The 150 mm thick wall of a gas fired furnace is constructed of fireclay brick (k=1.5 W/m.K) , tho=2600 kg/m3, and cp=1000 J/kg.K
1 answer:
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Answer:
w = 10.437 kips
deflection at 1/4 span 20.83\E ft
at mid span = 1.23\E ft
shear stress 7.3629 psi
Explanation:
area of cross section = 18*76
length of span = 32 ft
moment = 334 kips-ft
we know that
moment = load *eccentricity
334 = w * 32
w = 10.437 kips
deflection at 1/4 span
= 20.83\E ft
at mid span
shear stress
Answer:
Weight of cement = 10968 lb
Weight of sand = 18105.9 lb
Weight of gravel = 28203.55 lb
Weight of water = 5484 lb
Explanation:
Given:
Entrained air = 7.5%
Length, L = 40 ft
Width,w = 12 ft
thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft
Specific gravity of sand = 2.60
Specific gravity of gravel = 2.70
The volume will be:
40 * 12 * 0.5 = 240 ft³
We need to find the dry volume of concrete.
Dry volume = wet volume * 1.54 (concrete)
Dry volume will be = 240 * 1.54 = 360ft³
Due to the 7% entarained air content, the required volume will be:
V = 360 * (1 - 0.07)
V = 334.8 ft³
At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:
Total of ratio = 1+2+3 = 6
Their respective volume will be =
Volume of cement =
Volume of sand =
Volume of gravel =
To find the pounds needed the driveway, we have:
Weight = volume *specific gravity * density of water
Specific gravity of cement = 3.15
Weight of cement =
55.8 * 3.15 * 62.4 = 10968 pounds
Weight of sand =
111.6 * 2.60 * 62.4 = 18105.9 lb
Weight of gravel =
167.4 * 2.7 * 62.4 = 28203.55 lb
Given water to cement ratio of 0.50
Weight of water = 0.5 of weight of cement
= 1/2 * 10968 = 5484 lb
Answer:
Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22
Explanation:
Assuming a = 2b
Attached below is the required steps to the solution
The cutoff frequencies for the first ten modes of a rectangular waveguide listed in ascending order is :
Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22
