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Q1: From the following network data, determine the critical path, starting and finishing times and total and free floats. See Answer
Q2: 7:00 material for ceiling joists is very similar to floor joists. Calculate number of joists on each side of the center load-bearing wall. We will calculate the number of each even length joists needed. FORMULA for # of CEILING JOISTS (per span) # of Joist LF perpendicular to joist x space factor + 1 (starter) Ceiling Beams and Bracing Beams are placed in the ceiling joists to eliminate drops in the ceiling at doorways. The beams may be similar to methods and materials used for girders. A built-up beam needs to bear at least 2" on each wall. Joist hangers are needed for each joist on each side of the beam. A box beam is placed above the ceiling joists with plywood saddles holding the joists up. FORMULA for amount of BEAM MATERIAL Material for Beam - Calculated per LF of Beam - Strong backs and Stiffbacks Two types of bracing are used for ceiling joists. A strongback or stiffback provide more support to the joist than a rib band. A strongback helps keep the joists straight, especially when the plan has large rooms. They are placed at mid-span and are secured to the end walls. For residential construction a 2x4 is laid flat on top and nailed into each of the ceiling joist. A 2x6 is placed on edge and nailed into the edge of the 2x4. Each type of material and each span need to be calculated for the house. FORMULA for MATERIALS for STRONGBACKS (per span) Strongback Material = LF of Building (perpendicular to joists) Rib band The other ceiling joist bracing is a 1x4 nailed to the top edge of the joist. This keeps the joists spaced "oc" and from twisting and bowing. The rib band is also placed at mid-span and secured to the end walls. The rib band may also be called a "rat run". We calculate the even LF of material needed. FORMULA for RIB BAND MATERIAL (per span) Material for Rib band = L of Building (perpendicular to joists) Drywall Backing or Nailers A nailer is placed on top of walls when the joist is not next to the wall edge. This provides a nailing surface for edges of ceiling material. On top of 2x4 walls a 2x6 is commonly centered, leaving one (1") inch exposed on each side to nail or screw into. Depending on how the joists lay in relation to the walls provides a variety of situations that may develop for the backing. We will estimate an amount of 2x6 equal to two-thirds (2/3) the length of the interior 2x4 walls that run parallel to the ceiling joist. The remaining one-third (1/3) will come from scrap material. We will estimate the even LF of 2x6 needed. FORMULA for DRYWALL BACKING MATERIAL Material for Backing = LF of walls parallel to joist x .6666 5-8 (2x6) ==‒‒‒‒‒‒‒[[[///// Wall Sheathing Veneered panels, solid lumber, or non-wood products are common materials used for sheathing. Rated sheathing or OSB are structural sheathings that provide stiffness to the wall and are used under vinyl or steel sidings and brick veneers. The non-wood products (Black Jack, fiberboard, and rigid insulating board) usually provide more insulating value. The veneered panels are 4x8, while the non- wood products come in 4x8 or 4x9 sheets. Except for large openings, the sheathing is installed over doors and windows then cut out later. Evaluate deductions for large openings. Divide the SF by 32 for 4x8 sheets and by 36 for 4x9 sheets. Estimate the number of sheets for the gable ends and add them to what is needed for the walls. If different material is used as corner bracing be sure to make the deduction. We will calculate the number of each type and size of sheathing needed. FORMULA for # of SHEETS for GABLE END (# for one gable end) # of Sheets SF of gable end divided by SF per sheet FORMULA for # of SHEETS of WALL SHEATHING # of Sheets = = (4x9 sheets) OSP divided by 4 + (# for gable end area) - (deductions) B OR FORMULA for # of SHEETS of WALL SHEATHING (4x8 sheets) # of Sheets = OSP divided by 4 x 1.125 + (# for gable end area) - (deductions) 7.00 Moisture and Air Barrier Moisture barrier is placed directly under the siding to prevent water from penetrating into the sheathing. An air barrier is a membrane that limits infiltration of air through the wall. Either a moisture or a vapor barrier may also seal the wall against air infiltration. A vapor barrier is a membrane on the warm side of the wall that retards passage of water vapor from the warm inside air into the cooler wall where it could condense. This vapor barrier may