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jek_recluse [69]

3 years ago

15

A truss is subjected to three loads. The truss is supported by a roller at A and by a pin joint at B. What is most nearly the re

action force at B

1 answer:

slava [35]3 years ago

8 0

Answer:

Hello the diagram related to the question is missing attached is the diagram

Answer : 3833.33 KN

Explanation:

The most nearly reaction force at B

= ∑ Mb = 0 = 21Ay

= (2000 * 17.5 ) + ( 3000 * 10.5 ) + ( 4000 * 3.5 )

= 35000 + 31500 + 14000 = 80500

therefore Ay = 80500 / 21 = 3833.33 KN

x = 7/2 = 3.5m

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Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at

andriy [413]

Answer:

A) $v_2 = 2016.80 ft/s$

B) $\Delta s = 0.006 Btu/lbm R$

Explanation:

Given data:

P-1 = 100 lbf/in^2

$T_1 = 500$ degree f

$V_1 = 100 ft/s$

$P_2 = 40 lbf/inc^2$

effeciency = 80%

from steady flow enerfy equation

$h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}$

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

$H_1 = 1278.8 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

for P1 = 40 lbf/in^2

$H_1 = 1193.5 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

exit enthalapy h_2

$\eta = \frac{h_1 - h'_2}{ h_1 - h_2}$

$0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm$

from above equation

$1278.8 \times 25037 + \frac{100^2}{2} = 1197.77 \times 25037 + \frac{v_2^2}{2}$ [1 Btu/lbm = 25037 ft^2/s^2]

$v_2 = 2016.80 ft/s$

b) amount of entropy

$\Delta s = s_2 - s_1$

$s_1 = 1.708 Btu/lbm -R$

at $h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2$

$s_2 is 1.714 Btu/lbm -R$

$\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R$

6 0

3 years ago

Briefly explain what are the following AIDC technologies for industry automation:

aliya0001 [1]

I believe a is the answer enjoy

3 0

3 years ago

A 0.3 m3 rigid tank initially contains refrigerant-134a at 14°C. At this state, 55% of the mass is in the vapor phase, and the r

Oksanka [162]

Answer:

I am a girl want private photos

Explanation:

8 0

3 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago

What is the present value of the future receipts of $2,000, 5 years from now at 10% compounded annually?

Elden [556K]

Answer:

P = $ 766.28

Explanation:

present value = ?

Future value = $ 2000

time = 5 years

compounded annually at the rate of = 10 %

$A = P + P(1+\dfrac{r}{100})^t$

$2000 = P + P(1+\dfrac{10}{100})^5$

2000 = P + 1.61 P

2.61 P = 2000

P = $ 766.28

hence, the present value of amount invested to get the future value of $2000 is equal to P = $ 766.28

3 0

3 years ago