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1
- #1
pob11646
Structural
- Mar 8, 2009
- 35
I have got a question about the placement of minimum temperature and shrinkage steel, say for a 35" thick rectangular footing 14 feet long and 5 feet wide. Let's say, that all I need for my footing is minimum temperature and shrinkage steel.
Hence, Ast = 0.0018bh = 0.0018 * (14*12) in * 35 in = 10.58 in2.
Say, I am using #7 bars, thus, I will need 10.58 in2/0.60 in2 = 18 # 7 bars.
Method 1: Or, say I place bars in the top and bottom layers. For my bottom layer, do I need to place 14 #7 parallel to the short side, and 4 # 7 parallel to the long side. And do the same for the top layer.
Or, Method 2: I still place bars in both the top and bottom layer. For the reinforcment parallel to the do I just need to place a total of 18 #7 in both the top and bottom layers, say 9#7 in the bottom layer, and 9#7 in the top layer.
For Method 2, the Ast required parallel to the long side will be 0.0018 * (5*12) in * 35 in = 3.78 in2, or 7 # 7 bars. Do I place a total of 7#7 bars in both the top and bottom layers, say 4#7 in the bottom layer, and 3#7 in the top layer.
Method 3. Not to confuse matters, but can I place reinforcement in the bottom layer alone, say a total of 18# 7 bars, say 14#7 parallel to the short side, and 4#7 parallel to the long side.
Please advise whether Method 1 or Method 2 is more appropriate. And is Method 3 practical?
Hence, Ast = 0.0018bh = 0.0018 * (14*12) in * 35 in = 10.58 in2.
Say, I am using #7 bars, thus, I will need 10.58 in2/0.60 in2 = 18 # 7 bars.
Method 1: Or, say I place bars in the top and bottom layers. For my bottom layer, do I need to place 14 #7 parallel to the short side, and 4 # 7 parallel to the long side. And do the same for the top layer.
Or, Method 2: I still place bars in both the top and bottom layer. For the reinforcment parallel to the do I just need to place a total of 18 #7 in both the top and bottom layers, say 9#7 in the bottom layer, and 9#7 in the top layer.
For Method 2, the Ast required parallel to the long side will be 0.0018 * (5*12) in * 35 in = 3.78 in2, or 7 # 7 bars. Do I place a total of 7#7 bars in both the top and bottom layers, say 4#7 in the bottom layer, and 3#7 in the top layer.
Method 3. Not to confuse matters, but can I place reinforcement in the bottom layer alone, say a total of 18# 7 bars, say 14#7 parallel to the short side, and 4#7 parallel to the long side.
Please advise whether Method 1 or Method 2 is more appropriate. And is Method 3 practical?