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《数学建模》美赛优秀论文:03 A O A Time-Independent Model of Box Safety for Stunt Motorcyclists

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A Time-Independent model 233 A Time-Independent model of Box Safety for Stunt motorcyclists van corwin Sheel Ganatra Nikita rozenblyum Harvard University Cambridge, ma Advisor: Clifford h. Taubes Abstract We develop a knowledge of the workings of corrugated fiberboard and create an extensive time-independent model of motorcycle collision with one box, our Single-Box Model. We identify important factors in box-to-box and frictional interactions, as well as several extensions of the Single-Box Model. Taking into account such effects as cracking, buckling, and buckling under other boxes, we use the energy-dependent Dual-Impact Model to show that the pyramid"configuration of large 90-cm cubic boxes-a configuration of boxes in which every box is resting equally upon four others-is optimal for absorption of the most energy while maintaining a reasonable deceleration. We show how variations in height and weight affect the model and calculate a bound on the number of boxes needed General assumptions The temperature and weather are assumed to be"ideal conditions"-they do not affect the strength of the box The wind is negligible, because the combined weight of the motorcycle and the person is sufficiently large The ground on which the boxes are arranged is a rigid flat surface that can take any level of force All boxes are cubic, which makes for the greatest strength [Urbanik 19971 The UMAP Journal 24(3)(2003)233-250. Copyright 2003 by COMAP, Inc. Allrights reserved Permission to make digital or hard copies of part or all of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial dvantage and that copies bear this notice. Abstracting with credit is permitted, but copyrights for components of this work owned by others than COMAP must be honored. To to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP

A Time-Independent Model 233 A Time-Independent Model of Box Safety for Stunt Motorcyclists Ivan Corwin Sheel Ganatra Nikita Rozenblyum Harvard University Cambridge, MA Advisor: Clifford H. Taubes Abstract We develop a knowledge of the workings of corrugated fiberboard and create an extensive time-independent model of motorcycle collision with one box, our Single-Box Model. We identify important factors in box-to-box and frictional interactions, as well as several extensions of the Single-Box Model. Taking into account such effects as cracking, buckling, and buckling under other boxes, we use the energy-dependent Dual-Impact Model to show that the “pyramid” configuration of large 90-cm cubic boxes—a configuration of boxes in which every box is resting equally upon four others—is optimal for absorption of the most energy while maintaining a reasonable deceleration. We show how variations in height and weight affect the model and calculate a bound on the number of boxes needed. General Assumptions • The temperature and weather are assumed to be “ideal conditions”—they do not affect the strength of the box. • The wind is negligible, because the combined weight of the motorcycle and the person is sufficiently large. • The ground on which the boxes are arranged is a rigid flat surface that can take any level of force. • All boxes are cubic, which makes for the greatest strength [Urbanik 1997]. The UMAP Journal 24 (3) (2003) 233–250. c Copyright 2003 by COMAP, Inc. All rights reserved. Permission to make digital or hard copies of part or all of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice. Abstracting with credit is permitted, but copyrights for components of this work owned by others than COMAP must be honored. To copy otherwise, to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP

234 The UMAP Journal 24.3 (2003) Table 1 Variables Variable Definition V(t) Velocity as a function of time(vo is velocity of impact At Energy change due to the top of the box kg-cm2/52 Energy change due to the buckling of the box g-cm ANET Total energy including gravity for top ACD-EDGE Energy absorbed by CD springs in modelled buckling kg-cm-/s AMD- EDGE Energy absorbed by MD springs in modelled buckling kg-cm/s NET Total depression of the top △ r Change in a distance am DOWn Component of edge's depression in the x-direction r(t) ownward displacement of the top of the box( F Final depression before top failure oL Offset from top center. Motorcycle and stuntman combined mass PcD Tensile strength in the cross-machine direction PMd Tensile strength in the machine direction kg/(s-cm) Maximum strength as measured with the Edge Crush Test Maximum strength as measured with the Mullen Test Force in the cross-machine direction kg-cm/st FCDMAx Maximum force in the cross-machine direction kg-cm Force in the machine direction FmDMAx Maximum force in the machine direction kg-cm/s FUP Net dampening force the box exerts on the motorcycle FUPMAX Maximum dampening force the box exerts on the motorcycle kg-cm/s Force box exerts on the frame kg-cm/s Maximum force the box exerts on the frame before yielding g-cm/s2 FNI Total force k Depression at which MD tensile strength is exceeded aCD Depression at which CD tensile strength is exceeded ECt Depression at which a buckle occurs Depression at which a puncture occurs Velocity in the r-direction cm/s Velocity in the y-direction cm/s Initial vele Initial velocity in the y-direction cm/s Vfr Final velocity in the x-direction cm/s Final velocity in the y-direction Mass of boxes displaced Energy in the r-direction kg-cm2/s2 Energy in the y-direction kg-cm/s Energy in the z-direction m2/s2 Distance

