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The reason to hold the comprehensive knowledge of carpentry details and practices for the estimator of carpentry work.
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Explanation of Solution
The estimate for any project is to be prepared before the work and various factors influence and contribute to the cost of the work. So before making an estimate for any carpentry or any work, comprehensive and detailed knowledge of principal and practices of the project is required.
The drawing of the project contains information regarding designing, dimension and location of the project but very few details are provided on the drawings of carpentry work of a housing project. It is assumed by architecture that the construction of the building is done by the builder as per the standard codes of the building and construction of the particular area.
So the estimator should be well-known to the building codes and standard methods of framing in the area for accurately estimating the work. He should be also known to the conditions which may require extra work such as the requirement of additional studs, joist and reinforced joints for completion of the project.
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Chapter 7 Solutions
Fundamentals Of Construction Estimating
- Following is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) N60 1.5 6 3 8 4.5 9 6 8 7.5 9 13 14 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 10 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 12.2 kN/m³. Use the relationship given in the equation C'N = 1 σo/Pa 0.5 to calculate the corrected penetration numbers. (Round your answers to the nearest whole number.) Depth (m) N60 (N1) 60 1.5 6 4.5 3 8 9 6 8 7.5 13 9 14arrow_forwardFollowing is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) 1.5 N60 5 3 6 4.5 9 6 7 7.5 9 10 11 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 18 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 20.2 kN/m³. Using the equation N60 0.5 - {11} Dr = determine the average relative density of sand. (Enter your answer to three significant figures.) Average D₁ = %arrow_forwardThe cantilever beam shown below supports a uniform service (unfactored) dead loadof 1.5 kip/ft plus its own self weight, plus two unknown concentrated service(unfactored) live loads, as shown. The concrete has f’c = 6,000 psi and the steel yieldstrength is 60 ksi.a. Determine the design mopment capacity .b. Set-up the factored applied bending moment equation. .c. Calculate maximum safe concentrated live load that the beam may carry.arrow_forward
- A rectangular reinforced concrete beam 18 in. wide by 28 in. overall depth is to support a superimposed (additional to the self-weight) service dead load of 0.5 kip/ft and a service live load of 1.3 kip/ft. Reinforcing for positive moment is 60 ksi yield strength. f’c = 5,000 psi. Use 6#9 rebars a. Determine the design moment capacity . b. Set-up the factored applied bending moment . c. Determine the maximum simple span length on which this beam may be safely utilized.arrow_forward. . . . . . . . TUGAS-1 For a moist soil sample, the following are given: -Total Volume: V 1.2 m³ -Total mass: M = 2350 kg -Moisture Content: Wc = 8.6% -Spesific Gravity of Soil Solids : Gs = 2.71. Determine the following a. Moist Density (Y) b. Dry Density (yd) C. Void Ratio (e) e. f. g. Porosity (n) Degree of Saturation (Sr) Volume of water in the soil sample (Vw) Draw the three phase of the soil element complete with the number TUGAS-2 Mass (kg) Volum V Mac= V₁ = M = 2350 M₁ = ☐ Air Water Solid A saturated soil has a dry unit weight of 16.18 kN/m³. Its moisture content (WC) is 23%. Determine: a. Saturated unit weight, ysat b. Spesific gravity, Gs C. Void Ratio, e TUGAS-3 The dry density of a sand with a porosity of 0.387 is 1600 kg/m³. Determine the void ratio of the soil and the specific gravity of soil solids. POLIT V= POLITIarrow_forward5. What is the lightest WT shape that would be adequate for tension yielding under a design tensile demand of 1,215 kips? Assume that geometric constraints within the structure require you to select a WT9 section.arrow_forward
- a. Determine the effective area for the case shown in the figure below. Suppose that l = 6 in. For L5 × 5 × 5/8: A₁ = 5.90 in.², = 1.47 in. L5 × 5 × 5/8 Weld (Express your answer to three significant figures.) A₁ = in.² b. Determine the effective area for the case shown in the figure below. Suppose that l = 5 in. PL³/8 X 4 Weld (Express your answer to three significant figures.) Ae = in.2 2 c. Determine the effective area for the case shown in the figure below. -5" PL5/8 X 5 Weld (Express your answer to three significant figures.) Ae in.2 d. Determine the effective area for the case shown in the figure below. 2" 2" PL1/2 X 512 ос 3/4-in.-diam. bolts (Express your answer to three significant figures.) Ae = in.² e. Determine the effective area for the case shown in figure below. PL³/8 X 6 7/8-in.-diam. bolts (Express your answer to three significant figures.) Ae= in.2arrow_forwardA single-angle tension member of A36 steel must resist a dead load of 35 kips and a live load of 84 kips. The length of the member is 18 feet, and it will be connected with a single line of 1- inch-diameter bolts, as shown in the figure below. There will be four or more bolts in this line. For the steel Fy = 36 ksi and F₁ = 58 ksi. Try the tension members given in the table below. Tension member rz (in.) A, (in.²) L6 × 6 × 9.75 1.17 L 5 × 3 × 1/10 4.93 0.746 L5 × 3 × 16 5 2.56 0.758 L5 × 3 × 166 3.31 0.644 Bolt line a. Select a single-angle tension member to resist the loads. Use LRFD. A) L 6 × 6 × B) L 5 × 3 × C) L5 × 3 X D) L 5 × 3 × 6 -Select- V What is the required gross area? (Express your answer to three significant figures.) Ag = in.2 What is the required effective area? (Express your answer to three significant figures.) Ae = in.2 What is the minimum radius of gyration? (Express your answer to three significant figures.) 1min = in. b. Select a single-angle tension member to…arrow_forwardIn the connection shown in the figure below, the bolts are 15/8-inch in diameter, and A36 steel is used for all components: Fy = 36 ksi, Fu = 58 ksi. 21/2" 11½½" 3". -3"- 料 11/2" 11/2" О О t = 3/8 31/2" О О t = 7/16 Consider both the tension member and the gusset plate and compute the following: a. the design block shear strength of the connection (Express your answer to three significant figures.) Rn = kips b. the allowable block shear strength of the connection (Express your answer to three significant figures.) Rn/= kipsarrow_forward
- 13/2/2025 Concrete Technology Q/ 1:1.5:3/0.62 concrete mix by weight, is to be used in cold weather, cement content = 5 Sack/m³ concreting when the temperature is assumed to be (32 °F), Calculate the temperature of the heated mixing water (in ˚C) to meet specification requirements in similar conditions.arrow_forwardThe tension member shown in the figure below must resist a service dead load of 60 kips and a service live load of 45 kips. Does the member have enough strength? The steel is A588: Fy = 50 ksi, Fu = 70 ksi; and the bolts are 11/8 inches in diameter. Assume that Ae = An. О О PL 3/8 X 71/2 a. Use LRFD. Determine the design strength and the factored load. Make a conclusion about the member. (Express your answers to three significant figures.) Φι Ρη = kips kips Pu = -Select- b. Use ASD. Determine the allowable strength and required strength. Make a conclusion about the member. (Express your answers to three significant figures.) Ft Ae = Pa = -Select- kips kipsarrow_forwardDraw neatly top side frontarrow_forward
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