packaging materials, se2127 problems with solutions

Packaging Materials, SE2127Problems with Solutions

—————————————————–Compiled by Mikael S. Magnusson and Soren Ostlund

—————————————————–

Department of Solid Mechanics

Royal Institute of Technology - KTH, Stockholm, Sweden

2010

Department of Solid Mechanics - KTH

Contents

1 Preface 1

2 Problems 2

2.1 Laminate theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.2 Paper Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3 Box design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.4 Design of cushioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Solutions 18

3.1 Laminate theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2 Paper Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3 Box design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4 Design of cushioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Appendix 22

Packaging Materials, SE2127 ii


1 Preface

This collection of problems constitutes the first attempts to illustrate parts of the contentin the Packaging Materials course by some mechanics oriented problems. The majorityof problems have their origin in previous exams and the course Paper Physics that wastaught at KTH by Prof. Christer Fellers for many years.Other problems are inspired by material from Innventia AB, former STFI Packforsk. Allcontributions are gratefully acknowledged.

February 2010Mikael S. Magnusson and Soren Ostlund

Packaging Materials, SE2127 1


2 Problems

Note that the following collection of problems is currently under development.

2.1 Laminate theory

The problems in this chapter are all developed by Prof. Christer Fellers as part of thecourse in Paper Physics that he was responsible for at KTH during many years.

2.1.1

Calculate the bending stiffness for the following paperIs the answer Sb= 48,18 mNm? What happens if we make the tensile stiffness index twice

Table 1: Ply propertiesPly k Ew = E/ρ [MNm/kg] ρ [kg/m3] w [g/m2] E = Ew · ρ [MPa] t = w/ρ [mm]

1 7.5 800 80 6000 0.12 2,5 400 80 1000 0,23 4,0 500 50 2000 0,14 7,14 700 70 5000 0,1

as high in all plies? Would you like to do the calculation by hand?

2.1.2

Use the pulps in the plies in problem in section 2.1.1, assuming that the total grammageis 300 g/m2, and try to constructa) the stiffest 3-layer paperb) the stiffest 2-layer paperc) the stiffest 1-layer paperWhy did you choose a particular pulp in a specific layer? Which product is the stiffest?

2.1.3

The table below shows data for different materials. Calculate the bending stiffness assum-ing that the grammage is 100 g/m2.Which material has the highest bending stiffness?



Table 2: Material propertiesMaterial Elastic modulus Density Tensile Stiffness Index

E [MN/m2] ρ [kg/m3] Ew = E/ρ [MNm/kg]

Steel 210 000 7 800 25Aluminium 73 000 2 800 25

Glass 73 000 2 400 25Concrete 15 000 2 500 6

Carbon fibre composite 200 000 2 000 100Wood 14 000 500 25

Paper, linerboard in MD 15 600 700 22

2.1.4

The article (Figure 1) was published in 2006.Assume that you put an aluminium layer of 2 g/m2 on the paperboard in Problem 1.How much stiffer will the product be?Would it be better to put additional fibres on the product? Assume that aluminium isfive times more expensive than fibres.

2.1.5

Show the importance of symmetry for bending stiffness.Construct a three-ply board with the following data.Two surfaces, 40 g/m2, Ew = 7 MNm/kg, ρ = 700 kg/m3

Centre, 200 g/m2, Ew = 2 MNm/kg, ρ = 400 kg/m3

Calculate the bending stiffness.Change the grammage of the two surface plies to 20 and 60 g/m2 respectively.Calculate the bending stiffness.How important was the symmetry?

2.1.6

A paper has the following data.Density ρ = 400 [kg/m3]Tensile stiffness index Ew = Ex = 10 [MNm/kg]Tensile stiffness index Ew = Ey = 3 [MNm/kg]Hygroexpansion coefficient βRH

x = 0.02 [%strain/ %RH]Hygroexpansion coefficient βRH

y = 0.04 [%strain/ %RH]And the plies have properties according to Table 3 below.

Calculate the curl and twist.



Figure 1: Article from Iggesund



Table 3: Ply propertiesLayer Angle phi [degree] Grammage, w [g/m2] Change in moisture, ∆RH [%]

1 -5 100 102 5 100 10

2.1.7

Calculate the curl and twist for a material with properties according to the paper inSection 2.1.6 and Table 4..