234 The UMAP Journal 24.3 (2003) Table 1. Variables. Variable Definition Units l Box edge length cm V (t) Velocity as a function of time (v0 is velocity of impact) cm/s AT Energy change due to the top of the box kg-cm2/s2 AB Energy change due to the buckling of the box kg-cm2/s2 ANET Total energy including gravity for top kg-cm2/s2 ACD-EDGE Energy absorbed by CD springs in modelled buckling kg-cm2/s2 AMD-EDGE Energy absorbed by MD springs in modelled buckling kg-cm2/s2 xNET Total depression of the top cm ∆x Change in a distance cm xDOWN Component of edge’s depression in the z-direction cm x(t) Downward displacement of the top of the box (x0 = 0) cm xF Final depression before top failure. cm δL Offset from top center. cm M Motorcycle and stuntman combined mass kg PCD Tensile strength in the cross-machine direction kg/(s2cm) PMD Tensile strength in the machine direction kg/(s2cm) PECT Maximum strength as measured with the Edge Crush Test kg/s2 PML Maximum strength as measured with the Mullen Test kg/(s2cm) FCD Force in the cross-machine direction kg-cm/s2 FCDMAX Maximum force in the cross-machine direction kg-cm/s2 FMD Force in the machine direction kg-cm/s2 FMDMAX Maximum force in the machine direction kg-cm/s2 FUP Net dampening force the box exerts on the motorcycle kg-cm/s2 FUPMAX Maximum dampening force the box exerts on the motorcycle kg-cm/s2 FECT Force box exerts on the frame kg-cm/s2 FECTMAX Maximum force the box exerts on the frame before yielding kg-cm/s2 FNET Total force kg-cm/s2 xMD Depression at which MD tensile strength is exceeded cm xCD Depression at which CD tensile strength is exceeded cm xECT Depression at which a buckle occurs cm xML Depression at which a puncture occurs cm Vx Velocity in the x-direction cm/s Vy Velocity in the y-direction cm/s Vix Initial velocity in the x-direction cm/s Viy Initial velocity in the y-direction cm/s Vfx Final velocity in the x-direction cm/s Vfy Final velocity in the y-direction cm/s MB Mass of boxes displaced kg t Time s Ax Energy in the x-direction kg-cm2/s2 Ay Energy in the y-direction kg-cm2/s2 Az Energy in the z-direction kg-cm2/s2 d Distance cm

A Time-Independent model 235 Table 2 Constants Variable Definition Value EcD Young s modulus in the cross-machine direction 80000kg/(2-cm) E Sum of Emd and ecD 3800kg/(2-cm) Tire rect length in the machine direction 7cm front, 10cm back PcD Tensile strength in the cross-Machine direction 2000kg/(s2-cm) PMD Tensile strength in the machine direction 2500kg/(s2-cm) PECT Max strength as measured with the edge crush test 10000kg/ PML Max strength as measured with the Mullen test 25000kg/(s2-cm) Gravitational Constant 980cm/s2 Coefficient of kinetic friction Cardboard thickness peed variation 200 cm/s Angle variation away from y-axis leaving the ramp Mass of rider and motorcycle vi Initial velocity leaving ramp 1500cm/s 6 np angle of elevation Definitions and Key terms Buckling is the process by which a stiff plane develops a crack due to a stress exceeding the yield stres Compressive strength is the maximum force per unit area that a material can withstand, under compression, prior to yielding Corrugation is the style found in cardboard of sinusoidal waves of liner paper sandwiched between inner and outer papers. We use boxes with the most common corrugation, C-flute corrugation(see below for definition of a flute Cracking is a resulting state when the tensile force exceeds the tensile strength Cross-machine direction is the direction perpendicular to the sinusoidal wave of the corrugation Depression is when, due to a force, a section of a side or edge moves down wards ECT is the acronym for a common test of strength, the edge crush test . Fiberboard is the formal name for cardboard Flute is a single wavelength of a sinusoidal wave between the inner and outer portion papers that extends throughout the length of the cardboard A C-flute has a height of 0. 35 cm and there are 138 flutes per meter[Packaging glossary n d I

