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3 years ago

First step in solving frames in to solve support reactions when looking at the frame as a whole. a)- True b)-False

Engineering

1 answer:

drek231 [11]3 years ago

7 0

Answer:

Yes, it is true that for solving question related to frame we need to find out support reaction in the beginning.

Explanation:

Yes, it is true that for solving question-related to frame we need to find out support reaction in the beginning. The need for support reaction in the starting is just to help in finding out the moment at the desired location.

A support reaction is response force or forces related to system support. the support reaction in any rigid frame can be obtained from the free body diagram of every individual member of the given frame.

You might be interested in

How should you move your board through the planer? (Pick two choices.)

iragen [17]

Answer:

I would say do it at an even pace

Explanation:

Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast

4 0

3 years ago

At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t

nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

$PGP = \rho *\frac{\pi r^2}{2} *V*e \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}$

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0

3 years ago

Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1

babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0

3 years ago

Initially, a pump pressure of __________ pounds per square inch should be used to maintain a sprinkler or standpipe system.

jasenka [17]

Answer:

of 150 pounds per square inch

Explanation:

Note that the unit for measuring water pressure is called pounds per square inch (psi)

In the case of sprinklers and standpipe systems, a pressure of 150 pounds per square inch was used initially.

6 0

3 years ago

An airliner is flying at 34,000 ft cruise altitude on a standard day. Calculate the pressure difference between the cabin and th

nadya68 [22]

Answer:

$\Delta P=61,952.8\ lb/ft^2$

Explanation:

Given

Airline flying at 34,000 ft.

Cabin pressurized to an altitude 8,000 ft.

We know that at standard condition ,density of air

$\rho =0.074\ lb/ft^3$

We know that pressure difference

ΔP=ρ g ΔZ

Here ΔZ=34,000-8,000 ft

ΔZ=26,000 ft

$g= 32.2\ ft/s^2$

ΔP=0.074 x 32.2 x 26,000

$\Delta P=61,952.8\ lb/ft^2$

So pressure difference will be $\Delta P=61,952.8\ lb/ft^2$ .

7 0

3 years ago