thrust factors, Y ✔ Deep -groove, ball bearings

X = 0.56 for all values of Y

Example A bearing is to carry a radial load of 650 lb and a thrust load of 270 lb. Specify a suitable single-row, deep-groove ball bearing if the shaft rotates at 1150 rpm and the design life is 20,000 hrs.