be part of the insulation or a separate poly sheet. Coordinating the components is critical to avoid trapping water vapor in the wall cavity. Products, such as Tyvek serve as both a moisture and air barrier. Tyvek comes in rolls of 9'x195' or 1755 SF. The joints of the material need to be taped to help seal out the air and moisture. When estimating be sure to include the gable end areas. We will calculate the number of rolls needed. of cons br FORMULA for NUMBER of ROLLS of AIR BARRIER # of Rolls of Air Barrier = Exterior wall area divided by SF per Roll CEILING FRAMING Ceiling Joists Ceiling joists tie the outside walls together, support the attic area, the weight of the ceiling finish materials, and the tops of the interior walls. Generally the joists run parallel to the rafters. The span and the weight that the ceiling must support determine the size of the joists. Stub joists are necessary when a hip or low sloped gable roof is used on the house. If stub joists are used no additional material is required compared to regular joists. A 2x6 placed 16"oc is typical construction for ceiling joists. To produce smooth straight ceilings, 5/8" drywall should be used if the joists are 24"oc. To estimate for 8'1" high walls, cripple studs above the header are eliminated. For 2x4 stud walls, two 2x12's are builders will use a 2x12 header on all exterior and interior load-bearing walls. By using a 2x12 header nailed together with some ½" plywood strips sandwiched between to make the header 3 ½" thick. If the subfloor is not " material, you may have to calculate some 2" ply for the spacers. Additional material will need to be calculated for 2x6 walls. The maximum span for a solid stock 2x12 header is about 10'. If greater spans are needed you may have to use an alternate material (LVL, Paralam, etc.). We will estimate each opening separately. The actual length of the header is 3" longer than the rough opening width (Hdr L = RO + 3") or 5 ½" longer than the doors width. (Hdr = 3'5 " for 30 door) Pocket door RO = 2 x door + 1". Keeping in mind that the dimensional lumber comes in even lengths, use this formula for 2x4 walls. We will calculate the number needed for each size of even. length header material. (4- 2x12x14, 6 - 2x4x8, etc.) FORMULA for L of MATERIAL for HEADERS (2x4 stud walls) L of Header Material Opening width + 1'-0" (rd.off) x 2 Example: 2¹-8" door + 1'0" = 2'-6" door + 1'0" = 2-4" door + 1'0" = 3¹-8" (rd.up) to 4'-0" x 2 = 8'-0" 3'-6" (rd.up) to 4'-0" x 2 = 8'-0" 3'-4" (rd.down) to 3'-0" x 2 = 6'-0" 10/01 We will use a 4x6 (2 - 2x6's) header for non-load bearing walls. An optional chart can be used to determine the size of material for the headers. When truss rafters are used the load is carried to the exterior walls, and the interior walls become non-load bearing. Optional INTERIOR NON-LOAD BEARING WALL HEADER MATERIAL Max.opening for 2x4 Hdr. = 3'-6", 2x6 = 4'-6", 2x8 = 6'-0", 2x10 = 8'-0", 2x12 = 10'-0" Corner Bracing We will discuss two types of corner bracing for the walls. Both provide lateral support for the walls so they will not rack under heavy wind loads against the house. A rated sheathing or structural panel can be placed at each exterior 270° corner, if a non-structural sheathing is used for the rest of the wall sheathing. Two sheets would be placed at the corners, which are a min. wall length of 2'0". The 4x8 sheets will not cover the box header area, so a 1'x 4' piece of material will be calculated to cover this area (1' of an 8' sheet = .125 sheet). We will calculate the number of 4x8 sheets needed. FORMULA for # of SHEETS for CORNER BRACING # of Sheets # of 270° corners x 2 x 1.125 A let-in brace is required at each end of the wall, if possible, and at 25' intervals. The let-in brace should be placed from the top corner of the wall going towards the center at a 45° to 60° angle. Typically the brace is a continuous 1x4. Use a twelve (12') foot 1x4 for an 8'1" high wall. We will calculate the number of 1x4x12 needed for the 8' walls. FORMULA for LF of LET-IN BRACE LF of Let-In Brace = # of braces x LF per brace 5-6 IIII O 7:00 Contractors use one of several methods to estimate the quantity of studs, placed 16"oc. Several will simply estimate one stud per LF of wall. We will use the formula that adds one stud per corner to the LF of wall. Code stipulates that three (3) studs be placed in each corner. An estimator also needs to evaluate a small area where several studs would be located to see if additional ones are needed. (16" oc) FORMULA for WALL STUDS LF of Wall + 1 per Corner # of Studs = There are several optional formulas, which you might use, in special situations. The first one most closely counts the number of studs. When the walls have an abnormal number of doors and windows, such as a garage, you may want to use this formula. OPTION 1 FORMULA for # of WALL STUDS # of Studs = LF of Wall x space factor + 2 per corner, door, or window OPTION 2 for STUDS 16" oc # of Studs = OPTION 3 LF of wall x 1.1 for STUDS 24" oc # of Studs = LF of Wall x .85 OPTION 4 for STUDS 12" oc # of Studs LF of Wall x 1.35 = Gable End Studs The gable end studs fill in the area between the top of the wall and the gable end rafter. The wall sheathing and siding are fastened to them. The triangular shaped end will be converted to a rectangle for finding the length and number of studs, for one end. To find the number of studs multiply the span of the house, times the spacing factor. Because we are calculating a rectangular area divide this number by 2, then round down. To find the length of a gable end stud, multiply the slope number for the roof by the run, one half the span. The slope number for a 4-12 slope roof is "4". Convert the length to feet and order to the most convenient even LF increment. FORMULA for # of GABLE END STUDS (# for one gable end) # of Studs = Span x spacing factor divided by 2 FORMULA for LENGTH of GABLE END STUDS L of Studs (inches) Slope x run (feet) = Door and Window Headers All door or window openings in walls require some type of horizontal framing to support the load being placed above the wall. If the framing member is located in the wall it is called a header. A beam may be located in the ceiling joist to support this load. This will allow the ceiling height to remain the same as it passes through the opening. We will examine the beams when we estimate ceiling joist material. The header transfers the load of the ceiling and roof to the trimmers and then on down to the foundation walls. A load-bearing wall will carry larger loads, so larger headers are required. Most 5-5 . ♦ BRIDGING Bridging or blocking is installed between adjacent floor joist to provide lateral support. When bridging is used it helps distribute the load and allows the floor to work as one unit. Some types of bridging used are: prefab 1x3, 1x3 made from 1x6, solid, and metal. We will estimate the bridging being placed @ 8' oc. Cross Bridging For the cross bridging one set will consist of two (2) pieces. Cross bridging will be placed between joists spaced 16"oc (14 %2" space). For prefab sets calculate the number of 14 ½" spaces for the floor system. The remainder of openings will have solid blocking. Openings that would not be 142" would be located at the first and last space; and where the extra floor joists have been installed. When making the cross bridging from 1x6 material, 3.5 LF of 1x3 is needed for each set for 2x10, 16"oc joist. Divide the LF of 1x3 by 2 to convert to 1x6. FORMULA for # of SETS of CROSS BRIDGING (per row) # of sets = L perpendicular to joists x spacing factor # spaces containing solid bridging FORMULA for LF of CROSS BRIDGING LF of cross bridging = # of sets x 3'-6" An optional method for calculating LF of material for cross bridging is to take the length of the house times 2.5. The "2.5" is determined by multiplying the spacing factor times the length of one set (.75 x 3.3334 = 2.5). This optional method allows for no deductions for solid bridging. OPTION FORMULA for LF of CROSS BRIDGING LF of Cross Bridging = L of House x 2.5 Solid Bridging The same size material is used as the joists. When cross bridging is used then the solid blocks are placed in those joist spaces that are smaller than 14 " If only solid is used, then the formula places a block in each space. Order the material in a 4' increment if the LF of solid stock is longer than 18' and the joists are 16" oc. For 16"oc joist, we will use a 1'3" length for each block. The following formulas are using 16"oc joist. We will calculate the even LF of material needed. FORMULA for LF of SOLID BRIDGING (used with cross bridging) LF Solid Bridging # of spaces x 1'-3" = FORMULA for LF of SOLID BRIDGING LF of Solid Bridging = per row 7.00 L of House x 75 x 1'-3" 5-3 (used w/o cross bridging) . . •/n 4.0 PERNINCATATK #1 lan 1. Girder (6x10) LF of 2x10 2. LF of sill sealer 3. LF of 2x8 PT mudsill 30.0 4. LF of 2x10 box header 5. No of 2x10x14' floor joist 6. No of 2x10x12' floor joist 7. No of 2x10x10' floor joist M 10-10 30-0 Name 12-0 8. Number of sets of 1x3 bridging 9. LF of 2x10 solid bridging 10. No of sheets %" OSB