Sunday, November 24, 2013

Calculating The Weight of Iron

Practical formulation to calculate the weight of iron
In general formulation to calculate the weight of iron is :
Vb x Bjb = … .. Kg
where: Vb = volume of iron (m3)
Bjb = density of iron = 7850 (kg/m3)
Example:
1. Iron plate with a size (1m x 1m) with 1 mm thick plate, calculate the weight?
weight of iron = (1 x 1 x 0001) m3 x 7850 kg/m3 = 7.85 kg
(Note: 1 mm = 0.001 m)
2. Base plate with size (25 cm x 30cm) with 12 mm thick plate, calculate the weight?
heavy base plate = (0.25 x 0:30 x 0012) m3 x 7850 kg/m3 = 7065 kg
Up here is quite easy to understand right? … .. Well now how its formulation to calculate the weight of steel reinforcement for concrete?.
The answer:
Same way there is no difference, the point is the volume multiplied by the weight of iron.
Example:
1. Calculate the weight of steel reinforcement diameter 16 with a length of 12 meters?
Ø16-sectional area = 1 / 4 (π) d2 = 1 / 4 (3.14) (0,016) 2 = 0.00020096 m2
Ø16 volume = trunk cross-sectional area x length = 0.00020096 m 2 x 12 m = 0.002411 m3
iron weight Ø16 x 7850 kg/m3 = Volume m3 x 0.002411 = 7850 kg/m3 = 18.93 kg
easy enough right?, from the way I have described above, there is again a faster way to calculate the weight of steel reinforcement, namely by using the formulation:
Heavy iron bars = 0.006165 x d2 x L … (Kg)
where: d = diameter of reinforcement (mm)
L = length of reinforcing rod (m)
Example:
2. Calculate the weight of iron from the sample questions # 1, with the formulation of the above?
heavy iron Ø16 = 0.006165 x 162 x 12 = 18.93 kg
same right result, .. please you count yourself by trial and error is another measure of iron reinforcement, and I make sure that the two ways above will produce the same results, see for yourself … brow, God willing, must be the same.
Well … now the question is “where the origin of the formulation above number 0.006165?”.
The following are penjabarannya:
As I’ve described above, the formula to find the weight of iron is: Vb x Bjb
where Vb = Volume of iron and iron Bjb = density = 7850 kg/m3
So the heavy iron bars (round cross-section):
= Vb x 7850 kg/m3
= (1 / 4 x π x d2 x L) x 7850 kg/m3
= 1 / 4 x 3.1415 x d2 x L x 7850 kg/m3
because d = diameter of reinforcement specified in millimeters (mm), then we had the conversion to meters (m),
d2 = (d x d) … … … … … … … …. … … mm2
converted to meters (1 mm = 0.001 m)
= (X 0.001d 0.001d)
= (1x 10-6) d2 … … … … … … … m2
Thus,
= 1 / 4 x 3.1415 x (1x 10-6) d2 x L x 7850
D2 = 0.006165 x L
So the formulation to calculate the weight of iron is d2 = 0.006165 x L

Concrete Core Correction Factor Chart

Hi friends,
I was search in the web for correction factors chart for Obtaining and testing drilled cores of concrete (see my previous post Click Here),
But unfortunately i did not found it. So, i decided to create an Interpolation chart to determine correction factors for L/D ratio.
I hope it will help you guys...

   

Tuesday, November 19, 2013

Obtaining And Testing Drilled Cores of Concrete

Obtaining And Testing Drilled Cores of Concrete

ASTM Designation : C42 / C 42M

AASHTO Designation : T 24

Scope: 
This test method covers obtaining, preparing and testing core drilled from concrete for length or compressive or splitting strength.

Apparatus: 
Core Drill, for obtaining cylindrical core specimens with diamond impregnated bits attached to core barrel.

Sampling:
Samples of hardened concrete for use in the preparation of strength test specimens shall not be taken until the concrete strong enough to permit sample removal without disturbing the bond between the mortar and the coarse aggregate. when preparing strength test specimens from samples of hardened concrete, samples that show defects or samples that have been damaged in the process of removal shall not be used.
specimens containing embedded reinforcement shall not be used for determining splitting tensile strength and specimens for determining flexural strength shall not be used if reinforcement is embedded in the tensile portion of the specimen.

