General questions (1 Viewer)

Glyde · Oct 20, 2015

As some of you guys may know I'm doing engineering by distance education hence I won't have a teacher to answer my questions. Google is unspecific at times so i guess I'm relying on you guys

astroman · Oct 20, 2015

what do u want to know.

Glyde · Oct 20, 2015

Is there a difference between truss girder and box girder bridges ? Is a box girder just more specific ? If so what would each look like. What makes a truss girder bridge different from the rest ?

Kolmias · Nov 16, 2015

Glyde said:
Is there a difference between truss girder and box girder bridges ? Is a box girder just more specific ? If so what would each look like. What makes a truss girder bridge different from the rest ?

There's a thing called Google.

Anyway, a truss bridge is a bridge whose load-bearing superstructure is composed of a truss, a structure of connected elements forming triangular units.

A box girder bridge is a bridge in which the main beams comprise girders in the shape of a hollow box. The box girder normally comprises either prestressed concrete, structural steel, or a composite of steel and reinforced concrete.

Got it from Wikipedia.

Joplin · Feb 10, 2016

On the Excel Senior High School Engineering Studies (author by Peter M & Roger M) text book page 28, where I'm introduced to the Engineering Mechanics and Hydraulics, there's an example where the area is divided by 4.

I'll rewrite the work:
Tensile stress
Determine the Ultimate Tensile Stress (UTS) of a 15mm diameter rod if the maximum load it can withstand is 70 kN.

Load = 70k x 10^3 N
Diameter = 15 mm

Area = pi15^2 divide by 4

= 176.7 mm^2

...and then it continues with

UTS = L/A

But I want to know where the "divide by 4" comes from?? Is this textbook even reliable?

InteGrand · Feb 10, 2016

Joplin said:
On the Excel Senior High School Engineering Studies (author by Peter M & Roger M) text book page 28, where I'm introduced to the Engineering Mechanics and Hydraulics, there's an example where the area is divided by 4.

I'll rewrite the work:
Tensile stress
Determine the Ultimate Tensile Stress (UTS) of a 15mm diameter rod if the maximum load it can withstand is 70 kN.

Load = 70k x 10^3 N
Diameter = 15 mm

Area = pi15^2 divide by 4

= 176.7 mm^2

...and then it continues with

UTS = L/A

But I want to know where the "divide by 4" comes from?? Is this textbook even reliable?

$\noindent Yes, the divide by 4 is correct. This is because the area of a circle with diameter $d$ is given by $A=\frac{\pi d^2}{4}$. You probably know that if the radius is $r$, the area is $A=\pi r^2$. Since the radius is half the diameter, $r=\frac{d}{2}$. So$

$\begin{align*}A&= \pi \left(\frac{d}{2}\right)^2 \\ &= \pi \cdot \frac{d^2}{4}.\end{align*}$

Nailgun · Feb 10, 2016

InteGrand said:
$\noindent Yes, the divide by 4 is correct. This is because the area of a circle with diameter $d$ is given by $A=\frac{\pi d^2}{4}$. You probably know that if the radius is $r$, the area is $A=\pi r^2$. Since the radius is half the diameter, $r=\frac{d}{2}$. So$

$\begin{align*}A&= \pi \left(\frac{d}{2}\right)^2 \\ &= \pi \cdot \frac{d^2}{4}.\end{align*}$

I don't think I'll ever work out what exactly it is you do lol
Are you an engineer, mathematician, chemist, physicist, medical doctor - who knows lol

Joplin · Feb 11, 2016

Thanks. Wasn't sure if they could've just half the diameter instead of making it more complicated. But now I know a new (probably redudant) equation and that some maths here will try and be tricky. I'll be sure to ask again when I'm bewildered.

Joplin · May 22, 2016

Hai again. I'm irritated at a question and I'm not entirely sure how the conversion thing goes. I've got the numbers right but I don't got the numbers in the right places... See below:

It's from a book called Engineering Studies The Definitive Guide.
Q. Find the resistance of a copper wire used in an appliance that is 300mm long, and has a diameter of 2mm, if the resistivity of copper (at 20^oC) is 1.7x10^-10 Ohm metre.

The formula given is R = p x l/a

where
R is the Resistance (Ohm)
p is the resistivity (Ohm m)
l is the length (m)
a is the cross sectional area (m²)

Here's my work out:

R = (1.7x10^-8) x (0.3)/ pi(3.141...) x 0.001
= 0.00000162338 Ohm

And, the answer given was 1.62 Ohm m

I thought I got the conversion wrong but I'm really not sure now.

InteGrand · May 22, 2016

Joplin said:
Hai again. I'm irritated at a question and I'm not entirely sure how the conversion thing goes. I've got the numbers right but I don't got the numbers in the right places... See below:

It's from a book called Engineering Studies The Definitive Guide.
Q. Find the resistance of a copper wire used in an appliance that is 300mm long, and has a diameter of 2mm, if the resistivity of copper (at 20^oC) is 1.7x10^-10 Ohm metre.

The formula given is R = p x l/a

where
R is the Resistance (Ohm)
p is the resistivity (Ohm m)
l is the length (m)
a is the cross sectional area (m²)

Here's my work out:

R = (1.7x10^-8) x (0.3)/ pi(3.141...) x 0.001
= 0.00000162338 Ohm

And, the answer given was 1.62 Ohm m

I thought I got the conversion wrong but I'm really not sure now.

Here's what I get.

$\noindent We are given diameter $d=2\text{ mm}$, which in metres is $d=2\times 10^{-3} \text{ m}$. Hence the area of the cross-section is $A = \frac{\pi d^2}{4}$. We are given the length to be $\ell = 0.3 \text{ m}$. Also, the resistivity is $\rho = 1.7 \times 10^{-10} \, \Omega \, \text{m}$. Therefore, the resistance is (noting everything we have is now in SI units)$

$\begin{align*}R &= \frac{\rho \ell }{A}\\ &= \frac{1.7 \times 10^{-10} \times 0.3 }{\frac{\pi \left(2\times 10^{-3}\right)^2}{4}} \\ &= 0.0000162338\ldots \Omega \\ &\approx 1.62 \times 10^{-5} \Omega .\end{align*}$

Joplin · Oct 22, 2016

Someone help explain where does the longitudinal force exist on a body/subject.

Examples I'd like to know... A Cube, an airplane and a bridge. I'm just not really certain of the definition and prefer a picture, but I haven't been able to find much.

General questions (1 Viewer)

Glyde

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astroman

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Glyde

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Kolmias

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Joplin

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InteGrand

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Nailgun

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Joplin

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Joplin

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InteGrand

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Joplin

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