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velikii [3]
3 years ago
10

4. In the Hyatt Regency walkway case study, it is reported that Jack Gillum stamps the 42 shop drawings, including the revised S

hop Drawing 30 and Erection Drawing E-3 denoting the revised hanger connection detail, with his engineering review seal, authorizing construction. When stamping these drawings, what is Jack Gillum accepting? A one-word response is sufficient.
Engineering
1 answer:
mario62 [17]3 years ago
3 0

Answer:

Responsibility

Explanation:

By stamping the drawings that he was looking over, Jack Gillum conveys the fact that he is accepting responsibility for this work. The purpose of Gillum's stamp is to explain that such work has been under engineering review, and that it has fulfilled all the requirements that he watches our for. By putting his stamp in this work, Gillum accepts responsibility in case an error or a discrepancy is found in the drawings.

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The shear flow at the point depends on the value of Q for the portion of the upper flange to the right of the point. Calculate t
Ostrovityanka [42]

Q)The shear flow at a point is given by q = 1 of 2 An I-beam has a flange width b = 200 mm , height h = 200 mm , web thickness tw = 8 mm, and flange thickness t = 12 mm, For the given point The shear flow at the point depends on the value of Q for the portion of the upper flange to the right of the point. Calculate the value of Q. Express your answer with appropriate units to three significant figures.( <u>I have also attached the diagram for better understanding)</u>

Answer:

Q = 56400 mm^3

Explanation:

<u>I have given the explanation in the attached file below.</u>

6 0
3 years ago
Which is a drawback of solar energy
alexira [117]

Answer:

Drawbacks include that it is costlier than other clean energies. Another thing is that when it isnt sunny you wont get as much energy

8 0
3 years ago
A rear wheel drive car of mass 1000 kg is accelerating with a constant acceleration without slipping from 0 to 60 m/s in 1 min.
tester [92]

Answer:

500 N

Explanation:

Given;

Mass of the car, M = 1000 kg

initial speed of the car, u = 0 m/s

Final speed of the car, v = 60 m/s

Time, t = 1 min = 60 s

Now,

Force, F is given as:

F = Ma

where,

a is the acceleration

From the Newton's equation of motion, we have

v = u + at

on substituting the values, we get

60 = 0 + a × 60

or

a = 1 m/s²

Thus,

Force = 1000 × 1 = 1000 N

now,

this force will be equal to the friction force provided by the rear wheels

let the friction force on a single rear wheel be 'f'

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5 0
4 years ago
1. If a bolt is size 1/2" or larger, then its corresponding wrench size should be____ larger than the bolt size
Mrac [35]

Answer:

  1. C. 1/4"
  2. B. 3/16"

Explanation:

1. For hex bolts, lag bolts, and square bolts, the wrench size is 1/4" larger than the bolt size for 1/2" and 9/16" bolts. For 5/8" bolts and larger, the wrench size is <em>50% larger than the bolt size</em>.

__

2. For 7/16" bolts, the wrench size is 5/8", so is 3/16" larger than the bolt. This holds down to 1/4" bolts, where the wrench size may be 3/8" or 7/16".

3 0
4 years ago
In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri
ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

Here

  • \sigma_x is the normal stress which is 1750 psf
  • \sigma_y is 0
  • \tau_{xy} is the shear stress which is 800 psf

So the formula becomes

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}

Similarly, the minimum normal stress is given as

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}

The maximum shear stress is given as

\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}

5 0
3 years ago
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