Design of Mechanical Springs
July 22, 2022 | Author: Anonymous | Category: N/A
Short Description
Download Design of Mechanical Springs...
Description
ME-PC 321 MACHINE DESIGN 1 ENGR. LOUIE A. LARIOSA FACULTY – ME DEPARTMENT CTU – MAIN CAMPUS
CONTENTS 1. Analysis of Simple Stresses 2. Tolerance and Allowances 3. Variable Stress analysis 4. Shaft design 5. Keys and coupling Design 6. Design of Threaded Member and Power Screws 7. Design of Mechanical Springs 9. Gears
ME-PC 321
MACHINE DESIGN 1 DESIGN OF MECHANICAL SPRINGS
DESIGN OF MECHANICAL SPRINGS Spring is an elastic body (generally metal) that can be twisted, pulled, or stretched by some force.
It deflects when loaded (i.e., applied force ) and returns to its original shape when the load is removed. It is also termed as a resilient member, which stores energy once deflected and release the same as it recovers to its original shape. Materials for Springs: Oil-tempered spring wire, music wire, hard -drawn
spring wire, carbon steel, chrome-vanadium steel, chrome-silicon steel, stainless steel. Uses of Springs
1. To absorb energy or shock loads, as in autom automobile obile shock absorbers 2. To maintain contact between machine member members, s, as in valves and clutches 3. To act as source of energy, as in clocks 4. To serve as measuring device, as in spr springs ings scales
YPES ES OF SP SPRI RING NGS S T YP
T YP YPES ES O F EN END D S O F CO COIL IL S P RI RIN N GS
Plain Plain
Ground
Squared Square
Ground
Actual number of coils
Solid Length
Free Length
n
(n+1)d
np+d
n
nd
np
n+2
(n+3)d
np+3d
n+2
(n+2)d
np+2d
Where:
= = ℎ =
, = = = , =
SPRING RATE/SPRING SCALE F
DEFLECTION,
= …(Series) = = = ⋯ (Parallel) FORCE,
F
= = = … (Series) = … (Parallel)
( 1997, 1997, 2006) 2006) ℎ ℎ ℎ ℎ ℎ ℎ ℎ 100 . . ℎ 0.4 0.400 00 ℎ ℎ 0.64 . ℎ . EXAMPLE 1:
GIVEN:
SOLUTION:
= = = = 100 100 = 0.400 = = 0.64
= Where:
=
REQUIRED:
, , =?
Thus,
=
100 10 0 100 10 0 100 = 0.64 0.64 0.40 = . .
( 1998) 4 00 . 0.709 . ℎ . EXAMPLE 2:
GIVEN:
= 400 400 400 = = = = = = 100 100 4 4 = = = = 0.709
SOLUTION:
= = = = Where:
=
Thus, REQUIRED:
, , =?
=
100 10 0 = 0.709 = ..
STRESSES AND DEFLECTION OF COIL SPRING TORSIONAL SHEAR STRESS IN THE WIRE, = 8
Spring Factor,
=
IMPACT LOAD ON SPRING (FREE FALL),
0.615
(ℎ ( ℎ ) =
Where:
Where: ℎ , , =
4 − 1 4 − 4
SPRING DEFLECTION,
= 8
2
= / = = − = = 2 = = = ℎ
ℎ = ℎℎ
( 1997) ℎ ℎℎ ℎ 16 − ℎ ℎ ℎ ℎ ℎ ℎ 85 . ℎ 9.66 0.65 . ℎ ℎ ℎ .
EXAMPLE 3:
GIVEN:
Squared and ground ends = 16 = 84 = 9.66 9.66 = 0.65 0.65 REQUIRED:
ℎ , , =? , , =?
SOLUTION:
4 − 1 =
4 − 4
Where:
9.6 66 − 0.6 .65 5 − 9. = = = 13.86 = 0.65 Thus, 4(13 4(13.8 .86) 6) − 1
=
4(13 4(13.8 .86) 6) − 4
= . .
=
4 − 1 4 − 4
= 1.06 1.06 = .
0.615 0.615 =
0.615 13.86
( 1997) ℎ 20 ℎ ℎ 75 . . ℎ ℎ ℎ ℎ 79.84 , 101.6 9.525 .
EXAMPLE 4:
GIVEN:
Squared and ground ends = 20 = 75 = 735 735.7 .75 5 = 79 79.8 .84 4 = 79,840 79,840 = 101.6 101.6 = 9.525 9.525 REQUIRED:
, , =?
SOLUTION:
8 =
Where:
101. 1.6 6 − 9. 9.52 525 5 − 10 = = = 9.67 9.525 = − 2 = 20 20 − 2 = 18 18 (for square and ground ends)
=
Thus,
8 73 735.7 9.6 .67 7 (18) = 79,840 (9.525 (9.525 ) = ..
EXAMPLE 5:
( 1998)
ℎ ℎ 40 0
, 2.5 ℎ, ℎ, ℎ ℎ ℎ 8 ℎ ℎ . ℎ ℎ . = 750 , 1.25, 1.25, 57,296 = 10,800 10,800 , : : . . . ℎℎ .
