Technician A says lever action pushes a rod into the brake booster and master cylinder

For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?

juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= $4\sqrt{2}$ cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as $P = \dfrac{\pi ^2EI}{L_e^2}$

If area moment of inertia(I) is high then buckling load will be high.

We know that area moment of inertia(I)

For circular tube $I = \dfrac{\pi }{64}(D_o^4-D_i^4)$

For circular bar $I = \dfrac{\pi }{64}D^4$

Now by putting the values

For circular tube $I=62.83 cm^4$

For circular bar $I=50.26 cm^4$

So we can say that for same cross sectional area the area moment of inertia(I) is high for tube as compare to bar.So buckling load will be higher in tube as compare to bar.

3 0

3 years ago

Please help me fast, I don’t have time

Anna71 [15]

Answer: precision

Explanation: Because accuracy is where you keep on getting it right but precision is where you get closer and closer

5 0

3 years ago

A total of 245 kip force is applied to a set of 10 similar bolts. If the spring constant of each bolt is 0.4 Mlb/in and that of

zubka84 [21]

Answer: The net force in every bolt is 44.9 kip

Explanation:

Given that;

External load applied = 245 kip

number of bolts n = 10

External Load shared by each bolt (P_E) = 245/10 = 24.5 kip

spring constant of the bolt Kb = 0.4 Mlb/in

spring constant of members Kc = 1.6 Mlb/in

combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in

Initial pre load Pi = 40 kip

now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them

External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip

So Total net Force on each bolt Fb = P_Eb + Pi

Fb = 4.9 kip + 40 kip

Fb = 44.9 kip

Therefore the net force in every bolt is 44.9 kip

4 0

3 years ago

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage

faltersainse [42]

Answer:

a) 0.489

b) 54.42 kg/s

c) 247.36 kW/s

Explanation:

Note that all the initial enthalpy and entropy values were gotten from the tables.

See the attachment for calculations

4 0

4 years ago

Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

alexdok [17]

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number ( $Re$ ) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

$Re=\frac{\rho v D}{\mu}$

where $\rho$ is the density of the fluid, $\mu$ is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

$v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s$

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

$D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm$

8 0

3 years ago