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Ostrovityanka [42]

3 years ago

8

A. For a 200g load acting vertically downwards at point B’, determine the axial load in members A’B’, B’C’, B’D’, C’D’ and C’E’.

1 answer:

Nonamiya [84]3 years ago

7 0

Answer:

attached below

Explanation:

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Wind blows on the side of a fully enclosed hospital located on open flat terrain where V= 120 mi/h. Determine the external press

ss7ja [257]

Answer:

The external pressure is p = -21.9 psf or p = -8.85 psf

Explanation:

Given :

Velocity of wind, v = 120 mi / hr

$k_d = k_c =1$ (wind direction factor)

$k _{zt} = 1$ = topographical factor (for flat terrain)

$q_n$ = velocity pressure at height h

$\therefore q_n = 0.00256 k_z k_{zt} k_d v^2$

$q_n = 0.00256 \times k_z (1)(1)(120)^2$

$= 36.86 k_z$

But for height h = 30 ft, $k_z$ = 0.98 (from table)

$\therefore q_n = 36.86 \times 0.98$

= 36.16

Now, $\frac{L}{B}= \frac{200}{200} =1$ , so $C_p=-0.5$ (from table)

$p = q(G)(C_p)-q_n(GC_{pi})$

where, p = external pressure

G = 0.85 = gust factor (for typical rigid building)

$GC_{pi} = \pm 0.18$ (internal pressure co efficient)

Therefore putting the values,

$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$

p = -21.9 psf or p = -8.85 psf

4 0

4 years ago

A flux used for welding, brazing, or soldering prevents the formation of, dissolves, or helps remove?

OverLord2011 [107]

Answer:

Helps to remove Oxides formed during brazing.

Explanation:

Flux is needed to dissolve and remove oxides that may form during brazing. Prevent or inhibit the formation of oxide during the brazing process

8 0

2 years ago

a piping system has an internal air pressure of 1,500 kpa. In addition to being subject to the air pressure, the piping supports

Alik [6]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

5 0

4 years ago

The outer surface of an steel alloy is to be hardened by increasing its carbon content. The carbon is to be supplied from an ext

hoa [83]

Answer:

Diffusion time needed for 610 degree celcius is 81.91 min

Explanation:

Given data:

Diffusion heat temperature = 826 + 273 = 1099 K

Diffusion time t_1 = 10 min

carbon concentration = 0.83%

$D_O = 8.3\times 10^{-8} m^2/s$

$Q_d = 78.5 kJ/kg$

$T_2 = 610+273 = 883 K$

From Fick's formula we hvae

$\frac{ C_x - C_o}{C_z -C_o} = 1 -erf(\frac{x}{2\sqrt{Dt}}$

at specific concentration

$\frac{ C_x - C_o}{C_z -C_o} = constant$

then

$(\frac{x}{2\sqrt{Dt}} = constant$

$(\frac{x^2}{{Dt}} = constant$

Dt = constant

$D_1 t_1 = D_2 t_2$

$t_2 = \frac{D_1 t_1}{D_2}$

$D_1 at T_1$

$D_1 = D_o exp(\frac{Q_d}{R t_1})$

$= 8.3\times 10^{-8} exp(\frac{78.5\times 10^3}{8.314 \times 1099})$

$= 1.54\times 10^{-11} m^2/s$

similarly for D_2 for T_2

$D_2 = 1.88\times 10^{-12} m^2/s$

$t_2 = \frac{D_1 t_1}{D_2}$

$= \frac{1.54\times 10^{-11} \times 10}{1.88\times 10^{-11}}$

$t_2 = 81.91 min$

Diffusion time needed for 610 degree celcius is 81.91 min

5 0

3 years ago

The clock period of a ripple counter must be longer than the total propagation

iogann1982 [59]

Answer:

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5 0

3 years ago