How to Solve Mechanics Problems Involving Framed Structures

[PhanthomJay]

... :( i think that this problem is easy if you draw the proper free body diagram of each section, but i don't know how to draw the free body diagram of each section as there are too many inclined and different things that i never saw. I did only one hour lecture on bending beams and one hour seminar on bending beams. That's all. How can i solve such this question with one hour lecture and only the simplest on beams? From the other hand i believe that a person which know more on bending beams it is easy to solve this in 10 minutes. I can't be a god with only two hours cover on bending beams. I try many books and notes but nothing. I have some hours to do it but i don't think so as you understand what is my level on bending beams. :(

It is not easy. Drawing the free body diagams is essential, but it is only the beginning. Drawing the shear and moment diagrams is the difficult part, especially due to the load distribution which gives a shear diagram that has a quadratic curve in part for the horizontal beam. With a 1 hour lecture on simple beams, there is no way you can solve it. It would probably take me about 15 minutes to do it properly, but I've been doing it for 40 years. Only a student with more than a few hours lecture will be able to sokve it all corectly.

 

[mikex24]

yes i understand it.

 

[PhanthomJay]

You can't pass the module with a 1 hour lecture. Unless you studied it on your own for many many hours after that. I don't know what you were expecting.

 

[mikex24]

it is not just bending on beams. There are many others such as second moment of area, kinetics, kinematics...

 

[PhanthomJay]

it is not just bending on beams. There are many others such as second moment of area, kinetics, kinematics... but because this is an assignment and the lecturer thinks that the students are experts with one hour lecture and one hour seminar on bending beams.. :( .

This problem as written has nothing to do with bending of beams. It asks for the shear and moment diagrams. The bending stress is simple, it 's just Mc/I. It sounds to me like you enrolled in a course without having the necessary prerequisite courses on the basics.

 

[mikex24]

Hello. I did some work and i need some tips for the next stage:

 

[mikex24]

Hello. I did some work and i need some tips for the next stage:

I found the P1, P2, Mx, Qx

P1=20*2.15=43 N and it acts (2.15/2)+0.1=1.175 from the right end
P2=0.5*2.15*20=21.5 N and it acts (2.15/3)+0.1=0.816 from the right end
therefore P1+P2=Qx therefore Qx=64.5 N
so Mx=(43*1.175)+(21.5*0.816)= 68 Nm
therefore R=Qx=64.5 so Rcos15=62.3 and Rsin15=16.69

is that right? Cheers

 

[PhanthomJay]

I'm glad you did not give up. Those values are correct, but I'm not sure why you call it Qx and Mx; the value you calculated for Qx is the vertical upward internal reaction at the right end of the horizontal beam (call it Q), and the value for Mx (call it M_right) is the internal clockwise moment at the right end of the horozontal beam. You still need to calculate the moment at the support. Now you need to draw the shear and moment doagrams as requested iin the problem..

 

[mikex24]

yes... here is the trouble. how can i found the moment on the fixed support(i think that is the same as the upward but in different direction)?

 

[PhanthomJay]

l already gave you the equation for the moment at the fixed base support, in a sketch many posts ago. That is one way of doing it. But you need to calculate the values of x1 and x2, using basic geometry and trig.

 

[mikex24]

I did it. I found 64.5. I have to use the Rcos15 to find the x1 and x2?

 

[mikex24]

I use trigonometry and i found the x1 and x2 and i found the Mx at the fixed end of the framework which is : 134.805. ??

 

[mikex24]

What is the total plan and the next stages to solve this question?

 

[PhanthomJay]

What is the total plan and the next stages to solve this question?

Please tell me how you would go about drawing shear and moment diagrams for the more simple case of a simple beam supported at each end with a uniformly distributed load. You have to get back to the basics...as i tried to note before, if you don't understand the basic steps, you will find drawing the shear and moment diagrams for this problem quite difficult...and I'm not about to do it for you...we are here to assist, but you must do the work.....

 

[mikex24]

i know that we have to cut the beam in x values to find in each point of x the value of moment and force. ?
for A-B the Mb and Qb will zero
then if i start from the left of the beam
then for B-C part
i will cut the just after the first force which is the force of triangular
the problem i found there is that is that i can't find many unknows x to have quadtrative equation because always the unknows x is just one like 8x-3 and this is a straight linear equation. ??

 

[Joshsamuel117]

We've got a similar problem(the exact same) to this one for our assignment and we were wundering if the strength of the tubeing by workin out the external area then taking the internal area away from it to get the area of the piping and then divide the values of the forces we got from the wieght of the billboard. we did this but the values we got were extremely small compared to the yeild stress of the low carbon steel. the value we got was 0.2MNm^-2 but the yeild stress is 300MNm^-2 thanks for any help you can give

 

[PhanthomJay]

We've got a similar problem(the exact same) to this one for our assignment and we were wundering if the strength of the tubeing by workin out the external area then taking the internal area away from it to get the area of the piping and then divide the values of the forces we got from the wieght of the billboard. we did this but the values we got were extremely small compared to the yeild stress of the low carbon steel. the value we got was 0.2MNm^-2 but the yeild stress is 300MNm^-2 thanks for any help you can give

This is the 3rd separate post on this same problem, and I need to know up front where this assignment is coming from, and for what purpose it is being assigned, before we can provide additional assistance.

