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Memo to Tom Schreiner 9 February 21, 2006 <br />Bulkhead Cost Estimate Permit No. M-1983-141 <br />Design for Hydraulic Pressure Gradient, Secondary Bulkhead <br />Ungrouted Bulkhead <br />Low Pressure Grouted Bulkhead <br />L _ 66 psi =13.2 feet <br />Spsi / ft <br />66 psi <br />L= =1.6feet <br />41 pst / ft <br />Clearly, with an almost ten fold decrease in required bulkhead length, and the difficulty in concrete pouring <br />for bulkheads greater than 10 feet in length, low pressure grouting is a necessity for the proposed main <br />bulkhead. <br />Design for Concrete Shearon Mount Royale Adit No. 1 Perimeter <br />The required bulkhead length for the concrete shear component of design, assuming minimum bulkhead <br />concrete compressive strength is specified at 3000 psi, is calculated as follows: <br />L_ pxhxP _ 66x8x8 =1.2feet <br />2x~h+P~x f', 2x~8+8~x110 <br />where: L =bulkhead length (feet) <br />p =pressure head (psi) <br />h = adit height (feet) <br />f = adit width (feet) <br />f', =concrete shear strength (psi) <br />Therefore, the 1.6 foot minimum bulkhead length required for pressure gradient exceeds the bulkhead <br />length requires for concrete shear, and pressure gradient controls the design at this stage of the analysis. <br />Design for Plain Concrete Deep Beam Bending Stress <br />The Applicant has proposed that plain concrete (unreinforced) bulkheads be constructed. American <br />Concrete Institute codes can be used to determine the required length for a plain concrete bulkhead to resist <br />deep-beam bending stress. For the analysis, the dead or fluid load acting on the bulkhead is multiplied by <br />1.4 (ACI 318-89, Section 9.2.1) and the plain concrete bending strength reduction factor of 0.65 is applied <br />(ACI 318-77, Section 9.3.2). ACI directs that the design tensile bending strength be: <br />f, =5 f'~ (ACI318-77, Section 15.11.1) <br />f = 5 3000 = 273 psi, assuming that minimum 3000 psi compressive strength is specified <br />w =1.4 x p x 144in Z l ft z =1.4 x 66 x 144 =13306 pounds per foot <br />