1 10 100 102 10 100 10

2.1.8

Twist occurs if some layer in the paper is asymmetrical in relation to the MD, see Figure2. During exposure to moisture, such crookedness leads to a turning moment which twiststhe sheet. Twist may be expressed in a simple equation as follows;

Figure 2: Twist occurs, for a two-ply paper when one or two plies have an off-axis sym-metry. The definition of the off-axis angle is indicated.

k6 =3t [ϕ

THT (βTy − βT

x )− ϕBHB(βBx − βB

x )]In the equation above, β is the hygroexpansion coefficient for the bottom ply B and thetop ply T respectively.



H is the change in moisture and x and y are two perpendicular directions in the paperwhere 1 is MD and 2 is CD.Calculate the twist. Note that the angle shall be given in radians. Compare the calcula-tions to those made previously in Problem 2.1.6.

2.1.9

A paper has the following dataDensity ρ = 600 [kg/m3]Tensile stiffness index Ew = Ex = 8 [MNm/kg]Tensile stiffness index Ew = Ey = 2 [MNm/kg]Hygroexpansion coefficient βRH

x = 0.03 [%strain/ %RH]Hygroexpansion coefficient βRH

y = 0.04 [%strain/ %RH]And the plies have properties according to the table below. Calculate the curl and twistHow does the paper look like ?


1 0 100 102 90 100 10

Which bending moment do we have to apply to make the paper flat ?

Figure 3:

2.1.10

Use the data for the problem in Section 2.1.9For one of the directions, Calculate the effect of release of internal strains of -0.1% strainin layer 1.

2.1.11

For one of the directions, Use data for the problem in Section 2.1.9 to show how sensitivethe calculation of curl is to an increase of the tensile stiffness index by a factor 5.Guess first - then make the calculation.



2.1.12

For one of the directions, Use data for the problem in Section 2.1.9 to show how an increaseof grammage by a factor of 2 would influence curl. Guess first then make the calculation.

2.1.13

For one of the directions, Use data for the problem in Section 2.1.9 to show how an increaseof hygroexpansion coefficient by a factor of 2 influences the curl. Guess first then makethe calculation.

2.1.14

It is possible by laminate theory to calculate curl and twist if the properties of the pliesof the paper are known. Simplified equations have been derived in special cases, when thepaper is modelled as two plies with equal thickness, see Figure 4Compare the data in the problem in Section 2.1.9.Curl = 1

R = 32t(β

BHB − βTHT )β is the hygroexpansion coefficient for the bottom ply B and the top ply T respectively.H is the change in moisture. Curl thus depends on the difference in hygroexpansion, and

Figure 4: Paper is modelled as two plies with equal thickness

on the strain on the bottom and top plies. Curl decreases with increasing thickness.It can be shown that the tensile stiffness index of the layers plays a very small role

2.1.15

A paper has the following material propertiesX : Ew = 5 [MNm/kg]Y : Ew = 2 [MNm/kg]Angle + 6 °Density 600Grammage 200



and is exposed to the following forces N1

N2

N6

=

1000

[kN/m]

Calculate the strains.Draw a picture of the final shape in relation to the original shape.

2.1.16

The paper in the problem of Section 2.1.15 is exposed for the following strains ϵ1ϵ2ϵ6

=

0.10.30.2

[−]

Calculate the stresses. Draw a picture of the shape.

2.1.17

An article deals with the calculation of curl and twist of corrugated board.Look at Figure 5. Show that you can simulate the change in shape of a corrugated boardby using the laminate program.Suppose the RH change from 20 to 85 % RH.Is the twist direction true?Use the following flute type data.C flute: 3.6 mm

2.1.18

Look at the author in Figure 6. He claims that you can calculate the twist by measuringthe curl at +45 degrees and -45 degrees and subtract the values from each other.Make a calculation that shows if he is right or wrong.

2.1.19

A paper is flat at 8 % moisture content, but gets a moisture profile in an offset press of14 % on the top side and 8 % on the bottom side. Suggest a way of simulating the curltendency.



Figure 5:

Figure 6:

4



Figure 7:

2.2 Paper Mechanics

2.2.1

The figure shows the stress-strain curves in MD and CD for a linerboard determined using150 mm long and 15 mm wide test pieces. The material has a grammage/basis weight of200 g/m2 and a thickness of 0.25 mm.

