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 perpendicular 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 assume 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 motorcycle 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 crossmachine direction, with the analogous formula for the depression.