A Time-Independent Model 235 Table 2. Constants. Variable Definition Value EMD Young’s modulus in the machine direction 3000000 kg/(s2-cm) ECD Young’s modulus in the cross-machine direction 800000 kg/(s2-cm) E Sum of EMD and ECD 3800000 kg/(s2-cm) LMD Tire rect. length in the machine direction 7cm front, 10cm back LCD Tire rect. length in the cross-machine direction 10 cm PCD Tensile strength in the cross-Machine direction 2000 kg/(s2-cm) PMD Tensile strength in the machine direction 2500 kg/(s2-cm) PECT Max. strength as measured with the edge crush test 10000 kg/s2 PML Max. strength as measured with the Mullen test 25000 kg/(s2-cm) g Gravitational Constant 980 cm/s2 µ Coefficient of kinetic friction 0.4 w Cardboard thickness 0.5 cm δv Speed variation 200 cm/s δφ Angle variation away from y-axis leaving the ramp π/36 M Mass of rider and motorcycle 300 kg vi Initial velocity leaving ramp 1500 cm/s θ Ramp angle of elevation π/6 Definitions and Key Terms • Buckling is the process by which a stiff plane develops a crack due to a stress exceeding the yield stress. • Compressive strength is the maximum force per unit area that a material can withstand, under compression, prior to yielding. • Corrugation is the style found in cardboard of sinusoidal waves of liner paper sandwiched between inner and outer papers. We use boxes with the most common corrugation, C-flute corrugation (see below for definition of a flute). • Cracking is a resulting state when the tensile force exceeds the tensile strength. • Cross-machine direction is the direction perpendicular to the sinusoidal wave of the corrugation. • Depression is when, due to a force, a section of a side or edge moves down￾wards. • ECT is the acronym for a common test of strength, the edge crush test. • Fiberboard is the formal name for cardboard. • Flute is a single wavelength of a sinusoidal wave between the inner and outer portion papers that extends throughout the length of the cardboard. A C-flute has a height of 0.35 cm and there are 138 flutes per meter [Packaging glossary n.d.]

236 The UMAP Journal 24.3(2003) Machine direction is parallel to the direction of the sinusoidal waves Motorcycle We use the 1999 BMw R1200c model motorcycle for structural information; it has a 7. 0-cm front wheel and a 10-cm back wheel, with radii 46 and 38 cm. It is 3 m long and has 17 cm ground clearance. The weight is 220 kg(dry)and 257 kg(fueled)[Motorcycle specs n d I Mullen test is a common measurement of the maximum force that a piece of cardboard can stand before bursting or puncturing Puncturing is when a force causes an area to burst through the cardboard surface Pyramid configuration is a configuration of boxes in which each box is rest ing equally on four others Strain is the dimensionless ratio of elongation to entire length. Stress is the force per unit area to which a material is subject Tensile strength is the maximum force per unit area which a material can withstand while under tension prior to yielding Youngs modulus of elasticity is the value of stress divided by strain and relates to the ability of a material to stretch Developing the Single-Box Model Expectations Given sufficiently small in area, the surface plane either punctures or cracks before the frame bu Given sufficiently large impact area the frame should buckle and no punc turing occurs. During buckling, corners are more resistive to crushing than edges Preliminary Assumptions The force is exerted at or around the center of the top The top of the box faces upward. This assumption allows us to use ECT (Edge Crush Test)results. This orientation also ensures that the flutes along the side are oriented perpendicular to the ground so they serve as columns

