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Romashka [77]
3 years ago
6

A W18 X 119 is used as a compression member with one end fixed and the other end pinned. The length is 12 feet. What is the desi

gn compressive strength of A992 steel is used
Engineering
1 answer:
laiz [17]3 years ago
3 0

Answer:

156.0 ksi

Explanation:

The formula of compressive strength is  CS = F / A

where F is the force or load while A is the cross-sectional area

Given the inertia Ag = 35.1 in^4 , which is less than 40, the steel has a short column.

E = 29000 ksi

L = 12ft

r = 2.69 in

therefore;

CS = π^2 E / (kL^2/r)^2

   = π^2 29000 / (0.8 x 12^2 / 2.69)^2

  = 286292.756 / 1834.4089 = 156 ksi

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This came off my car earlier today. What is it and what does it do?
patriot [66]

Answer:

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Explanation:

that's just whut I wold do.

6 0
3 years ago
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The speed of a vehicle is reduced with a constant acceleration from 72km/h to 18
Radda [10]

Answer:

The correct answer will be "1477.84 N".

Explanation:

Given that,

Mass,

m = 1.6 mg

or,

   = 1600 kg

Initial velocity,

u = 72 km/h

  = 72\times \frac{5}{18} \ m/s

  = 20 \ m/s

Final velocity,

v = 18 km/h

  = 18\times \frac{5}{18}

  = 5 \ m/s

Covered distance,

s = 250 m

By using the below relation, we get

⇒  v^2=u^2+2as

On putting the values, we get

⇒  (5)^2=(20)^2+2\times a\times 250

⇒      a=-0.75 \ m/s^2 (shows the deceleration)

Slope will be given as 1 in 25, then

⇒  Sin \theta=\frac{1}{25}

           \theta=2.3^{\circ}

hence,

As we know,

⇒  \Sigma F=ma

or,

⇒  Braking \ force+350-mgSin\theta=ma

⇒  Braking \ force=ma+mgSin\theta-350

On substituting all the values, we get

⇒                           =1600(0.75+1600\times 9.81 Sin(2.3^{\circ})-350

⇒                           =1477.84 \ N

7 0
3 years ago
Which one of these is NOT considered a skill?
Ugo [173]

Answer:

number 2 people skills

Explanation:

i think people skills does not make cense because you can not control what people think

4 0
4 years ago
Read 2 more answers
A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500
lukranit [14]

Answer:

The resulting strain is 1.39\times 10^{-3}.

Explanation:

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm

Force, F = 44,500 N

Th elastic modulus of Cu to be 110 GPa

The resulting strain is given by the formula as follows :

\epsilon=\dfrac{F}{AE}

E is elastic modulus of Cu is are of cross section

\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}

So, the resulting strain is 1.39\times 10^{-3}.

6 0
4 years ago
Investigating how slime molds reproduce is an example of applied research.<br> True<br> False
azamat

Answer:

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Explanation:

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