How to Solve Mechanics Problems Involving Framed Structures

[mikex24]

Homework Statement

A flat, solid trapezoidal-shaped metallic bill board plate is welded to the inverted
L-shaped framework, the details of which are shown in the figure below. The frame
is to be made of standard hollow tubing of low carbon steel with typical yield stress
of 300 MNm-2. (I attached a photo of the diagram)

I have to :
a. Draw the free-body diagrams of each section of the framework(forces and moments)
b. Determine the shear forces and bending moments in the frame and draw
diagrams to show their distribution.

If anyone can help me to solve it i would appreciate it. Cheers and i hope a kind person help me or give the idea how to solve it. Thanks.

 

[PhanthomJay]

mikex24

I'd like to first welcome you to PF.

I'm pretty kind-hearted I think, but you've got to show some effort before we can help. Those are the rules.

Thanks, please show something...

 

[mikex24]

mikex24

I'd like to first welcome you to PF.

I'm pretty kind-hearted I think, but you've got to show some effort before we can help. Those are the rules.

Thanks, please show something...

Thanks for the reply and thanks for the quick response .. I never get in touch with inclined and combo problems like this one.. I understand that i have to find shear stresses and moments and that i have to do two different shapes. But there is and an inclined frame which changes the order of the solution. :(

 

[PhanthomJay]

The problem gets a bit messy with that trapezoidal load distribution, but nevertheless, the first step is to find the support reactions at the fixed end support. Can you do this? The trapezoidal loading consists of a uniform load plus a triangularly distributed load. You should find the resultant for each when determining the support reactions.

 

[mikex24]

Thank you very much. Yes, i think. There is one moment on the fixed and and a shear force on the frame which it has two components. One on the X axis and one on the Y axis. Cheers

 

[PhanthomJay]

Thank you very much. Yes, i think. There is one moment on the fixed and and a shear force on the frame which it has two components. One on the X axis and one on the Y axis. Cheers

At the support, yes, there is a fixed end moment which you can calculate by summing moments of the loading force about it. But in terms of the force, R, at that support, and with the assumption that the x-axis is horizontal and the y-axis is vertical, then what's the force in the X direction using Newton 1 in that direction? The force, R, can be split into components parallel to, and perpendicular to, the sloping member.

 

[mikex24]

xm.. because i am student and this is the first time i am facing such problems it is little difficult to understand. I think that i get the forces on the fixed end. i attached a pic. Cheers

 

[pongo38]

There are two alternative definitions of shear force, and both could be helpful. If you were to draw the M diagram first, then the shear force can be obtained from the rate of change of M. Alternatively use the definition for shear force at a section being the algebraic sum of forces parallel to that section, and on one side of it. This is a statically determinate problem, and therefore everything is self-checking: the reaction components can be checked, and so can all the internal resultants as you develop the problem to its conclusion. You won't have to ask if it's right because you can do all the checks yourself. I think it would have been helpful if you had also been asked for the normal force (axial force) diagram, as that brings completeness to the problem.

 

[PhanthomJay]

xm.. because i am student and this is the first time i am facing such problems it is little difficult to understand. I think that i get the forces on the fixed end. i attached a pic. CheersView attachment 24272

You have identified the moment at the fixed end support, but its direction is wrong. Can you calculate its value? Regarding the reaction force at the end, you have not correctly shown the components. Since the frame is subject to vertical loading only, what must be the magnitude and direction of the reaction force at the support? Can there be a horizontal component to it in light of the fact that there are no horizontal forces applied? Once you identify the reaction force and its direction, you can then break it into componnents parallel to and perpendicular to the sloping member. Then proceed with the moment and shear diagrams in accordance with pongo38 suggestions. You'll probably have to rotate your paper 85 degrees ccw to draw the moment and shear diagrams for that piece (or tilt your head ). You should be thoroughly familiar with drawing shear and moment diagrams for simple cases (like a horizontal cantilever with uniform load distribution), before attempting to solve this problem.

 

[mikex24]

thank you very much for any help but i can't understand without a sketch. Cheers and thanks for your time.

 

[PhanthomJay]

thank you very much for any help but i can't understand without a sketch. Cheers and thanks for your time.

maybe this helps, I don't know...you'll have to calculate P1, P2, x1, x2, M, R, and its components...and that's just the beginning...


http://img28.imageshack.us/img28/5542/invertedslantedlframe.jpg

 

[mikex24]

on the Bill Board which there is the P1 and P2 there isn't any upward force on the beam? Cheers

 

[PhanthomJay]

on the Bill Board which there is the P1 and P2 there isn't any upward force on the beam? Cheers

P1 and P2 are the applied loads on the frame...P1 represents the weight from the rectangular piece of the billboad and P2 represents the weight from the traingular piece..these are downward loads on the beam (gravity acts down). In terms of upward loads, there must be upward reaction loads on the frame for equilibrium (R = P1 +P2, up), and also, for equilibrium, internal upward loads on the horizontal beam, whch you can find by taking a free body diagram of that horizontal beam. For example, if you cut it free from the slanted member at the right far right end of the horizontal beam, there must be an upward shear force equal to P1 + P2 at that cut, (plus a bending moment). If that is what you mean ...continue.

 

[mikex24]

cheers mate i understand this but on free body diagram i think i have to show only the forces and how can i show the equal to the P1 and P2 shear force (or on free body diagram i have to part he body and show my forces?)?