236 The UMAP Journal 24.3 (2003) • Machine direction is parallel to the direction of the sinusoidal waves. • Motorcycle We use the 1999 BMW R1200C model motorcycle for structural information; it has a 7.0-cm front wheel and a 10-cm back wheel, with radii 46 and 38 cm. It is 3 m long and has 17 cm ground clearance. The weight is 220 kg (dry) and 257 kg (fueled) [Motorcycle specs n.d.]. • Mullen test is a common measurement of the maximum force that a piece of cardboard can stand before bursting or puncturing. • Puncturing is when a force causes an area to burst through the cardboard surface. • Pyramid configuration is a configuration of boxes in which each box is rest￾ing equally on four others. • Strain is the dimensionless ratio of elongation to entire length. • Stress is the force per unit area to which a material is subject. • Tensile strength is the maximum force per unit area which a material can withstand while under tension prior to yielding. • Young’s modulus of elasticity is the value of stress divided by strain and relates to the ability of a material to stretch. Developing the Single-Box Model Expectations • Given sufficiently small impact area, the surface plane either punctures or cracks before the frame buckles. • Given sufficiently large impact area, the frame should buckle and no punc￾turing occurs. • During buckling, corners are more resistive to crushing than edges. Preliminary Assumptions • The force is exerted at or around the center of the top. • The top of the box faces upward. This assumption allows us to use ECT (Edge Crush Test) results. This orientation also ensures that the flutes along the side are oriented perpendicular to the ground, so they serve as columns

A Time-Independent model The Conceptual Single-Box Model During the puncturing or cracking of the top of the box The frame stays rigid The cardboard can be modeled as two springs with spring constants equal to the length of board times its modulus of elasticity [Urbanik 1999](Figure 1) Cross-Machine Direction (CD) Machine Direction (MD) Figure 1. Our model for a tire hitting the top of one box. We treat the portions of the box directly vertical or horizontal from the edges of the motorcycle tire(the rectangle in the middle)as ideal springs, neglecting the effect of the rest of the box. The surface of the motorcycle tire that strikes the box is approximated as a rectangle with dimensions LMD(the wheel's width) and Lcd( the length of the wheel in contact with the cardboard surface). We neglect the spin and tread of the tire The part of the spring beneath the tire does not undergo any tension. In re- ality, this is not the case; but with this assumption, cracking and puncturing occur along the edges of the tire. The force still comes from the rigid frame, and the springs have the same constant; therefore, we believe this assump tion affects only the position of the cracking and puncturing, not when it occurs or how much energy is dissipated There is no torque on the box during this first process. This assumption can be made since the force is at the center of the top The top of the cardboard box can fail in several ways If the resistive upwards force from the top, FUP, exceeds PML. LMD. LCD (the Mullen maximum allowable pressure over this area), then puncturing occurs

A Time-Independent Model 237 The Conceptual Single-Box Model During the puncturing or cracking of the top of the box: • The frame stays rigid. • The cardboard can be modeled as two springs with spring constants equal to the length of board times its modulus of elasticity [Urbanik 1999] (Figure 1). L L Machine Direction (MD) Cross-Machine Direction (CD) l MD CD EMD E CD Figure 1. Our model for a tire hitting the top of one box. We treat the portions of the box directly vertical or horizontal from the edges of the motorcycle tire (the rectangle in the middle) as ideal springs, neglecting the effect of the rest of the box. • The surface of the motorcycle tire that strikes the box is approximated as a rectangle with dimensions LMD (the wheel’s width) and LCD (the length of the wheel in contact with the cardboard surface). We neglect the spin and tread of the tire. • The part of the spring beneath the tire does not undergo any tension. In re￾ality, this is not the case; but with this assumption, cracking and puncturing occur along the edges of the tire. The force still comes from the rigid frame, and the springs have the same constant; therefore, we believe this assump￾tion affects only the position of the cracking and puncturing, not when it occurs or how much energy is dissipated. • There is no torque on the box during this first process. This assumption can be made since the force is at the center of the top. The top of the cardboard box can fail in several ways: • If the resistive upwards force from the top, FUP, exceeds PML · LMD · LCD (the Mullen maximum allowable pressure over this area), then puncturing occurs