subfloor 13. Total Cost Floor Framing lan 11. No of quart tubes adhesive 12. No of sheets 3/8" plywood underlayment 1| Page Estimating for Light Construction DDRT 1350 12-0 11'-0" #2 Mariah G ELI 22-0 3'-0" Estimate the plan and use the checklist to fill out a quantity take off form. Girder-Built-up 6x8 Post-4 Lally columns Sill Sealer-6" Mudsill-2x8 PT Bridging-1x3 prefab & 2x10 solid Subfloor-glued %" OSB T&G Underlayment - ½/2" plywood 10-10 26-0 Name Floor Framing lan 2 | Page Estimating for Light Construction DDRT 1350 #3 Savannah 2018, 1²0.C FLOOR JOIST 17.P 23:10, 1678.C FLOOR JOIST 10 PULL 4-4 189 SUBFLOOR 12-165 Name Floor Framing lan 3 | Page Estimating for Light Construction DDRT 1350See Answer
Q3: Cost Analysis - 0401-533 Assignment 2: Cost Estimating, Uncertainty Analysis and Cost Planning Course: Question 1: 1) Perform quantity takeoff for the foundation structure provided in the following figure and estimate the quantities of: 1. Excavation 2. Concrete including the concrete slab 3. Gravel 2) Design crew for excavation (labors, excavators, hauling trucks... etc.) to excavate the above quantity and to haul it to a dumping location 15 km far from the construction site. Estimate productivity of crew and cost of the excavation process 3) Evaluate uncertainty in the excavation productivity and estimate its impact on schedule and budget. Note: Include a list of all assumptions that you have made at the beginning of the analysis including any missing data in the drawing 45° 3' 0" 1' 0" typical 32' 0" Plan view Width 32' 0" रहे Length 85' 0" 4" concrete slab 4" gravel Typical section view 85' 0" 3' 0" 10' 0" 1'0" 3' 0" 3' 0" 45° Question 2: Given the bridge abutment shown below, estimate its construction cost. Data for this problem should be extracted from both: 1) UAE construction data and 2) Published data such as the RS Means Unit Price Estimating Methods (available in the University library). 1) Perform quantity takeoff of the formwork and concrete needed to build the abutment 2) Design minimum reinforcing steel mesh for the abutment (provide the reinforcing steel drawings) and do quantity takeoff of steel (Select appropriate design loads and specifications of the material) 3) Use resource numeration to design crews (labour and power tools) to build the formwork, to place concrete and to fix the reinforcing steel. 4) Specify productivity of the different crews 5) Estimate cost of the materials 6) Estimate cost per unit of work (i.e. cost per ton of fixing steel, cost per m³ of placing concrete... etc. 7) Study the market trends in term of adding overhead, contingency and profit and prepare a bid proposal to build the abutment package. 30' 50' 1' 30' 8′See Answer
Q4:1. (15 Points) I want to build a vacation home in the mountains near Winter Park and I need your help with a cost estimate (and donations). The homeowner's association (HOA) has some strict guidelines. The minimum square footage for a single-story home is 1,800sf. The minimum square footage for a 2-story home is 2,400sf. The square footage is measured above ground, basements are not included in the calculation. It must have high-end materials on the exterior to make it seem we're fancy people. There must be multiple roof pitches. Interior finishes are not a concern of the HOA. Dreaming about this home, I want a 2-car detached garage for snowmobiles, dirt bikes and a ski- tuning workshop. I want 4 bedrooms and 3 baths. I would like two big wood stoves (comparable in price to a fireplace). The kitchen should be fancy for hosting spring break parties. I would like an unfinished basement for now in hopes that I'll finish it later. The land is a separate cost and will run $200k. I will also need a well and septic system which cost around $20k-$25k total. The location factor in Grand County is 1.35. I have an estimating guide from 2016 which gives us square footage pricing, but assume an inflation rate of 3%/year (compounded annually). Hypothetically, my budget is $400k. 1a. I really want a 2-story home to provide better views of the mountains. Using the RS Means estimating guides (saved on Canvas), perform a cost estimate for this home. Show all of your calculations and list all of your assumptions. Am I over budget? If so, by how much? 1b. It seems that a 1-story home is more realistic. Perform another cost estimate for a single-story home. Be very aggressive in your pricing and try to get me under budget. If you need to make changes to my dream home, be sure to state your assumptions. You cannot change the HOA rules, however. Show all of your calculations and list all of your assumptions. Am I over budget still? What is your conclusion regarding my dream home in Winter Park?See Answer