 

[mikex24]

This is an assignment of a UK university. As you can there are many person with difficulties on this assignment as we teached only 2 hours on this module for the beams..

 

[PhanthomJay]

This is an assignment of a UK university man. As you can there are many person with difficulties on this assignment as we teached only 2 hours on this module for the beams..

If this is a homework assignment, we can offer help, but not solutions, and even then, it is assumed that the 'UK university man' does not object to your seeking outside help. If this is a take home exam that determines whether you pass or fail a course, for example, then we cannot provide help, per Forum rules. That would be , in a sense, cheating. That is why I am being a bit cautious here, and I need you and the others to be up front and honest about this.

 

[mikex24]

PhantomJay how can i find the induced stresses on this tube? it has thickness t=2.6mm and radious=21.2mm. I try to find them by torsion theory. the problem is that i don't know the shear modulus of low carbon steel.

 

[PhanthomJay]

PhantomJay how can i find the induced stresses on this tube? it has thickness t=2.6mm and radious=21.2mm. I try to find them by torsion theory. the problem is that i don't know the shear modulus of low carbon steel.

There are no torsional stresses, and you don't need any modulus to solve for the induced stresses. With all due respect, especially since i know this problem is eating away at you, you need a lot more background, like a semester or 2 or more of prerequisite courses. You can't expect to become an engineer overnight...it takes 4 long and hard years...

 

[PhanthomJay]

We've got a similar problem(the exact same) to this one for our assignment and we were wundering if the strength of the tubeing by workin out the external area then taking the internal area away from it to get the area of the piping and then divide the values of the forces we got from the wieght of the billboard. we did this but the values we got were extremely small compared to the yeild stress of the low carbon steel. the value we got was 0.2MNm^-2 but the yeild stress is 300MNm^-2 thanks for any help you can give

Hey Josh, you may be looking at axial stresses in the slanted member, or perhaps shear stresses, I don't know, but in any case, these stresses pale in comparison to the max bending stress in the frame. Focus on the bending stress.

 

[D44]

For this same problem I have calculated a reaction force at the rhs of the horizontal part of 64.5N. So this would also have to be the same as the reaction force of the ground on the stand. I understand how to calculate the resultant force and distance of the trapezoid (hence calculating the reaction force). I also understand how to calculate the x values that phanthomjay suggested on previous posts. I just don't know how to calculate the shear forces and bending moments. I have tried several different ways and I'm getting a few different results. I'm just not exactly sure what the correct way is.

As for the rest...I'll worry about that after...

If anyone could help, it'd be much appreciated

Cheers

 

[jobz1]

i have got the same problem
i don't know how to solve when it comes to shear force and moment.. i mean when u cut it into pieces.
pls help

 

[PhanthomJay]

This problem seems to be a popular one. This response is for you and jobz1

For this same problem I have calculated a reaction force at the rhs of the horizontal part of 64.5N. So this would also have to be the same as the reaction force of the ground on the stand.

yes, correct, 64.5 N up

I understand how to calculate the resultant force and distance of the trapezoid (hence calculating the reaction force). I also understand how to calculate the x values that phanthomjay suggested on previous posts. I just don't know how to calculate the shear forces and bending moments. I have tried several different ways and I'm getting a few different results. I'm just not exactly sure what the correct way is.

As for the rest...I'll worry about that after...

If anyone could help, it'd be much appreciated

Cheers

Start with the horizontal member first. Draw a free body diagram of that member, isolating it (cutting it) just to the left of the knee. The vertical (shear) force at that cut is 64.5 N, as you have noted. That is the shear at that point. There is also a moment at that point. Can you calculate the moment at that point? It acts opposite and equal to the moment about that point from the trapezoidal load. The moment from the trapezoidal load can be found by breaking that load into a uniformly distributed load and a triangularly distributed load, and summing the moments of the resutants of each about that point.

 

[D44]

I think I have calculated the moment about the rhs point. I have this as 68.112Nm. Does this look about right?

When calculating shear forces and bending moments, do I still need to keep the rectangle and triangle separate? This part is what's causing trouble!

 

[PhanthomJay]

I think I have calculated the moment about the rhs point. I have this as 68.112Nm. Does this look about right?

Yes, clockwise on the beam and counterclockwise on the joint. Round it off to 68.1 N-m

When calculating shear forces and bending moments, do I still need to keep the rectangle and triangle separate? This part is what's causing trouble!

It's easiest to do it this way, since it helps to fing the load at a given point and the cg of the load. Start with the shear diagram, working left to right. There's nothing going on between 0 and 0.1 m, so the shear is 0 in this portion. Now look at the next section between 0.1 m and 2.15 m. Call the 0.1 m mark the '0' point along the x axis. Now draw a free body diagram of the beam around the left end that cuts through the distributed loading at a distance 'x' from the '0' point. What do you get as a total (shear) load at that cut, as a function of x? What do you get for the bending moment at that cut? You should proceed in the same manner as you did when calculating the shear and moment at the knee, except you will have your answer in terms of x.

