Question

3. Design a concrete mix using BS 8328 for the following information: a. QC = Good b. % defects = 5% c. Cement type = SRPC d. 28 days Fc = 33 MPa e. Slump = 75 mm f. Coarse aggregate = natural gravel g. NMS = 40 mm h. FM of sand = 2.5 i. Exposure condition = moderate j. Concrete cover = 35 mm k. Specific gravity of sand and gravel = 2.7

Answer 1

sol Given Qe = good el defects - 50% Cement- lype -- SR PC 25 days Fe= 33 mpa glamp - 75mm ccasse aggregate = natugal ggavell NMS.= Homm ec -35mm specific graui key of sand e gravels 27 Now : foraet me Target mean strength = fm fetkas fc230 K2 1.642. ( f8 54, defects) S3 6,1 . fm a 30+. (1.64x61). - 240 N/mm w/c ratio from table for SRPC cement w/c=0.5 .. uncrushed aggregate is yampa

For Hompa strength WLC Yalio is = 0.58 permissible limic is 0.5 0.5830.5 so w/C =0.5 Now water content from table of w/c ratio aggregate size - 20mm skimp - 75mm water demand for natural fine aggregate = 195btes Wates demand for crushed aggregate = 225.llas .. 23X195 + bax225 • 205 kg/m3 WIC=0.5 cement content = 205 X2 - Hlo kg/m3 4105350 (As per data book table.)

water content =205 kgims , Aggregate size=20mm specific geavity = 2.7 from goaph betweele euel density & Water content wel density = 2375.49% Total weight of aggregate - wat density - (cement twater). 100 -2375-CH10+205) - 1760 kg/m3 given is fine aggregato 50% fine aggregates passes through Goousille fine aggregati :10 -41% so weight of fine aggregate - 4141760 = 721.6kglms weight of coars aggregate = 59 *7.60 = 1038.4kgim? classify into fractions ieglm3 cement fine course water 410 7216 1038.4 205 1.1.76 2.54 0.5 for Ibag cement so 88 127 25 100

if aggregate abscbs water content (moisture) ondjustments need to be done : if Ic/ moisline content is added. 100 : 921.6 + 2 * 721.6 - 738kgim? half percent absåbed by GA. . 1038.4* -idx1038.4 . 1028 kgoms Now Design proportions cement F A C:A water 410 738 1028 205 1:18 12:51 : 0.5 to Ibag ,50 90 05:5 25 coment 50