What is civil engineering

A circular bar is 800mm in length and 32mm in diameter. The bar is made from a material with a modulus of elasticity E = 150 GPa

Yuki888 [10]

Answer:

For any material if ∈ is the axial strain then the lateral strain is given by -μ∈ is the lateral strain in the object

Where,

μ is the poisson's ratio of the material

The longitudinal strain is calculated as follows

$\varepsilon _{axial}=\frac{\Delta length}{Length_{original}}\\\\\therefore \varepsilon _{axial}=\frac{0.7}{800}=8.75\times 10^{-4}$

Thus the lateral strain becomes

$\varepsilon _{lateral}=-\mu\varepsilon _{axial}\\\\\varepsilon _{lateral}=-0.27\times 8.75\times 10^{-4}=-2.36\times 10^{-4}$

now by definition of lateral strain we have

$\varepsilon _{lateral}=\frac{\Delta diameter}{diameter_{original}}\\\\\Rightarrow \Delta Diameter=-2.36\times 10^{-4}\times 32=-7.56\times 10^{-3}\\\\D_{f}-D_{i}=-7.56\times 10^{-3}\\\\D_{f}=32-7.56\times 10^{-3}=31.992mm$

By hookes law the stress developed due to the given strain is given by

$\sigma =\varepsilon _{axial}E$

Applying values we get

$\sigma =8.75\times 10^{-4}\times 150\times 10^{9}\\\\\sigma =131.25MPa$

Thus the force is calculated as

$Force=\sigma \times Area\\\\Force=131.25\times 10^{6}\times \frac{\pi (32\times 10^{-3})^{4}}{4}\\\\Force=105.55kN$

5 0

3 years ago

provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki

Vaselesa [24]

Answer:

hello some parts of your question is missing attached below is the missing part ( the required fig and table )

answer : The solar collector surface area = 7133 m^2

Explanation:

Given data :

Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2

percentage of solar power absorbed by refrigerant = 60%

Determine the solar collector surface area

The solar collector surface area = 7133 m^2

attached below is a detailed solution of the problem

8 0

3 years ago

A pressurized 2-m-diameter tank of water has a 10-cm-diameter orifice at the bottom where water discharges to the atmosphere. Th

alukav5142 [94]

Answer:

A fluid is defined as a material that deforms continuously and permanently under the

application of a shearing stress.

• The pressure at a point in a fluid is independent of the orientation of the surface

passing through the point; the pressure is isotropic.

• The force due to a pressure p acting on one side of a small element of surface dA

defined by a unit normal vector n is given by −pndA.

• Pressure is transmitted through a fluid at the speed of sound.

• The units we use depend on whatever system we have chosen, and they include quantities

like feet, seconds, newtons and pascals. In contrast, a dimension is a more

abstract notion, and it is the term used to describe concepts such as mass, length and

time.

• The specific gravity (SG) of a solid or liquid is the ratio of its density to that of water

at the same temperature.

• A Newtonian fluid is one where the viscous stress is proportional to the rate of strain

(velocity gradient). The constant of proportionality is the viscosity, µ, which is a

property of the fluid, and depends on temperature.

• At the boundary between a solid and a fluid, the fluid and solid velocities are equal;

this is called the “no-slip condition.” As a consequence, for large Reynolds numbers

(>> 1), boundary layers form close to the solid boundary. In the boundary layer,

large velocity gradients are found, and so viscous effects are important.

• At the interface between two fluids, surface tension may become important. Surface

tension leads to the formation of a meniscus, drops and bubbles, and the capillary rise

observed in small tubes, because surface tension can resist pressure differences across

the interface.

5 0

4 years ago

Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon al

Ahat [919]

Answer:

the carburizing time necessary to achieve a carbon concentration is 31.657 hours

Explanation:

Given the data in the question;

To determine the carburizing time necessary to achieve the given carbon concentration, we will be using the following equation:

(Cs - Cx) / (Cs - C0) = ERF( x / 2√Dt)

where Cs is Concentration of carbon at surface = 0.90

Cx is Concentration of carbon at distance x = 0.30 ; x in this case is 4 mm = ( 0.004 m )

C0 is Initial concentration of carbon = 0.10

ERF() = Error function at the given value

D = Diffusion of Carbon into steel

t = Time necessary to achieve given carbon concentration ,

so

(Cs - Cx) / (Cs - C0) = (0.9 - 0.3) / (0.9 - 0.1)

= 0.6 / 0.8

= 0.75

now, ERF(z) = 0.75; using ERF table, we can say;

Z ~ 0.81; which means ( x / 2√Dt) = 0.81

Now, Using the table of diffusion data

D = 5.35 × 10⁻¹¹ m²/sec at (1100°C) or 1373 K

now we calculate the carbonizing time by using the following equation;

z = (x/2√Dt)

t is carbonizing time

so we we substitute in our values

0.81 = ( 0.004 / 2 × √5.35 × 10⁻¹¹ × √t)

0.81 = 0.004 / 1.4628 × 10⁻⁵ × √t

0.81 × 1.4628 × 10⁻⁵ × √t = 0.004

1.184868 × 10⁻⁵ × √t = 0.004

√t = 0.004 / 1.184868 × 10⁻⁵

√t = 337.5903

t = ( 337.5903)²

t = 113967.21 seconds

we convert to hours

t = 113967.21 / 3600

t = 31.657 hours

Therefore, the carburizing time necessary to achieve a carbon concentration is 31.657 hours

7 0

3 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

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What is civil engineering​

What is civil engineering