Marks

Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e

marshall27 [118]

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

$\Delta x=\frac{PL^3}{3EI}$

Hence we can write

$P=\frac{3EI\cdot \Delta x}{L^3}$

Comparing with the standard spring equation $F=kx$ we find the cantilever analogous to spring with $k=\frac{3EI}{L^3}$

Now the angular frequency of a spring is given by

$\omega =\sqrt{\frac{k}{m}}$

where

'm' is the mass of the load

Thus applying values we get

$\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}$

$\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec$

8 0

3 years ago

Read 2 more answers

A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain. The h

sertanlavr [38]

Answer:

Explanation:

Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)

fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)

vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)

BFFS = 50+5, BFFS =55 (given) fLW= 6.6

TLC=6+3=9 fLC= 0.65

fM= 0.0

fA= 1.0

FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)

Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)

After: fA= 3.0

FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)

Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln

8 0

3 years ago

A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr

Hoochie [10]

Answer:

$\Delta m = 102.25\,kg$

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

$m_{o} = \rho_{air}\cdot V_{tank}$

$m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{o} = 31.25\,kg$

The final mass inside the rigid tank is:

$m_{f} = \rho \cdot V_{tank}$

$m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{f}= 133.5\,kg$

The supplied air mass is:

$\Delta m = m_{f}-m_{o}$

$\Delta m = 133.5\,kg-31.25\,kg$

$\Delta m = 102.25\,kg$

4 0

3 years ago

A dc shunt generator rated at 85 kW produces a voltage of 280 V. The brush voltage drop is 2.5 V, and the armature and field res

dalvyx [7]

Answer:

(a) Calculate the field, armature, and load currents versus load = 306A

(b) Determine the terminal voltage at no-load and at rated load conditions = 300V

(c) Calculate the voltage regulation of the generator. Use the no-load voltage as the base value = 6.67%

(d) Plot the terminal voltage as a function of the load. Determine the load that corresponds to a 5% voltage drop using the no-load voltage as the base = 294.74V

Explanation:

CHECK THE ATTACHED FILES FOR DETAILED EXPLANATION.

3 0

3 years ago

What position would be the lowest paid in a technology company?

murzikaleks [220]

I think the answers is b

3 0

3 years ago