Calculate the following fundamental mechanical properties:a) The MD elastic modulus in [Pa]b) The CD elastic modulus in [Pa]c) The MD tensile stiffness in [N/m]d) The CD tensile stiffness in [N/m]e) The MD tensile stiffness index in [Nm/kg]f) The CD tensile stiffness index in [Nm/kg]g) The MD yield stress in [Pa]h) The CD yield stress in [Pa]i) The MD ultimate stress in [Pa]j) The CD ultimate stress in [Pa]k) The MD tensile strength in [N/m]l) The CD tensile strength in [N/m]



m) The MD tensile strength index in [Nm/kg]n) The CD tensile strength index in [Nm/kg]o) The MD ultimate strainp) The CD ultimate strain

2.2.2

A test-piece of the material in Problems 1 and 2 is cut at an angle 30° to MD according tothe figure. Calculate all in-plane strains in the test-piece when it is loaded with a stress σin the 30°-direction.

Figure 8:

2.2.3

A popular method for determination of the bending stiffness of paper-based materials isfour-point bending, as illustrated in the figure below.a) Discuss why this method is particularly useful for corrugated board and multi-plypaperboard.b) Derive the relation S = ∆F

∆δl1l2216 for determination of the bending stiffness S as function

of applied load and measured deflection ∆δ. Hint use to use the elementary beam solutionsprovided in the Appendix.



Figure 9:

2.2.4

The newsprint used in a roll printing press is expected to contain edge cracks no longerthan 4 mm. Calculate the ratio between the maximum allowable line load when considerthe edge cracks and when the influence of the edge cracks is neglected. Assume that linearelastic fracture mechanics (LEFM) is valid. It is known that the tensile strength of thenewsprint is 40 MN/m2. The stress intensity factor is given by where σ is the tensile stressin the web and a is the length of the crack. The fracture toughnessKIc = 2MPa

√m and

the average thickness of the newsprint is 20 µm.

Figure 10:



2.2.5

Consider a paperboard material with stress-strain curve in MD according to Figure 11.The thickness of the paperboard is 0.48 mm. A frequently used method for measurement

Figure 11:

of the bending stiffness of paperboard is the two-point method illustrated in Figure 11in Appendix. Here, the force, F , is determined that is required to deform a test-piece ofwidth b = 38 mm and length, L = 50 mm, to a deflection, δ, corresponding to a certainangle, θ, as illustrated in the Figure ?? in Appendix. From a colleague you learn that anangle of 15° is frequently used in paperboard testing, and you determine the force to 1.1N.a) Calculate the bending stiffness per unit width of the paperboard using this force.b) When applying this value in box compression analysis you suspect this value not to bereally accurate. What could be wrong? Would you expect a higher or a lower value?c) Estimate a more correct value of the bending stiffness per unit width considering thestress-strain curve in Figure 11.



2.3 Box design

2.3.1

You are the chief designer in an old-fashioned corrugated board mill that does not havethe latest computer softwares for design of corrugated board boxes. One day a customeris requesting a new container (Box B) with dimensions according to Figure 12. Each boxshould contain a product with a mass of 10 kg, and in the packaging system five suchboxes will be stacked on top of each other. Design a box that is strong enough in termsof type of corrugated board, according to Table 7 in the Appendix, and choice of flutingand liner, according to Table 8 in the Appendix. For simplicity a symmetric single typeof board with the same type of material for the top and botton liner. Furthermore, thecustomer will use the boxes in a varying climate that requires a safety factor of 6 againstbox failure.The box dimensions are: L = 300mm, H = 300mm and W = 200mm.

Figure 12:

Since the mill does not have any computer softwares, you have to rely on McKee’s formulaPc = βF 0.75

c S0.25Z0.5

wherePc is the Box compression strength,Fc is the Compressive strength of plane panel (ECT),

S is the geometric mean of MD and CD bending stiffnesses√

SbMDS

bCD,

Z is the perimeter of the box, andβ is an empirical constant. From experience you know that another box (Box A) producedby your company using Flute 1 and White Top Liner 1 has strength and stiffness dataaccording to Table 8 in the Appendix.In the analysis you need to estimate the compressive strength and the bending stiffnessesof the panels. Carry out the analysis using as simple mathematical expressions as possible.If necessary derive your own equations in order to obtain an applicable design. Your man-ager will not accept NO answer. One useful equation is the bending stiffness of corrugated



board when the influence of the corrugated medium is neglected.Sb = Eliner

bt2

2In this equation, t is the thickness of the corrugated board, Eliner

b is the tensile stiffnessof the liner and Sb is the bending stiffness (per unit width in the MD or CD). The Box Ahad the following properties

Table 6: Properties of Box AParameter Value

Length 0.3 mWidth 0.3 mHeight 0.2 mECT 6.58·103 N/mBCT 2285 NSbMD 6.2 Nm

SbCD 2.69 Nm

2.3.2

Design a RSC (0201) corrugated board box with dimensions L = 400mm, H = 300mm andW = 350mm that should contain a product with a mass of 20 kg, and in the packagingsystem five such boxes will be stacked on top of each other. Since the boxes will besubjected to a varying climate, a high safety factor of 8 is required. Carry out the designusing the Billerud Box Design tool.

