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Brut [27]
4 years ago
14

Use Routh's stability criterion to determine how many roots with positive real parts the following equations have:

Engineering
1 answer:
Pavlova-9 [17]4 years ago
3 0

Answer:

a) no roots not in LHP

b) 2 roots not in LHP

c) 2 roots not in the LHP

d) 2 roots not in the LHP

e) 2 roots not in LHP

Explanation:

a) s^4 + 8s^3 + 32s^2 + 80s + 100 = 0\\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:32\:\:\:\:\:\:100\\s^3:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\\s^2:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:100\\s^1:\:\:\:80-\frac{800}{22} =43.6\\s^0:\:\:\:100

No roots not in the LHP

b) s^5 + 10s^4 + 30s^3 + 80s^2+344s + 480 =0 \\\\s^5:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:30\:\:\:\:\:\:344\\s^4:\:\:\:10\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\:\:\:\:\:\:480\\s^3:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:296\\s^2:\:\:\:-545\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:480\\s^1:\:\:\:490\\s^0:\:\:\:480

2 roots not in the LHP

c) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:7\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^3:\:\:\:2\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-2\\s^2:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^1:\:\:\:-4\\s^0:\:\:\:8

There are roots in the RHP (not all coefficients are greater than 0).

2 roots not in the LHP

d) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^3:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:20\\s^2:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:78\\s^1:\:\:\:-58\\s^0:\:\:\:78

There are two sign changes in the first column of the Routh array.

2 roots not in the LHP

e) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^3:\:\:\:4\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:12\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: new \:\:row \\s^2:\:\:\:3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^1:\:\:\:12-\frac{100}{3}=-21.3 \\s^0:\:\:\:25

2 roots not in LHP

check:

 a (s) = 0  ⇒

 s^2 = -3 \limits^+_- 4j = 5e^{j(\pi \limits^+_- 0.92)}\\\\s = \sqrt5 e^{j( \frac{\pi}{2} \limits^+_-  0.46)+n\pi j},\:\:\:\:\: n= 0, 1\\

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Varvara68 [4.7K]

Answer:

Explanation:

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8 0
3 years ago
A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected
Pavlova-9 [17]

Answer:

the required diameter of the rod is 9.77 mm

Explanation:

Given:

Length = 1.5 m

Tension(P) = 3 kN = 3 × 10³ N

Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa

E = 70 GPa = 70 × 10⁹ Pa

δ = 1 mm = 1 × 10⁻³ m

The required diameter(d)  = ?

a) for stress

The stress equation is given by:

S = \frac{P}{A}

A is the area = πd²/4 = (3.14 × d²)/4

S = \frac{P}{(\frac{3.14*d^{2} }{4}) }

S = \frac{4P}{{3.14*d^{2} } }

3.14*S*{d^{2}} = {4P}

{d^{2}} =\frac{4P}{3.14*S}

d=  \sqrt{\frac{4P}{3.14*S} }

Substituting the values, we get

d=  \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }

d=  \sqrt{\frac{12000 }{125600000  } }

d=  \sqrt{9.55*10^{-5}  }

d = (9.77 × 10⁻³) m

d = 9.77 mm

b) for deformation

δ = (P×L) / (A×E)

A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063

d² = (4 × A) / π = (0.000063 × 4) / 3.14

d² = 0.0000819

d = 9.05 × 10⁻³ m = 9.05 mm

We use the larger value of diameter = 9.77 mm

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Answer:

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Explanation:

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Answer:

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