 

[D44]

I understand what you said about zero shear force in the 0.1m section. When drawing the free body diagram for the cut, I still need to include this 0.1m section though, right?

I've had a go at this bit already and got 3 possibilities:
Qx=-4.65(x-0.1)^2
Qx=-4.44(x-0.1)^2
Qx=-4.26(x-0.1)^2
When x=0.1, Qx=0 and when x=2.25, Qx=20.52N?

For the rectangular section:
Qx=-20(x-0.1)
When x=0.1, Qx=0 and when x=2.25, Qx=-43N?
For the bending moments I have 0Nm at 0.1m from the left, -46.23Nm at the 2.25m point and -50.53Nm at the very rhs?

Cheers

 

[PhanthomJay]

I understand what you said about zero shear force in the 0.1m section. When drawing the free body diagram for the cut, I still need to include this 0.1m section though, right?

well you could, and did, but it just gets in the way and makes calculations more difficult than they need be. Just cut out that piece and chuck it. Pretend its not there. It's at the free end of the beam with no loading, so it doesn't do anything.

I've had a go at this bit already and got 3 possibilities:
Qx=-4.65(x-0.1)^2
Qx=-4.44(x-0.1)^2
Qx=-4.26(x-0.1)^2
When x=0.1, Qx=0 and when x=2.25, Qx=20.52N?

Your first possibility is correct for the triangular piece, in which case the shear (Qx?) in this section at x = 2.25 is 21.5 N under this load. I don't know where your other possibilities came from, since you did not show your work. Also, you can get rid of that 0.1 and just say Qx = -4.65x^2, where x is 0 at the start and 2.15 m at the end of the distributed loading , instead of 0.1 and 2.25 m.
, respectively.

For the rectangular section:
Qx=-20(x-0.1)
When x=0.1, Qx=0 and when x=2.25, Qx=-43N?

Yes, good. Now add up the shear from each loading to get 64.5 N at your x =2.25 m, whch agrees with the shear you got earlier at the knee (the shear is constant over the last 0.1 m). In between (your) x =0.1 and x= 2.25 , the shear is Qx= (-4.65(x-0.1)^2) + (-20(x-0.1)). The slope of the shear diagram at a given point is the negative of the loading intensity (N/m) at that point.

For the bending moments I have 0Nm at 0.1m from the left, -46.23Nm at the 2.25m point and -50.53Nm at the very rhs?

Cheers

Something's amiss here because the moment at the very rhs is -68.1 which we established earlier. Please show your work, as you are making me work hard for my money

 

[D44]

Hmmm... but then surely I already knew the shear forces of both the triangle and the rectangle when I calculated the resultant force in order to work out the 64.5N force as the reaction force at the right hand side?

The moment of the triangle I got as:
Mx=-(4.65*(x-0.1)^2)*((x-0.1)/3)
At 2.25, Mx=-13.35Nm
At 0.1, Mx=0

As for the rectangle...
Mx= -10(x-0.1)^2

 

[D44]

Ooops, Mx = -4.65x^3 for the triangle

 

[D44]

Sorry

For the rectangle I have between 0.1 and 2.25m:
Mx=-10(x-0.1)^2
at 0.1 Mx = 0
at 2.25 Mx = -46.23Nm

Then between 2.25m and the rhs I have:
Mx = -43(x-1.175)
at 2.25m Mx = -46.23Nm
at 2.35m Mx = -50.5Nm

I now understand that I can remove the 0.1 from my equations. I'm just keeping them in for consistency on here :)

 

[D44]

Right...

I've been looking at it some more...

Rectangular section:
Mx= -43*(x-1.175)
at x=2.25m, Mx=-46.23Nm
at x=2.35m, Mx=-50.5

Triangular section:
Mx= -21.5*(x-((2/3)*2.15))
at x=2.25m, Mx=-15.48Nm
at x=2.35m, Mx=-17.63Nm

Looking any better?

 

[PhanthomJay]

Hmmm... but then surely I already knew the shear forces of both the triangle and the rectangle when I calculated the resultant force in order to work out the 64.5N force as the reaction force at the right hand side?

The moment of the triangle I got as:
Mx=-(4.65*(x-0.1)^2)*((x-0.1)/3)
At 2.25, Mx=-13.35Nm

math error...Mx = - 15.4

At 0.1, Mx=0
As for the rectangle...
Mx= -10(x-0.1)^2

yes...

 

[PhanthomJay]

Right...

I've been looking at it some more...

Rectangular section:
Mx= -43*(x-1.175)
at x=2.25m, Mx=-46.23Nm
at x=2.35m, Mx=-50.5

Triangular section:
Mx= -21.5*(x-((2/3)*2.15))
at x=2.25m, Mx=-15.48Nm
at x=2.35m, Mx=-17.63Nm

Looking any better?

Yes, that looks about right...good work. Now you have to (roughly) draw the shear and moment diagrams... watch your curvatures...then proceed to the slanted member.