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aleksandrvk [35]
3 years ago
5

You talk with the owner and he likes the idea of using two large glulam beams as shown to carry the joist loads. Design the glul

am beams using stress class 20F- 1.5E glulam material (see Table 5A in the wood supplement handout) and assume 5²-1/8" wide by 25²-1/2" deep glulam beam. Check bending and shear stresses, do not check deflection.
If this beam does not work, what are your options to make it work without changing the stress class? Select one of these options and try to make it work. If your second attempt still fails, show your work and move on to Problem 3 after you price your more expensive answer. Pricing for ghulam is about $4 per inch of depth per linear foot. For example: For a 25.5" deep beam, price will be $4*25.5" per linear foot = $102/linear foot of beam length. Price out your final answer for use in the last problem.
Engineering
1 answer:
bezimeni [28]3 years ago
7 0

Answer:

People also ask

What type of soil has equal parts of sand silt and clay?

What soil horizon contains mostly clay silt and sand?

E Horizon - This eluviation (leaching) layer is light in color; this layer is beneath the A Horizon and above the B Horizon. It is made up mostly of sand and silt, having lost most of its minerals and clay as water drips through the soil (in the process of eluviation).

You might be interested in
Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres
natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}

\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}

\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}

G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

h₃=2105KJ/Kg

h₄=957.14KJ/Kg

The net power developed:

P_{net}=P_{Tur}-P_{Comp}=G_{m}((h_{3}-h_{4})-(h_{2}-h_{1}))

The rate of heat:

Q=G_{m}(h_{3}-h_{2})

The thermal efficiency:

\mu_{ther}=\frac{P_{net}}{Q}

3 0
3 years ago
Challenge:
goldenfox [79]

Answer:

what?

Explanation:

8 0
3 years ago
9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show the sample space & n(s)
zzz [600]

Answer:

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

7 0
3 years ago
1 A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is expo
Alexeev081 [22]

Answer:

a) q' = 351.22 W/m

b) q'_total = 1845.56 W / m

c) q'_loss = 254.12 W/m

Explanation:

Given:-

- The diameter of the steam line, d = 100 mm

- The surface emissivity of steam line, ε = 0.8

- The temperature of the steam, Th = 150°C

- The ambient air temperature, T∞ = 20°C

Find:-

(a) Calculate the rate of heat loss per unit length for a calm day.

Solution:-

- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.

- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.

- The net heat loss per unit length due to radiation is given by:

                     q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]      

Where,

          σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)

          Ts: The absolute pipe surface temperature = 150 + 273 = 423 K

          T∞:The absolute ambient air temperature = 20 + 273 = 293 K

Therefore,

                    q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]    

                    q' = (1.4251*10^-8)* [ 24645536240 ]    

                    q' = 351.22 W / m   ... Answer

Find:-

(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.

Solution:-

- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.

- We will first evaluate the following properties of air at T∞ = 20°C = 293 K

                  Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s

                  Thermal conductivity ( k ) = 0.025596

                  Prandtl number ( Pr ) = 0.71559

- Determine the flow conditions by evaluating the Reynold's number:

                 Re = U∞*d / v

                      = ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )

                      = 52941.56574   ... ( Turbulent conditions )

- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):

           Nu_D = 0.3 + \frac{0.62*Re_D^\frac{1}{2}*Pr^\frac{1}{3}  }{[ 1 + (\frac{0.4}{Pr})^\frac{2}{3} ]^\frac{1}{4}  }*[ 1 + (\frac{Re_D}{282,000})^\frac{5}{8} ]^\frac{4}{5}    \\\\Nu_D = 0.3 + \frac{0.62*(52941.56574)^\frac{1}{2}*(0.71559)^\frac{1}{3}  }{[ 1 + (\frac{0.4}{0.71559})^\frac{2}{3} ]^\frac{1}{4}  }*[ 1 + (\frac{52941.56574}{282,000})^\frac{5}{8} ]^\frac{4}{5}  \\\\

           Nu_D = 0.3 + \frac{127.59828 }{ 1.13824  }*1.27251  = 142.95013

- The averaged heat transfer coefficient ( h ) for the flow of air would be:

            h = Nu_D*\frac{k}{d} \\\\h = 143*\frac{0.025596}{0.1} \\\\h = 36.58951 W/m^2K

- The heat loss per unit length due to convection heat transfer is given by:

           q'_convec = h*( π*d )* [ Ts - T∞ ]

           q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]

           q'_convec = 11.49493* 130

           q'_convec = 1494.3409 W / m

- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):

           q'_total = q_a + q_convec

           q'_total = 351.22 + 1494.34009

           q'_total = 1845.56 W / m  ... Answer

Find:-

For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)

Solution:-

- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.

- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.

- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":

             q'_t = 2*\pi*k \frac{T_h - T_o}{Ln ( \frac{r_2}{r_1} )}  

             q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]

             

Where,

             T_o = T∞ = 20°C

            T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C

             r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m

             r_1 = d/2 = 0.1 / 2 = 0.05 m

- The heat loss per unit length would be:

            q'_loss = q'_rad - q'_cond

- Compute the individual heat losses:

            q'_t = 2*\pi*0.08 \frac{150 - 85}{Ln ( \frac{0.07}{0.05} )}\\\\q'_t = 0.50265* \frac{65}{0.33647}\\\\q'_t = 97.10 W/m

Therefore,

             q'_loss = 351.22 - 97.10

            q'_loss = 254.12 W / m   .... Answer

- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.

6 0
3 years ago
How are engine bearings lubricated?
Musya8 [376]

Answer:

Engine bearings are lubricated by <u>motor oils</u> constantly supplied in sufficient amounts to the bearings surfaces. Lubricated friction is characterized by the presence of a thin film of the pressurized lubricant

Explanation:

7 0
3 years ago
Read 2 more answers
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