2.3.3

Which type of corrugated board is for example required if each box should contain a loadof 20 kg and the box dimensions are are: L = 500mm, H = 400mm and W = 400mm..Four boxes are stacked on top of each other and a safety factor of 8 is required.

2.3.4

Design exercise with Korsnas Optipack. a) Consider a A2120 box with dimension L = 100mm, W = 100 mm and H = 150 mm made from Korsnas White 265 g/m2. What is theBCT-value according to an analysis using Korsnas Optipack when the box is loaded inthe vertical direction, and the fibre orientation is along the box?

b) What would the BCT-value be for a box of the same dimensions, but made fromKorsnas White 320? Use Grangard´s formula and compare with Korsnas Optipack.



2.4 Design of cushioning

2.4.1

Consider a loud speaker according to Figure 13. Design cushioning using a material havinga cushioning factor according to Figure 14.

Figure 13: Dimensions and shock resistance of loud speaker

Figure 14: Cushioning factor

The values wmin and wmax defines the damping materials optimal application interval.Within this interval, a common approximation of the cushioning factor isC = K = 1.2Cmin = 1.2 · 2.7 = 3.2The necessary thickness, T of the corner protectors is given by T = Ch

G

and the necessary volume is given by V = Mghw where, G is the damping factor i g, M is



the mass, h is the height of the fall, w is the cushion materials energy absorption capacityper unit volume and g is the gravity acceleration.

2.4.2

You have been requested to design the cushioning of a package containing a flat screenTV as shown in Figure 15. The cushioning should consist of four pieces of the dampeningmaterial mounted at the bottom and top corners of the TV-screen. By drop testing theallowable failure retardation (negative acceleration) of the TV-screen has been determinedto 25g, where g ≈ 10[m/s2] is the gravity constant. The dampening factor of the cushioningmaterial is given by the graph in Figure 16. The weight of the screen is 14 kg. Determinethe thickness and area, A, of each of the four pieces of dampening material in order toavoid damage in case the screen is dropped vertically from a height of 0.6 m. Note thatonly the load-carrying area need to be considered, i.e. the shaded area in Figure 15.Useful formulas:c = aT

h , where a is the allowable retardation/g, h is the drop height and T is the thicknessof dampening material.W = mgh

V , where W is the impact energy(strain energy) per unit volume, m is the mass,g the gravity constant, V is the volume of dampening material.

Figure 15:



Figure 16: Cushioning factor, c

3 Solutions

The work of including solutions to the exercises have just begun and will be furtherdeveloped in upcoming revisions.

3.1 Laminate theory



3.2 Paper Mechanics

3.2.1 Solution to 2.2.1

Figure 17: Stress-Strain curve with calculations

Thickness, t = 2.5 · 10−2 mGrammage, w = 0.2 kg/m2

a) EMD = 79 · 106/0.01 = 7.9 · 109 = 7.9 GPab) ECD = 21 · 106/0.01 = 2.1 · 109 = 2.1 GPac) The MD tensile stiffness = EMDt = 1.975 MN/md) The CD tensile stiffness = ECDt = 0.525 MN/me) The MD tensile stiffness index = EMDt/w = 9.875 MNm/kgf) The CD tensile stiffness index = ECDt/w = 2.625 MNm/kgg) σy,MD = 41 MPah) σy,CD = 12 MPai) σb,MD = 72 MPaj) σb,CD = 25 MPak) The MD tensile strength = σb,MDt = 18 kN/ml) The CD tensile strength = σb,CDt = 6.25 kN/mm) The MD tensile strength = σb,MDtw = 90 kNm/kgn) The CD tensile strength = σb,MDtw = 31.25 kNm/kgo) ϵb,MD = 0.017 or 1.7%p) ϵb,CD = 0.036 or 3.6%



3.3 Box design


With a packaging system of five boxes per stack results in a load of in the bottom box ofm = 4 · 10 = 40kgwith a safety factor, s = 6 the required Box-Compression-Strength of box B is calculatedasPcBreq. = mgs = 2357,where g is the gravity acceleration.Using Mackee’s formula for box A and assuming that the β is the same for Box B as forBox A, givesβ = PcA

F 0.75cA S0.25

A Z0.5A

= 2

The specific bending stiffness is approximated asSbMD = Eliner

bt2

2, where the geometric mean of the bending stiffnesses

S =√

SbMDS

bCD is used.

