Answer:
The annualized cost is:
$299,272.
Explanation:
a) Data and Calculations:
Year 0 Capital cost of road construction = $6 million
Year 10 Major improvements cost = $17 million
Year 25 Replacement and upgrade cost = $29 million
Year 40 Federal government one-time tax credit = $12 million
Period of project analysis = 50 years
Cost of capital (discount rate) = 10%
Annualized cost at present value costs:
Amount spent Discount Factor Present value
$6 million 1 $6,000,000
$17 million 0.386 6,562,000
$29 million 0.092 2,668,000
($12 million) 0.0222 (266,400)
Total cost $14,963,600
Annualized cost = $14,963,600/50 = $299,272
b) The annualized cost for the road construction project, which is the annualized value of the net present costs of $14,963,600, is divided by 50. Before obtaining the net present costs, the cash outflows, including the tax credit, are discounted to their present values, using the discount rate of 10%. And then, the average of the cost is obtained by dividing the total net present cost into 50 years.
Which tool is used for cutting bricks and other masonry materials with precision?
Answer:
A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.
Explanation:
Answer:
Masonry Saw
Explanation:
Masonry Saws can be used to cut brick, ceramic, tile and or stone.
A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maximum shear stress is 450 N/cm2 when the maximum bending stress is 1500 N/cm2. Calculate the value of the centrally applied point load W and the span L. The overall height of the I section is 29 cm.
Answer:
W = 11,416.6879 N
L ≈ 64.417 cm
Explanation:
The maximum shear stress, [tex]\tau_{max}[/tex], is given by the following formula;
[tex]\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )[/tex]
[tex]t_w[/tex] = 1 cm = 0.01
h = 29 cm = 0.29 m
[tex]h_w[/tex] = 25 cm = 0.25 m
b = 15 cm = 0.15 m
[tex]I_c[/tex] = The centroidal moment of inertia
[tex]I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )[/tex]
[tex]I_c[/tex] = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴
Substituting the known values gives;
[tex]I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}[/tex]
[tex]I_c[/tex] = 1.2257083[tex]\bar 3[/tex] × 10⁻⁴ m⁴
From which we have;
[tex]4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )[/tex]
Which gives;
W = 11,416.6879 N
[tex]\sigma _{b.max} = \dfrac{M_c}{I_c}[/tex]
[tex]\sigma _{b.max}[/tex] = 1500 N/cm² = 15,000,000 N/m²
[tex]M_c[/tex] = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²
From Which we have;
[tex]M_{max} = \dfrac{W \cdot L}{4}[/tex]
[tex]L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417[/tex]
L ≈ 0.64417 m ≈ 64.417 cm.
The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect the total quantity of bricks used in the wall
True or False
Answer:
false
Explanation: