MACHINE ELEMENTS Formula Sheet PDF

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MACHINE ELEMENTS I (Prof. Dr. Özgen Ü. Çolak) g : Stress amplitude m : Mean stress, m = ( ü + a ) /2 Kt: Theoretical stress concentration factor max = Kt . n smooth

Notch factor: K ç

Notch sensitivity : q

notch

Static equivalent stress : ,

/

) 2 3(

(

eş

ü

/

m=

Kç 1 Kt 1 Dynamic equivalent stress:~ eş

)2

K bK y Ak

SHAFT AND AXLE Strength control of shafts : i) Torsional stress = Mb / W b 3

(

~) 2

~) 2

3(

tg ,

S

d

g = ( ü - a ) /2

k

Kç

Kb K y

;

D

Kç

em .

16M b m

Md (Mb): Torsional moment, Md = 9550P/n dır. ( Md [Nm], P [kW] : power, n [rpm] ) ii) If there is only static bending (static axle) : Me / We m Me :Bending moment [Nmm] Min. shaft diameter, d

32.M e

em

=

Ak / S

m

~ Me = ii) If there is only dynamic bending (rotating axle) : We ~ 32.M e d Allowable stress: em = eD.Kb.Ky / Kç

m

m

Bending section modulus: W e = .d3 / 32, Torsional section modulus: W b = .d3 / 16 Deformation Control d 2y Me In bending: ymax / L 1 /3000. Euler-Bernouilli equation: 2 EI dx 4 M bL I In torsion: , Polar moment of inertia: p m 32 GI p Variable cross sections:

Mb G

I pi

m

, for torsional length, L=1 m,

Vibration control of shafts: Critical angular speed in bending: k , m: mass of the part on the shaft, m Critical angular speed in torsion: For a disc, Im

em

( 1/ 4)

k= (48.E.I) / l3 , Critical speed in bending: n kr

30

k m

k Im : Mass moment of inertia of the machine part on the shaft Im

mr 2 / 2 , Critical speed in torsion : n kr

30

k Im

KEYS - Geometry based keys: Parallel keys Surface pressure; Md Circumferential force; Fç (d/2) Between shaft and key: d: shaft diameter Fç Shear stress in the keys ; m Between hub and key: b

Fç pem t1 L Fç p pem t2 L

p

pem =

Ak / S

;

Tapered keys tg = 1 / 100 b : key width, h : key height, L : key length Press and disassembly forces: Fçk = Fön tg 2 + tg ( + 1 ) and Fçz = Fön - tg 2 + tg ( - 1 ) For self-locking 2 : Friction angle

Total friction moment Ms = 2 . Fs ( d / 2 ) = FN d Ms S.Md FN= Fön Surface pressure control of key: p = FN / (b.L) pem

PRESS FIT CONNECTIONS Press force: Fpres =

bpd

Friction moment: Ms = (1/2)

b p d2

p min

Ms > Md .

2.Ms .d 2 .b.

Conical press fits: Friction force: Fs= b p do/Cos , do: average diameter Required press force Fpr d b p (tg + ), Force for disassemble 1 d1 d 2 Conicity: k b

Fçöz = d b p (tg - )

BOLT AND NUT CONNECTIONS tg = h / ( .d), : helix angle, h: pitch Force for assembly, Fh = Fö.tg ( + ),

Force for disassemble: Fçz = Fö.tg (

Total moment: Mstop =M1 + M2 Msık-çö = Fö. [ sRs + r2tg ( )], Rs : Average radius under nut, For every types of thread profile: = / Cos /2 1-Connections without preload F 4F 0,6. m , m 2 A1 1 2- Tightening with preload Fön 4.Fön , Tension stress: A1 d21

k

-

)

tg =

, d1 : minor diameter

torsional stress:

=

M Wb

Fö .r2tg( ) , 3 d1 / 16

2 ş

3.

3. Connections with preload

Fö

k p x , kc, kp: stiffness of bolt and parts k ş

k cx

A .E c , kv lş

A1 .E c lv

1 kc

,

Ap.Ep

2 .(D 2 d delik ) , D = s + k.(lp / 2) , , Ap 4 lp k : 0,25 for cast iron, 0,2 for steel.

kp

Fmax

Fz

x

.kc

kc kc

kp

.Fiş

Fö

Fz ,

1

1 .F kp /kc ) iş

Fa

Fö

Fk ,

Fmax

Fiş

Fk ,

Fiş

Fa

Fz

1 kv

1 kş

m

Maximum stress Fmax 4.Fmax max A1 d2

m

,

m

0,6.

k

1

If external load is dynamic, Fmax , max min A1 Mean stress:

Fg

Fö

Fö , A1

.

Fiş

= max

Max. stress:

A1

ü

Surface pressure for screw thread 4F p= p em , pem=0.25 Ak z. d2 - d 21 ) 4F z H = z.h , H: Height of nut 2 d - d21 ).p em

(if external load is static)

Fmax

=

m+

min or

2

Fz 2 A1

g

Shear for screw thread: F = em= em/2 m , zA A= d1h (for triangle profile) A= d1h/2 (for square profile)

WELDING

Weld throat ( a) and weld lenght ( k ):

( em)kay = V1.V2.( em)som ( em)kay = V1.V2 ( em)som V1 : weld factor V2 : quality factor V2 = 1 (for 1. Quality), V2 = 0.8 (for 2. Quality), V2 = 0.5 (for 3. Quality),

a = min (Butt joint), a = 0.7. min (T joint)  k =  - 2.a Cross section for weld seam, Ak =

( a.  k )

For rectangle cross section 1 1 We 2. a.l k 2 .a.l k 2 3 6 For circular cross section, D d 2.a, 1 D4 d4 1 D4 d4 . , Wb . 32 D 16 D For square cross section B.H3 b.h3 We H h 2.a, B b 2.a 6.H We

SPRINGS 1. Helical spring : Total stress ,

max

K.

8.F.D 1 ) 3 (1 d 2D / d

ax

8.F.D d3

.

m

K : Wahl faktor

To find out wire diameter (d): τmax= τem . 8.F.n D 3 ( ) Spring stiffness: k G.d d G : Shear modulus, n : number of active coils

Deflection : x

Allowable stress:

/S

m

F x

G.d 1 8.n (D / d)3

0 ise ) , τD: Fatique strength , S: Safety factor

(

2. LEAF SPRINGS: Bending stress in single leaf, F.l 3 3.E.I Spring stiffness : k= Fmax / ymax

Maximum deflection : x

6.F.l b. 4.F.l 3 b. .E

Uniform strength cantilever beams Deflection: y

K1

4F.l 3

3

K1: f (b' / b)

b. .E

Multileaf springs: Benging stress in a single leaf 6.F.l n.b. n: Number of layers Deflection: y K2= 12/(2+n /n)

K2

F.l3 n.b. 3 .E n : main leaf number, n: total leaf number....