Likewise, if the work requires 2. The result simply mean that in one 6 meters long steel bar we could get 4. We will not consider the fractional value of. Instead, we will only use the whole value 4. Divide the result of dtep 3 by 4. Find the total cut bars for 30 footings. The following rules will help in making the right choice. Find the total length of pes. Determine the net length of one reinforcing cut bar x 1. Divide 6. Divide by 6. If the result in dividing the length of one steel bar by ,the length of one cut bar is a whole number exact value use m.
If the result has a fractional value, use the first method. The above solution is correct because in dividing 6.
Estimating the Tie Wire was initially discussed in Section by 1. Applying the same principles of 5. Meaning, there is no extra cut in one 6. In step 5, dividing by 6. Therefore, I , 1. Looking at the plan of Figure there are: all cuts are exactly the same without excesses.
The net length of one cut bar is 1. Solve for the total ties of the 24 footings: number of bars in 30 footings is pieces.
Select a commercial length of steel bar which is divisible 3. Using a 30 centimeters long ties, multiply: H by 1. Try 6. Convert this leng h to kilogram. One kilogram jf No. Divide: 3. This simply mean that in a 6. Divide the lli. Looking at the plan of Figure There are 36 steel bar to the additional length I.
I wire a. Additional length for beam depth and floor thickness if 2. Solve for the total ties of 30 footings. Distance from floor to footing slab. Provisions for splices of succeeding floors. If one tie is 30 cm. From the following figure, list down the main reinforcement 4, Convert this length to kilograms. Using the value of 53 from the footing to the second floor using 20 mm if mere are 10 meters per kilogram, divide: columns in the plan. The m Ii or vertical reinforcement, b. The lateral ties or c.
SpirAl tip. Identify the bars with hook and bend for adjustment of their order length. Determine the total length of the main steel bar reinforce- ments. Slab Slab Beam Beam a. Bend at the base footing Height from ground line to beam Depth of beam Dowel for second floor 20 x 20 mm Select a 6. Multiply by number of bars in one post x'10 post. Order: 80 pes. Verify the plan if the span or distance of the column Tied column has reinforcement 'conSisting of vertical bars where the beam is resting indicates the following condi- held In a position by lateral reinforcement calied lateral ties.
Take note that "the lesser th splice the Ie.. Determine the spacing distance of the lateral ties. Ties should not be less than No. The lowest value is Therefore, adopt 30 centimeters shall not exceed the following: spacing of the lateral ties 1. Designation Inches mm. The least dimension of the column 2 Y- '. Determine the spacing of- the lateral ties. TIES 16 x 20 mm. Find how many 1. By trial division we have: The least dimension of the column is 50 cm.
Adopt 32 cm. Determine the number of lateral ties in one column. Iong steel bar. CommeBU: 32cm. Step 7 is very important treca use wtihout these trial divi- sions, cutting could be done on a 6. In cases where results of the trial divisions does not give an exad quotient, it becomes the estimator's choice to FIGURE decide which length to use that will not produce exces- sive waste.
TaKe note that this 22 is the spacing distance between lateral ties. What we are after is the number of ties in one '3. In this example, we have learned. Add one to get the number of lateral ties. Solve for the total lateral ties In the 26 cohllmns. Find the number of main. Detennlne the length of one Lateral Tie. By inspection, ties. Find the total intersections in the 26 columns: 1.
Or 32 cm. This 4, is the total numbl:! Or 48 cm. The least dimension of the column is 40 cm. Adopt 32 em. One kilogram of No. Find the total lateral ties in th e 30 columns. By Inspection, the length of the lateral ties are : tie wire in kilograms. By trial divi- main reinf. The results dictate that we use 6. Convert to kilograms.
The x - en- Spacing per M. The table will guide the estimator in selecting rei nforcing 15 6. To use the table, consider the following example: 25 4.
Find the 40 2. If the length of each tie is 40 cm. By Inspection, there are two types of lateral ties. Convert to kilograms, divide by 53 m. Find the total length of the 20 columns. Problem Exercise 7. From the following figure, find the number of 10 mm lateral ties and tie wire if there are 36 columns with cross sectional di- 3. Refer to Tatile Under spacing of lateral ties at 30 cm.
Multiply: m. Refer again to Table Under length of ties along centimeters, 4 and 6 pi'eces could pe derived from a 5. For the 80 cm. If there are 16 beams of the The procedures adopted in estimating the number of stirrups same design, find the materials required for the stirrups.
By'direCt counting, there are 17 stimJps at 99 cm. For a 1. For easy handling use 6. By inspection the length of one stinup is cm. Refer' to Table , along cm. If we chose 6. If we chose 7. The spiral reinforce. Find the number of 10 mm b'ilr spirals. Refer to Table 3- spacers under the following considerations 9.
For a 50 cm. That, the clear spacing between the spirals should not 3. Multiply: 3. Finding the Tie Wire 7. Find the number of vertical bars per column :: 12 pieces. Under 50 cm. Total Ties for 14 cplumn at 7. TotallengtlJ 9f the wire at. Convert to' kilograms. Divide by 5j, , 5.
The sup- tennlne the number of 12 mm. The principal reinforcement runs in one direction parallel to the slab span and perpendicular to the A. One method used in finding the number of steel bars for a one-way reinforced concrete slab is either by the direct Given Data: counting or by the area method.
Find the number of main reinforcements at. This Find the number of cut bars in. Using a 6. Find the main reinforcement; add step 2 and step 5. Across the main reinforcement, divide by. Along the 1. Since there are ,2 sides at 1. Add the results of step I and step 3. Detennine the floor area. Refer to Table , along b ,r spacing at 15 crn. I '"'! I "- T-rii -l 3. Bars 7. Solve for the Area of the floor: 7. Refer to Table along For tie wire, refer to Table S4 sq.