Q5:5. (15 Points) QUANTITY TAKEOFF (not a full cost estimate) - You are building twenty-two (22) of the 2'-0"x 2'-0"x 0'- 10"h spread footings shown below. Using the material schedule at the bottom of the page, estimate the material quantities listed below. Show all calculations and list all assumptions in the space provided. Don't forget to include waste. 2"-3" 2'-6" 2 x 12 Concrete Form 1'-6" Plan View 2"-3" Side View (2) #4 rebar each way top and bottom Section A-A Concrete 2 x 4 Stakes 2 x 12 Concrete Form 2 x 4 Stakes 2'-6" Side View 49-1 キミュラ 2'-6" 0'-10". Concrete エニラ 2"-3" 2-3" Section A-A 2 x 12 Concrete Form 2x4 Stakes (2) #4 rebar each way top and bottom/nMaterial Concrete Rebar Forms Stakes Nails Specification Concrete Rebar Forms Stakes Nails 4,000 psi #4 (0.668 lbs/lf) 2 x 12 x length Oak, 2 x 4 x 18 in. 3½ in. common Frequency of Use Once Once Once Once Once Units Waste, % cy lbs bf each lbs 3 10 10 25 10 Hint: a 2 X 4 at 8' length = 5.33 board feet (convert two dimensions to feet, the third to inches) Required quantity takeoff items - Material Quantity Comments 2" clearance from ends to forms Board Foot = 1" XI'XI' Total length is 18". Board Foot = 1" XI'XI' 1 lb/100 bf of total lumber (forms plus stakes) Note: Presentation is important and your grade may be adjusted +/- 10% depending on your work productSee Answer
Q6:1. Determine the drywall, joint compound and tape needed to cover four 8' high by 12' long walls. The drywall will be run horizontally and is available in 4' widths and 12' lengths. To finish 100 ft2 of drywall, 0.4 boxes of joint compound and 0.13 rolls of tape are needed (1 box of joint compound can cover approximately 250 ft² and a roll of tape can cover the seams in a surface of approximately 770 ft2). Round your answers up.See Answer
Q7:2. If one gallon of paint covers 400 ft2, how many gallons are required to paint the walls and the ceilings of two rooms that each measure 15' long by 12' wide by 7' high, with two coats (ignore the window and door openings)? Round your answer up.See Answer
Q8:3. How many square yards of solid colored carpet is needed for a 12'-6" x 15'-3" room? If the carpet is only available in 12' wide rolls and is only sold in whole foot lengths, what minimum length of roll will be required in whole feet?See Answer
Q9:5. Determine the quantity (in units of tons) of Base Course, Hot mix asphalt (HMA) and Prime Coat needed to pave a 1,200' by 1,200' parking lot. The Base Course must be 6" thick, the layers of HMA must be 0.25" thick (combined total) and 0.15 gallons of Prime Coat can cover 1 square yard. The density of the Base Course is 2.1 tons per cubic yard (2.1 T/CY). The density of the HMA is 2.0 tons per cubic yard (2.0 T/CY). The density of the Prime Coat is 8.1 pounds per gallon (8.1 lbs/GAL).See Answer
Q10:EXERCISES 1. Determine an OM estimate for a 2,500-pound capacity hydraulic passenger elevator for a two-story office building. Use commonly available estimating references. 2. Determine a reasonable unit price for HVAC of a two-story fire station. What kind of system is this? cnic 3. How does the cost of the HVAC for the fire station compare with a two-story medical office building both in unit price and percentage of total building cost?See Answer
Q11:2. The following bar chart shows the activities to complete a project in 6 months (each cell represents 1 month). The total project cost is $700. Month 1 Month 2 Month 3 Activity Cost $100 ABCDEFGHIKL $30 $40 $80 $20 $50 $70 $30 $200 $60 $20 Month 4 Month 5 Month 6 After 2 months, Activities A, B, half of C, and half of D are completed for $200. For this point in time (2 months after the start date). a. What are the values of the budgeted cost of work performed, the budgeted cost of work scheduled, and the actual cost of work performed? b. What is the status of the project in terms of budget (over or under) and the schedule (ahead or behind)? (Give each answer in terms of a percent variance).See Answer
