
#1
Feb2406, 08:25 AM

P: 69

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/janez10meh1.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
Feb2406, 11:09 AM

HW Helper
P: 2,280

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
Feb2406, 11:20 AM

P: 69





#5
Feb2406, 12:00 PM

P: 69

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. 



#7
Feb2506, 05:49 AM

P: 69

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? 



#8
Feb2606, 08:23 AM

P: 69

Please take a look here, i have 90% of the task, just without an extrems :( http://www.sciencechatforum.com/bull...pic.php?t=1051




#9
Feb2606, 10:19 AM

P: 96

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: BxFx=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+ByFy=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. 



#10
Feb2606, 10:24 AM

P: 69

haynewp i have already figured out what you typed. I have numbers and graph which is showed here http://www.sciencechatforum.com/bull...pic.php?t=1051
but somehow I have to know how to get extrem for diagonal force, extrem for axis force and extrem for moment. 



#11
Feb2606, 10:51 AM

P: 96

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
Feb2606, 11:03 AM

P: 69

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/v3...ii/Graph34.jpg I would mark extrem of M as 4.68 kNm 



#13
Feb2606, 11:23 AM

P: 96

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
Feb2606, 11:27 AM

P: 69

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
Feb2606, 02:20 PM

P: 69

Hmm does anyone know this?




#16
Feb2606, 02:58 PM

P: 96

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*8m9.67kN*4m4.68kn*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. 



#17
Feb2606, 03:05 PM

P: 69

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
Feb2606, 03:20 PM

P: 96

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.



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