Packaging Materials, SE2127 Problems with Solutions

Packaging Materials, SE2127 Problems with Solutions Compiled by Mikael S. Magnusson and Sören Östlund Department of Solid Mechanics Royal Institute of Technology - KTH, Stockholm, Sweden 2010

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 content in the Packaging Materials course by some mechanics oriented problems. The majority of problems have their origin in previous exams and the course Paper Physics that was taught at KTH by Prof. Christer Fellers for many years. Other problems are inspired by material from Innventia AB, former STFI Packforsk. All contributions are gratefully acknowledged. February 2010 Mikael S. Magnusson and Sören Östlund 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 the course in Paper Physics that he was responsible for at KTH during many years. 2.1.1 Calculate the bending stiffness for the following paper Is the answer S b = 48,18 mnm? What happens if we make the tensile stiffness index twice Table 1: Ply properties Ply k E w = E/ρ [MNm/kg] ρ [kg/m 3 ] w [g/m 2 ] E = E w ρ [MPa] t = w/ρ [mm] 1 7.5 800 80 6000 0.1 2 2,5 400 80 1000 0,2 3 4,0 500 50 2000 0,1 4 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 grammage is 300 g/m 2, and try to construct a) the stiffest 3-layer paper b) the stiffest 2-layer paper c) the stiffest 1-layer paper Why 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 assuming that the grammage is 100 g/m 2. Which material has the highest bending stiffness? Packaging Materials, SE2127 2

Table 2: Material properties Material Elastic modulus Density Tensile Stiffness Index E [MN/m 2 ] ρ [kg/m 3 ] E w = E/ρ [MNm/kg] Steel 210 000 7 800 25 Aluminium 73 000 2 800 25 Glass 73 000 2 400 25 Concrete 15 000 2 500 6 Carbon fibre composite 200 000 2 000 100 Wood 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/m 2 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 is five 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/m 2, E w = 7 MNm/kg, ρ = 700 kg/m 3 Centre, 200 g/m 2, E w = 2 MNm/kg, ρ = 400 kg/m 3 Calculate the bending stiffness. Change the grammage of the two surface plies to 20 and 60 g/m 2 respectively. Calculate the bending stiffness. How important was the symmetry? 2.1.6 A paper has the following data. Density ρ = 400 [kg/m 3 ] Tensile stiffness index E w = E x = 10 [MNm/kg] Tensile stiffness index E w = E y = 3 [MNm/kg] = 0.02 [%strain/ %RH] Hygroexpansion coefficient βy RH = 0.04 [%strain/ %RH] And the plies have properties according to Table 3 below. Hygroexpansion coefficient β RH x Calculate the curl and twist. Packaging Materials, SE2127 3

Figure 1: Article from Iggesund Packaging Materials, SE2127 4

Table 3: Ply properties Layer Angle phi [degree] Grammage, w [g/m 2 ] Change in moisture, RH [%] 1-5 100 10 2 5 100 10 2.1.7 Calculate the curl and twist for a material with properties according to the paper in Section 2.1.6 and Table 4.. Table 4: Ply properties Layer Angle phi [degree] Grammage, w [g/m 2 ] Change in moisture, RH [%] 1 10 100 10 2 10 100 10 2.1.8 Twist occurs if some layer in the paper is asymmetrical in relation to the MD, see Figure 2. During exposure to moisture, such crookedness leads to a turning moment which twists the 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 symmetry. The definition of the off-axis angle is indicated. k 6 = 3 t [ϕt H T (β T y β T x ) ϕ B H B (β B x β B x )] In the equation above, β is the hygroexpansion coefficient for the bottom ply B and the top ply T respectively. Packaging Materials, SE2127 5

H is the change in moisture and x and y are two perpendicular directions in the paper where 1 is MD and 2 is CD. Calculate the twist. Note that the angle shall be given in radians. Compare the calculations to those made previously in Problem 2.1.6. 2.1.9 A paper has the following data Density ρ = 600 [kg/m 3 ] Tensile stiffness index E w = E x = 8 [MNm/kg] Tensile stiffness index E w = E y = 2 [MNm/kg] Hygroexpansion coefficient βx RH = 0.03 [%strain/ %RH] Hygroexpansion coefficient βy RH = 0.04 [%strain/ %RH] And the plies have properties according to the table below. Calculate the curl and twist How does the paper look like? Table 5: Ply properties Layer Angle phi [degree] Grammage, w [g/m 2 ] Change in moisture, RH [%] 1 0 100 10 2 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 2.1.9 For one of the directions, Calculate the effect of release of internal strains of -0.1% strain in layer 1. 2.1.11 For one of the directions, Use data for the problem in Section 2.1.9 to show how sensitive the calculation of curl is to an increase of the tensile stiffness index by a factor 5. Guess first - then make the calculation. Packaging Materials, SE2127 6

