Example 1 Evaluate ESWL at depth 3", 30", and 50" based on equal stress if dual spacing S = 20", single wheel load = 750

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Example 1 Evaluate Eswl At Depth 3 30 And 50 Based On Equal Stress If Dual Spacing S 20 Single Wheel Load 750 1 (25.81 KiB) Viewed 39 times

Example 1 Evaluate ESWL at depth 3", 30", and 50" based on equal stress if dual spacing S = 20", single wheel load = 7500 lb, and tire pressure = 90 psi. P a= = 5.2 inch P₂ X R d = S-2a = 9.6 inch d/2 = 4.8 inch 2 S=40 inch 1. At Z=3 inch <d/2→ ESWL = P = 7500 lb 2. At Z = 50 inch > 2S → ESWL = 2P = 15000 lb 3. At Z = 30 inch (d/2<Z<2S) → P< ESWL<2P: me=mi + m₂ Z/a-30/5.2 5.8 r/a - zero Z/a-30/5.2 5.8 r/a S/a 3.85 mi m; m₁-4.5% ₂1.8% m₂m₂ + m₂-6.3 Z/a-30/ag-5ag-6. ESWL = P = P₁ n (ag)² ESWL = P = 90 π (6)² =