# Reinforced Concrete QTO Design Stage 1 Preconstruction Stage 2: Procurement Conceptual Planning Stage3: Construction Stage 4: Project Close-out.

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Reinforced Concrete QTO Design Stage 1 Preconstruction Stage 2: Procurement Conceptual Planning Stage3: Construction Stage 4: Project Close-out

Construction Drawings The goal is to compute the amount of concrete, concrete forming, and reinforcing. Refer to the example in figures 1, 2. The structure is one floor building of RC foundation, columns, exterior wall, and beams. The roof is metal deck. The floor is 8” RC slab. Five drawings: –Slab on grade foundation plan –Interior Column. –Exterior wall detail –Footing detail –Beam detail.

Concrete Volume Basically computed by multiplying the the width, length, and height of net dimensions. In our example, compute the perimeter at the center of the wall first. In simple cases, multiply the sum of the vertical and the horizontal dimensions by two. Check for any exceptions (recess). Do an adjustment to the perimeter to find out the perimeter along the centerline if necessary. A value that will be used again in the estimate is labeled: super. Refer to figure 3 for the QTO. Neglect the volume occupied by the reinforcing steel.

The Homework Not all the area to be taken off Columns and walls extend from the top of the footings to the bottom of first floor slab. Information about the first floor slab can be obtained from the “structural details” drawing of the first floor. Note that not all the columns within the wall have separate footings. Only the ones at the corners do. The square shape around the columns is the edge of the concrete slab. Take the change in wall thickness into account.

Concrete Forming Construction Drawings do not specify the location or quantity of formwork. Concrete forming is usually taken off in square feet of concrete contact area (SFCA), regardless of the fact that forming may extend beyond the contact area. Computation of SFCA is carried out in a less accurate manner. This is because of the uncertainties involved such as: which system will be used, the need to cut and shape the boards. Also, the cost of forming material is less than the cost of concrete. SFCA is rounded to the nearest 5 feet of contact area.

Quantities of different elements are calculated separately. This is due to the varying degrees of difficulty involved. Depending on the type of the soil, footings may and may not need forming work; they do not in the homework. Auxiliary hardware is not computed in details. Their cost is included in the unit price of the SFCA. Refer to figure 5 for an example of concrete QTO. Notice that the dimensions used to compute the concrete QTO are used again.

SFCA for slabs the QTO includes the amount of forms and the supporting system. Scaffolding (falsework) consists of temporary structural members utilized to hold forms in place. Scaffolding can be estimated by two methods: –Taking off every scaffold member. –including the cost of scaffolding in the price, which is used to estimate the slab forms that are supported by the scaffolding, since both costs are usually related. This method is more commonly used. Refer to figure 6 for an example of a T-beam slab.

Special Cases In certain cases, the QTO is estimated using the length of the forming instead of the SFCA. Refer to Figure 7 for two examples: the keyway and the brick ledge concrete section. The cost in both cases is not related to the contact area, but is related to the length of the keyway or the ledge.

Reinforcing Steel Reinforcement (reinforcing bars or rebar) QTO is done considering every reinforcing bar from the drawings The estimate of the quantity is obtained in terms of weight for each size used. Quantity and lengths of rebar vary from one member to another. Refer to the attached table for dimensions and unit weights for reinforcing bars. From the drawings, the estimator can identify the sizes, the lengths, and the distribution of the bars.

Methods for QTO Three Methods: Bar size approach: Search for all bars of specific sizes identified in all the members. Member approach: Take-off the reinforcing bars in the members, one member at a time. Combined method: take every member. The reinforcing bars are sorted out as they are taken off the drawings. The most commonly used method.

Detailed Computations The procedure is to visit each member and compute the total length of each size. When all the members are visited, the total length of each size is obtained and multiplied by the weight per unit length. The total weight is finally computed by adding up all the weights of all the different sizes. Refer to figure 8 for an example of concrete Reinforcing QTO.

Additional Considerations Splices: –When the required bars are too long and needed to be continuos (walls). –to join two bars together, they must overlap and be tied together using tie wire. the length of the splice depends on the diameter of the bars. –how many? a contractor’ decision. –the estimator must develop a rule of thumb. For example, add 10% to the rebar quantity to account for overlaps.

Dowel Bars: –reinforcing bars that serve to transfer load from one member to another or to connect two members. – they usually exist between two concrete pours. –their length and size are usually well defined in the drawings

Project: Class Example Location:UW EngineerK.A Classification: Concrete Reinforcing Concrete cover is assumed 0.25ft NumberLength#4#5#6#8#9 Reinf. Continuous Footings Perimeter 8152 1216 Transverse 1522.5 3800.5 cover Dowels 3041.5 456 no cover needed Reif. Isolated Footings Horizontal 4*8 322.5 80 2X0.25 ft cover

Reinf. Slab Number Le ngt h#4#5#6#8#9 East-west 2340 920 Assumed that the extra East-west 1214 168 bar is the longer length North-south 2622 572 in both directions North-south 1534 510 no need for cover Reinf. Foundation Wall Vertical 2X152 (two lines of rebarr))304 11. 42 3471.68 ht is 11.67-0.25, no slab on top Ties NOT GIVEN13X2152 3952 Assum #6@1' cts about 13 in 11.67 ft Reinf. Columns Vertical2X81611 176 Ties#4@9' cts294117

Reinf. Beams NumberLength#4#5#6#8#9 Horizont al 498 392 Stirrups #4@ 6"cts1962.33 457 Total Length 117176486550262684 Weight/foot 0.6681.0431.5022.673.4 Total Weight 78.378183.577307.713418.69125.6 Total Weight for Project in Pounds 30113.84 Total Weight for Project in Tons 15.1 Add 10% Allowance for splices 16.6

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