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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:

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If you are setting up a race car. What is the cross weight? Does it matter?
lara31 [8.8K]

Answer:

cross-weight is used to tighten it up.

Explanation:

and yes this is important because Cross-weight percentage compares the diagonal weight totals to the car's total weight.

hope  this help

(mark this answer as an brainliest answer)

7 0
4 years ago
A steam turbine receives steam at 1.5MPa and 220oC, and exhausts at 50kPa, 0.75 dry. Neglecting heat losses and changes in kinet
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4 0
3 years ago
A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl
bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1.  Steady operating conditions exist.

2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

PP_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The  isentropic  efficiency, n_{T} = \frac{h_{3}-h_{4}  }{h_{3}- h_{4s} }

==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

b. Pump

h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

7 0
4 years ago
Read 2 more answers
A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f
KIM [24]

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) \frac{dm}{2}     ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = \frac{L}{\pi dm}      ...............2

here lead L = n × p

so tan(α) = \frac{2\times 2}{\pi 19}

α = 3.83°  

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and  now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) \frac{19\times 10^{-3}}{2}  

T = 1.59 Nm

so

power = \frac{2\pi N \ T }{60}     .................3

put here value

power = \frac{2\pi \times 300\times 1.59}{60}

power = 49.95 W

and

as φ > α

so it is self locking screw

 

8 0
3 years ago
**Please Help. ASAP**
natima [27]

Answer:

The answer is below

Explanation:

1)

\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at

2)

\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)

3)

x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}

4)

x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}

5)

x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}

6)

E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\  \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}

7)

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\  \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}

8)

ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\

4 0
3 years ago
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