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Chp.12 Cont. – Examples to design Footings. Example Design a square footing to support a 18 in. square column tied interior column reinforced with 8 #9.

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Presentation on theme: "Chp.12 Cont. – Examples to design Footings. Example Design a square footing to support a 18 in. square column tied interior column reinforced with 8 #9."— Presentation transcript:

1 Chp.12 Cont. – Examples to design Footings

2 Example Design a square footing to support a 18 in. square column tied interior column reinforced with 8 #9 bars. The column carries an unfactored axial dead load of 245 k and an axial live load of 200 k. The base of the footing is 4 ft. below final grade and allowable soil pressure is 5 k/ft 2 Use f c = 3 ksi and f y = 60 ksi

3 Example 1 Assume a depth of footing. (2 ft or 24 in.) The weight of concrete and the soil are:

4 Example 1 The effective soil pressure is given as:

5 Example 1 Calculate the size of the footing:

6 Example 1 Calculate net upward pressure:

7 Example 1 Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering.

8 Example 1 Calculate perimeter for two-way shear or punch out shear. The column is 18 in. square.

9 Example 1 Calculate the shear V u The shape parameter

10 Example 1 Calculate d value from the shear capacity according to chose the largest value of d  s is 40 for interior, 30 for edge and 20 for corner column

11 Example 1 The depth of the footing can be calculated by using two way shear

12 Example 1 The second equation b o is dependent on d so use the assumed values and you will find that d is smaller and  = 40 Actual (d = in.) b o = in

13 Example 1 The depth of the footing can be calculated by using one-way shear

14 Example 1 The depth of the footing can be calculated by using one-way shear The footing is 19.5 in. > 13.9 in. so it will work.

15 Example 1 Calculate the bending moment of the footing at the edge of the column

16 Example 1 Calculate R u for the footing to find  of the footing.

17 Example 1 From R u for the footing the  value can be found.

18 Example 1 Compute the area of steel needed The minimum amount of steel for shrinkage is The minimum amount of steel for flexure is

19 Example 1 Use a #7 bar (0.60 in 2 ) Compute the number of bars need Determine the spacing between bars

20 Example 1 Check the bearing stress. The bearing strength N 1, at the base of the column, 18 in x 18 in.,  0.7 The bearing strength, N 2, at the top of the footing is

21 Example 1 The bearing strength, N 2, at the top of the footing is

22 Example 1 P u =683 k < N 1, bearing stress is adequate. The minimum area of dowels is required. Use minimum number of bars is 4, so use 4 # 8 bars placed at the four corners of the column.

23 Example 1 The development length of the dowels in compression from ACI Code for compression. The minimum l d, which has to be greater than 8 in., is

24 Example 1 Therefore, use 4#8 dowels in the corners of the column extending 19 in. into the column and the footing. Note that l d is less than the given d = 19.5 in., which is sufficient development length.

25 Example 1 The development length, l d for the #7 bars for the reinforcement of the footing. There is adequate development length provided.

26 Example 1 - Final Design

27 Example 2 Design a footing to support a 18 in. square column tied interior column reinforced with 8 #9 bars. The column carries an unfactored axial dead load of 245 k and an axial live load of 200 k. The base of the footing is 4 ft. below final grade and allowable soil pressure is 5 k/ft 2 Use f c = 3 ksi and f y = 60 ksi. Limit one side of the footing to 8.5 ft.

28 Example 2 Assume a depth of footing. (2 ft or 24 in.) The weight of concrete and the soil are:

29 Example 2 The effective soil pressure is given as:

30 Example 2 Calculate the size of the footing:

31 Example 2 Calculate net upward pressure:

32 Example 2 Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering.

33 Example 2 V u =150.7 k in short direction The depth of the footing can be calculated by using the one-way shear (long direction)

34 Example 2 The depth of the footing can be calculated by using one-way shear design The footing is 19.5 in. > 18.8 in. so it will work.

35 Example 2 Calculate perimeter for two-way shear or punch out shear. The column is 18 in. square.

36 Example 2 Calculate the shear V u The shape parameter

37 Example 2 Calculate d from the shear capacity according to chose the largest value of d.  s is 40 for interior, 30 for edge and 20 for corner column

38 Example 2 The depth of the footing can be calculated for the two way shear

39 Example 2 The third equation b o is dependent on d so use the assumed values and you will find that d is smaller and  = 40 Actual (d = in.) b o = in

40 Example 2 The depth of the footing can be calculated by using the two way shear

41 Example 2 Calculate the bending moment of the footing at the edge of the column (long direction)

42 Example 2 Calculate R u for the footing to find  of the footing.

43 Example 2 Use the R u for the footing to find .

44 Example 2 Compute the amount of steel needed The minimum amount of steel for shrinkage is The minimum amount of steel for flexure is

45 Example 2 Use A s =8.36 in 2 with #8 bars (0.79 in 2 ). Compute the number of bars need Determine the spacing between bars

46 Example 2 Calculate the bending moment of the footing at the edge of the column for short length

47 Example 2 Calculate R u for the footing to find  of the footing.

48 Example 2 Use R u for the footing to find .

49 Example 2 Compute the amount of steel needed The minimum amount of steel for shrinkage is The minimum amount of steel for flexure is

50 Example 2 Use A s =9.36 in 2 with #6 bar (0.44 in 2 ) Compute the number of bars need Calculate the reinforcement bandwidth

51 Example 2 The number of bars in the 8.5 ft band is 0.83(22)=19 bars. So place 19 bars in 8.5 ft section and 2 bars in each in (12ft -8.5ft)/2 =1.75 ft of the band.

52 Example 2 Determine the spacing between bars for the band of 8.5 ft Determine the spacing between bars outside the band

53 Example 2 Check the bearing stress. The bearing strength N 1, at the base of the column, 18 in x 18 in.,  0.7 The bearing strength, N 2, at the top of the footing is

54 Example 2 The bearing strength, N 2, at the top of the footing is

55 Example 2 P u =683 k < N 1, bearing stress is adequate. The minimum area of dowels is required. Use minimum number of bars is 4, so use 4 # 8 bars placed at the four corners of the column.

56 Example 2 The development length of the dowels in compression from ACI Code for compression. The minimum l d, which has to be greater than 8 in., is

57 Example 2 Therefore, use 4#8 dowels in the corners of the column extending 19 in. into the column and the footing. Note that l d is less than the given d = 19.5 in., which is sufficient development length.

58 Example 2 The development length, l d for the #8 bars There is adequate development length provided.

59 Example 2 The development length, l d for the #6 bars There is adequate development length provided.

60 Example 2 - Final design 12 #823 #6


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