hollow circular shaft

[logistic_guy]

here is the question

The hollow circular shaft is subjected to an internal torque of $\displaystyle T = 10$ kN$\displaystyle \ \cdot$ m. Determine the shear stress developed at points $\displaystyle A$ and $\displaystyle B$. Represent each state of stress on a volume element.




my attemb
the solution depend on the polar moment of inertia
i don't know how to derive it for circular cross section

 

[khansaheb]

here is the question

The hollow circular shaft is subjected to an internal torque of $\displaystyle T = 10$ kN$\displaystyle \ \cdot$ m. Determine the shear stress developed at points $\displaystyle A$ and $\displaystyle B$. Represent each state of stress on a volume element.

View attachment 38959


my attemb
the solution depend on the polar moment of inertia
i don't know how to derive it for circular cross section

Use Google or Wikipedia or your textbook and tell us about your findings....

 

[logistic_guy]

Use Google or Wikipedia or your textbook and tell us about your findings....

i'm very bad in calculus but i'll try my best

the polar moment of inertia is $\displaystyle J$
$\displaystyle dJ = r^2 dA$ so $\displaystyle J = \int dJ = \int r^2 dA$
i don't know how to solve this integration

 

[khansaheb]

Can you derive the expression for the "area of a circle" with radius "r" - using calculus?

 

[logistic_guy]

Can you derive the expression for the "area of a circle" with radius "r" - using calculus?

yes, very easy
area $\displaystyle = \int dA = A + c = \pi r^2$
i assume $\displaystyle c = 0$

 

[topsquark]

yes, very easy
area $\displaystyle = \int dA = A + c = \pi r^2$
i assume $\displaystyle c = 0$

Hint: Use $y = \sqrt{r^2 - x^2}$. (What shape is this?)

-Dan

 

[khansaheb]

Can you derive the expression for the "area of a circle" with radius "r" - using calculus?

You have to do "double" integration. Read you calculus (II) text book.

 

[logistic_guy]

Hint: Use $y = \sqrt{r^2 - x^2}$. (What shape is this?)

-Dan

i've to cheat. it's the top part of the circle

but how to use it? i don't have $\displaystyle y$ in the integration

 

[logistic_guy]

You have to do "double" integration. Read you calculus (II) text book.

 

[topsquark]

I agree!

Okay, let's take it in baby steps.

I gave you y.

You said the graph of y is a semi-circle. Good!

Now: How do you find the area between a curve and the x-axis?

Hmmmm... Calc I again?

-Dan

 

[logistic_guy]

I agree!

Okay, let's take it in baby steps.

there's no baby steps in mathematics. we're humans, so after time pass we forget even the simplest idea
i don't remember what's my dinner last nighthow will i remember everything in calculus from three years ago?

Now: How do you find the area between a curve and the x-axis?

by integration but setting up the integration isn't always easy
i need to see the curve by eyes to be able to integrate it

Hmmmm... Calc I again?

my weaknessi'm very good in calculus III

 

[topsquark]

there's no baby steps in mathematics. we're humans, so after time pass we forget even the simplest idea
i don't remember what's my dinner last nighthow will i remember everything in calculus from three years ago?

By going back and reviewing what you've forgotten. It's a process called "learning."

by integration but setting up the integration isn't always easy
i need to see the curve by eyes to be able to integrate it

Then graph it if you must. What is
$\displaystyle \int_{-r}^r dy = \int_{-r}^r \sqrt{r^2 - x^2}~dx$

my weaknessi'm very good in calculus III

You've got a lot of weaknesses. And calc III is one of them, as you proved in another post.

My advice? Put aside your problem set for the moment and review Calc I - III. Then review any other field of study you are deficient in. (Differential equations, complex Calculus, Physics, etc.)

