# Mechanic - Static (force, moments)

1. Feb 24, 2006

### Andreii

Hi

I would like to ask for some help about the construction which I show in this post. I have to get maximum (in + and - is the same number) of internal axis force, internal cross/diagonal force, internal moment. All those three numbers i can get from the graph, example of this graph is here: http://server2.uploadit.org/files/janez10-meh1.JPG

Now what is my problem is this task - It starts already at the beginning - with setting the reactions :(. I setted sum of x axis and i didn't have any bigger problems with this (α is 30 degree): Ax - F*Cos α = 0 and from that I can get that Ax is 7.36 kN.
But the problem comes in the next step - how to get Ay or/and B (there is no Bx, so I can mark By as B) for suma of y axis, and how to set up moments? I got for y axis this: Ay + B - F*SIN α = 0. Now I have to get somehow this B and Ay. For moments, I have to count length too... How do I set reactions for moment in my example? When I come to fields, there are two of them, right?
Construction:

2. Feb 24, 2006

### Pyrrhus

Take Moment about the support on the left (where you have your moment). Remember Couples have the same moment about every points and they are also free vectors.

3. Feb 24, 2006

### Andreii

I know this too but I can't help with this to setting up a reactions and both fields.

4. Feb 24, 2006

### Pyrrhus

5. Feb 24, 2006

### Andreii

I have already done that in my first post when i built a topic.
I got alone sum of X axis:

Ax - F*Cos α = 0
Ax - 8.5 cod 30 = 0
Ax = 7.36 kN

Now I got alone setted reaction for Y axis too:

Ay + B - F Sin α = 0

But here it stops because later (after I will be done with both fields) I will prolly need Ay and B for the graph but I don't know how to get Ay and B. Perhaps from sum of moments? Im not sure how to set the sum of moments too. Here I need to count length somehow.

6. Feb 24, 2006

### Pyrrhus

i meant you moment equation attempt.

7. Feb 25, 2006

### Andreii

Here are all steps:

Reactions:

Sum in x axis: Ax – F * COS α = 0
Ax = 7.36 kN
Sum in y axis: Ay + B – F * SIN α = 0
9.67 + B – 8.5 * SIN α = 0
B = - 9.67 + 8.5 * SIN 30
B = - 9.67 + 4.25
B = -5.42 kN
Sum of moments: M - L1 * Ay + F * sin α * (L1+L2) = 0
Ay = [M + F * sin α * (L1+L2)]/L1
Ay = [4,68kNm + 8,5kN * sin 30° * 8m ]/4m=
Ay = 9,67kN

1. Field: Moment is turned in the same line as clock is moving, N (axis force) in horizontal right side, T (cross/diagonal force) vertical downward:

Sum in x axis: N=0
Sum in y axis: -T + 9.67 = 0
T = 9.67 kN
Sum of moments: M – 9.67*x = 0
M = 9.67x

M0 = 0
M4 = 38.7 kN/m

2. Field: Moment is turned in the same line as clock is moving, N (axis force) horizontal in left side, T (cross/diagonal force) horizontal upward:

Sum in x axis: N + F * COS α = 0
N = -F * COS α
N = -7.4 kN/m
Sum in y axis: T – F * SIN α = 0
T = F * SIN α
T = 4.25 kN
Sum of moments: M + 4.25*x = 0
M = -4.25x

M0 = 0
M4 = -17

Extrems: Nextrem, Textrem, Mextrem = ?
How do I get those 3 numbers of extrems from the graph?

-------------------
I tried also this:
-------------------

Here are all steps:

Reactions:

Sum in x axis: Ax – F * COS α = 0
Ax = 7.36 kN
Sum in y axis: Ay + B - F*SIN α = 0
Ay + B = F*SIN α
B = - 1.17
Sum of moments: M - L1*Ay + F*L2*SIN α = 0
4.68 + 34*0.5 - 4*Ay = 0
21.68 - 4*Ay = 0
Ay = 5.42 kN

1. Field: Moment is turned in the same line as clock is moving, N (axis force) in horizontal right side, T (cross/diagonal force) vertical downward:

Sum in x axis: N = 0
Sum in y axis: -T + 6.6 = 0
T = 6.6 kN
Sum of moments: M - 6.6x = 0
M = 6.6x
M0 = 0
M4 = 26.4 kNm

2. Field: Moment is turned in the same line as clock is moving, N (axis force) horizontal in left side, T (cross/diagonal force) horizontal upward:

Sum in x axis: N + F * COS α = 0
N = -F * COS α
N = -7.4 kN/m
Sum in y axis: T – F * SIN α = 0
T = F * SIN α
T = 4.25 kN
Sum of moments: M + 4.25*x = 0
M = -4.25x

M0 = 0
M4 = -17

Extrems: Nextrem, Textrem, Mextrem = ?
How do I get those 3 numbers of extrems from the graph?

