Maximum weight a bar and cables can hold before breaking

[Alison A.]

Homework Statement

A weight W is supported by bar AO and cables AB, AC, and AD. Point A lies in the xy plane, if the bar and cables have the strengths listed below, determine the largest value W may have.

Hint given: A FBD of joint A is recommendedto being this problem. Bar AO should be assumed to carry load along its length. Unit vectors point from A to B and from A to C are needed to describe the direction of cable tensions. Since cables are required to be in tension, it should be clear from the figure that AO will be in compression. It is usually not obvious which structural member first reaches its limit load. Loads carried by each member should be obtained in terms of W. Individual failure criteria can then be applied to each member to find W inducing failure in that member. The largest weight that may be supported is the minimum of this set of failure loads.

Homework Equations

Unit vector equation
ΣF=0, ΣF_x=0, ΣF_y=0,
Basic trig

The Attempt at a Solution

First I drew out the FBD which was basically exactly the same as the picture given so I didn't find it necessary to post it. Then I found the point coordinates to be:
A (15,48,0) in
B (15,0,36) in
C (0,48,36) in
O (0,0,0) in

Like the hint suggests, I found the unit vector of AB and AC to be (0,-4/5,3/5) and (-5/13,0,12/3) respectively.
Then I multiplied those by the strengths given making the, equal to (0,-1568,1176) for AB and (-538.5,0,1292) for AC. The problem statement says the force of AO is already in it's direction so AO would be (29400,94080,0).

From here I don't really know how to go about "Loads carried by each member should be obtained in terms of W." I just need a good push in the right direction.

Thank you so much for any help, I really appreciate it!

 

[SteamKing]

You might want to check the coordinates for point A.

What you have to do in order to check the strength of each member is to calculate the loads in OA, AB, and AC using the equations of equilibrium, assuming that the load is W applied at point D. These loads will be in terms of W. Once you have calculated the loads, you then compare them with the allowable strengths listed in the table to find out what the maximum value of W can be without overloading any part of this structure.

 

[Alison A.]

You might want to check the coordinates for point A.

What you have to do in order to check the strength of each member is to calculate the loads in OA, AB, and AC using the equations of equilibrium, assuming that the load is W applied at point D. These loads will be in terms of W. Once you have calculated the loads, you then compare them with the allowable strengths listed in the table to find out what the maximum value of W can be without overloading any part of this structure.

My bad, on my actual work I have written out my A coordinate is (15,48,0) I just had a typo. So when I'm finding the components of all the cables, D is the only one that will have the vertical W force? How does that make every other cable in terms of W?

 

[SteamKing]

My bad, on my actual work I have written out my A coordinate is (15,48,0) I just had a typo. So when I'm finding the components of all the cables, D is the only one that will have the vertical W force? How does that make every other cable in terms of W?

Because W is the only load applied to this structure. All of the loads in the cables and the bar OA are produced by this one load; hence, it stands to reason that these loads will also be proportional to the magnitude of W.

Like I said in my previous post, you'll have to solve for the loads in each member using the equations of static equilibrium. The cables will be in tension, because they don't work if they're not, but the bar OA will probably be in compression.

 

[Alison A.]

Let me see if I got this right, I found the unit vectors and I know the tension/compression of all of them because of the table given.

AB = (0, -1568, 1176)
AC = (-538.5, 0, 1292)
OA = (29400, 94080, 0)

So, from theses I can find
ΣF_x = 0 - 538.5 + 29400 = 28861.5
ΣF_y = -1568 + 0 + 94080 = 92512
ΣF_z = 1176 + 1292 + 0 = 2468

I know w is in the z-direction, but I'm not really sure where to apply it from here.

EDIT: Do I find the magnitudes of AB, AC and OA? I attempted that and tried to put in the lowest number for the answer and that didn't work.

 

[SteamKing]

Let me see if I got this right, I found the unit vectors and I know the tension/compression of all of them because of the table given.

AB = (0, -1568, 1176)
AC = (-538.5, 0, 1292)
OA = (29400, 94080, 0)

No, these are not the forces created by the weight W. You are misinterpreting the maximum allowable forces from the given table as the actual forces in the structure.

So, from theses I can find
ΣF_x = 0 - 538.5 + 29400 = 28861.5
ΣF_y = -1568 + 0 + 94080 = 92512
ΣF_z = 1176 + 1292 + 0 = 2468

I know w is in the z-direction, but I'm not really sure where to apply it from here.

EDIT: Do I find the magnitudes of AB, AC and OA? I attempted that and tried to put in the lowest number for the answer and that didn't work.

