All categories

3 years ago

Six forces act on a beam that forms part of a building's

Engineering

1 answer:

IrinaVladis [17]3 years ago

7 0

Answer:

Explanation:

write each force in terms of magnitude and directions

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let

FA = FA sin (110) + FA cos (110)

FB = 20 sin (270) + 20 cos (270)

FC = 16 sin (140) + 16 cos (140)

FD = 9 sin (40) + 9 cos (40)

FE = 20 sin (270) + 20 cos (270)

FG = FG sin (50) + FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110) + 0 + 16 sin (140) + 9 sin (40) + 0 + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110) + 0 + 10.285 + 5.785 + 0 + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110) + 16.070 + FG sin (50) = 0

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

You might be interested in

A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis

kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

$V_1=0.3m^3,T_1=298 K$

$V_2=0.1m^3,T_1=1273 K$

Mass of air=1 kg

For polytropic process $pv^n=C$ ,n is the polytropic constant.

$Tv^{n-1}=C$

$T_1v^{n-1}_1=T_2v^{n-1}_2$

$298\times .3^{n-1}_1=1273\times .1^{n-1}_2$

n=2.32

Work in polytropic process given as

w= $\dfrac{P_1V_1-P_2V_2}{n-1}$

w= $mR\dfrac{T_1-T_2}{n-1}$

Now by putting the values

w= $1\times 0.287\dfrac{289-1273}{2.32-1}$

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

5 0

3 years ago

The following is a list of metals and alloys:

viktelen [127]

Answer:

A) Gray cast iron

B) Aluminum

C) Titanium alloy

D) Tool steel

E) Titanium alloy

F) magnesium

G) Tungsten

5 0

3 years ago

Jnjn freeeeeeeeeeeeeeeeeeeeeeeeeeeeeee pointtttttttttt

kvasek [131]

Answer:

thx

Explanation:

5 0

3 years ago

Read 2 more answers

1. The construction process begins with which of the following stages?

Firdavs [7]

Answer:

c) site preparation

Explanation:

A construction process can be defined as a series of important physical events (processes) that must be accomplished during the execution of a construction project.

Generally, in the construction of any physical asset such as offices, hospitals, schools, stadiums etc, the first step of the construction process is site preparation. Site preparation refers to processes such as clearing, blasting, levelling, landfilling, surveying, cutting, excavating and demolition of all unwanted objects on a piece of land, so as to make it ready for use.

This ultimately implies that, site preparation should be the first task to be accomplished in the construction process.

Hence, the construction process typically begins with site preparation before other activities such as the laying of foundation can be done.

Additionally, construction costs can be defined as the overall costs associated with the development of a built asset, project or property. The construction costs is classified into two (2) main categories and these are; capital and operational costs.

7 0

3 years ago

A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The

Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

s1 = s-P1 = 0.6492 KJ/kg.K

v1 = v-P1 = 0.001010 m^3 / kg

P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

$w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}$

- From the following relation we can determine ( h2 ) as follows:

h2 = h1 + wp

h2 = 191.81 + 8.0699

h2 = 199.88 KJ/kg

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

P3 = 8 MPa

T3 = ? ( assume fluid exist in the saturated vapor phase )

h3 = hg-P3 = 2758.7 KJ/kg

s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

$q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}$

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

P4 = 10 KPa

s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

- Compute the quality of the mixture at condenser inlet by the following relation:

$x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947$

- Determine the isentropic ( h4s ) at this state as follows:

$h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}$

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

$h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\$

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

$w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}$

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

$W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}$

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

$n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31$

Answer: The thermal efficiency of the cycle is 0.31

7 0

3 years ago