Last edited: Feb 25, 2006
8. Feb 26, 2006

### Andreii

9. Feb 26, 2006

### haynewp

This is a simple beam problem so I will just give you a procedure to solve it and ones like it.

I will call the reaction at the left end of the beam "A" which will have components "Ax" and "Ay" since it is a pinned condition and has an applied moment "Ma". I will call the other reaction "B" which will also have components "Bx" and "By".

1) Assign a sign convention to the problem. I will use forces to the right as positive, forces up as positive and counter clockwise moments as positive.

2) You need to resolve the force at the right end of the beam into x and y components, Fx=-7.36N and Fy=-4.25kN

3) Set up the equilibrium equations, assume unknowns are acting in the POSITIVE directions:

Horiz:
Bx-Fx=0, therefore Bx=7.36kN to the right
and Ax=0 since Bx cannot displace and Fx is at the opposite end.

Rotational:
Sum moments about reaction "A", remember "By" direction is up and CCW is positive. This gives -Fy*(L1+L2)+By*(L1)-Ma=0, solve for By.

Vertical:
Ay+By-Fy=0, solve for Ay.

If any results for the reactions are negative when solved in the equations, then the direction is opposite of the way you assumed (positive).

4) Draw the axial, shear and bending moment diagrams, find maximum and minimum forces and moments.

I will verify your reactions is you follow this procedure to get them.

Last edited: Feb 26, 2006
10. Feb 26, 2006

### Andreii

11. Feb 26, 2006

### haynewp

Your N and T diagrams are correct, which you can get that the max axial force is 7.36kN and the max absolute shear force is 5.42kN (which is what I assume you mean when you say diagonal force).

Your moment diagram is still incorrect.

12. Feb 26, 2006

### Andreii

Hmm but if I change moment diagram to get extrem of M then that means I must change formulas for first and/or for second field too, or not?

For first field for moment: M1 = M + B * x
For second field for moment: M2 = M + B * (L1+x) + Ay * x

I think this formulas are right. I changed in the graph for M now, new graph is here: http://img.photobucket.com/albums/v309/Andreii/Graph34.jpg

I would mark extrem of M as 4.68 kNm

13. Feb 26, 2006

### haynewp

Your moment formulas are not right. There is no way you can have a moment at the right end since there is no reaction and no applied moment there. The moment at the right end must equal zero. The moment diagram will be a linear decrease from the right end to reaction B, then linear increase from B to the left end A. At A, the moment diagram value must equal the applied moment 4.68kN*m.

14. Feb 26, 2006

### Andreii

Hmm then in the first graph, the beginning of moment (4.68) is right but it goes to the end of x axis (including the second field) to 0.
How do I get this M (moment) extrem?

15. Feb 26, 2006

### Andreii

Hmm does anyone know this?

16. Feb 26, 2006

### haynewp

It's easy. Sum moments about the right end, it checks out as being equal to zero based on the reactions you have already found.

5.42kN*8m-9.67kN*4m-4.68kn*m+Mrightend=0, therefore Mrightend=0.

It's obvious because there is no applied moment there and there is no moment resisting reaction there either, and no internal moments being carried since it is at the beam end.

Last edited: Feb 26, 2006
17. Feb 26, 2006

### Andreii

yeah you got that from formula but extrem - from formula or from graph (maximum) of moment can't be 0, otherwise it wouldn't be showed on the picture.

18. Feb 26, 2006

### haynewp

Showed on what picture? Your moment diagram is wrong. The right end of the diagram should be at point zero. You need to get your equations right first, then plot your moment diagram.

19. Feb 26, 2006

### Andreii

5.42kN*8m-9.67kN*4m-4.68knm+Mrightend=0 is sum of moments for second field, right? from this i can see that M2 = 0 (with M2 i mean moment of second field). what is it for first one then? the formula for first one then? You said it have to be 0, so M1 = M + B*x is wrong too because i get 4.68 (x length is 0) and -17 (x length is 4)

20. Feb 26, 2006

### haynewp

No, you misunderstand. That equation I gave is summing moments about the right end to prove to you the moment is zero there, it is not the equation for the entire second field. The second field moment equation is a function of the distance. That is why I said there is a linear decrease in the moment from the right end to support B. You need to sit down with your teacher and go through this, you have spent way too much time on this simple problem. I may post the correct moment diagram when I get a chance, but my effort here is beginning to seem futile.