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Anuta_ua [19.1K]

4 years ago

6

You will only have two attempts to answer this question correctly. Assuming you determine the required section modulus of a wide

flange beam is 200 in3, determine the lightest beam possible that will satisfy this condition.

1 answer:

Setler79 [48]4 years ago

3 0

Answer:

W18 * 106

Explanation:

Section modulus of wide flange = 200 m^3

<u>Determine the value of the lightest beam possible </u>

The lightest beam possible that will satisfy the given condition will have a section modulus ≥ 200m^3 ( note: it will also be the nearest to 200 in^3 )

From Beam Table ; The Lightest beam with its section modulus( 204 in^3) > 200in^3 is W18 * 106

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The basic concept of feedback control is that an error must exist before some corrective action can be made?

weqwewe [10]

Answer:

The correct answer is True.

Explanation:

The feedback control system implies that to make a feedback, there must first be an error, otherwise there will be nothing to correct.

This system works so that there is an output that is controlled through a signal.

This signal will be feedback and it will signal an error which will be detected by a controller that will allow entry into the system.

In basic words, this system processes signals, samples them in the form of an output, and re-enters them feedback to detect the error signal.

7 0

3 years ago

Calculate total hole mobility if the hole mobility due to lattice scattering is 50 cm2 /Vsec and the hole mobility due to ionize

Ad libitum [116K]

Answer:

The total hole mobility is 41.67 cm²/V s

Explanation:

Data given by the exercise:

hole mobility due to lattice scattering = ul = 50 cm²/V s

hole mobility due to ionized impurity = ui = 250 cm²/V s

The total mobility is equal:

$\frac{1}{u} =\frac{1}{ul} +\frac{1}{ui} \\\frac{1}{u}=\frac{1}{50} +\frac{1}{250} \\u=41.67cm^{2} /Vs$

5 0

3 years ago

Read 2 more answers

A soil weighs 2,520 lbs/CY in its in situ condition, 1,970 lb/CY in its loose condition after excavation, and 3,025 lbs/CY in an

azamat

Answer:

load factor = 0.782

Shrink Factor = 0.833

no of truck is 62500

Explanation:

given data

soil weighs in situ condition = 2,520 lbs/CY

soil weighs in loose condition = 1,970 lb/CY

soil weighs in embanked state = 3,025 lbs/CY

average volume = 16 LCY

soil from a borrow pit = 1 million CCY

solution

first we get here Load Factor that is express as

load factor = $\frac{1,970}{2550}$

load factor = 0.782

and Shrink Factor will be as

Shrink Factor = $\frac{2520}{3025}$

Shrink Factor = 0.833

and

no of truck will be

no of truck = $\frac{1000000}{16}$

no of truck is 62500

6 0

4 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago

The principal component in glass manufacturing is_____

Oksanka [162]

Answer:

Silica is the principal component in glass.

Explanation:

Step1

Glass is the amorphous solid and transparent. Glass products have many of the shapes and design that are available in market.

Step2

Natural quartz is the primary source of glass in sand. Silica is the principal component in approximately all glass. Lime stone, soda ash and aluminum oxide are added in the galas for desired properties depending upon application. So, silica is the principal component in glass.

5 0

4 years ago