For some reason, you are resisting my repeated suggestions that you find the forces in the structure due to the applied weight W, whose maximum magnitude is what you are supposed to be finding. You must use the equations of static equilibrium in order to do this. This is not a simple plug-and-chug problem, which you assume it to be.

 

[Alison A.]

No, these are not the forces created by the weight W. You are misinterpreting the maximum allowable forces from the given table as the actual forces in the structure.For some reason, you are resisting my repeated suggestions that you find the forces in the structure due to the applied weight W, whose maximum magnitude is what you are supposed to be finding. You must use the equations of static equilibrium in order to do this. This is not a simple plug-and-chug problem, which you assume it to be.

...I don't know what you are referring to when you say equations of static equilibrium. I get they all have W acting on them but we don't know W so I don't understand how I
"find the forces in the structure due to the applied weight W". How are the loads of the structures not the magnitudes. I don't get it.

 

[SteamKing]

...I don't know what you are referring to when you say equations of static equilibrium. I get they all have W acting on them but we don't know W so I don't understand how I
"find the forces in the structure due to the applied weight W". How are the loads of the structures not the magnitudes. I don't get it.

The equations of static equilibrium are:

ΣM = 0, ΣF_x = 0, ΣF_y = 0, ΣF_z = 0

Thus, there should be no net force and no net moment acting at point A.

If you are not comfortable with using W for the load on the structure, then assume for the purposes of this calculation that W = 1 lb or 100 lbs or 1000 lbs, whichever value is more convenient. The forces in the structure you obtain from your calculations can then be scaled by this assumed value to find out their proportion to the applied load W. Then, you may use the values from the table to find out what the actual maximum magnitude of W is such that the structure is not overloaded.

 

[Alison A.]

The equations of static equilibrium are:

ΣM = 0, ΣF_x = 0, ΣF_y = 0, ΣF_z = 0

Thus, there should be no net force and no net moment acting at point A.

If you are not comfortable with using W for the load on the structure, then assume for the purposes of this calculation that W = 1 lb or 100 lbs or 1000 lbs, whichever value is more convenient. The forces in the structure you obtain from your calculations can then be scaled by this assumed value to find out their proportion to the applied load W. Then, you may use the values from the table to find out what the actual maximum magnitude of W is such that the structure is not overloaded.

Alright, but I did take the sum of all the components and you told me I wasn't listening to you. I'm assume I take the sum without adding the strengths.

That would be
ΣF_x = 0 - 5/13 - 15
ΣF_y = -4/5 + 0 - 48
ΣF_z = 3/5 + 12/3 + 0

Making ΣF_M = ( -200/13, -244/5, 23/5)

Is this along the line of what you are talking about? I am having trouble seeing where W is going to be applied here, I know it scales the the applied load, but is that found by the all the components?

 

[SteamKing]

Alright, but I did take the sum of all the components and you told me I wasn't listening to you. I'm assume I take the sum without adding the strengths.

That would be
ΣF_x = 0 - 5/13 - 15
ΣF_y = -4/5 + 0 - 48
ΣF_z = 3/5 + 12/3 + 0

Making ΣF_M = ( -200/13, -244/5, 23/5)

Is this along the line of what you are talking about? I am having trouble seeing where W is going to be applied here, I know it scales the the applied load, but is that found by the all the components?

I have no idea what these calculations represent. They certainly are not the components of the forces in the cables AB and AC or the bar OA.

You should start by calculating the unit vectors for cables AB and AC and the bar OA.

Like the hint suggests, I found the unit vector of AB and AC to be (0,-4/5,3/5) and (-5/13,0,12/3) respectively.

The unit vector for AB looks OK, but you should check the unit vector for AC. You also should calculate the unit vector for OA. Remember to check the lengths of the unit vectors to see if each equals 1 exactly.

 

[Alison A.]

The unit vector for OA is (-0.2983, -0.9545, 0)
AB = (0, -4/5, 3/5)
AC = (-5/13, 0, 12/13)

I still got the same thing for AC, AC = (0-15, 48-48, 36-0) = (-15, 0, 36) and the magnitude of this is 39. I reduced the fraction and got (-5/13, 0, 12/13) for the unit vector

 

[SteamKing]

The unit vector for OA is (-0.2983, -0.9545, 0)
AB = (0, -4/5, 3/5)
AC = (-5/13, 0, 12/13)

I still got the same thing for AC, AC = (0-15, 48-48, 36-0) = (-15, 0, 36) and the magnitude of this is 39. I reduced the fraction and got (-5/13, 0, 12/13) for the unit vector

You still need to calculate a unit vector for the bar OA.

 

[Alison A.]

You still need to calculate a unit vector for the bar OA.

I did... (-0.2983, -0.9545, 0)

 

[SteamKing]

I did... (-0.2983, -0.9545, 0)

Now, the forces in the cables, call them F_AB and F_BC, along with the force in the bar, F_OA, are all unknown at this point.

