17 Constr onst r ucti uct i on Mat Materi erial alss
17–1
A sample of wood with dimensions 3 in. 4 in. 12 in. has a dry density of 0.35 g/cm3. (a) Calculate the number of gallons of water that must be absorbed by the sample to contain 120% water. (b) Calculate the density after the wood absorbs this amount of water. Solution:
V 3
4
12
dry weight
0.35
@120% water (a) water
144 in.3
2359.7
2359.7 cm3 825.9 g
weight of water weight of dry wood
100
1.2 2 1 825.9 825.9 2 991 g 2.183 lb 1 1.2
2.183 lb 2 1 7.48 7.48 gal/ft3 2 62.4 62.4 lb/ft3 0.262 gal 1 2.183
(b) If the volume volume remains remains the same, same, then density 17–2
825.9 g of dry wood
991 g of water 3
2359.7 cm
0.77 g/cm3
The density of a sample of oak is 0.90 g/cm 3. Calculate (a) the density of completely dry oak and (b) the percent water in the original sample. Solution:
r 12% water
0.68 g/cm3
Table 17–1 2 1 Table
(a) Therefo Therefore, re, in 100 100 cm3 of wood at 12% H2O, there are 68 g. 12% water
dry weight
green weight
dry weight
dry weight 68 1.12 1.12
68
dry weight
dry weight
100
60.71 g
191
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Instructor ’s Solution Manual
(b) When the density is 0.90 g/cm3, there are 90 g of green wood per 100 cm3. The water is therefore 90 60.71 g, or 29.29 g. %H 2O 17–3
90 g
60.71 g
60.71 g
100
48.2%
Boards of maple 1 in. thick, 6 in. wide, and 16 ft. long are used as the flooring for a 60 ft 60 ft hall. The boards were cut from logs with a tangential-longitudinal cut. The floor is laid when the boards have a moisture content of 12%. After some particularly humid days, the moisture content in the boards increases to 45%. Determine the dimensional change in the flooring parallel to the boards and perpendicular to the boards. What will happen to the floor? How can this problem be corrected? Solution:
Perpendicular: ctangential 0.00353 in./in. # %H 2O for maple ¢ x xo 3 c 1 M f M i 2 4
Over a 60 ft span:
6 3 0.00353 1 45
¢ x
12 2 4
0.699 in. in 6 in.
1 60 ft 2 1 12 in./ft 2 1 0.699 in. 2 6 in.
83.9 in.
The floor will therefore buckle due to the large amount of expansion of the boards perpendicular to the flooring. Parallel: For most woods, only about a 0.2% change in dimensions occurs longitudinally. Thus the total change in the length of the boards will be about ¢ y 17–4
1 0.002 2 1 60 ft 2 1 12 in./ft 2 1.44 in.
A wall 30 feet long is built using radial-longitudinal cuts of 5-inch wide pine, with the boards arranged in a vertical fashion. The wood contains a moisture content of 55% when the wall is built; however the humidity level in the room is maintained to give 45% moisture in the wood. Determine the dimensional changes in the wood boards and estimate the size of the gaps that will be produced as a consequence of these changes. Solution:
ctangential 0.00141 in./in. # %H 2O for pine ¢ x
1 30 ft 2 1 12 in./ft 2 3 1 0.00141 in./in. # %H 2O 2 1 45 55 2 4
5.076
in.
The total number of boards in the width of the wall is: # of boards
1 30 ft 2 1 12 in./ft 2 5 in./board
72 boards
Therefore there are 71 gaps between the boards. The average width of the gaps is: gap 17–5
5.076 in. 71 gaps
0.0715 in.
