Types and Bearing Capacity: for assistance with this exercise, refer to the read
ID: 1760128 • Letter: T
Question
Types and Bearing Capacity: for assistance with this exercise, refer to the reading ackboard ("Fundamentals of Building Construction", Wiley, Ch apter 2, Foundations) Figures 2.3 and 2.6 and the accompanying text. a. How wide must a wall footing be if the load is 8700 pounds per foot of wall and the footing rests on a sandy gravel? Show calculations and make a sketch. Assume that it is 18" thick. How large must a square column footing be to support a load of 74,000 pounds if it stands on silty gravel? Show calculations. Make a sketch of the footing, assuming that it is 18" thick. b.Explanation / Answer
a.
Please refer Safe bearing capacity of soils in below link
http://www.civil-estimation.com/blog/safe-bearing-capacity-of-rocks-soils-as-per-is-1256/
Sandy Gravel = 1.5 kg/cm2 = ((1.5*9.81)/0.0929) N/m2 = 158.40 N/m2
Given Data –
8700 pounds load = 38.70 KN/m, Thickness of footing = 18 inch = 457 mm
Assumptions considered :
M20 grade concrete and Fe 415 steel reinforcement
Step 1: Size of the footing
Given axial load = 38.70 kN/m and safe bearing capacity of soil qc = 158.4 kN/m2 at a depth of 1 m below the ground level. Assuming the self weight of footing and backfill as 10 per cent, the area of the base required = 38.7(1.1)/158.4 = 0.269 m2. Provide width of 0.269m for every one metre to get the required area of 0.269 m2.
Step 2: Thickness of footing slab based on shear
18” thick = 457mm
Step 3: Checking for the moment
The critical section for the bending moment is marked, where the bending moment (factored) is,
Mu = (1)(0.269)(1000)(1000)/2 = 134.5 kNm/m
The capacity of the section = (2.76)(1000)(457)(457) = 576.40 kNm/m > 134.5 kNm/m. Hence, o.k.
M/bd2 = 134.5(106)/1000(457)(457) = 0.7644 N/mm2
Table 2 of SP-16 gives p = 0.2282 per cent. Accordingly, Ast = 0.2282(1000)(457)/100 = 1042.87 mm2/m. Provide 16 mm diameter bars @ 100 mm c/c which gives 1134 mm2/m > 1042.87 mm2/m.
Step 4: Development length
The development length of 16 mm bars = 47.01(16) = 752.1 mm
Length available = 900 – 50 = 850 > 542.88
Step 5: Distribution reinforcement
Minimum reinforcement @ 0.12 per cent should be provided longitudinally.
Ast = 0.12(2200)(360)/100 = 950.4 mm2
Provide 13 bars of 16 mm diameter (area = 1073 mm2). The spacing = (2200 – 50 – 16)/12 = 172.33 mm c/c, say @ 175 mm c/c.
Step 6: Transfer of loads at wall-footing base
Bearing stress = 0.45 fck (A1/A2)1/2 = 0.45(20)(1) = 9 N/mm2
The forces that can be transferred = 9(1000)(269)(10-3) kN = 2421 kN >> factored load of 38.7(1.5) = 58.05 kN. Hence, o.k.
b.
Please refer Safe bearing capacity of soils in below link
http://www.civil-estimation.com/blog/safe-bearing-capacity-of-rocks-soils-as-per-is-1256/
Silty Gravel = 2.5 kg/cm2 = ((2.5*9.81)/0.0929) N/m2 = 263.98 N/m2
Given Data –
74000 pounds load = 333.60 KN/m, Thickness of footing = 18 inch = 457 mm
Assumptions considered :
M20 grade concrete and Fe 415 steel reinforcement
Step 1: Size of the footing
Given axial load = 333.6 kN/m and safe bearing capacity of soil qc = 263.98 kN/m2 at a depth of 1 m below the ground level. Assuming the self weight of footing and backfill as 10 per cent, the area of the base required = 333.6(1.1)/263.98 = 1.39 m2. Provide width of 1.18m for every one metre to get the required area of 1.39 m2.
Step 2: Thickness of footing slab based on shear
18” thick = 457mm
Depth required 510mm
Step 3: Checking for the moment
The critical section for the bending moment is marked, where the bending moment (factored) is,
Mu = (1)(1.39)(1000)(1000)/2 = 695 kNm/m
The capacity of the section = (2.76)(1000)(510)(510) = 718 kNm/m > 695 kNm/m.
Hence, o.k.
M/bd2 = 695(106)/1000(510)(510) = 2.672 N/mm2
Table 2 of SP-16 gives p = 0.36 per cent. Accordingly, Ast = 0.36(1000)(510)/100 = 1062.00 mm2/m. Provide 16 mm diameter bars @ 100 mm c/c which gives 1134 mm2/m > 1042.87 mm2/m.
Step 4: Development length
The development length of 16 mm bars = 47.01(16) = 752.1 mm
Length available = 900 – 50 = 850 > 542.88
Step 5: Distribution reinforcement
Minimum reinforcement @ 0.12 per cent should be provided longitudinally.
Ast = 0.12(1180)(360)/100 = 509.76 mm2
Provide 10 bars of 16 mm diameter (area = 524.12 mm2). The spacing = (2200 – 50 – 16)/12 = 172.33 mm c/c, say @ 175 mm c/c.
Step 6: Transfer of loads at wall-footing base
Bearing stress = 0.45 fck (A1/A2)1/2 = 0.45(20)(1) = 9 N/mm2
The forces that can be transferred = 9(1000)(1180)(10-3) kN = 10620 kN >> factored load of 333.6(1.5) = 500.4 kN. Hence, o.k.
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