ℎ ℎ ℎℎ
GIVEN:
Squared and ground ends = 400 ℎ = 2.5 2.5 ℎ ℎ = 8 ℎ ℎ = 750 750
= 57,29 1.25 6 (( = 57,296 = 750 ))
= 10 10,8 ,800 00 REQUIRED: . , , ℎ ℎ . , , . ℎ ℎℎ ℎ ℎ ℎ,, . ℎℎ ℎℎ ,,
From:
SOLUTION: From,
= 2 = 7.8125 2 0.5
8
=
8( )
= 8(1.25 .25)( )(750 750 )( )(2.5 2.5 ) 8(1 57,296 = = . . From:
=
= . . From: =
8
Where,
=
8
Also,
Where:
750 = 1.875 1.875 = = Τ 400 2.5 0.5 0.5 2.5 =6 = = = 0.5 Thus,
8 75 750 6 1.8 1.875 75 = (10,800,000 Τ )(0 )(0.5 .5 ) = 7.8125 = 2 = 7. 7.81 8125 25 2 = 9. 9.81 81
= − = 8 − 4. 4.91 91 = 3.09 3.09 Thus,
= 400
3. 3.09 09 = 1,237.5 1,237.5
Therefore:
=
8(1.25)(1 8(1.2 5)(1,237. ,237.5 5 )(3 ) (0.5 (0 .5 )
= .
,. ,.
EXAMPLE 6: ℎ ℎ 3 ℎ ℎ ℎ 10 0 . ℎ ℎ ℎ ℎ ℎ ℎ . ℎ 30.48 ℎ ℎ ℎ ℎ ℎℎ . 8,
3,868 Τ 808,720 Τ . : . . .
GIVEN:
ℎ = 3 = 30 300 = 100 100 = 30.48 30.48 =8 = 3,868
= 808,720 REQUIRED: . . .
From:
SOLUTION: From,
= 8 = 8 3.67 3.67
8 =
= . .
From:
Where:
=
; = 8 8= 4 − 1 0.615
=
8
8(2,168.50 8(2,16 8.50 )(8) 808,720 (3.67 ) (3.67 = . . 30 30.4 .48 8 =
4(8 4(8) − 1
0.615
= 4 − 4 = 4(8 4(8) − 4 8 = 1.18 ℎ = 2 30.4 30.48 8 100 100 30 300 0 30.4 30.48 8 = 2 = 2,168.50 2,168.50 Thus,
1.18 2,1 2,168. 68.50 50 ((8) 8) 3,868 3,86 8 / = 8 1.18 = . .
The leaf spring is used mainly for vehicle suspension and in one form consists of a stack of slightly curved narrow plates of equal width and varying length clamped together, with the shorter plates in the centre to form a semielliptical shape.
STRESSES AND DEFLECTION OF LEAF SPRING SIMPLE CANTILEVER (QUARTER ELLIPTICAL LEAF SPRING)
6 , , = 6 = , , =
Where:
SEMI-ELLIPTICAL SEMI-ELLIPTICA L LEAF SPRING
= ℎ = ℎ ℎ ℎ ℎ = ℎ ℎ ℎ ℎ
, , = , , = , , =
3
2 = (without extra full length leaf)
(+ )
= = ℎ = ′ = − ℎ ℎ
=
(with extra full length leaf) +
EXAMPLE 7:
ℎ 60 12 ℎ . ℎ ℎ ℎ 700 700 8,000 ℎ . = 200 . : . . .
GIVEN:
=6 = 60 60 = 12 12 = 700 700 = 8,00 8,000 0 = 200 = 2 200, 00,000 000
REQUIRED: . . .
SOLUTION: For simple cantilever spring
= =
6 6(8,00 6(8 ,000 0 )( )(70 700 0 ))
6( 6(60 60 )( )(12 12 )) = . . From,
= =
6
6(8,00 6(8 ,000 0 )(7 )(700 00 )
6(6 (60 0 ) 12 = ..
200,000
EXAMPLE 8:
− ℎ 20 . ℎ 4 ℎ ℎ. ℎ ℎ 1.225 55 . 90 360. . ℎ ℎ 360 . = 200 . : . GIVEN:
− = 4 = 55 55 = 1.225 1.225 = 1,225 1,225 = 20 = 20, 20,00 000 0
SOLUTION: For semi-elliptical spring Deflection of a spring without extra full length leaf
3
= 200 = 2 200, 00,000 000
= 4 8 360 360 (1, (1,22 225 5 ) = =
= 360 = 90 90
4(200, 000 0 )(9 )(90 0 ) 4 = . . 4(2 00,00
=
From, REQUIRED: . ℎ ℎ .
=
Where:
= = =
= 5 = 5, 5,0 000
3(5,00 3(5 ,000 0 )( )(1,2 1,225 25 ) = 8 = (9 (90 0 )) 8( 8(55 55 )) 7.50 7.50 200,000 3
= .
EXAMPLE 9:
− 1. 5 , ℎ ℎ − ℎ , 60 ℎ . 8, 00 0 ℎ 70 . . = 200 200 . . : . ℎ ℎ . .
GIVEN:
− = 10 ′ = 1 = 60 60 = 70 70 = 1.5 1.5 = 1,50 1,500 0 = 8,00 8,000 0
= 200 = 2 200, 00,000 000
REQUIRED: . ℎ ℎ . .
SOLUTION: For semi-elliptical spring with extra full-length leaves =
=
3 = ′ ′ 4 2 2 2
3(8,000 3(8,00 0 )(1,50 )(1,500 0 ) 3 = 1 ′ 10 60 200,000 4 2 (70 (70 ) 4 2 10
= . . From, = ′ 2 2 ′
=
2 2 2 2 0. 0.1 1 200, 200,00 000 0 (10. (10.47 47 )( )(70 70 ) ) = 1, 1,50 500 0
View more...
Comments
