Question

esioRS 31-5) are based on the problem statement below: A concrete plant used the following materials with the bulk specific gravity (BSG) and the weight below: Materials Cement Sand Stone Water BSG 3.15 2.63 2.62 1.00 Wt. 5,500 lb 9,290 lb 1,7930 lb 284 gallons (weight of water: 8.33 lb/gallons) 51) The volume of cement ft3 a) 29.78, b) 27.98; e) 28.97 fts 52) The volume of sand a) 56.61; b) 65.61; c) 66.15 a) 106.97; b) 109.67;e) 107.76 a) 91.37; b) 37.91;c) 79.13 a) 35085.7; b) 30585.7; c) 38505.8 a) 217.33; b) 233.71; c) 232.17 ft 53) The volume of stone 54) The volume of waterf lbs 55) The total weight of the concrete ft 56) The volume of the fresh concrete lbs/ ft3 57) The unit weight of the concrete produced a) 153.1;b) 151.1; c) 152.1 Questions 58-63 are based on the following problem statement Given the following batch weights, calculate w/c ratio: %abs of CA %abs of FA 0.8% 0.2% Moist CA Moist FA Cement Water 20 lbs, М.С. of CA-|.896; 14 lbs; M.C. of FA-1.8%; 8 lbs 4 lbs 58) The free moisture content in CA is a) 1.096; b) 2.6%; c)-1.0% 59) The free moisture content in FA is a) 2.0%; b) 1.6%; c)-1.6%

Answer 1

Solutions

51. Volume of cement:

$\because Volume =\frac{Weight}{Unit\ weight}$

$Unit\ weight\ of\ cement =G\gamma _{w} = 3.15\times 62.4\ lb/ft^{3}$

$\therefore \ Volume\ of\ cement=\frac{5500\ lb}{3.15\times 62.4\ lb/ft^{3}} = 27.98\ ft^{3}$

52. Volume of sand

similar to above

$Volume\ of\ sand=\frac{9290\ lb}{2.63\times 62.4\ lb/ft^{3}} = 56.608\ ft^{3}$

53. Volume of stone

$Volume\ of\ stone=\frac{17930\ lb}{2.62\times 62.4\ lb/ft^{3}} = 109.67\ ft^{3}$

54. Volume of water

$Volume\ of\ water=\frac{(284\ gallons\times 8.33\ lb/gallon)}{1.0\times 62.4\ lb/ft^{3}} = 37.91\ ft^{3}$

55. Total weight of concrete

W_total = 5500 + 9290 + 17930 + (284x8.33)

W_total = 35085.72 lbs

56. Volume of fresh concrete

V = 27.98 + 56.61 + 109.67 + 37.91 = 232.17 ft³

57. Unit weight of concrete

$\gamma _{c}= \frac{Total\ weight\ of\ concrete}{Volume\ of\ concrete}$

$\gamma _{c}= \frac{35085.72\ lbs}{232.17\ ft^{3}}$

$\gamma _{c}= 151.12\ lb/ft^{3}$

58. Free moisture content in CA = Total moisture content - Absorption factor

= 1.8 - 0.8

= 1.0 %

59. Free moisture content in FA = Total moisture content - Absorption factor

= 1.8 - 0.2

= 1.6 %