238 The UMAP Journal 24.3 (2003) If the force FMD on the machine direction spring exceeds PMD W. 3LCD-l)/2 (the tensile strength in the machine direction times cross-sectional area per- pendicular to the force), then a crack occurs in the cross-machine direction We assume that this force that causes cracking is evenly distributed over a section larger than the actual edge of the tire rectangle, because of the solid nature of cardboard If the force FcD on the cross machine direction spring exceeds PcD w(3LMD 2)/2, then a crack occurs in the machine direction, for the same reasons as above If the force on the edges exceeds Pect 4l(the compression strength of the edges times the total edge length), then buckling occurs first. Here we as- sume that even though we model the top as two springs, the force is evenl distributed over the edges, taking advantage of the high spring constant and solid behavior of cardboard There are several ways in which boxes can fail Puncture: The wheel enlarges the hole, and only when the hull of the mo- torcycle hits the edge does buckling occur Crack: The tire does not break through the material, and buckling eventually occurs. Buckling: Can occur without a puncture or crack first We assume that at most one puncture or crack occurs per box, followed inevitably by buckling Calculations for the box Top [EDITOR'S NOTE: We do not give all the details of the following calculations. Refer to Figure 2. We solve for FCD, getting LCD FC DLMD a(t) We apply the results to solve for FMD and combine them to get Fup. After obtaining the vertical component of FcD V(=)2+o)P- ( eCDL =2g)2+[r(t)2

238 The UMAP Journal 24.3 (2003) • If the force FMD on the machine direction spring exceeds PMD ·w·(3LCD−l)/2 (the tensile strength in the machine direction times cross-sectional area per￾pendicular to the force), then a crack occurs in the cross-machine direction. We assume that this force that causes cracking is evenly distributed over a section larger than the actual edge of the tire rectangle, because of the solid nature of cardboard. • If the forceFCD on the cross machine direction spring exceedsPCD·w·(3LMD− l)/2, then a crack occurs in the machine direction, for the same reasons as above. • If the force on the edges exceeds PECT · 4l (the compression strength of the edges times the total edge length), then buckling occurs first. Here we as￾sume that even though we model the top as two springs, the force is evenly distributed over the edges, taking advantage of the high spring constant and solid behavior of cardboard. There are several ways in which boxes can fail: • Puncture: The wheel enlarges the hole, and only when the hull of the mo￾torcycle hits the edge does buckling occur. • Crack: The tire does not break through the material, and buckling eventually occurs. • Buckling: Can occur without a puncture or crack first. We assume that at most one puncture or crack occurs per box, followed inevitably by buckling. Calculations for the Box Top [EDITOR’S NOTE: We do not give all the details of the following calculations.] Refer to Figure 2. We solve for FCD, getting FCD = ECDLMD   l − LCD 2 2 + [x(t)]2 − l − LCD 2   . (1) We apply the results to solve for FMD and combine them to get FUP. After obtaining the vertical component of FCD, x(t)ECDLMD    l−LCD 2 2 + [x(t)]2 − l−LCD  2  l−LCD 2 2 + [x(t)]2   ,

A Time-Independent Model 239 e 2x(t) x(t) Figure 2. Side view of the depression of the motorcycle tire into the top of the box and the analogue by symmetry for FMD, we have FUP, namely FuP=x()2EcDLMD 1 (=2)2+{r() EMD LMD1 (4=2m)2+( The force FuP is the resistive force that the top exerts on the motorcycles heel. Balancing the force and taking into effect gravity (in the form of the normal force), we get the force equation of the motion of the motorcycles wheel on the box prior to puncture, crack, or buckle FUP +mg We use this expression to calculate the energy as a function of depression into the box. We use our initial force calculation to determine the level of depression and the type of failure that the top incurs. This depression is the minimum depression for which any failure occurs If the force FcD on the cross-machine direction spring(contributed by both sides of the spring) exceeds 2PCDLmDw, then a crack occurs in the machine direction. Solving for the depression, we get PcDw 2 PCU (- LCD D CD Likewise, if the force FMD on the machine direction spring(contributed by both sides of the spring)exceeds 2PMD LCDw, then a crack occurs in the cross- machine direction, with the analogous formula for the depression