Q12: 7:00 material for ceiling joists is very similar to floor joists. Calculate number of joists on each side of the center load-bearing wall. We will calculate the number of each even length joists needed. FORMULA for # of CEILING JOISTS (per span) # of Joist = Ceiling Beams and Bracing LF perpendicular to joist x space factor + 1 (starter) Beams are placed in the ceiling joists to eliminate drops in the ceiling at doorways. The beams may be similar to methods and materials used for girders. A built-up beam needs to bear at least 2" on each wall. Joist hangers are needed for each joist on each side of the beam. A box beam is placed above the ceiling joists with plywood saddles holding the joists up. FORMULA for amount of BEAM MATERIAL Material for Beam = Calculated per LF of Beam - Strong backs and Stiffbacks Two types of bracing are used for ceiling joists. A strongback or stiffback provide more support to the joist than a rib band. A strongback helps keep the joists straight, especially when the plan has large rooms. They are placed at mid-span and are secured to the end walls. For residential construction a 2x4 is laid flat on top and nailed into each of the ceiling joist. A 2x6 is placed on edge and nailed into the edge of the 2x4. Each type of material and each span need to be calculated for the house. Rib band FORMULA for MATERIALS for STRONGBACKS (per span) Strongback Material = LF of Building (perpendicular to joists) The other ceiling joist bracing is a 1x4 nailed to the top edge of the joist. This keeps the joists spaced "oc" and from twisting and bowing. The rib band is also placed at mid-span and secured to the end walls. The rib band may also be called a "rat run". We calculate the even LF of material needed. FORMULA for RIB BAND MATERIAL Material for Rib band = Drywall Backing or Nailers (per span) L of Building (perpendicular to joists) A nailer is placed on top of walls when the joist is not next to the wall edge. This provides a nailing surface for edges of ceiling material. On top of 2x4 walls a 2x6 is commonly centered, leaving one (1") inch exposed on each side to nail or screw into. Depending on how the joists lay in relation to the walls provides a variety of situations that may develop for the backing. We will estimate an amount of 2x6 equal to two-thirds (2/3) the length of the interior 2x4 walls that run parallel to the ceiling joist. The remaining one-third (1/3) will come from scrap material. We will estimate the even LF of 2x6 needed. FORMULA for DRYWALL BACKING MATERIAL Material for Backing = LF of walls parallel to joist x .6666 (2x6) 5-8 HI 7.00 Wall Sheathing Veneered panels, solid lumber, or non-wood products are common materials used for sheathing. Rated sheathing or OSB are structural sheathings that provide stiffness to the wall and are used under vinyl or steel sidings and brick veneers. The non-wood products (Black Jack, fiberboard, and rigid insulating board) usually provide more insulating value. The veneered panels are 4x8, while the non- wood products come in 4x8 or 4x9 sheets. Except for large openings, the sheathing is installed over doors and windows then cut out later. Evaluate deductions for large openings. Divide the SF by 32 for 4x8 sheets and by 36 for 4x9 sheets. Estimate the number of sheets for the gable ends and add them to what is needed for the walls. If different material is used as corner bracing be sure to make the deduction. We will calculate the number of each type and size of sheathing needed. FORMULA for # of SHEETS for GABLE END (# for one gable end) # of Sheets = SF of gable end divided by SF per sheet FORMULA for # of SHEETS of WALL SHEATHING # of Sheets = (4x9 sheets) OSP divided by 4+ (# for gable end area) - (deductions) OR FORMULA for # of SHEETS of WALL SHEATHING (4x8 sheets) # of Sheets = OSP divided by 4 x 1.125 + (# for gable end area) - (deductions) Moisture and Air Barrier Moisture barrier is placed directly under the siding to prevent water from penetrating into the sheathing. An air barrier is a membrane that limits infiltration of air through the wall. Either a moisture or a vapor barrier may also seal the wall against air infiltration. A vapor barrier is a membrane on the warm side of the wall that retards passage of water vapor from the warm inside air into the cooler wall where it could condense. This vapor barrier may be a part of the insulation or a separate poly sheet. Coordinating the components is critical to avoid trapping water vapor in the wall cavity. Products, such as Tyvek serve as both a moisture and air barrier. Tyvek comes in rolls of 9'x195' or 1755 SF. The joints of the material need to be taped to help seal out the air and moisture. When estimating be sure to include the gable end areas. We will calculate the number of rolls needed. ня FORMULA for NUMBER of ROLLS of AIR BARRIER # of Rolls of Air Barrier Exterior wall area divided by SF per Roll = CEILING FRAMING leasa grilion Ceiling Joists Ceiling joists tie the outside walls