Core Drilling: 
A core specimen taken perpendicular to horizontal surface shall be located, when possible, so that its axis is perpendicular to the bed of the concrete as originally placed and not near formed joints or obvious edges of a unit of deposit. A specimen taken perpendicular to vertical surface with a batter, shall be taken from near the middle of a unit of deposit when possible and not near formed joints or obvious edges of a unit of deposit.

Length of Drilled Cores:
Core specimens drilled through a structure for the purpose of measuring structural dimensions shall have a
diameter of at least 3.75 in. [95 mm]. For cores which are not intended for measuring structural
dimensions, measure the longest and shortest lengths on the cut surface along lines parallel to the core axis. Record the average length to the nearest 1⁄4 in. [5 mm].

Cores for Compressive Strength:
Test Specimen—The nominal diameter of core specimens for the determination of compressive strength shall be at least 3.75 in. [95 mm]. Core diameters less than 3.75 in.[95mm] are permitted when it is impossible to obtain cores with length to diameter (L/D) ratio > 1 for compressive strength evaluations in cases other than load bearing situations. For concrete with nominal maximum aggregate size greater
than 11⁄2 in. [37.5 mm], the nominal diameter should preferably be at least three times the nominal maximum size of the coarse aggregate and must be at least twice the nominal maximum size of the coarse aggregate. The preferred length of the capped specimen is between 1.9 and 2.1 times the diameter. If the ratio
of the length to the diameter of the core specimen exceeds 2.1, reduce the length of the specimen so that the ratio is between 2.1 and 1.9. Core specimens with length-to-diameter ratios less than 1.8 require corrections to the measured compressive strength. A core having a maximum length of less than 95 % of its diameter before capping or a length less than its diameter after capping shall not be tested.
End Preparation:
The ends of core specimens to be tested in compression shall be essentially smooth, perpendicular
to the longitudinal axis, and of the same diameter as the body of the specimen. If necessary, saw the ends of the specimens until the following requirements are met:
Projections:
 if any, shall not extend more than 0.2 in.[5 mm] above the end surfaces,
The end surfaces shall not depart from perpendicularity to the longitudinal axis by more than 0.5°, and
The diameters of the ends shall not depart more than 0.1 in. [2.5 mm] from the mean diameter of the specimen.
Moisture Conditioning:
Test specimens shall be tested in a moisture condition representative of the in-place concrete or as directed by the specifying authority.
Compressive strength test results are usually used for the evaluation of the in-place concrete strength; therefore, the cores shall be conditioned in a moisture condition most representative of the in-place strength. If the concrete service condition is dry, the cores can be tested in either an “as received condition” after allowing the drilling moisture to evaporate or tested in a “dry condition” where the cores are air dried in a temperature range of 60 to 80°F [16 to 27°C] at a relative humidity less than 60 % for seven days, as directed by the specifying authority.
The following procedure is used to bring the cores to the “as received condition.” After drilling, transport the cores to the testing laboratory within 24 h. Dry the cores for 12 to 24 h in air at a temperature between 60 and 80° [16 to 27°C] and at less than 50 % relative humidity. Cap or grind the cores, and
test them within 48 h of receipt.

Capping:
The ends of the cores shall conform to the tolerance requirements of Test Method C 39. The ends shall be
sawed or ground to tolerance or capped in accordance with capping procedures for hardened concrete specimens of Practice C 617.

Calculation:
Calculate the compressive strength of each specimen using the computed cross-sectional area based
on the average diameter of the specimen.
If the ratio of length to diameter (L/D) of the specimen is 1.75 or less, correct the result obtained in 7.7 by multiplying by the appropriate correction factors shown in the following

       Ratio of Length to Diameter (L/D)                          Strength Correction Factor
                            1.75                                                                 0.98
                            1.50                                                                 0.96
                            1.25                                                                 0.93
                            1.00                                                                 0.87
Use interpolation to determine correction factors for L/D values not given in the table.



Drilling Core at R.C. Box Culvert Bottom Slab


                                           

                                             Drilling Core at R.C. Box Culvert Bottom Slab


Drilling Core at R.C. Box Culvert Bottom Slab


Drilling Core at R.C. Box Culvert Bottom Slab


Drilling Core at R.C. Box Culvert Bottom Slab

For clear reference please see AASHTO - T24 and ASTM - C42