A Time-Independent Model 239 FUP CD _____ l-L 2 CD _____ l-L 2 l x(t) F CD F CD LCD CD _____ l-L 2 +∆LCD x(t) θ Figure 2. Side view of the depression of the motorcycle tire into the top of the box. and the analogue by symmetry for FMD, we have FUP, namely: FUP = x(t)  2ECDLMD  1 − l−LCD  2  l−LCD 2 2 + [x(t)]2   + 2EMDLMD  1 − l−LMD  2  l−LCD 2 2 + [x(t)]2    . (2) The force FUP is the resistive force that the top exerts on the motorcycle’s wheel. Balancing the force and taking into effect gravity (in the form of the normal force), we get the force equation of the motion of the motorcycle’s wheel on the box prior to puncture, crack, or buckle: FNET = FUP + mg. Weuse this expression to calculate the energy as a function of depression into the box. We use our initial force calculation to determine the level of depression and the type of failure that the top incurs. This depression is the minimum depression for which any failure occurs. If the force FCD on the cross-machine direction spring (contributed by both sides of the spring) exceeds 2PCDLMDw, then a crack occurs in the machine direction. Solving for the depression, we get xCD = PCDw ECD 2 + PCDw ECD (l − LCD) Likewise, if the force FMD on the machine direction spring (contributed by both sides of the spring) exceeds 2PMDLCDw, then a crack occurs in the cross￾machine direction, with the analogous formula for the depression.

240 The UMAP Journal 24.3 (2003) If the resistive upwards force FUP from the top exceeds PML LMDLCD, punc turing occurs. Similarly, if the net force on the edges, 2FCD+ 2FMD, exceeds 4PECTL, buckling occurs first. We find the respective depressions in the next section We can use the r-position in energy calculations to determine the new speed of the motorcycle Energy is a distance integral of net force, so using (2)we can find energy Ar absorbed by the tor A()=/6FNEd=EDL(2-(-Lm)√(=m)+n2+如m2 +EMDLcD(22-(-LMD)v(2mD)2+22 + mg Extensions of Top Model Testing the model shows that not all the force and dissipative energy comes from the top of the box springs. We make the following further assumptions Since the deflection is small compared to the edge lengths, a I-LCM/2 The boxes have two layers coming together at the top that are corrugated in different directions, hence the cardboard on the top really is of width 2 Since the flutes in one top piece are perpendicular to the flutes of the other the resulting combined modulus of elasticity is the sum of the two original values in both directions. This means that we can define e= emd+ ecd and modify all equations accordingly. The equations for the forces Fcd (1)and Fmd are of the form f(x)=k(a2+x2) with a=(L-LcD)/2. For small deflections a, such a function can be approxi mated well by its second-degree Taylor expansion around x=0 f(a) The resulting equations are FCD= 2ELMD FMD= 2ELCD m=0(2E40b+280-L0

240 The UMAP Journal 24.3 (2003) If the resistive upwards force FUP from the top exceeds PMLLMDLCD, punc￾turing occurs. Similarly, if the net force on the edges, 2FCD + 2FMD, exceeds 4PECT l, buckling occurs first. We find the respective depressions in the next section. We can use the x-position in energy calculations to determine the new speed of the motorcycle. Energy is a distance integral of net force, so using (2) we can find energy AT absorbed by the top: AT (x) =  x 0 FNET ds = ECDLMD x2 − (l − LCD)  l−LCD 2 2 + x2 + (l−LCD)2 2 +EMDLCD x2 − (l − LMD)  l−LMD 2 2 + x2 + (l−LMD)2 2 + mgx. Extensions of Top Model Testing the model shows that not all the force and dissipative energy comes from the top of the box springs. We make the following further assumptions: • Since the deflection is small compared to the edge lengths, x l − LCM/2 and x l/2. • The boxes have two layers coming together at the top that are corrugated in different directions, hence the cardboard on the top really is of width 2w. • Since the flutes in one top piece are perpendicular to the flutes of the other, the resulting combined modulus of elasticity is the sum of the two original values in both directions. This means that we can define E = EMD + ECD and modify all equations accordingly. The equations for the forces FCD (1) and FMD are of the form f(x) = k(a2 + x2) 1/2 − a, with a = (l − LCD)/2. For small deflections x, such a function can be approxi￾mated well by its second-degree Taylor expansion around x = 0: f(x) ≈ x2 2a . The resulting equations are FCD = 2ELMD x2 l − LCD , FMD = 2ELCD x2 l − LMD , FUP = x(t) 2ELMD x(t) l − LCD + 2ELCD x(t) l − LMD

