Mechanics of Materials: find combined loading

[SoylentBlue]

Homework Statement

This is not a homework assignment. I am studying civil engineering by myself.

Relevant Equations

Sum of the moments when you cut a beam, as shown above.

TL;DR Summary: A frame with a triangular distributed load is pin-connected to a 2-force member. Find the combined stress at point E on the frame.

I am stuck at determining the value of M at the cut. The book shows the value at 8.25KN-meter, but I cannot see how they arrived at that number.
Thank you in advance. I am trying to learn this on my own, so the Internet is my professor right now.

 

[erobz]

Begin by writing the shear $V$ as a function of $x$. To do this use the definition:

$$ \int dV = -\int w(x)~dx $$

Be careful writing $w(x)$. The value of the distributed load at distance $x$ from $A$ is that of a line passing through the origin. Write the equation of that line, and integrate on the left from $V_o$ to $V$ ( note initial conditions of $V$ are not zero), and on the right from $0$ to $x$.

Once you have obtained $V(x)$, integrate once more to obtain $M(x)$.

 

[erobz]

Homework Statement: This is not a homework assignment. I am studying civil engineering by myself.
Relevant Equations: Sum of the moments when you cut a beam, as shown above.

TL;DR Summary: A frame with a triangular distributed load is pin-connected to a 2-force member. Find the combined stress at point E on the frame.

View attachment 335991View attachment 335992

I am stuck at determining the value of M at the cut. The book shows the value at 8.25KN-meter, but I cannot see how they arrived at that number.
Thank you in advance. I am trying to learn this on my own, so the Internet is my professor right now.

What you are currently trying to do, you have incorrectly calculated the effective load. What is the value of the distributed load evaluated at $x = 1.5 ~\rm{m}$ ( i.e. $w( 1.5 ~ \rm{m} ) ~\rm{ \frac{[kN]}{[m]}}$)?

 

[SoylentBlue]

Begin by writing the shear $V$ as a function of $x$. To do this use the definition:

$$ \int dV = -\int w(x)~dx $$

Be careful writing $w(x)$. The value of the distributed load at distance $x$ from $A$ is that of a line passing through the origin. Write the equation of that line, and integrate on the left from $V_o$ to $V$ ( note initial conditions of $V$ are not zero), and on the right from $0$ to $x$.

Once you have obtained $V(x)$, integrate once more to obtain $M(x)$.

OK, so the equation for the load is simply y=4x.
To obtain shear, integrate y=4x to yield y=2x².
At a value of x=1.5, the shear is 4.5KN, which agrees with the book's answer key.
Is this correct so far?
Integrate again and you get y=2x³/3.
At a value of 1.5, this yields a moment of 2.25 KNmeters.

I'm still a little foggy on how these pieces fit together. In a statics video using a single triangular distributed load, the instructor mentioned several times that the moment curve is the integral of the shear curve. But when he solved for the moment he did not use those relationships. Instead he cut the beam and examined the left side; he used M_cut - (distance to origin)*A_y (where A_y is the upward force at the origin) plus 0.5 (the centroid of the triangle)*concentrated load.
The book's answer key shows a triangular distributed load of 2kN/m at 1.5 meters. I'm not sure how we jumped from 4kN/m to 2kN/m using the shear-moment diagram and the intregrals.
Thank you again...all help is appreciated!! :^)

 

[SoylentBlue]

OK, for anyone else who is stuck here, I found the answer in another youtube video:

The equation of our line is y=mx+b. b=0. m is the slope, which they call 4KN/3m when you examine the left-most 3 meters of the load (back to that in a, ahem, moment. Sorry, I just couldn't resist making that pun).
So the equation of our line is y= 4/3x.
V=6 - integral of 4/3xdx = 6- 4/3 x²/2 = 6 - 2/3 x²
For x = 1.5, V = 6 -2/3(1.5)² = 4.5, which agrees with the book's answer key.
To get the moment, we integrate again: M = integral of 2/3x²dx = 2/3 x³/3 = 2/9x³. For x=1.5, we get a moment of 0.75, which, when subtracted from the moment of 9kNmeters, yields 8.25 kNmeters, which agrees with the book's answer key.

And yes, I realize I drew the moment diagram incorrectly; it should look like an upside-down U.

So, back to that determination of the slope. When I see 4KN/m, I assume 4KN rise per meter of run, or a slope of 4. They show a triangle 3 meters wide and 4 KN tall, and call the slope 4/3. Can anyone comment on this please? Thank you.

 

[erobz]

They show a peak distributed load value of 4 kN/m @ x = 3m. It is not 4kN load at x=3m.

 

[Lnewqban]

...
So, back to that determination of the slope. When I see 4KN/m, I assume 4KN rise per meter of run, or a slope of 4. They show a triangle 3 meters wide and 4 KN tall, and call the slope 4/3. Can anyone comment on this please? Thank you.

Please, see:
https://mathalino.com/reviewer/mech...tion-to-problem-417-shear-and-moment-diagrams

https://mathalino.com/reviewer/mech...o-problem-443-relationship-between-load-shear

 

[SoylentBlue]

They show a peak distributed load value of 4 kN/m @ x = 3m. It is not 4kN load at x=3m.

Oh, thank you!!! That was a subtlety that I had missed. And it obviously had a profound effect on the calculations. Thank you again.