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第六章 包装缓冲与防振包装设计 ( GB8166-87 ) PowerPoint PPT Presentation


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第六章 包装缓冲与防振包装设计 ( GB8166-87 ). 第 1 节 缓冲与防振包装设计的六步法. 一、一般要求 减小传递到产品上的冲击、振动等外力 分散作用在产品上的应力 保护产品的表面及凸起部分 防止产品的相互接触 防止产品在包装容器内移动. 表面保护. 二、缓冲包装设计要解决的问题. 选择恰当的缓冲材料并确定其形状和大小; 对指定的材料确定其形状和大小; 比较若干不同品种的材料的缓冲效果; 判断现有包装结构的合理性和可靠性;

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第六章 包装缓冲与防振包装设计 ( GB8166-87 )

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GB8166-87


1




.

  • .

    1

    2

    3

    4)

    5

    6


2

++


1.

2.

3.

4.

5.

6.

7.Ph6.2~8.0

8.

9.

10.

11.


1

=0

=0.5


1.1

k

E

AF

T

1.2

k0x0

dbxdbF


1.3

k0

F0F

1.4

k0

>0<0

1.5


2

+

2.1

2.1.1

(1) E

(2) E

E2<E<E1

E>E2

E<E1


2.1.2

k

(1)

1

2

12

+=1

112aabbcc

2

3


2

(2)xb

2.2

A=A1+A2


2.2.1

EA=E1A1+E2A2

E1>E2

E

E

E2<E<E1


2.2.2

1

+=1

112aabbcc

2

3


2

1 t2

2


[] 5cm24cm

F1=2x1+0.12 x13F2=3x2+0.32 x23

13cm222cm25cm2

5cm2


1=F1/A 2=F2/A

A=5cm2

A1=3cm2A2=2cm2

x1=x2=x


3

FE

3.1

1

2

db

db

1

3


th,tanh

coth


3.2

()

( )

e

3.2.1

1

2 3

123mm/min

C


CC-

CC-

CEC-E

0.014

0.020

0.033

0.035

1000.050

1500.060

2000.080

0.03

0.03

0.31

0.42

0.07 0.08 0.087


3.2.2

1

2

Gm

ATH

Gm

Gm

Gmst

HATGmst

TAHGmst

GmstC

WH=

Gmst Cm


3.2.3


4

4.1

4.2

Tr

,

0.08-0.20

0.10-0.50

0.02-0.16

0.11-0.23

0.008-0.016


4.3

,,

,

,

,

20%

24

,


4.4

,

,

,


4.5

-30

-20

-50

-40

-20


4.6

4.7

4.8

PH6.0-8.0


4.9


5

5.1

5.1.1

6-5

PE,PP,PS,PVC

PU,

2


3

0.5mm)

0.25mm)

>0.4g/cm ,<1.5

=0.1-0.4g/cm ,

1.5-9

<0.1g/cm >9

>700kPa)

70-700kPa)

<70kPa)

3

3

23 50% RH


5.1.2 EPS()

2

EPS

EPS

EPS


5.1.3 EPE()

EPS


5.1.4 PU()


5.1.5 EVA(-) PEF()

EVA

PEF


5.1.6 EPDM() CR()

EPDM 130C

CR

CR

EPDM


5.1.7 EPP()

EPEEPS


5.1.8


5.1.9

30%


4.2 4.2.1

1/3


4.2.2


4.2.3


4.2.4

EPS


4.2.6 Pillowpack


3

  • .

    1.

    2.

    3.

    4.


.

1.

2.

3.

4.,



.

1.

2.

3.


4

.


.

  • a.

    h=300W-1/2

    hcm

    WKg>16kg

    b.

    GB4857.584

    JISI0202

    MILP116H


a.

b.

MILHDBK304

MTS

NDS

c.