A Time-Independent model 241 We deal briefly with the position of the rectangle on the box. The ect de- flection should not depend on the position of the rectangle, since ECT depends only on the net force. The other three deflections become less as the force moves away from the center, so we can model them with a standard linear decrease factor of 1-oL(/2), where dL is the distance radially from the center Now we solve equations for a, taking the d factor into account, as well FECT=2(FCD+FMD), getting rcD=(1-2) L-LcD) MmD=(1-242)sm=m, 26L UPMAX FECTMAX ML CECT ELM To obtain energy, we integrate the Taylor-expanded version of FuP +mg Ar()=2E r mg F, where F=min. cD, IMD, ML, ECTI Now we can substitute the maximum values of the respective forces and solve for the T-values. We use the same values of Fup= PMl Lmd Lcd and FECT= 4PECTL. For the cracking forces, we double their values to take into account both springs on the side of the tire. So this gives FCD= 2(PCD+ PDW LCD, FMD= 2(PcD+ RMD)wLMD We make one last change. Energy is absorbed not only from the elasticity of the top but also from that of the edges. We determine the average force on this edge spring and the change in the depression that occurs for this spring. It is reasonable to use average force, because this is over a small dis tance, and a parabolic function is reasonably linear in such an interval. Using modulus of elasticity E and first-order Taylor approximations, we estimate the displacement and energy absorbed We also assume that the edge does not fail before the top. The is reasonable because an edge has greater structural integrity. So, assume that we are dealing with edge on which FcD acts. To find th average force, integrate FcD with respect to distance and divide by distance Denote by F the depression at which failure occurs in our original top model The average force is FCD= ELMD To solve for A c observe that if the force is centered at the edge then

A Time-Independent Model 241 We deal briefly with the position of the rectangle on the box. The ECT de- flection should not depend on the position of the rectangle, since ECT depends only on the net force. The other three deflections become less as the force moves away from the center, so we can model them with a standard linear decrease factor of 1 − δL(l/2), where δL is the distance radially from the center. Now we solve equations for x, taking the δ factor into account, as well as FECT = 2(FCD + FMD), getting xCD =  1 − 2δL l  FCDMAX(l−LCD) 2ELMD , xMD =  1 − 2δL l  FMDMAX(l−LMD) 2ELCD , xML = 1 − 2δL l  FUPMAX 2ELMD l−LCD + 2ELCD l−LMD , xECT =  FECTMAX 2ELMD l−LCD + 2ELCD l−LMD . To obtain energy, we integrate the Taylor-expanded version of FUP + mg: AT (x)=2E x3 F 3 LMD l − LCD + LCD l − LMD + mgxF , where xF = min{xCD, xMD, xML, xECT}. Now we can substitute the maximum values of the respective forces and solve for the x-values. We use the same values of FUP = PMLLMDLCD and FECT = 4PECTl. For the cracking forces, we double their values to take into account both springs on the side of the tire. So this gives FCD = 2(PCD + PMD)wLCD, FMD = 2(PCD + PMD)wLMD. We make one last change. Energy is absorbed not only from the elasticity of the top but also from that of the edges. We determine the average force on this edge spring and the change in the depression x that occurs for this spring. It is reasonable to use average force, because this is over a small dis￾tance, and a parabolic function is reasonably linear in such an interval. Using modulus of elasticity E and first-order Taylor approximations, we estimate the displacement and energy absorbed. We also assume that the edge does not fail before the top. The is reasonable because an edge has greater structural integrity. So, assume that we are dealing with edge on which FCD acts. To find the average force, integrate FCD with respect to distance and divide by distance. Denote by xF the depression at which failure occurs in our original top model. The average force is FCD = ELMD x2 F 3(l − LCD) . To solve for ∆x, observe that if the force is centered at the edge, then ∆x =  l 2 2 + x2 CD-EDGE − l 2 .