The perimeter of the box isZ = 2w + 2L.And lastly, since we don’t have value of the ECT test for other box panels than that ofBox A, we could assume that Fc is directly proportional to the thickness, t and thereforuse the value of Box A asFcB = FcA

tBtA

Now, Mackee’s formula for the Box-Compression-Strength for Box B is calculated as

PcB = β(FcAtBtA)0.75(

√Eliner

MB ElinerCD

t2B2 )0.25

√2w + 2L

Assuming a type A board and White Top Liner 2 in Table 7 and Table TA2 in theAppendix givesPcB = 2950 > PcBreq.

Note that in this approximation the fluting’s contribution to the bending stiffnesses areneglected.



3.4 Design of cushioning


The cushioning factor is approximated by c = K = 3.2 according to the standard, theheight of the fall, h = 0.4 and the dampening factor is given as G = 30g, therefore, thethickness of the corner protection, T is calculated asT = ch

G = 0.0427mThe least amount of material needed is when the cushioning material is at it’s optimalperformance, that is when w = wmax = 155kJ/m3, henceVmin = Mgh

wmax= 0.3041 · 10−3 m3

and w = wmin = 32kJ/m3 givesVmax = Mgh

wmin= 1.5 · 10−3 m3

The dampening surface area can then be calculated asAmin = Vmin

T = 0.00712 m2

andAmax = Vmax

T = 0.0351 m2

To obtain a low cost design, the least amount material is choosen. Assume a load casewhere the box falls with at least one face parallell to the fall direction, 4 surfaces of thecorner protection will damp the fall, see Figure 18 Assume that the cut-out corner of the

Figure 18: Principle sketch of one of the four corner protectors with the dampening surfaceshaded

cushioning material is square, and it’s side has the length, B, then

4B2 = Amin ⇒ B =√

Amin4 = 0.0422 m

The cushioning material could then be chosen as a cube with sides equal to T +B and acut-out cube with sides B. Note that the cushioning area is often chosen slightly largerto obtain some safety factor if the height of fall is larger than h.



4 Appendix

Following are some useful formulas and tables used in the exercises

Two-Point BendingFor two-point bending (as seen in Figure 19) in a linear elastic material, the relationbetween the force, F and the end-displacement, δ is given byδ = FL3

3EI ,where the moment of inertia for a rectangular cross section is given byI = bt3

12 .The relation between the end-deflection angle θ and the applied force is calculated from

θ = FL2

2EIThe maximum bending normal stress for a beam of rectangular cross-section isσ = FLt

2I ,where E is Young’s modulus, b is the width and t is the thickness of the test piece.

Figure 19:



Additional tablesFollowing are tables with material properties used in the problems above.

Table 7: Corrugated board material propertiesType Thickness [t/mm] Wave length of fluting pattern [l/mm] Take-up factor, f

A 4.7 8.6 1.53B 2.6 6.1 1.36C 3.6 7.3 1.45D 6.0 11.5 1.5E 1.2 3.2 1.29F 0.8 2.3 1.25

Table 8: Fluting and liner material propertiesQuality Grammage SCT Tensile Stiffness MD Tensile Stiffness CD Thickness

[g/m2] [kN/m] [kN/m] [kN/m] [mm]

Flute 1 127 2.6 1400 420 0.21Flute 2 140 2.9 1500 500 0.23Flute 3 150 3.2 1600 550 0.24Flute 4 175 3.8 1850 620 0.28Flute 5 200 4.3 1800 750 0.315

White Top Liner 1 140 2.6 1400 510 0.17White Top Liner 2 175 3.3 1700 620 0.22White Top Liner 3 200 3.8 2000 720 0.25

White Liner 1 110 2.1 1100 400 0.135White Liner 2 120 2.3 1250 450 0.145White Liner 3 135 2.5 1400 500 0.165White Liner 4 170 3.2 1700 600 0.215White Liner 5 190 3.8 2000 700 0.24



Solutions for simple beam theory

Figure 20:



Figure 21:


packaging materials, se2127 problems with solutions

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