By multiplying the magnitude of each force by the corresponding unit vector for the member, you can obtain the components of each force. These components are then used to write the equations of equilibrium for this structure. You can write a single force equation each for the x-, y-, and z- directions. You can also write one moment equation.

 

[Alison A.]

Now, the forces in the cables, call them F_AB and F_BC, along with the force in the bar, F_OA, are all unknown at this point.

By multiplying the magnitude of each force by the corresponding unit vector for the member, you can obtain the components of each force. These components are then used to write the equations of equilibrium for this structure. You can write a single force equation each for the x-, y-, and z- directions. You can also write one moment equation.

Are the magnitudes the same as the strengths given?
OA = 1960(-0.2983, -0.9545, 0) = (-584.668, -1870.82, 0) (1960, because AO is in compression)
AB = 1960(0, -4/5, 3/5) = (0, -1568, 1176)
AC = 1400(-5/13, 0, 12/13) = (-538.462, 0, 1292.31)

From these
ΣF_x = -584.668 + 0 + -538.462 = -1123.13
ΣF_y = -0.9545 - 1568 + 0 = -1568.9545
ΣF_z = 0 + 1176 + 1292.31 = 2468.31

ΣF_m = (-1123.13, -1567.9545, 2468.31)

 

[SteamKing]

Are the magnitudes the same as the strengths given?
OA = 1960(-0.2983, -0.9545, 0) = (-584.668, -1870.82, 0) (1960, because AO is in compression)
AB = 1960(0, -4/5, 3/5) = (0, -1568, 1176)
AC = 1400(-5/13, 0, 12/13) = (-538.462, 0, 1292.31)

From these
ΣF_x = -584.668 + 0 + -538.462 = -1123.13
ΣF_y = -0.9545 - 1568 + 0 = -1568.9545
ΣF_z = 0 + 1176 + 1292.31 = 2468.31

ΣF_m = (-1123.13, -1567.9545, 2468.31)

No, they are not. You must find the forces in the members before you can use the loads from the tables.

The forces in the members will be some factor times W for each member.

 

[Alison A.]

No, they are not. You must find the forces in the members before you can use the loads from the tables.

The forces in the members will be some factor times W for each member.

I know that the magnitudes of the unit vectors are going to be 1 so it must be from the original coordinates.

AB = (0, -48, 36) has a magnitude of 60
AC = (-15, 0, 36) has a magnitude of 39
OA = (-15, -48, 0) has a magnitude of 50.289

but this seems a little redundant.. I don't know what else the magnitudes could be

 

[SteamKing]

I know that the magnitudes of the unit vectors are going to be 1 so it must be from the original coordinates.

AB = (0, -48, 36) has a magnitude of 60
AC = (-15, 0, 36) has a magnitude of 39
OA = (-15, -48, 0) has a magnitude of 50.289

but this seems a little redundant.. I don't know what else the magnitudes could be

We need to find the magnitudes of the forces, not the lengths of the members.

Now, the forces in the cables, call them F_AB and F_BC, along with the force in the bar, F_OA, are all unknown at this point.

You must use the equations of static equilibrium to find the magnitudes of these forces.

 

[Alison A.]

OA = (-0.2983, -0.9545, 0)
AB = (0, -4/5, 3/5)
AC = (-5/13, 0, 12/13)

ΣF_x = -0.2983 + 0 - 5/13, Mag_x = 0.4867
ΣF_y = -0.9545 - 4/5 + 0, Mag_y = 1.2454
ΣF_z = 0 + 3/5 + 12/13, Mag_z = 1.1009

 

[SteamKing]

OA = (-0.2983, -0.9545, 0)
AB = (0, -4/5, 3/5)
AC = (-5/13, 0, 12/13)

ΣF_x = -0.2983 + 0 - 5/13, Mag_x = 0.4867
ΣF_y = -0.9545 - 4/5 + 0, Mag_y = 1.2454
ΣF_z = 0 + 3/5 + 12/13, Mag_z = 1.1009

You don't seem to be grasping the concept here.

The unit vectors for each member are as follows:
OA: (-0.2983, -0.9545, 0.0000)
AB: ( 0.0000, -0.8000, 0.6000)
AC: (-0.3846, 0.0000, 0.9231)

Multiply each member's unit vector by the corresponding magnitude of the unknown force:
F_OA = F_OA * (-0.2983, -0.9545, 0.0000)
F_AB = F_AB * ( 0.0000, -0.8000, 0.6000)
F_AC = F_AC * (-0.3846, 0.0000, 0.9231)

The vector for the weight W is simply W = W * (0, 0, -1)

If you want to write the equations of equilibrium, then you use the force vector components defined above.

