TM 5-809-1/AFM 88-3, Chap. 15
APPENDIX C
DESIGN EXAMPLES
C-1.
Example 1: Concrete slab thickness for interior loads.
a. Problem. The floor slab for a warehouse will be designed based on the following information:
Traffic
Type of Traffic
Average Daily Volume
Category
5-axle trucks
50
I
4 axles, 5 kips each
1 axle, 10 kips
15-kip forklift or trucks
15
II
Stationary Live Load
1,200 pounds per square inch
Interior Wall Load
1,400 pounds per linear foot
Material properties
Concrete flexural strength = 650 pounds per square inch
Modulus of subgrade reaction, k = 150 pounds per cubic inch
b. Solution.
(1) Floor slab thicknesses h should be determined by using equivalent forklift truck axle load
below.
Equivalent
Forklift
Number
Maximum
Truck Axle
of
Average
Operations
Design
Load, kips
Axles
Daily Volume
Per Day
Index
5
4
50
200
4
10
1
50
50
4
15
1
15
15
7
Matching the axle loads and maximum operations per day in table 5-1, the design index for each axle-load
group is selected as shown in the far right column in the above-mentioned table. Design index 7 is selected
for the design. From figure 5-1, usingk = 150 pounds per cubic inch and 650 pounds per square inch flexural
strength, slab thickness equal to 6.7 inches, and round to 7 inches should be selected.
(2) One must check for adequacy of 7 inch slab for stationary live load, w = 1 200 pounds per
square foot. Table 3-1 should be entered using 650 pounds per square inch flexural strength concrete and 7
inch slab thickness; allowable stationary live load is selected, w =1,109 pounds per square inch. The w is
adjusted based on k = 150 pounds per cubic inch.
150
' 1,358lb/ft 2 > 1,200lb/ft 2
w ' 1,109
100
(3) Thickness, tc, of thickened floor slab supporting interior wall weighing 1,400 pounds per linear
foot should be determined by entering table 3-2 using 650 pounds per square inch flexural strength concrete
and wall load p = 1,400 pounds per linear foot. Thus, tc equals 10 inches, and tc is adjusted based on k = 150
pounds per square inch.
C-1
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