chunk_38.json•1.54 kB
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"text": "1475 DB =TB-3.5 19.5” \n1490 PN=0.2/FY 0.0033333 \n1500 U =PN*FY*/(1.78*FC) 0.039215 \n1510 V =DC/H 0.2692 \n1520 J = 10.8*FY*PN 2.159784 \n1530 MU=1/500*KA*W1*LF 0.121716 Inch Kips \n1540 A1 =MU/J \n1550 B1 = -(V^2-U*V^2) -0.069626 \n1560 C1 = 2*U*DT**V 0.2639169 \n1570 D1 = -(DT^2+2*V*DT) -162.98 \n1575 REM A1,B1,C1 and D1 are coefficients of \nthe General Cubic Equation AY1\n3\n + B1Y\n2\n + \nC1Y + D1 = 0 To solve for Y we restore \ngeneral cubic equation program no 1 \n \n1580 PRINT” TAB(1);A1;”Y CUBE”;TAB(10);B1;” \nY Square”;TAB(15);C1;”Y”;TAB(22);”=0” \n \n ***Note computer prints on the monitor \nscreen the third degree equation AY1\n3\n + \nB1Y\n2\n + C1Y + D1 = 0 \n \n1590 PRINT “Copy coefficients of Y cube, Y \nsquare and y and constant D1 for data \nstatement then type continue to resume \n \n\n269 \n \nrunning” \n1600 BREAK \n1610 DATA Put value of A1,B1,C1 & D1 \n1620 GOSUB 2400 \n1630 PRINT “Location of minimum steel \nreinforcements is valid at a distance \nY=”;”Feet from top of wall”1640 \n \n1640 REM To find required area of steel",
"metadata": {
"book_id": 34880,
"title": "BASIC-STRUCTURAL-388457483-Retaining-Wall-Design-Analytical-and-Computer-Methods-By-Ben-David-CE",
"authors": "Ben David",
"project": "basic_structural",
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