2.1.12 For one of the directions, Use data for the problem in Section 2.1.9 to show how an increase of 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 increase of hygroexpansion coefficient by a factor of 2 influences the curl. Guess first then make the calculation. 2.1.14 It is possible by laminate theory to calculate curl and twist if the properties of the plies of the paper are known. Simplified equations have been derived in special cases, when the paper is modelled as two plies with equal thickness, see Figure 4 Compare the data in the problem in Section 2.1.9. Curl = 1 R = 3 2t (βb H B β T H T ) 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 properties X : E w = 5 [MNm/kg] Y : E w = 2 [MNm/kg] Angle + 6 Density 600 Grammage 200 Packaging Materials, SE2127 7

and is exposed to the following forces N 1 10 N 2 N 6 = 0 0 [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 ϵ 1 0.1 ϵ 2 ϵ 6 = 0.3 [ ] 0.2 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 board by 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 measuring the 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 of 14 % on the top side and 8 % on the bottom side. Suggest a way of simulating the curl tendency. Packaging Materials, SE2127 8

Figure 5: Figure 6: 4 Packaging Materials, SE2127 9

Figure 7: 2.2 Paper Mechanics 2.2.1 The figure shows the stress-strain curves in MD and CD for a linerboard determined using 150 mm long and 15 mm wide test pieces. The material has a grammage/basis weight of 200 g/m 2 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] Packaging Materials, SE2127 10

m) The MD tensile strength index in [Nm/kg] n) The CD tensile strength index in [Nm/kg] o) The MD ultimate strain p) 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 to the figure. Calculate all in-plane strains in the test-piece when it is loaded with a stress σ in the 30 -direction. Figure 8: 2.2.3 A popular method for determination of the bending stiffness of paper-based materials is four-point bending, as illustrated in the figure below. a) Discuss why this method is particularly useful for corrugated board and multi-ply paperboard. b) Derive the relation S = F l 1 l 2 2 δ 16 for determination of the bending stiffness S as function of applied load and measured deflection δ. Hint use to use the elementary beam solutions provided in the Appendix. Packaging Materials, SE2127 11

Figure 9: 2.2.4 The newsprint used in a roll printing press is expected to contain edge cracks no longer than 4 mm. Calculate the ratio between the maximum allowable line load when consider the edge cracks and when the influence of the edge cracks is neglected. Assume that linear elastic fracture mechanics (LEFM) is valid. It is known that the tensile strength of the newsprint is 40 MN/m 2. The stress intensity factor is given by where σ is the tensile stress in the web and a is the length of the crack. The fracture toughnessk Ic = 2MP a m and the average thickness of the newsprint is 20 µm. Figure 10: Packaging Materials, SE2127 12

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 11 in Appendix. Here, the force, F, is determined that is required to deform a test-piece of width b = 38 mm and length, L = 50 mm, to a deflection, δ, corresponding to a certain angle, θ, as illustrated in the Figure?? in Appendix. From a colleague you learn that an angle of 15 is frequently used in paperboard testing, and you determine the force to 1.1 N. 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 be really 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 the stress-strain curve in Figure 11. Packaging Materials, SE2127 13

2.3 Box design 2.3.1 You are the chief designer in an old-fashioned corrugated board mill that does not have the latest computer softwares for design of corrugated board boxes. One day a customer is requesting a new container (Box B) with dimensions according to Figure 12. Each box should contain a product with a mass of 10 kg, and in the packaging system five such boxes will be stacked on top of each other. Design a box that is strong enough in terms of type of corrugated board, according to Table 7 in the Appendix, and choice of fluting and liner, according to Table 8 in the Appendix. For simplicity a symmetric single type of board with the same type of material for the top and botton liner. Furthermore, the customer will use the boxes in a varying climate that requires a safety factor of 6 against box 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 formula P c = βfc 0.75 S 0.25 Z 0.5 where P c is the Box compression strength, F c is the Compressive strength of plane panel (ECT), S is the geometric mean of MD and CD bending stiffnesses SMD b Sb CD, Z is the perimeter of the box, and β is an empirical constant. From experience you know that another box (Box A) produced by your company using Flute 1 and White Top Liner 1 has strength and stiffness data according to Table 8 in the Appendix. In the analysis you need to estimate the compressive strength and the bending stiffnesses of 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 manager will not accept NO answer. One useful equation is the bending stiffness of corrugated Packaging Materials, SE2127 14

