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4 years ago

Primary Creep: slope (creep rate) decreases with time

Engineering

1 answer:

Igoryamba4 years ago

5 0

Answer:

true

Explanation:

Creep is known as the time dependent deformation of structure due to constant load acting on the body.

Creep is generally seen at high temperature.

Due to creep the length of the structure increases which is not fit for serviceability purpose.

When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.

When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.

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The net potential energy EN between two adjacent ions, is sometimes represented by the expression

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3 years ago

Complete the given statement with the most appropriate word.

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3 years ago

Consider a 8-m-long, 8-m-wide, and 2-m-high aboveground swimming pool that is filled with water to the rim. (a) Determine the hy

Stolb23 [73]

Answer:

The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.

Explanation:

To calculate force on the walls of swimming pool whose dimensions are given as 8-m-long, 8-m-wide, and 2-m-high. We know that formula for hydrostatic force is $\text {hydrostatic force}=\text {pressure} \times \text {area,}=\rho g h \times(l \times h)$

$\equiv \rho g h^{2} l$ , we know ρ=density of fluid=1000 $g / c m^{3}$ ,

g=acceleration due to gravity=9.81 $m / s^{2}$ , h=height of the pool=2 m and l=length of the pool=8 m.

hydrostatic force on each wall= $1000 \times 9.81 \times 2^{2} \times 8$ = 313920 N.

The distance at which hydrostatic force is acted is half of the height of the swimming pool.

At 1 m from the ground this hydrostatic force is acted on each wall.

The force is quadruple if the height of the walls of the pool is doubled this is because, the height is doubled and taken as h=4 m and substitute in the equation = $\rho g h^{2} l$ = $1000 \times 9.81 \times 4^{2} \times 8$ = 1255680 N. This is 4 times 313920 N.

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3 years ago

A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress = 50 MPa

amplitude stress = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first maximum stress and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

$\sigma _{m}$ = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2

50 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........1

and

225 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........2

from eqution 1 and 2 we get maximum and minimum stress

$\sigma _{maximum}$ = 275 MPa ............3

and $\sigma _{minimum}$ = -175 MPa ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio = $\sigma _{minimum}$ / $\sigma _{maximum}$

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = $\sigma _{maximum}$ - $\sigma _{minimum}$

magnitude = 275 - (-175) = 450 MPa

3 0

3 years ago