Design of Rectangular Short Column With Four Side Equal Reinforcement
PiscesaCol 1.05

     A short column with bending in X-Axis direction will be design using PiscesaCol 1.05 and PCACOL V3.00, then the result of each application will be compared to ensure the result from PiscesaCol 1.05 can be used safely in design. Rectangular Column with width 500 mm and height 500 mm, the load applied in columns consist of uniaxial bending in X Axis about 260 kNm and Axial Load consist of 450 kN with Concrete Compressive Strength (f'c) 40 MPa and Reinforcement Steel Yield Strength 400 MPa. Each step in using PiscesaCol 1.05 will be explained step by step below :

Step 1
Open PiscesaCol 1.05 programs from start menu button. Click Start >> Program Files >> PiscesaCol 1.05 >> PiscesaCol 1.05.

Step 2
After opening the program, the main window of PiscesaCol 1.05 program will be shown and can be seen in the picture below, every part of the main window has it own function which will be discussed later. Main Menu show the sub menu of File, Input, Solve, Design and View. Column Section and Reinforcement Over View will show the shape and reinforcement position of current defined column. Information of Material, Section and Reinforcement Properties will show the information that had inputed before. Interaction Diagram and Loading Position Graphics will show the current capacity according to the code selected. Information Sheet of Analysis, Loading and Design Result will show the point of current analysis by value.

Step 3
Ok... now we'll start with define the general information about the current column. Click Input >> General Information. The picture shown below is the General Information Window of column which include the Codes were used and question about Slenderness Effect to be included in analysis. For beginning lets type "Test" in Project Text Box, "K500x500" in Column Text Box and "Your Name" in Engineer Text Box. Second let's pick up SNI 2847-2002 which is similiar with ACI 318-1999 except for the reduction factor or ACI 318-2002 (Unified Design Theory). Neglect slenderness effect for this time. This two method (Both of Limit State Method and Unified Design Provision) will be explained briefly in other topics for now on let's keep the SNI 2847-2002 which is familiar in Indonesia. After all field was filled then proceed with click OK button.

Step 4
Define Material Properties of column, click Input >> Material Properties. The Picture shown below is Material Properties Window of column which includes the major Characteristic of Compressive Strength of Concrete (f'c) which the value is 40 MPa and Steel Reinforcement Yield Strength (fy) which value is 400 MPa. Modulus Elasticity of Concrete will be calculated automatically with normal Modulus Elasticity of Concrete formula where Ec = 4700*(f'c)^0.5. Maximum Stress f'c will also be calculated automatically with strength reduction of axial concrete compressive strength 0.85. Beta1 was calculated using the formula B1=0.85-((f'c-30)/28), B1>=0.65. Modulus Elasticity of Steel Reinforcement Es = 200000 MPa and Ultimate Strain at Yield of Steel Reinforcement will be calculated automatically with formula es=fy/Es. After all field was filled proceed with OK button.

Step 5
Design of short rectangular column in this case will be somewhat different with other programs since in this programs the design was based on graphics design as many reinforced concrete books used. Click Design >> Graphics Design. The Picture shown below is Design Graphics Manual Column Window now let's configure the width and height of the column where the width is 500 mm and height is 500 mm. Then let's configure myu for the column where myu is the length of top center reinforcement to bottom center reinforcement (d-d') divided by total height, in this case lets fill 0.8 for myu value, it's means that (d-d')/h = 0.8.

Step 6
Since the type of rectangular reinforcement is not determine yet so we can use four type's of rectangular reinforcement position such as Two Side Equal Along X-Axis, Two Side Equal Along Y-Axis, Four Side Equal and Circular Equal. For this case we will use Four Side Equal. After all field was filled then proceed with calculate button.

Step 7
Now to calculate the reinforcement needed we need to know the X and Y diagram value which can be calculated as follows :
Check Small Axial Load on Column
if Pu < 0.1 f'c Ag then
     Phi = 0.8 - (0.15*Pu / (0.1f'cAg)) >= 0.65
     [Phi = 0.8 - (0.15*450000 / (0.1*40*500*500)) = 0.7325] This value of Phi were used !!!
else
     Phi = 0.8
end if

Check X and Y Value
X = Phi * Mn / (Ag * h), Mu = 260 kNm then Phi = 0.7325, Mn = Mu / 0.7325 = 260 / 0.7325 = 355 kNm. X = 0.7325 * 355 * 1000000 / ((500*500) * 500) = 2.08.
Y = Phi * Pn / Ag, Pu = 450 kN then Phi = 0.7325, Pn = Pu / 0.7325 = 614 kN, Y = 0.7325 * 614 * 1000 / (500*500) = 1.8.
Then we will plot the point to the design graphics to get the reinforcement ratio in percent where the first small diagram interaction capacity is with 1% reinforcement capacity and the next diagram interaction capacity is 2% and so on. The reinforcement ratio achived approximately about 1.3 % from the Picture below.

Step 8
Now let's calculate the reinforcement needed in column to ensure the capacity provided is enough to carry the load. Try with D19 reinforcement where As = 283 mm2, then number of reinforcement required n = 1.3 * 500 * 500 /( 100*283) = 11.48 pieces, Dekking = ((500-500*[myu=0.8])/2)-Stirrups-0.5*D19 = 30.5 mm and Stirrups = 10 mm since it was Four Side Equal then the number of reinforcement required n = 12 pieces.

Step 9
Let's check the new capacity of design column using the same way in this link but with customized number of reinforcement required n = 12 pieces, Dekking = 30.5 mm and Stirrups = 10mm. The Picture below shown the new capacity of design column.

Step 10
This conclude this tutorial. If you have any question please feel free to use the forum or email me... ^^, Any adviced and critics will be helpfull for the implementation of ITS Column Analysis 2009. Thanks...

Regards,

Bambang Piscesa