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3 years ago

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(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced

with tension steel As=5 in2 . The beam supports a uniformly distributed dead load (including its own weight) DL=2.2 kips/ft and a uniformly distributed live load LL=1.8 kips/ft. The properties of the materials are as follows: f’c=4000 psi, steel fy=60,000 psi. Calculate the long-term deflections in the beam after five years.

1 answer:

Nat2105 [25]3 years ago

5 0

Answer:

Zx = 176In³

Explanation:

See attached image file

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An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [

crimeas [40]

Answer:

Modulus of resilience will be $3216942.308j/m^3$

Explanation:

We have given yield strength $\sigma _y=818MPa$

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience $=\frac{\sigma _y^2}{2E}$ , here $\sigma _y$ is yield strength and E is elastic modulus

Modulus of resilience $=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3$

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3 years ago

Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?

zhenek [66]

Answer:

it is indeed C

Explanation:

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3 years ago

Read 2 more answers

Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25

Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

$y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x$

$y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}$

$dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx$

$dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)$

$dy = \frac{1}{2\pi}$

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

#SPJ1

4 0

2 years ago

A round bar of chromium steel, (ρ= 7833 kg/m, k =48.9 W/m-K, c =0.115 KJ/kg-K, α=3.91 ×10^-6 m^2/s) emerges from a heat treatmen

Lerok [7]

Answer:

Q = 424523.22 kw

Explanation:

$\rho =7833 kg/m$

k = 48.9 W/m - K

c = 0.115 KJ/kg- K

$\alpha = 3.91*10^{-6} m^2/s$

$T_s = 285 degree celcius$

T_∞ = 35 degree celcius

velocity of air stream = 15 m/s

D = 40 cm

L = 200 cm

mass flow rate $\dot m = \rho AV = 7833 *\frac{\pi}{4} 0.4^2*15$

$\dot m = 14764.85 kg/s$

$A_s = \pi DL = \pi 0.4*2 = 2.513 m^2$

$Q = \dot m C \Delta T = h A_s \Delta T$

$\dot m C \Delta T = h A_s \Delta T$

solving for h

$h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}$

h = 675.6 kw/m^2K

$Q = h A_s\Delta T$

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

7 0

3 years ago

Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.

Novosadov [1.4K]

Answer:

i) SF: $v(x) = \frac{(w_0* x )^2}{2L}$

ii) BM : $= \frac{(w_0*x)^3}{6L}$

Explanation:

Let's take,

$\frac{y}{w_0} = \frac{x}{L}$

Making y the subject of formula, we have :

$y = \frac{x}{L} * w_0$

For shear force (SF), we have:

This is the area of the diagram.

$v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0$

$= \frac{(w_0* x )^2}{2L}$

The shear force equation =

$v(x) = \frac{(w_0* x )^2}{2L}$

For bending moment (BM):

$BM = v(x) * \frac{x}{3}$

$= \frac{(w_0* x )^2}{2L} * \frac{x}{3}$

$= \frac{(w_0*x)^3}{6L}$

The bending moment equation =

$= \frac{(w_0*x)^3}{6L}$

5 0

4 years ago