MSE 220
HOMEWORK SOLUTIONS ASSIGNMENT #2
CH. 2
AGGREGATES
Problems: 1, 3, 6, 10, 13, 16, 21, 22, 23
1.
100 lb. of loose gravel can fill a 1-ft3 box. Its specific gravity is 2.68. What is the
dry rodded weight?
The “unit weight” or “bulk density”, aka “dry-rodded weight” is the mass of aggregate
per unit bulk volume. The mass of the sample is determined in oven-dry condition (after
driving away all moisture) and the volume is the solid volume of particles plus volume of
voids.
The volume, including voids, is 1 ft3. The sample mass is 100 lb. Therefore, the bulk
density is 100 pcf.
3.
A sample of aggregate has a moisture content of 10%. Its absorption capacity is
4%. What is the weight of free water in a 100-lb sample of this aggregate?
This sample has an excess moisture content of (10% - 4%) or 6%. Therefore, 6% of 100
lb is free water, or 6 lbs.
6.
How is fineness modulus measured, and what does it represent?
Fineness modulus is determined by doing a sieve analysis on a sample of aggregate. Add
the values of the percentages coarser than each full sieve in the set, and divide this
number by 100. See, for example, table 2.4 on p. 59 of Somayaji. This number tells you
the location of the average size of the aggregate, counted up from the bottom pan. For
example, if the fineness modulus is 2.6, the average aggregate size lies between the
aggregate in the 2nd and 3rd pans from the bottom of the stack.
10. The specific gravity and bulk density of an aggregate are 2.6 and 95 pcf, respectively.
Find the loose (bulk) volume of a 190-lb sample.
Bulk density = Mass/Volume (inc. pores) = 95 pcf.
Therefore, Volume (w/pores) = Mass/bulk density = 190 lb./95 lb/ft3 = 2 ft3
13.
What causes bulking?
Bulking is defined as the percent increase in volume of an aggregate over its dry-rodded
volume, caused by the pulling apart of the particles due to surface tension (moisture
content). The particles grow with increasing moisture content, up to a maximum at a
value of 5-8% MC.
13.
Plot typical gradation curves for two samples, one with a fineness modulus of 2.2,
and the other of 3.2
Sieve No.
¾” (19-mm)
3/8” (9.5-mm)
#4 (4.75-mm)
#8 (2.36-mm)
Log(sieve size)
1.279
0.978
0.677
0.354
Cumulative
Fineness Modulus
% coarser than (2.2)
0
50
75
95
220
2.2
% coarser than (3.2)
30.5
90
99.6
99.9
320
3.2
120
% Coarser than
100
80
Fineness 2.2
60
Fineness 3.2
40
20
0
0
0.5
1
1.5
Log(sieve size)
21.
A sample of gravel weighs 420 g. Its oven dry weight is 400 g. What is the
moisture content?
MC = (original sample weight – oven-dry weight)/oven-dry weight x 100
MC = (420 g – 400 g)/400 g = 5%
22.
A sample of fine aggregate with a specific gravity of 2.6 has a bulk density of 105
pct. What is the void content?
Void content (%) = (SG x W-B)/(SG x W)
Here, SG = specific gravity = 2.6
B = bulk density = 105 pcf
W = density of water = 62.4 pcf
Void content, % = (2.6 x 62.4 – 105)/(2.6 x 62.4) = 35.3%
23. The maximum aggregate size for a wall is 1/5 the wall thickness, thus Size, max =
1/5 x 5 in. = 1 in.