board when the influence of the corrugated medium is neglected. S b = Eb liner t 2 2 In this equation, t is the thickness of the corrugated board, Eb liner is the tensile stiffness of the liner and S b is the bending stiffness (per unit width in the MD or CD). The Box A had the following properties Table 6: Properties of Box A Parameter Value Length 0.3 m Width 0.3 m Height 0.2 m ECT 6.58 10 3 N/m BCT 2285 N SMD b 6.2 Nm 2.69 Nm S b CD 2.3.2 Design a RSC (0201) corrugated board box with dimensions L = 400mm, H = 300mm and W = 350mm that should contain a product with a mass of 20 kg, and in the packaging system five such boxes will be stacked on top of each other. Since the boxes will be subjected to a varying climate, a high safety factor of 8 is required. Carry out the design using the Billerud Box Design tool. 2.3.3 Which type of corrugated board is for example required if each box should contain a load of 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 Korsnäs Optipack. a) Consider a A2120 box with dimension L = 100 mm, W = 100 mm and H = 150 mm made from Korsnäs White 265 g/m 2. What is the BCT-value according to an analysis using Korsnäs Optipack when the box is loaded in the 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 from Korsnäs White 320? Use Grangård s formula and compare with Korsnäs Optipack. Packaging Materials, SE2127 15

2.4 Design of cushioning 2.4.1 Consider a loud speaker according to Figure 13. Design cushioning using a material having a cushioning factor according to Figure 14. Figure 13: Dimensions and shock resistance of loud speaker Figure 14: Cushioning factor The values w min and w max defines the damping materials optimal application interval. Within this interval, a common approximation of the cushioning factor is C = K = 1.2C min = 1.2 2.7 = 3.2 The necessary thickness, T of the corner protectors is given by T = Ch G and the necessary volume is given by V = Mgh w Packaging Materials, SE2127 16 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 capacity per 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 screen TV as shown in Figure 15. The cushioning should consist of four pieces of the dampening material mounted at the bottom and top corners of the TV-screen. By drop testing the allowable failure retardation (negative acceleration) of the TV-screen has been determined to 25g, where g 10[m/s 2 ] is the gravity constant. The dampening factor of the cushioning material is given by the graph in Figure 16. The weight of the screen is 14 kg. Determine the thickness and area, A, of each of the four pieces of dampening material in order to avoid damage in case the screen is dropped vertically from a height of 0.6 m. Note that only 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 thickness of 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: Packaging Materials, SE2127 17

Figure 16: Cushioning factor, c 3 Solutions The work of including solutions to the exercises have just begun and will be further developed in upcoming revisions. 3.1 Laminate theory Packaging Materials, SE2127 18

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 m Grammage, w = 0.2 kg/m 2 a) E MD = 79 10 6 /0.01 = 7.9 10 9 = 7.9 GPa b) E CD = 21 10 6 /0.01 = 2.1 10 9 = 2.1 GPa c) The MD tensile stiffness = E MD t = 1.975 MN/m d) The CD tensile stiffness = E CD t = 0.525 MN/m e) The MD tensile stiffness index = E MD t/w = 9.875 MNm/kg f) The CD tensile stiffness index = E CD t/w = 2.625 MNm/kg g) σ y,md = 41 MPa h) σ y,cd = 12 MPa i) σ b,md = 72 MPa j) σ b,cd = 25 MPa k) The MD tensile strength = σ b,md t = 18 kn/m l) The CD tensile strength = σ b,cd t = 6.25 kn/m m) The MD tensile strength = σ b,md tw = 90 knm/kg n) The CD tensile strength = σ b,md tw = 31.25 knm/kg o) ϵ b,md = 0.017 or 1.7% p) ϵ b,cd = 0.036 or 3.6% Packaging Materials, SE2127 19