 

[PhanthomJay]

cheers mate i understand this but on free body diagram i think i have to show only the forces and how can i show the equal to the P1 and P2 shear force (or on free body diagram i have to part he body and show my forces?)?

Free body diagrams must include all forces and all moments acting on the object. I chose the FBD of the top horiz beam as an example; If you took the FBD thru the appied billboard load, the forces and moments become a bit more dificult ot calculate (using Newton 1), since you cannot simply use the resultant load P1 or P2 when looking internally within that dostributed load area..
You should 'google' on Shear and mmoment diagrams.

 

[mikex24]

can anyone help me as i have no more time to solve it? I have to hand it until Monday and i don't know how can i sovle it. Cheers

 

[pongo38]

Can you get the shear force and bending moment just to the left of the 'knee' of the frame? If you can do this on a FBD of the horizontal beam, then draw a FBD for the sloping member, and at the top, you put the shear and moment from the beam, but with arrows reversed. Then resolve the force parallel and perpendicular to the sloping member. Then draw the M and V force diagram for the sloping member. The end reactions at the bottom should then agree with your global approach to the reactions as suggested by Jay. Good luck.

 

[mikex24]

PhantomJay do you get my msg? Cheers

 

[PhanthomJay]

PhantomJay do you get my msg? Cheers

Yes, you've been given quite a few tips...now it's time for you to show some work , and someone can provide further assistance.

 

[mikex24]

yes. thank you very much and i hope someone assist me .. :(

 

[PhanthomJay]

yes. thank you very much and i hope someone assist me .. :(

Mikex24: You must show some attempt at this on your own...start by calculating P1 and P2, so that you cam determine the shears and moments just to the left of the knee, and/or the shears and moments at the support...

 

[mikex24]

i am able to find the shear forces on a normal beam which there are loads and there are two reactions on the two end of that beam but it is difficult to this one as it is an inclined frame and there is only loads and many planes. Also i try to find the forces but if the forces is only P1, P2, shear force and the moment how an equilibrium takes place?? all of the has direction down and which forces is upward? Also what is the aim of the other two forces on the down end of the frame for this stage? Also how can i determine the x for each force (the forces the aren't stable ) ? If i had a correct FBD it will be moore easier to solve it. sorry for my english.

 

[PhanthomJay]

i am able to find the shear forces on a normal beam which there are loads and there are two reactions on the two end of that beam but it is difficult to this one as it is an inclined frame and there is only loads and many planes. Also i try to find the forces but if the forces is only P1, P2, shear force and the moment how an equilibrium takes place?? all of the has direction down and which forces is upward? Also what is the aim of the other two forces on the down end of the frame for this stage? Also how can i determine the x for each force (the forces the aren't stable ) ? sorry for my english.

OK why don't you start with the top horizontal member. Cut it loose from the frame. in a free body diagram (Pretend that you cut it with a saw just left of the knee joint, and put it aside.). Now when you cut it at the knee, there are generally 3 unknown internal forces or moments at that cut...a horizontal force, a vertical force, and a moment. Pretend that the vertical force at that cut is a support if it helps, and now you should be able to solve for the upward value of this force, using sum of forces in vertical direction = 0. You can also solve for the unknown moment by using sum of moments about that cut = 0. There won't be any horizontal force at that cut because there are no horizontal forces applied. You must of course calculate P1 and P2, which you apparently have not done yet.

 

[mikex24]

very very difficult to understand these forces. i know that i have 2 downward forces one for the plate and one for the weight of the horizontal plate. also i know that if i cut the horizontal beam there are one addition force(shear force downward) and a moment. so P1+P2+QX=0 this is for the equilibrium of the shear force and for the moments i have to calculate the x of each force. is that right?

 

[PhanthomJay]

very very difficult to understand these forces. i know that i have 2 downward forces one for the plate and one for the weight of the horizontal plate. also i know that if i cut the horizontal beam there are one addition force(shear force downward) and a moment. so P1+P2+QX=0 this is for the equilibrium of the shear force and for the moments i have to calculate the x of each force. is that right?

If you let Q be the unknown vertical force at the cut end, then P1 + P2 + Q = 0. Sove for Q. It will have a negative value, which means it points upward. But you still have not calculated P1 and P2 yet, why are you holding back? Now for the moment equation, yes, you need to find the x values between P1 and the right end of the horizontal member, and also the x value between P2 and that end.

 

[mikex24]

how P1 P2 can be calculated if they don't have a stable point...?

 

[mikex24]

difficult.

 

[PhanthomJay]

i try this but i don't know if it right. Q/40=x/2.15 => Q=40*(x/2.15) therefore i found the P2 which P2=0.5*(Q*x) => P2=9.30x^2... is that right?

I guess your definition of the stable point is the point of the resultant force of the weight distribution. You should know for a uniform load (the rectangle part) that P1 is just (w)L, where in this case, w is 20 N/m and L is 2.15 m. Thus P1 = 43 N, and it acts at the center of gravity of the load, which is halfway in between. Thus, P1 acts 2.15/2 + 0.1 = 1.18 m to the left of the right end of the horizontal member. Now try P2, a little more difficult.

 

[mikex24]

yes but why you calculate the P1 only for the rectangle part? what about triangle part and what about P2?

 

[PhanthomJay]

yes but why you calculate the P1 only for the rectangle part? what about triangle part and what about P2?

P2 is the triangular part.