Cantilevered and line-loaded steel beam

Kalle · Sunday, 3:03 PM

I found the internal force Vxy 25 +6 =31kn. , bending moment Mmin -qb2/2 , M min- 14 kNm. delta =M/W I can't find Mmax and tau= V/As. So I need help to find Mmax and As wall area, how to find.

Lnewqban · Sunday, 3:12 PM

Welcome, Kalle!
Could you show us a diagram of the beam and loads distribution?

haruspex · Monday, 3:43 AM

That's very hard to read, and mostly blank space. Please post a better shot.

Kalle · Monday, 4:25 AM

Lnewqban said:

Welcome, Kalle!
Could you show us a diagram of the beam and loads distribution?

Kalle · Monday, 4:25 AM

haruspex said:

That's very hard to read, and mostly blank space. Please post a better shot.

haruspex · Monday, 5:41 AM

It would help if you were to explain exactly what you have done so far (all the steps).
E.g., I do not know how you are defining "internal force" or how you found Mmin.

What is the support force at B?
What is the bending moment at a point x to the right of B?
What is the bending moment at a point x to the left of B?

Kalle · Monday, 7:08 AM

1.first I found the effect of the linear load on the total length of the beam q=0, 5 x 12 `= 6kNm
2.Total vertical force Vxy =25 +6=31kN.
3.I find Mz = ´- q(b)2/2 Mz `=12x 0,5 squared =1, 5kNm linear load, Mz point load -12,5kNm .Mz min total -1,5- 12,5 =-14kNm. Mz max don't know
4.I found the tension delta =M/W max is missing, then can't count. And can't find a suitable cross section for IPE.
4.I found tau =V/As , As can't find.

haruspex · Monday, 3:57 PM

Kalle said:

1.first I found the effect of the linear load on the total length of the beam q=0, 5 x 12 `= 6kNm
2.Total vertical force Vxy =25 +6=31kN.

5x12=6?
I think you are misinterpreting " load across beam 12kNm".
The units kNm make no sense, so I read this as a distributed load of 12kN/m, for a total 5m x 12kN/m = 60kN.

Kalle said:

3.I find Mz =

What axis are you taking moments about?

Kalle said:

´- q(b)2/2 Mz `=12x 0,5 squared =1, 5kNm linear load, Mz point load -12,5kNm .

This is very hard to parse. I can't tell where one mathematical statement ends and the next one starts. Use separate lines.
What is "q(b)2/2"?

Kalle said:

Mz min total -1,5- 12,5 =-14kNm.

How can you find a minimum without first finding the general formula?
Please try to answer my questions in post #6.

Kalle · Tuesday, 2:02 AM

5x12=6?
I think you are misinterpreting " load across beam 12kNm".
The units kNm make no sense, so I read this as a distributed load of 12kN/m, for a total 5m x 12kN/m = 60kN.
Yes, the load over the beam is 5 x 12 =60 kN, and the point load on the right 25kN . The total vertical force is 85 kN. I need to find Vxy ?
vertical axis of the cantilever beam
Mzmin= − q(b) squared divided by 2.
M z min is the minimum bending moment,
q is the magnitude of the distributed load (linear load),
b is the beam height.
How can you find a minimum without first finding the general formula? Mzmin= − q(b) squared divided by 2.

haruspex · Tuesday, 3:12 AM

Kalle said:

need to find Vxy ?

You have not defined it. What is it?
You referred to it as "internal force", but I don’t know what that means either.

Kalle said:

Mzmin= − q(b) squared divided by 2.

What is "b"? Shouldn’t P be involved?

I say again, please try to answer my questions in post #6.

Kalle · Tuesday, 6:14 AM

E.g., I do not know how you are defining "internal force" - Vxy is the internal force, denoted V. how you found Mmin. ------- the moment of compression Mmin was found by the formula - q(linear load) multiplied by b(beam length) squared and divided by 2. What is the bending moment at a point x to the right of B? - 12,5 kNm What is the bending moment at a point x to the left of B? 37,5 kNm

haruspex · Tuesday, 3:18 PM

Kalle said:

Vxy is the internal force, denoted V

Telling me the internal force is the internal force gets me nowhere. What is meant here by "internal force"?

Kalle said:

the moment of compression Mmin was found by the formula - q(linear load) multiplied by b(beam length) squared and divided by 2

I would think that formula only applies in particular circumstances, e.g. for the case of a uniform load between two supports, and nothing else. That does not apply here.

Kalle said:

What is the bending moment at a point x to the right of B? - 12,5 kNm What is the bending moment at a point x to the left of B? 37,5 kNm

You have not understood my questions.
If you consider a point of the beam at a position x to the right of B, what is the sum of moments about that point due to forces to the right of it? It will be a formula involving x, and it will contain a term due to the point load P as well as a term from that portion of the distributed load to the right of x.

Kalle · Tuesday, 11:36 PM

What is meant here by "internal force"? -In mechanics, internal forces are often analyzed in structures like bridges or buildings, where the forces between different parts of the structure determine its stability and behavior.
lI would think that formula only applies in particular circumstances, e.g. for the case of a uniform load between two supports, and nothing else. That does not apply here.
Please suggest which formula I should use then.
Please give instructions on how and what to find.

haruspex · 2024-03-20T01:45:21-0500

Kalle said:

internal forces are often analyzed in structures like bridges or buildings, where the forces between different parts of the structure determine its stability and behavior

Sure, but you referred to "the" internal force. Which one? And internal to what system?
You named it Vxy. Why that name? What do x and y refer to here?

Kalle said:

Please suggest which formula I should use then.
Please give instructions on how and what to find.

Seems you do not know how to answer my questions, so I'll help you.
It will be clearer if I avoid numbers, so I'll call the point where P is applied C, and use ##\rho## for the force density of the distributed load
Consider a point D at distance x to the right of point B.
The point load at C exerts a torque ##-P(BC-x)## about D.
An element of the distributed load between ##y## and ##y+dy## to the right of x has weight ##\rho dy## and exerts a torque ##-y\rho dy## about D.
The distributed load over BC therefore exerts a torque ##\int_0^{BC-x}\rho y.dy=-\frac 12(BC-x)^2\rho## about D.
The total torque from forces to the right of x is therefore ##-P(BC-x)-\frac 12(BC-x)^2\rho##.
To find the minimum or maximum within BC, differentiate wrt x, etc.

Now, can you do the same for a point distance x to the left of B? It is a bit more complicated because you have to include the normal force from the support at B.

Kalle · 2024-03-21T01:22:23-0500

I'm trying to solve it step by step and figure it out.
I attach a table of the formulae that are required to solve.

Something is still wrong

haruspex · 2024-03-21T01:58:09-0500

Kalle said:

I'm trying to solve it step by step and figure it out.
I attach a table of the formulae that are required to solve.

Something is still wrong

That table is a dreadful way to solve such problems. Far better to understand the principles. But that does require some basic calculus.

Do you understand how I did it post #14?

Kalle · 2024-03-21T02:02:47-0500

I tried to understand, but since I have to keep track of the formulas, it gets confusing.
We would only be required to work with the formulas.

haruspex · 2024-03-21T03:11:11-0500

Kalle said:

I tried to understand, but since I have to keep track of the formulas, it gets confusing.
We would only be required to work with the formulas.

Since the formulas cannot cover every case specifically, you need some rules on how to combine the formulas in any given case.
Can you see two examples in the table which, in some sense, should add up to the case in this thread?

Kalle · 2024-03-21T04:23:12-0500

yes, with the help of these ready-made helpers I will try to solve it.

Cantilevered and line-loaded steel beam

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