From Figure , detennine the number of 12 mm steel 2. Using a 7. Solve for the number of main reinforcement. There are two cross run of main remforcement, multiply: Centimeters 6. For additional cut bars at 1. Divide Find the Temperature bars across the 1.
Multiply by 4 sides at 2 layers, one atihe bottom and one Length of Steel Bars at the bend bars. Summary of steel bars from step 2, 3 and 5.
Total length of one ring plus. Refer to Table , along 10 cm bar spacing and 30 cm. Find the total number of ring at 15 cm. Order' 37 kilograms of No. Add one to get the number of ring: 6. Find the number of shrinkage and temperature bars at say. From the following Figure, determine the quantity of 10 mm steel bars and the tie wire required.
Summary of the Reinforcements Reinf. Ring 8 pes. Total number of ring multiplied by the number of shrinkage and temperature bars. Convert this length to kilograms. Divide by Solve for the circumference of the circle at midpoint of the I concrete. From the following circular column, determine the spiral. A flood control project requires 80 and 50 pieces con-. Number of main reinforcement 16 mm d Concrete using class A mixture.
From the following figure, find the main and lateral ties steel reinforcement for 20,columns each with a height of 15 me- ters. Determine the required reinforcement and tie wire of a one way reinforced concrete slab using 12 mm steel bars de- signed as follows:.
Main reinforcement 12 mm at 5 inches on center. Temperature bars 12 ITU11 at 10 inches on center. For accuracy of the lateral ties cut length make a full scale drawing then measure the actual length. Remember the additional length for hook and bend. A two way reinforced concrete slab will be reinforced by 12 mm steel bars space at 7 Inches and the temperature bars at 14 inches on center.
WOOD 5. Wood is that fibrous substance which composes the trunk and the brancttes of a tree that lies between the pith and the bark. The versatility of using wood In every construction has Wood, because of its strength, light in weight, durability and ease of fastening be- come one of the most important building materials.
Rough Lumber - is the term applied to unplaned or un- dressed lumber. In short, those lumber with rough surfaces. S2s and S4s - are dressed lumber wherein the number con- notes the smooth sides. For instance, S2S means lumber hav- ing two smooth sides and S4s with four sides. Slab - is a kind of rough lumber cut tangent to the anri4al rings running through the ,full length of the log containing at least one flat surface.
With Respect to Leaves - is either: smallest dimension. Needle shape b. Yellow c. Red Flitch - iss thick piece of lumber. Brown e. Black, etc. Fine Grained - when the annual rings are small, the grain or marking , which separates the adjacent rings is said to be fine grained.
When large, it is called Coarse Grained. Ii' Modular Rays Straight Grained - Is a term used when the direction of the fibers are nearly parallel with the side and edges of the board. Wood is Classified According to: ; 1. Mode of Growth Crooked Grain Cro.. Graln Straight Grain a. Indigenous - are those trees that grows from the. These kind of trees has a soft center core ,and are not preferred for lumbering.
With Respect to the Grain a. Straight b. Exogenous - are those outward growing trees pre-' b. Cross ferred for lumbering. Coarse a. Soft b. Hard 6. Grained c. Figllr8d or Marked 1. Plain or Bastard Sawing 2. Radial Cambined RadiaLand Tangential b.. Tangential c. Quarter Tangential d. And the most common defects in wood are: Quarter 'Tangential Tangential Plain or 1.
Heart Shakes are radical cracks in wood origi- Star Shake nating from the heart of the logs. Heart shake is commonly found in old trees. Wind Shakes or Cup Shakes. Star Shakes are composed of several heart shakes radiating from the center of the log in a star like manner. Knots - occurs at the starting point of a limb or branch of the tree. Due To Deterioration Lumbering is the term applied to the operations performed in a. Logging is the ber due to the presence of moisture. Sawing on the other hand is the op- b.
Wet Rot - takes place sometime in the growth of eration of cutting logs into commerCial sizes of lumber. This 1. External Process. Experiments proved that wood im- wood. Internal process. A chemical compound is impreg- nated at a prescribed pressure to permeate the ' wood ting and decay.
It reduces warping but become brittle and less elastic. Despite the adoption of the Mettie System 1. One board foot simply mean, one square foot by of wood although the period involved is relatively longer. The width and thickness of commercial lumber are expressed in 2. The Artificial Seasoning is a process where lumbers are inches while the length is in feet 6f even numbers. Under this method, lumber undergoes a quick Board foot is found by dividing the product of the thickness, drying process.
By Forced Air Drying Find the total board feet of 5 pieces 2" x 6" x ft. By Kiln Drying 3. Finding the board foot of a commercial size lumber is as simple as the above illustration. But the question is how to find The Common Causes of Decay in Wood are: the net board foot of a round log or a standing tree knowing its diameter and height? Alternate moisture and dryness following formula:. Fungi or molds 3. Insects and worms Bd. On the other hand, the meter length method is mul- tiplying the width in inches by the tength in meter times the agreed unit price per meter run.
How much will it cost to slice a 6" x 6" x 3. In solving board foot of iumber, convert first all measure- 12 ments from Metric to English. Multiply by the unit price say P 2. By board foot or 2. Just construcfion file title and download max fajardo simplified construction estimate pdf will popup Our goal is to provide high-quality PDF documents, Mobile apps, video, TV streams, music, software or any other files uploaded on shared hosts for free!
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