We have been asked to prepare 100 yd 3 of normal concrete using a volume ratio of cement-sand-coarse aggregate of 1 : 2 : 4. The water-cement ratio (by weight) is to be 0.5. The sand contains 6 wt% water and the coarse aggregate contains 3 wt% water. No entrained air is expected. (a) Determine the number of sacks of cement that must be ordered, the tons of sand and aggregate required, and the amount of water
CHAPTER 17
Constructi on Materials
193
needed. (b) Calculate the total weight of the concrete per cubic yard. (c) What is the weight ratio of cement-sand-coarse aggregate? Solution:
First we can determine the volume of each material on a “sack ” basis, keeping in mind the 1: 2 : 4 volume ratio of solids and the 0.5 watercement weight ratio: cement
1 1 sack 2 1 94 lb/sack 2 190 lb/ft3 0.495 ft3 /sack
sand
1 2 2 1 0.495 ft3 /sack 2
0.990 ft3 /sack
aggregate
1 4 2 1 0.495 ft3 /sack 2
1.980 ft3 /sack
water
1 0.5 2 1 94 lb 2 62.4 lb/ft3 2
0.753 ft3 /sack
total volume of materials/sack 4.218 ft3 /sack In 100 yd3, or 1 100 yd3 2 1 27 ft3 /yd3 2 : cement
2700 ft3 4.218 ft3 /sack 640 sacks
sand
1 640 sacks 2 1 0.990 ft3 /sack 2 1 160 lb/ft3 2
101,376 lb
1 640 sacks 2 1 1.980 ft3 /sack 2 1 170 lb/ft3 2
215,424 lb
1 640 sacks 2 1 0.753 ft3 /sack 2 1 62.4 lb/ft3 2 30,072 lb or 1 640 sacks 2 1 0.753 ft3 /sack 2 1 7.48 gal/ft3 2 3,605 gal
aggregate water
50.7 tons 107.7 tons
But we must make adjustments for the water that is already present in the sand and aggregate. There is 6% water in the sand and 3% water in the aggregate. We can either multiply the dry sand by 1.06, or divide the dry sand by 0.94, to obtain the amount of wet sand that we need to order. wet sand
1 101,376 lb 2 1 1.06 2 107,459 lb 53.7 tons
water in sand
107,459
wet aggregate
1 215,424 lb 2 1 1.03 2 221,887 lb 110.9 tons
water in aggregate
221,887
101,376
215,424
6083 lb
6463 lb
The actual amount of water that should be added to the concrete mix is: water
30,072
gal water
1 17,526 lb 2 1 7.48 gal/ft3 2 62.4 lb/ft3
2101 gal
6083
6463
17,526 lb
Therefore: (a) The ingredients of the concrete mix are: 640 sacks of cement 53.7 tons of sand 110.9 tons of aggregate 2101 gal of water (b) The total weight per yd3 is: wt /yd3
1 640 sacks 2 1 94 lb/sack 2 107,459 221,887 17,526 100 yd3 4070 lb/yd
3
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Instructor ’s Solution Manual
(c) The cement-sand-aggregate ratio, on a weight basis, is: ratio
1 640 sacks 2 1 94 lb/sack 2 : 107,847 lb : 221,887 lb
60,160 :107,847 : 221,887 1:1.79 : 3.69
17–6
We plan to prepare 10 yd 3 of concrete using a 1 : 2.5 : 4.5 weight ratio of cementsand-coarse aggregate. The water-cement ratio (by weight) is 0.45. The sand contains 3 wt% water, the coarse aggregate contains 2 wt% water, and 5% entrained air is expected. Determine the number of sacks of cement, tons of sand, and coarse aggregate, and gallons of water required. Solution:
First, we can determine the volume of each material required, using the 1 : 2.5 : 4.5 ratio to determine the weights per sack of cement and dividing by the density to determine the volume. Per sack of cement: 94 lb/sack 190 lb/ft3
0.495 ft3 /sack
sand:
1 2.5 2 1 94 lb/sack 2 160 lb/ft3
1.469 ft3 /sack
aggregate:
1 4.5 2 1 94 lb/sack 2 170 lb/ft3
2.488 ft3 /sack
cement:
1 0.45 2 1 94 lb/sack 2 62.4 lb/ft3 0.678 ft3 /sack
water:
Volume per sack 5.130 ft3 /sack But 5% of the concrete is expected to be entrained air. The volume of air “ x” per sack of cement is: x 1 5.130
x 2
0.05
or x
0.27 ft3
Therefore the total volume of concrete per sack is: Volume of concrete
5.130
0.27
5.400 ft3 /sack
In 10 yd3 270 ft3: cement
270 ft3 5.400 ft3 /sack 50 sacks
sand
1 50 sacks 2 1 1.469 ft3 /sack 2 1 160 lb/ft3 2
11,752 lb
aggregate
1 50 sacks 2 1 2.488 ft3 /sack 2 1 170 lb/ft3 2
21,148 lb
water
1 50 sacks 2 1 0.678 ft3 /sack 2 1 62.4 lb/ft3 2 2,115 lb
But we must also adjust for the water present in the wet sand (3%) and wet aggregate (2%). For example, to find the amount of wet sand, we could either multiply the dry sand by 1.03 or divide by 0.97: wet sand
11,752 lb 0.97
12,115 lb;
H2O
363 lb
wet aggregate
21,148 lb 0.98
21,580 lb;
H2O
432 lb
Therefore, the ingredients for the concrete mix include: cement
50 sacks
sand
12,115 lb
6.06 tons
aggregate
21,580 lb
10.8 tons
water
2115 363 432 1320 lb 1 1320 lb 2 1 7.48 gal/ft3 2 62.4 lb/ft3
158 gal