[G]<20g

30g<[G]<250g

400g<[G]<600g

[G]>1000g

Cm

Cs

C

CE

(rf)


1.Cm

T: cm W: , N A: , cm2 G: H:, cm


AA=

T C

1

2

3

Cm


1 100N80g60cm

0.031g/cm3

W=100NG=80gH=60cm

m =3.6 105PaC=3.6

T=CH/ G=3.6 60/80=2.7 cm

A=WG 104/ m

=100 80 104/(3.6 105)=222 cm2


2 100N30g2000cm260cm

W=100NG=30gH=60cm

A=2000 cm2

m=WG/A 104=100 30/2000104 =0.15105 Pa

m =0.15105Pa0.08C=3.5

T=CH/G=3.560/30=7 cm

0.08

AT=20007 (cm3)


3: 90N50g750cm260cm0.075g/cm3

W=90NG=50gH=60cm,

C=5.7

m=0.17105 Pa

A=WG104/m =9050104/(0.17105 ) =2648 cm2

C=3.3

m=0.7105 Pa

A=WG104/m

=9050104/(0.7105)643 cm2

T=CH/G=3.360/50=3.96 cm


2.Gs


Gs

T

1

2HG GsT

1H

2

3T

4


200N60g50cmx40cm,50cm6-6

W=200NG=60gH=50cm

1

b, Gm=60gB

B


D

EE

T=4cmA=285Cm2A=500cm2

A=1/4A

(2)

Gm=60g

T=4cmCDD


.

Pa

1.()

Pa

2.

Amin/1.33T2>1 Amin>1.33T2

Tcm

Acm2


3.

1


2

ldb

Ae= 3lbd/

l=b=d

Ae=3l2/ = l2

:

1.73


3

Ae=L2(l+b+d)/

Ae= L2

4

Ae=4L2(l+b+d)/

14K

Ae=KL2(l+b+d)/


4

Ae= L2L2 L2

L2Ae=4 L24L24 L2


4.

Tc=T1+Cr

Tccm

Cr%

Tcm

5.


1

2T=5cmAB0.015105Pa0.05105PaB

[] W=200NG=85g90cm10%

34

L2=A/4=400/4=100cm2

4 Amin>1.33T2 L2=100>1.33T2=44


5

K=1 Ae=K L2=173(cm2);

s=W/Ae104=200/173104=0.116105(Pa)

B

Ae=W/s104=200/(0.05105)104=400(cm2)

L2=400,L2=400/ =231.3(cm2)

L=15.2cm

A=4L2=4231.3=925cm2

s=W/A104=200/925104=0.0216105Pa

6

Tc=T1+Cr=51+10%=5.5cm

5.5cm15.2cm


.C

ATV=AT

A=WG/ T=CH/G V=HWC/

WHC

V=WHC -C/ 2

V=0C=C/

C- C0 C0 C0

100C

2C0

3

4


50N50cm40g

1

CC=2.4,

10KPa

A=WG/ =5040/10103104=2000cm2

T=CH/G=2.450/40=3(cm)

V=AT=20003=6000(cm3)

2C02.6

11.5KP

A=WG/ 104=5040/11500104=1739.1(cm2)

T=CH/G=2.650/40=3.25(cm)

VAT=1739.13.25=5652(cm2)


5

.

1>,

2

3

4


.

  • .

    1

    2

    3


.

1.

(1)


2

  • ()

()E

A T

(fn)


2.

0.08-0.20

0.10-0.50

0.02-0.16

0.11-0.23

0.008-0.016

3.

()


.

(1)

WAl04(Pa)

(2) )

(3)

(4)

(5) ,

(6)


300N3030cm225Hz00245g10cm

(1)W/A104=300/(3030) 104 =3.33(kPa)


T(Trf)7Hz2.57Hz1g 2.5l2.5(g)45g(2)25Hz0.55g0.25=0.250.55=0.14(g)

Tr = = 1.04

Tr = =25


6

.

1

2

3

4

5

.


1.

>

>,


2.

  • T>d(1E)

    T--

    E--

    d--


(629)


3.

(1),

(2)


(3)

(4)14


(5)

(6)

(7)

(8)


7

Ae

baT

P=WG

dxWG(EAy)dy

x

Ae=T/

A

Ae=


630E

aby

Ae

(i=1,2,3n)


8

1.CAD


2.


1 -

2

3 --

4 103-

1172310--

21210--

5

6


7.

8.

9.

10.

11.

12.-

13.-

14.

15.


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