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Citrus2011 [14]
3 years ago
15

A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. D

etermine the minimum required wire diameter assuming a factor of safety 5 and that the yield strength of the steel is 900MPa.
Engineering
1 answer:
zhuklara [117]3 years ago
7 0

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

d = \sqrt{4*P / (\pi*\sigma)}

d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm

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For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about ____
7nadin3 [17]

Answer:

One

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___

<h3>what is supported mounted?</h3>
  • A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.

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4 0
2 years ago
Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio
SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

a)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

b)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

c)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0
4 years ago
When buttons or switches are pressed by humans for arbitrary periods of time, we need to convert a signal level to a pulse. In t
ddd [48]

Answer:

The FSM uses the states along with the generation at the P output on each of the positive edges of the CLK. The memory stores the previous state in the machine and the decoder generates a P output based on the previous state.

Explanation:

The code is in the image.

6 0
3 years ago
1. A thin plate of a ceramic material with E = 225 GPa is loaded in tension, developing a stress of 450 MPa. Is the specimen lik
mina [271]

Answer:

fracture will occur as the value is less than E/10 (= 22.5)

Explanation:

If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}

=  2\times 750 (\frac{\frac{0.2mm}{2}}{0.001 mm}})^{1/2}

                 = 15 GPa

fracture will occur as the value is less than E/10 = 22.5

7 0
3 years ago
A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is th
tigry1 [53]

Answer:

Q = 65.388 KJ

Explanation:

To calculate the heat required for the given process Q, we recall the energy balance equation.

Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)

We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.

Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ  = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg

Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)

                   u₁ = 0.001043m³/kg + 0.123 . ( 1.6720m³/kg - 0.001043m³/kg)

                   u₁ = 0.2066m³/kg

Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.

Therefore, V = 14L . 1m² / 1000L = 0.014 m³

Hence, m = V / u₁

                 0.014m³ / 0.2066 m³/kg

              m = 0. 0677 kg

Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.

Therefore, u₁ = vₐ + x₁ . vₐₙ

                   u₁ = 419.06 kj / kg + 0.123  .  2087.0 kj/kg

                    u₁ = 675.76 kj/kg

For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ =  0.39248m³/kg

Hence, x₂ = u₂ - vₙ / uₐ

x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg

x₂ = 0.524

Moving forward to calculate the final internal energy u₂, we have :

u₂ = vₙ + x₂ . vₙₐ

u₂ = 631.66 kj/kg + 0.524  . 1927.4 kj/kg

u₂ = 1641.62 kj/kg

We now return to equation (1) to plug in the values generated thus far

Q = m (u₂ - u₁)

0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)

Q = 65.388KJ

7 0
3 years ago
Read 2 more answers
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