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3 years ago

8

Consider a W21x93. Determine the moment capacity of the beam. Assume the compression flange is not laterally braced and that the

unbraced length is 15 feet. Assume Cb

1 answer:

OLga [1]3 years ago

8 0

Answer:

The answer is "828.75"

Explanation:

Please find the correct question:

For W21x93 BEAM,

$Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3$

For A992 STREL,

$F_y= 50\ ks$

Check for complete section:

$\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}$

Design the strength of beam = $\phi_b Z_x F_y\\\\$

$=0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\$

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3 years ago

Differentiate between isohyetal method and arithmetical average method of rainfall

prohojiy [21]

Answer:

Explained below

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3 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

The mathematical relation between stress to strain on the plastic region of deformation is

$\sigma _T =K\epsilon^n_T$

Where K is a constant

n is known as the strain hardening exponent

This constant K can be obtained as follows

$K = \frac{\sigma_T}{(\epsilon_T)^n}$

No substituting $345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n$ from the question we have

$K = \frac{345}{(0.02)^{0.22}}$

$= 815.82MPa$

Making $\epsilon_T$ the subject from the equation above

$\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }$

$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

$\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

$=0.0443$

From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

$l_i = l_o e^{\epsilon_T}$

Where

$l_i$ is the instantaneous length

$l_o$ is the original length

$Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T$

$l_i = 470 * e^{0.0443}$

$=491.28mm$

We can also obtain the elongated length mathematically as follows

$Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

$Elongated \ Length = 491.28 - 470$

$=21.29mm$

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3 years ago

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Luba_88 [7]

Answer:

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Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

(4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

__

<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)

These will be equal when ...

y = y

12 -0.5x = 8 -0.25x

4 = 0.25x . . . . . . . . . . add 0.5x -8

16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

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2 years ago