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Advocard [28]
3 years ago
7

Ultimate tensile strength is: (a) The stress at 0.2% strain (b) The stress at the onset of plastic deformation (c) The stress at

the end of elastic deformation (d) The stress at fracture (e) The maximum stress in a stress-strain curve
Engineering
1 answer:
MissTica3 years ago
4 0

Answer:

By definition the ultimate tensile strength is the maximum stress in the stress-strain deformation. The stress at 0.2% strain, the stress at the onset of plastic deformation, the stress at the end of the elastic deformation and the stress at the fracture correspond, by definition, to other points of the stress-strain curve.

Explanation:

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A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates wi
____ [38]

Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

V_{max}=\omega A

now by putting the values

V_{max}=\omega A

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

6 0
4 years ago
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
An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50
netineya [11]

Answer:

a)Ek=115759.26J

b)23.15kW

Explanation:

kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation

Ek=0.5mv^{2}

for the first part

m=1200Kg

V=50km/h=13.89m/s

solving for Ek

Ek=0.5(1200)(13.89)^2

Ek=115759.26J

For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.

m=1200kg

V=100km/h=27.78m/s

Ek=0.5(1200)(27.78)^2=462963J\\

taking into account all of the above the following equation is inferred

ΔE=\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15}  =23146.916W=23.15kW

3 0
3 years ago
You are designing a 200mm long x 100mm wide x 50mm deep rectangular housing, with a wall width of 1.5 mm, and 1 degree draft. Yo
Nimfa-mama [501]
365gpa I don’t even know what it is but this is with 10 point and I need it
6 0
2 years ago
Can you give me examples of reciprocating motion?
tensa zangetsu [6.8K]

Answer:

Other examples of reciprocating motion include piston pumps and compressors, roller pressure and tensioning systems, material testing devices, and insertion machines.

Explanation:

6 0
3 years ago
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