-Dan

 

[logistic_guy]

By going back and reviewing what you've forgotten. It's a process called "learning."

i know
but sometimes it's embarassing to read baby steps as you call them

Then graph it if you must. What is
$\displaystyle \int_{-r}^r dy = \int_{-r}^r \sqrt{r^2 - x^2}~dx$

i can't do this integration
but i know to solve the beginning $\displaystyle \int_{-r}^{r} dy = y\big|_{-r}^{r}$

You've got a lot of weaknesses. And calc III is one of them, as you proved in another post.

you're wrong

My advice? Put aside your problem set for the moment and review Calc I - III. Then review any other field of study you are deficient in. (Differential equations, complex Calculus, Physics, etc.)

i'm very good in all of them except calculus I and geometry

 

[khansaheb]

i know
but sometimes it's embarassing to read baby steps as you call them



i can't do this integration
but i know to solve the beginning $\displaystyle \int_{-r}^{r} dy = y\big|_{-r}^{r}$


you're wrong


i'm very good in all of them except calculus I and geometry

You cannot be very good at calc II and III without MASTERING calc I AND geometry.
We know - you feel that reviewing BASICS - like calc I AND GEOMETRY is waste of time - but you have to climb the stairs - from the bottom....

 

[topsquark]

i know
but sometimes it's embarassing to read baby steps as you call them

More embarrassing than what you are doing now? You have to have control of the basics before you move to the more advanced material. That's not me preaching, it's just the way it is. Else you run into having to ask for help with basic problems, as you are now.

i can't do this integration
but i know to solve the beginning $\displaystyle \int_{-r}^{r} dy = y\big|_{-r}^{r}$

THEN GO BACK AND LOOK IT UP!! It's a simple integration and it's probably in your text. Hint: Use a trig substitution.

By the way, I posted the wrong integral. It was a typo. The integral is
$\displaystyle \int_{-r}^r y(x)~dx \neq y \left |_{-r}^r \right .$

The integral with the square root is correct.

i'm very good in all of them except calculus I and geometry

If you were then this problem would have been trivial for you.

-Dan

 

[logistic_guy]

By the way, I posted the wrong integral. It was a typo.

don't worry teacher. even me as advance engineering i make silly mistakes

THEN GO BACK AND LOOK IT UP!! It's a simple integration and it's probably in your text. Hint: Use a trig substitution.

i'll try

$\displaystyle x = \sin u$
$\displaystyle dx = \cos u du$

$\displaystyle \int_{-r}^{r}\sqrt{r^2 - x^2}dx = \int_{\sin^{-1}-r}^{\sin^{-1}r}\sqrt{r^2 - \sin^2 u} \cos u du$

is my analize correct?

 

[topsquark]

don't worry teacher. even me as advance engineering i make silly mistakes

Your mistakes aren't silly. They are very sad and indicate that you have learned very little of any of this material.

i'll try

$\displaystyle x = \sin u$
$\displaystyle dx = \cos u du$

$\displaystyle \int_{-r}^{r}\sqrt{r^2 - x^2}dx = \int_{\sin^{-1}-r}^{\sin^{-1}r}\sqrt{r^2 - \sin^2 u} \cos u du$

is my analize correct?

Like I said. This is probably right in your text.

Correction: x = r sin(u).

Try again.

-Dan

 

[logistic_guy]

Your mistakes aren't silly. They are very sad and indicate that you have learned very little of any of this material.

Like I said. This is probably right in your text.

Correction: x = r sin(u).

Try again.

-Dan

new idea but i think i understand it

$\displaystyle x = r\sin u$
$\displaystyle dx = r\cos u du$

$\displaystyle \int_{-r}^{r}\sqrt{r^2 - x^2}dx= \int_{\sin^{-1}-1}^{\sin^{-1}1}\sqrt{r^2 - (r\sin u)^2} r\cos u du$

i think it get more complicated than the original

 

[khansaheb]

don't worry teacher. even me as advance engineering i make silly mistakes


i'll try

$\displaystyle x = \sin u$
$\displaystyle dx = \cos u du$

$\displaystyle \int_{-r}^{r}\sqrt{r^2 - x^2}dx = \int_{\sin^{-1}-r}^{\sin^{-1}r}\sqrt{r^2 - \sin^2 u} \cos u du$

is my analize correct?

No.... Try
x = r * sin(I)

 

[topsquark]

new idea but i think i understand it

$\displaystyle x = r\sin u$
$\displaystyle dx = r\cos u du$

$\displaystyle \int_{-r}^{r}\sqrt{r^2 - x^2}dx= \int_{\sin^{-1}-1}^{\sin^{-1}1}\sqrt{r^2 - (r\sin u)^2} r\cos u du$

i think it get more complicated than the original

New idea?!

You still haven't done a review, have you?

-Dan