together, support the attic area, the weight of the ceiling finish materials, and the tops of the interior walls. Generally the joists run parallel to the rafters. The span and the weight that the ceiling must support determine the size of the joists. Stub joists are necessary when a hip or low sloped gable roof is used on the house. If stub joists are used no additional material is required compared to regular joists. A 2x6 placed 16" oc is typical construction for ceiling joists. To produce smooth straight ceilings, 5/8" drywall should be used if the joists are 24" oc. To estimate 10/01 builders will use a 2x12 header on all exterior and interior load-bearing walls. By using a 2x12 header for 8'1" high walls, cripple studs above the header are eliminated. For 2x4 stud walls, two 2x12's are the subfloor is not ½" material, you may have to calculate some ½" ply for the spacers. Additional nailed together with some ½" plywood strips sandwiched between to make the header 3 ½" thick. If material will need to be calculated for 2x6 walls. The maximum span for a solid stock 2x12 header is about 10'. If greater spans are needed you may have to use an alternate material (LVL, Paralam, etc.). We will estimate each opening separately. The actual length of the header is 3" longer than the rough opening width (Hdr L = RO + 3") or 5 ½" longer than the doors width. (Hdr = 35 ½" for 3° door) Pocket door RO = 2 x door + 1". Keeping in mind that the dimensional lumber comes in even lengths, use this formula for 2x4 walls. We will calculate the number needed for each size of even length header material. (4 - 2x12x14, 6 - 2x4x8, etc.) FORMULA for L of MATERIAL for HEADERS (2x4 stud walls) L of Header Material == Opening width + 1'-0" (rd.off) x 2 Example: 2'-8" door + 1'0" 2'-6" door + 1'0" = 2'-4" door + 1'0": = 3'-8" (rd.up) to 4'-0" 3'-6" (rd.up) to 4'-0" x 2 = 8'-0" x 2 = 8'-0" x 2 = 6'-0" = 3'-4" (rd.down) to 3'-0" We will use a 4x6 (2 - 2x6's) header for non-load bearing walls. An optional chart can be used to determine the size of material for the headers. When truss rafters are used the load is carried to the exterior walls, and the interior walls become non-load bearing. Optional INTERIOR NON-LOAD BEARING WALL HEADER MATERIAL Max.opening for 2x4 Hdr. =3'-6", 2x6 = 4'-6", 2x8 = 6'-0", 2x10 = 8'-0", 2x12 = 10'-0" Corner Bracing We will discuss two types of corner bracing for the walls. Both provide lateral support for the walls so they will not rack under heavy wind loads against the house. A rated sheathing or structural panel can be placed at each exterior 270° corner, if a non-structural sheathing is used for the rest of the wall sheathing. Two sheets would be placed at the corners, which are a min. wall length of 2'0". The 4x8 sheets will not cover the box header area, so a 1'x 4' piece of material will be calculated to cover this area (1' of an 8' sheet = .125 sheet). We will calculate the number of 4x8 sheets needed. FORMULA for # of SHEETS for CORNER BRACING # of Sheets = # of 270° corners x 2 x 1.125 A let-in brace is required at each end of the wall, if possible, and at 25' intervals. The let-in brace should be placed from the top corner of the wall going towards the center at a 45° to 60° angle. Typically the brace is a continuous 1x4. Use a twelve (12') foot 1x4 for an 8'1" high wall. We will calculate the number of 1x4x12 needed for the 8' walls. FORMULA for LF of LET-IN BRACE LF of Let-In Brace = # of braces x LF per brace 5-6 D 7:00 Contractors use one of several methods to estimate the quantity of studs, placed 16" oc. Several will simply estimate one stud per LF of wall. We will use the formula that adds one stud per corner to the LF of wall. Code stipulates that three (3) studs be placed in each corner. An estimator also needs to evaluate a small area where several studs would be located to see if additional ones are needed. FORMULA for WALL STUDS # of Studs = (16"oc) LF of Wall +1 per Corner There are several optional formulas, which you might use, in special situations. The first one most closely counts the number of studs. When the walls have an abnormal number of doors and windows, such as a garage, you may want to use this formula. OPTION 1 FORMULA for # of WALL STUDS # of Studs = LF of Wall x space factor + 2 per corner, door, or window OPTION 2 for STUDS 16" oc # of Studs = LF of wall x 1.1 1 Gable End Studs OPTION 3 for STUDS 24" oc # of Studs = LF of Wall x .85 OPTION 4 for STUDS 12" oc # of Studs = LF of Wall x 1.35 The gable end studs fill in the area between the top of the wall and the gable