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3 years ago

Explain the difference between statically determinate beam and statically in determinate beam with sketch.

Engineering

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Answer

Statically determinate beams are those beams in which the unknown reaction forces are equal or less than the equilibrium equation.

As shown in figure 1 in which reaction forces are 3 and we have 3 equilibrium equation so beam is determinate.

Statically indeterminate beams are those beams in which unknown reaction force are more than the equilibrium equation.

As shown in figure 2 in which reaction forces are 6 and we have 3 equilibrium equation so beam is indeterminate.

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the AADT for a section of suburban freeway is 150000 veh/day. Assuming this is an urban radial facility, what range of direction

igomit [66]

Answer:

design hour volumes will be 4000 to 6000

Explanation:

given data

AADT = 150000 veh/day

solution

we get here design hour volumes that is express as

design hour volumes = AADT × k × D ..............1

here k is factor and its range is 8 to 12 % for urban

and D is directional distribution i.e traffic equal divided by the direction

so here design hour volumes will be 4000 to 6000

7 0

3 years ago

Note that common skills are listed toward the top, and less common skills are listed toward the bottom.

Sladkaya [172]

Answer:

BDEG

Explanation:

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3 years ago

For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta

Crank

Answer:

The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

$c=\sqrt(\gamma*R*T)$

Where R is the gas's particular constant defined in terms of Cp and Cv:

$R=Cp-Cv$

For particular values given:

$R=840.4 \frac{J}{Kg-K}- 651.5 \frac{J}{Kg-K}$

$R=188.9 \frac{J}{Kg-K}$

The gamma undimensional constant is also expressed as a function of Cv and Cp:

$\gamma=Cp/Cv$

$\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}$

$\gamma=1.29$

And the variable T is the temperature in Kelvin. Thus for the known temperature:

$c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)$

$c=\sqrt(91867.73 \frac{J}{Kg})$

The Jules unit can expressing by:

$J=N.m=\frac{Kg.m}{s^2}* m$

$J=\frac{Kg.m^2}{s^2}$

Replacing the new units for the speed of the sound:

$c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})$

$c=\sqrt(91867.73 \frac{m^2}{s^2})$

$c=313 m/s$

3 0

3 years ago

Read 2 more answers

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0

3 years ago

Read 2 more answers

If an imbalance occurs, the _

pochemuha

A. AFGI is the answer for this question.

7 0

3 years ago