3.3 Box design 3.3.1 Solution to 2.3.1 With a packaging system of five boxes per stack results in a load of in the bottom box of m = 4 10 = 40kg with a safety factor, s = 6 the required Box-Compression-Strength of box B is calculated as P cbreq. = 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 for Box A, gives β = P ca FcA 0.75S0.25 A Z0.5 A = 2 The specific bending stiffness is approximated as SMD b = Eliner t 2 b 2, where the geometric mean of the bending stiffnesses S = SMD b Sb CD is used. The perimeter of the box is Z = 2w + 2L. And lastly, since we don t have value of the ECT test for other box panels than that of Box A, we could assume that F c is directly proportional to the thickness, t and therefor use the value of Box A as t F cb = F B ca ta Now, Mackee s formula for the Box-Compression-Strength for Box B is calculated as t P cb = β(f B ca ta ) 0.75 ( EMB linereliner t 2 B CD 2 ) 0.25 2w + 2L Assuming a type A board and White Top Liner 2 in Table 7 and Table TA2 in the Appendix gives P cb = 2950 > P cbreq. Note that in this approximation the fluting s contribution to the bending stiffnesses are neglected. Packaging Materials, SE2127 20

3.4 Design of cushioning 3.4.1 Solution to 2.4.1 The cushioning factor is approximated by c = K = 3.2 according to the standard, the height of the fall, h = 0.4 and the dampening factor is given as G = 30g, therefore, the thickness of the corner protection, T is calculated as T = ch G = 0.0427m The least amount of material needed is when the cushioning material is at it s optimal performance, that is when w = w max = 155kJ/m 3, hence V min = Mgh w max = 0.3041 10 3 m 3 and w = w min = 32kJ/m 3 gives V max = Mgh w min = 1.5 10 3 m 3 The dampening surface area can then be calculated as A min = V min T = 0.00712 m 2 and A max = V max T = 0.0351 m 2 To obtain a low cost design, the least amount material is choosen. Assume a load case where the box falls with at least one face parallell to the fall direction, 4 surfaces of the corner 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 surface shaded cushioning material is square, and it s side has the length, B, then 4B 2 A = A min B = min 4 = 0.0422 m The cushioning material could then be chosen as a cube with sides equal to T + B and a cut-out cube with sides B. Note that the cushioning area is often chosen slightly larger to obtain some safety factor if the height of fall is larger than h. Packaging Materials, SE2127 21

4 Appendix Following are some useful formulas and tables used in the exercises Two-Point Bending For two-point bending (as seen in Figure 19) in a linear elastic material, the relation between the force, F and the end-displacement, δ is given by δ = F L3 3EI, where the moment of inertia for a rectangular cross section is given by I = bt3 12. The relation between the end-deflection angle θ and the applied force is calculated from θ = F L2 2EI The maximum bending normal stress for a beam of rectangular cross-section is σ = F Lt 2I, where E is Young s modulus, b is the width and t is the thickness of the test piece. Figure 19: Packaging Materials, SE2127 22

Additional tables Following are tables with material properties used in the problems above. Table 7: Corrugated board material properties Type Thickness [t/mm] Wave length of fluting pattern [l/mm] Take-up factor, f A 4.7 8.6 1.53 B 2.6 6.1 1.36 C 3.6 7.3 1.45 D 6.0 11.5 1.5 E 1.2 3.2 1.29 F 0.8 2.3 1.25 Table 8: Fluting and liner material properties Quality Grammage SCT Tensile Stiffness MD Tensile Stiffness CD Thickness [g/m 2 ] [kn/m] [kn/m] [kn/m] [mm] Flute 1 127 2.6 1400 420 0.21 Flute 2 140 2.9 1500 500 0.23 Flute 3 150 3.2 1600 550 0.24 Flute 4 175 3.8 1850 620 0.28 Flute 5 200 4.3 1800 750 0.315 White Top Liner 1 140 2.6 1400 510 0.17 White Top Liner 2 175 3.3 1700 620 0.22 White Top Liner 3 200 3.8 2000 720 0.25 White Liner 1 110 2.1 1100 400 0.135 White Liner 2 120 2.3 1250 450 0.145 White Liner 3 135 2.5 1400 500 0.165 White Liner 4 170 3.2 1700 600 0.215 White Liner 5 190 3.8 2000 700 0.24 Packaging Materials, SE2127 23

Solutions for simple beam theory Figure 20: Packaging Materials, SE2127 24

Figure 21: Packaging Materials, SE2127 25