end rafter. The wall sheathing and siding are fastened to them. The triangular shaped end will be converted to a rectangle for finding the length and number of studs, for one end. To find the number of studs multiply the span of the house, times the spacing factor. Because we are calculating a rectangular area divide this number by 2, then round down. To find the length of a gable end stud, multiply the slope number for the roof by the run, one half the span. The slope number for a 4-12 slope roof is "4". Convert the length to feet and order to the most convenient even LF increment. FORMULA for # of GABLE END STUDS (# for one gable end) # of Studs = Span x spacing factor divided by 2 FORMULA for LENGTH of GABLE END STUDS L of Studs (inches) == Slope x run (feet) Door and Window Headers All door or window openings in walls require some type of horizontal framing to support the load being placed above the wall. If the framing member is located in the wall it is called a header. A beam may be located in the ceiling joist to support this load. This will allow the ceiling height to remain the same as it passes through the opening. We will examine the beams when we estimate ceiling joist material. The header transfers the load of the ceiling and roof to the trimmers and then on down to the foundation walls. A load-bearing wall will carry larger loads, so larger headers are required. Most 5-5 1 BRIDGING 7.00 Bridging or blocking is installed between adjacent floor joist to provide lateral support. When bridging is used it helps distribute the load and allows the floor to work as one unit. Some types of bridging used are: prefab 1x3, 1x3 made from 1x6, solid, and metal. We will estimate the bridging being placed @ 8' oc. Cross Bridging For the cross bridging one set will consist of two (2) pieces. Cross bridging will be placed between joists spaced 16" oc (14 ½" space). For prefab sets calculate the number of 14 1/2" spaces for the floor system. The remainder of openings will have solid blocking. Openings that would not be 14 1/2" would be located at the first and last space; and where the extra floor joists have been installed. When making the cross bridging from 1x6 material, 3.5 LF of 1x3 is needed for each set for 2x10, 16"oc joist. Divide the LF of 1x3 by 2 to convert to 1x6. FORMULA for # of SETS of CROSS BRIDGING (per row) # of sets = L perpendicular to joists x spacing factor - #spaces containing solid bridging THE FORMULA for LF of CROSS BRIDGING LF of cross bridging = # of sets x 3'-6" An optional method for calculating LF of material for cross bridging is to take the length of the house times 2.5. The "2.5" is determined by multiplying the spacing factor times the length of one set (.75 x 3.3334 = 2.5). This optional method allows for no deductions for solid bridging. OPTION FORMULA for LF of CROSS BRIDGING per row Solid Bridging LF of Cross Bridging = L of House x 2.5 The same size material is used as the joists. When cross bridging is used then the solid blocks are placed in those joist spaces that are smaller than 14 ½" If only solid is used, then the formula places a block in each space. Order the material in a 4' increment if the LF of solid stock is longer than 18 and the joists are 16" oc. For 16" oc joist, we will use a 1'3" length for each block. The following formulas are using 16"oc joist. We will calculate the even LF of material needed. FORMULA for LF of SOLID BRIDGING (used with cross bridging) LF Solid Bridging = # of spaces x 1'-3" (used w/o cross bridging) FORMULA for LF of SOLID BRIDGING LF of Solid Bridging = L of House x .75 x 1'-3" 5-3See Answer
Q13:Part 1. Nominal and Effective Interest Rates (present equations you used and provide sufficient calculations for each problem) 1. A credit card company charges a nominal 18% annual interest, compounded monthly. a. What is the effective annual interest rate? b. Re-calculate the effective interest rate if it is compounded daily. 2. Deposits of $300 are made quarterly into an account that pays 12% interest per year, compounded monthly. What is the effective quarterly interest rate, and how much money will be in the account at the end of 5 years? (Compounding periods more often than cash flows)See Answer
Q14: Draw the activity-on-node (AON) network diagrams for the following project: See Answer
Q15: 3-10. A 24,000-square-foot warehouse constructed for S1,200,000. If the cost-capacity factor is 0.88, what is the estimated cost to construct a32,000-square-foot warehouse? (See Answer

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