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2 years ago

Select three types of lines that engineers use to help represent the shape of a design in a sketch.

Engineering

1 answer:

Vikki [24]2 years ago

7 0

Hidden lines

Used to describe the in shown lines (like diagonals inside cubes)

Extension lines:-

Used to explain the expansion of structures like building

Object lines

Used to describe the structure of objects and the lining to show borders

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The locations and type of electrical device required on an architectural plan are referred to as

Vikki [24]

The locations and type of electrical devices required on an architectural plan are referred to as electrical symbols. The use of detailed electrical plan ensures that all electrical equipment and wiring is installed exactly as intended. If the electrical plans are insufficient or hazy, the installation is left to the electrician’s discretion.

Electrical plans are prepared using a CAD layered floor plan. Designers should not rely on electricians to design the electrical system, only to install it. Conversely, designers do not plan the position of every wire, only the position and relationship of all fixtures, devices, switches, and controls. This is done with the use of electrical symbols.

Hundreds of electrical symbols are used on floor plans to describe what and where electrical elements will be installed. On simple plans, electrical symbols are often included as a separate layer on the floor plan. For larger or more complex structures, a separate plan is prepared.

The locations and type of electrical device required on an architectural plan are referred to as electrical symbols.

Learn more:

https://brainly.in/question/40648976

5 0

2 years ago

Read 2 more answers

A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when

lianna [129]

Answer:

1. $3.4^{o}$

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

= (50)(1000/3600) m/s

= 13.89 m/s

The acceleration due to gravity, $g = 9.81 m/s^{2}$

The frictional force, f = μsN ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

ΣF = mg sinθ-f = 0 ...... (2)

Substitute the equation (1) in equation (2), we get

ΣF = mgsinθ-μsN = 0

mgsinθ-μsmgcosθ =0

μs = sinθ/cosθ

tanθ = μs

θ = tan-1( μs) = tan-1(0.06) = $3.4^{o}$

(b)The vertical component of the force is

N cosθ = fsinθ+mg

N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma (Centripetal acceleration, $a = \frac {v^{2}}{r}$

Nsinθ+fcosθ = $m(\frac {v^{2}}{r})$

Nsinθ+μsNcosθ = $m(\frac {v^{2}}{r})$

N[sinθ+μs cosθ] = $m(\frac {v^{2}}{r})$ ...... (4)

Dividing the equation (4) with equation (3),

[sinθ+μscosθ]/[cosθ- μs sinθ] = $\frac {v^{2}}{rg}$

cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] = $\frac {v^{2}}{rg}$

[tanθ+μs]/[1-μs tanθ] = $\frac {v^{2}}{rg}$

From part (1), tanθ = μs

Then the above equation becomes

$\frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

$\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

Therefore, the minimum radius of the curvature of the curve is

$r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g}$

= $\frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}$

= $\frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}$

= 163.3 m

5 0

3 years ago

A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in

Ierofanga [76]

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, $\sigma=\frac {F}{A}$

where F is applied force and A is original area

Also, engineering strain, $\epsilon=\frac {\triangle l}{l}$ where l is original area and $\triangle l$ is elongation

We also know that Hooke's law states that $E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}$

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

$\triangle l= 0.1 mm= 0.1\times 10^{-3} m$

By substitution we obtain

$E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa$

5 0

3 years ago

What are the characteristic features of stress corrosion cracks?

Oduvanchick [21]

Answer and Explanation:

The crack formation growth that takes place in an environment corrosive.

Stress corrosion cracks can be defined as the spontaneous failures of the metal alloy as a result of the combined action of corrosion and high tensile stress.

Some of the characteristic features of stress corrosion cracks are:

These occur at high temperatures.
Occurrence of failures in metals mechanically.
Occurrence of sudden and unexpected failures under tensile stress.
The rate of work hardening of the metal alloy is high.
Time
An environment that is specific for stress corrosion cracking.

5 0

3 years ago

Developed an automated program in any language which take 12 dependent variable and corresponding independent variables and show

otez555 [7]

Answer:

Explanation:

var generator = new Random(1);

// Now the nextGaussian() function returns a normal distribution of random numbers with the following parameters: a mean of zero and a standard deviation of one

var draw = function() {

var num = generator.nextGaussian();

var standardDeviation = 60;

var mean = 2003;

// Multiply by the standard deviation and add the mean.

var x = standardDeviation * num + mean;

noStroke();

fill(214, 159, 214, 10);

ellipse(x, 200, 16, 16); };

Hope this will be helpful

3 0

2 years ago