242 The UMAP Journal 24.3 (2003) Taking a first-order Taylorapproximation gives A c= aCD-EDGE/L Using Hookes law and FCD=EA r, we solve for the depression of this spring CD-EDGE TCD-EDGE=1/ LMDI 3(l The corresponding energy is AcD-EDGE- FCD L= ELMD-CD-EDGE(3(L-LcD))+mg.cD-EDGE At this point, we assume that even though we have treated the two spring groups(the top and the edge) as separate systems--despite the fact that they move and affect each other dynamically--we can add the depression values r and energy absorption results A. This simplifying assumption is valid because the depressions are small enough that the approximations are the same as in the dynamic case. To obtain MD-EDGE and AMD-EDGE, interchange CD and MD So, reviewing, we have F= min(aCD, IMD, IML, EcT)and total depression NET =F+2ccD-EDGE +2TMD-EDGe and ANET AF+2AcD-EDGE+ 2AMD-EDGE We calculate A ACD-EDGE ID-EDGE NET Finally, we take into account the fact that as the mass falls, it gains energy from gravity is countered by the energy gained. This fact leads to the important equation for Ar(the energy change during the top failures NET- mg(NET Single-Crack Buckling We deal with energy dissipation for single-crack buckling, which occurs when a single side develops a crack from the top to a side. We model this as two sets of springs(Figure 3 Once the top(side Ilm)and a single side(side I) become weak, all corners but the two adjacent to the crack(C1, C2)remain strong. The only nonrigid corners, free to move, are C1 and C2. We assume that we are in the elastic range of the cardboard, so we can model this situation with two springs connected to the adjacent corners. We have springs connecting C4 and C1, Cs and C1, O and c2, c6 and e apply the same methods as used in modeling the top to determine the energy as a function of how much the edges move in. [ EDITOR'S NOTE: We do not give all the details. The total force exerted by the box is FNET=2EMD o( )+2ECD0(√P+2-)

242 The UMAP Journal 24.3 (2003) Taking a first-order Taylor approximation gives ∆x = x2 CD-EDGE/l.Using Hooke’s law and FCD = E∆x, we solve for the depression of this spring xCD-EDGE: xCD-EDGE =  LMDl x2 F 3(l − LCD) . The corresponding energy is ACD−EDGE = FCD∆x = ELMDx2 CD-EDGE x2 F 3(l − LCD) + mgxCD-EDGE. At this point, we assume that even though we have treated the two spring groups (the top and the edge) as separate systems—despite the fact that they move and affect each other dynamically—we can add the depression values x and energy absorption results A. This simplifying assumption is valid because the depressions are small enough that the approximations are the same as in the dynamic case. To obtain xMD-EDGE and AMD-EDGE, interchange CD and MD. So, reviewing, we have xF = min(xCD, xMD, xML, xECT) and total depression xNET = xF + 2xCD-EDGE + 2xMD-EDGE and ANET = AF + 2ACD-EDGE + 2AMD-EDGE. We calculate FNET = AF + 2ACD-EDGE + 2AMD-EDGE xNET . Finally, we take into account the fact that as the mass falls, it gains energy from gravity is countered by the energy gained. This fact leads to the important equation for AT (the energy change during the top failures): AT = ANET − mg(xNET). Single-Crack Buckling We deal with energy dissipation for single-crack buckling, which occurs when a single side develops a crack from the top to a side. We model this as two sets of springs (Figure 3). Once the top (side III) and a single side (side I) become weak, all corners but the two adjacent to the crack (C1, C2) remain strong. The only nonrigid corners, free to move, are C1 and C2. We assume that we are in the elastic range of the cardboard, so we can model this situation with two springs connected to the adjacent corners. We have springs connecting C4 and C1, C5 and C1, C3 and C2, C6 and C2. We apply the same methods as used in modeling the top to determine the energy as a function of how much the edges move in. [EDITOR’S NOTE: We do not give all the details.] The total force exerted by the box is FNET = 2EMD l 10 l2 + x2 − l + 2ECD l 10 l2 + x2 − l

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