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3 years ago

Why we use the lagrange Equation?

Engineering

1 answer:

PIT_PIT [208]3 years ago

3 0

Answer:

It is used for solving optimization problems in which,given some functional,one seeks the function minimizing or minimizing it.

Explanation:

Please i need a Brainliest.

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Could I please get help with this

alex41 [277]

Answer:

1. $I_{xc}$ = 7.161458 $\overline 3$ in.⁴

$I_{yc}$ = 36.661458 $\overline 3$ in.⁴

Iₓ = 28.6458 $\overline 3$ in.⁴

$I_y$ = 138.6548 $\overline 3$ in.⁴

2. $I_{xc}$ = 114. $\overline 3$ in.⁴

$I_{yc}$ = 37. $\overline 3$ in.⁴

Iₓ = 457. $\overline 3$ in.⁴

$I_y$ = 149. $\overline 3$ in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Explanation:

1. The height of the beam having a rectangular cross section is h = 2.5 in.

The breadth of the beam, is = 5.5 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

$I_{xc}$ = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458 $\overline 3$

$I_{xc}$ = 7.161458 $\overline 3$ in.⁴

$I_{yc}$ = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458 $\overline 3$

$I_{yc}$ = 36.661458 $\overline 3$ in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458 $\overline 3$

Iₓ = 28.6458 $\overline 3$ in.⁴

$I_y$ = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548 $\overline 3$

$I_y$ = 138.6548 $\overline 3$ in.⁴

2. The height of the beam having a rectangular cross section is h = 7 in.

The breadth of the beam, b = 4 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

$I_{xc}$ = b·h³/12 = 4 × 7³/12 = 114. $\overline 3$

$I_{xc}$ = 114. $\overline 3$ in.⁴

$I_{yc}$ = h·b³/12 = 7 × 4³/12 = 37. $\overline 3$

$I_{yc}$ = 37. $\overline 3$ in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 4 × 7³/3 = 457. $\overline 3$

Iₓ = 457. $\overline 3$ in.⁴

$I_y$ = h·b³/3 = 2.5 × 5.5³/3 = 149. $\overline 3$

$I_y$ = 149. $\overline 3$ in.⁴

3. The deflection, $\delta _{max}$ , of a simply supported beam having a point load at the center is given as follows;

$\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}$

The given parameters of the beam are;

The length of the beam, L = 22 ft. = 264 in.

The applied load at the center, W = 750 lbs

The modulus of elasticity for Cedar = 10,000,000 psi

The height of the wood, h = 3 in.

The breadth of the wood, b = 5 in.

The moment of inertia of the wood, $I_{xc}$ = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

$\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552$

The maximum deflection of the beam, $\delta _{max}$ = 2.55552 inches

5 0

3 years ago

सत्य से अधिक उपयोगी एवं आज्ञापालन से श्रेष्ठ क्या है ?<br>answer fast plz

PilotLPTM [1.2K]

Answer:

I hope this helps you

Explanation:

किसी की जान बचाने के लिए झूठ बोलना सत्य से अधिक उपयोगी है । अन्यायपूर्ण आज्ञा का विरोध आज्ञापालन से श्रेष्ठ है ।

5 0

2 years ago

The convection heat transfer coefficient for a clothed person standing in moving air is expressed as h 5 14.8V0.69 for 0.15 , V

Rom4ik [11]

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7 0

3 years ago

Read 2 more answers

A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a

nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$

$T_{1}$ = 20+460 = 480°R

$T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}$ = 533.25°R

Pressure at the inlet of compressor at isentropic condition

$P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}$

$P_{a}$ = $(10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}$ = 14.45 psia

$P_{2}= 8P_{a} = 8(14.45) = 115.6 psia$

4 0

3 years ago

Read 2 more answers

What is the maximum thermal efficiency possible for a power cycle operating between 600P'c and 110°C? a). 47% b). 56% c). 63% d)

Anastasy [175]

Answer:

(b) 56%

Explanation:

the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature

here we have given T₁ (Higher temperature)= 600+273=873

lower temperature T₂=110+273=383

Efficiency of power cycle is given by =1- $\frac{T2}{T1}$

=1- $\frac{383}{873}$

=1-0.43871

=.56

=56%

5 0

3 years ago

Why we use the lagrange Equation? ​

Why we use the lagrange Equation?