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3 years ago

A pin-supported structure has unrestrained rotations at the support locations.a) True b) False

Engineering

1 answer:

antoniya [11.8K]3 years ago

6 0

Answer:

a)True

Explanation:

Yes it is true a pin support can not resist the rotation motion . It can resist only lateral or we can say that only linear motion of structure and can not resit angular moment of motion about hinge or pin joint.On the other hand a fixed support can resist linear as well rotation motion of structure.

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An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$

From the question the, the mass flow rate at the inlet $m_{1}}$ is 2Kg/s $\\$

Since 5% flow is delivered into the feedwater heating, $\\$

$m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s$ $\\$

Also for the outlet 3 the remaining 95% will flow out. Hence

$m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s$ $\\$

Now, from $m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$ we substitute values

$W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}$

$W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)$

$\\$

$W_{out} = 2285.19 kW$ .

Hence the power produced is 2285kW

7 0

3 years ago

Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of th

sleet_krkn [62]

Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, $\sigma$ is given by

$\sigma=\frac {F}{A}$ where F is force and A is area and area is given by $\frac {\pi d^{2}}{4}$ hence

$\sigma=\frac {4F}{\pi d^{2}}$ where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

$\sigma=\frac {4*9970 N}{\pi 0.01m^{2}}=126941982.6 N/m^{2}$

$\sigma \approx 127 Mpa$

From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

$\epsilon=\frac {\triangle l}{l}$ where $\epsilon$ is the strain, $\triangle l$ is elongation and l is original length and making elongation the subject

$\triangle l= \epsilon \times l$ and substituting strain with 0.0015 and length l with 100.6 mm then

$\triangle l=0.0015\times 100.6=0.1509 mm$

(b)

Lateral strain is given by

$\epsilon_{lat}=\frac {\triangle d}{d}$ and substituting $-v\epsilon$ for $\epsilon_{lat}$ where v is poisson ratio then

$-v\epsilon=\frac {\triangle d}{d}$ and making $\triangle d$ the subject then

$\triangle d=-vd\epsilon$ and substituting 0.35 for v, 0.0015 for strain and 10 mm for d

$\triangle d=-(0.35)*10*0.0015=-0.00525 mm$ and the negative sign indicates decrease in diameter

8 0

3 years ago

Noooo plssschvwekjshdjkshdjkshdjksahdk

sashaice [31]

Huh? Kffjkdkdodjenxkrkdkre

4 0

3 years ago

Read 2 more answers

The pressure at the bottom of an 18 ft deep storage tank for gasoline is how much greater than at the top? Express your answer i

Julli [10]

5.85 psig

Using a specific gravity of 0.75 as an average for red\automobile gasoline.

Water at standard conditions (60 degF) is 2.31 feet = 1 psig

80/2.31 then multiply x .75 to compensate for specific gravity of water being 1.0

4 0

3 years ago

A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu

suter [353]

Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

$emf= N\frac{d \phi}{dt}$

where;

ΔФ is change in magnetic flux

ΔФ = BA = B x πr²

ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²

$emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V$

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

6 0

3 years ago