For example:

∑ F_x = -0.2983 * F_OA - 0.3846 * F_AC = 0

I'll leave it to you to write the equations for ∑ F_y and ∑ F_z

 

[Alison A.]

You don't seem to be grasping the concept here.

The unit vectors for each member are as follows:
OA: (-0.2983, -0.9545, 0.0000)
AB: ( 0.0000, -0.8000, 0.6000)
AC: (-0.3846, 0.0000, 0.9231)

Multiply each member's unit vector by the corresponding magnitude of the unknown force:
F_OA = F_OA * (-0.2983, -0.9545, 0.0000)
F_AB = F_AB * ( 0.0000, -0.8000, 0.6000)
F_AC = F_AC * (-0.3846, 0.0000, 0.9231)

The vector for the weight W is simply W = W * (0, 0, -1)

If you want to write the equations of equilibrium, then you use the force vector components defined above.

For example:

∑ F_x = -0.2983 * F_OA - 0.3846 * F_AC = 0

I'll leave it to you to write the equations for ∑ F_y and ∑ F_z

∑ F_x = -0.2983 * F_OA - 0.3846 * F_AC = 0
∑ F_y = -0.9545 * F_OA - 0.8000 * F_AB = 0
∑ F_z = 0.6000 * F_AB + 0.9231 * F_AC = 0

EDIT: ∑ F_z = 0.6000 * F_AB + 0.9231 * F_AC - W = 0

There isn't enough information here to find the forces since there isn't one equation that has a value for x, y and z.

 

[SteamKing]

∑ F_x = -0.2983 * F_OA - 0.3846 * F_AC = 0
∑ F_y = -0.9545 * F_OA - 0.8000 * F_AB = 0
∑ F_z = 0.6000 * F_AB + 0.9231 * F_AC = 0

EDIT: ∑ F_z = 0.6000 * F_AB + 0.9231 * F_AC - W = 0

There isn't enough information here to find the forces since there isn't one equation that has a value for x, y and z.

Well, we haven't finished writing all of our equations of equilibrium. These are just the force equations.

There is still one equation to be written for the sum of the moments about point O, ∑ M_O = 0.

I don't know if you've learned how to calculate the moment of a force vector using the cross product,

or M = r × F.

 

[Alison A.]

∑ F_x = -0.2983 * F_OA - 0.3846 * F_AC = 0
∑ F_y = -0.9545 * F_OA - 0.8000 * F_AB = 0
∑ F_z = 0.6000 * F_AB + 0.9231 * F_AC = 0

EDIT: ∑ F_z = 0.6000 * F_AB + 0.9231 * F_AC - W = 0

Three equations, three unknowns

Well, we haven't finished writing all of our equations of equilibrium. These are just the force equations.

There is still one equation to be written for the sum of the moments about point O, ∑ M_O = 0.

I don't know if you've learned how to calculate the moment of a force vector using the cross product,

or M = r × F.

We did nothing like this for the examples my professor did in lecture for this homework, but I do know how to take the cross product. What is r in this case? I know it's the radius vector. And F is all the components we just found added together?

 

[SteamKing]

Three equations, three unknownsWe did nothing like this for the examples my professor did in lecture for this homework, but I do know how to take the cross product. What is r in this case? I know it's the radius vector. And F is all the components we just found added together?

You're close.

At point A, all of the forces in this structure created by the weight W come together. Since we have only one moment equation available, it's probably best to take Point O as the reference point for the radius vector r to each force.

Normally, if we were dealing with several known forces acting at point A, we would just find the resultant force R at this point and then calculate the moment vector as M = r × R. Since the forces on this structure are unknown, except for the one due to the weight W, it is easier if we compute r × F for each force at point A, with the same r being used in each calculation, that is vector OA. Once the individual moment vectors are calculated, their individual components can then be summed in a fashion similar to what was done with the force vectors.

 

[Alison A.]

You're close.

At point A, all of the forces in this structure created by the weight W come together. Since we have only one moment equation available, it's probably best to take Point O as the reference point for the radius vector r to each force.

Normally, if we were dealing with several known forces acting at point A, we would just find the resultant force R at this point and then calculate the moment vector as M = r × R. Since the forces on this structure are unknown, except for the one due to the weight W, it is easier if we compute r × F for each force at point A, with the same r being used in each calculation, that is vector OA. Once the individual moment vectors are calculated, their individual components can then be summed in a fashion similar to what was done with the force vectors.

Ok, so
OA x AB = (-0.5727, 0.1790, 0.2386)
OA x AC = (-0.8811, 0.2753, -0.3671)
and we already know OA x OA is going to be (0, 0, 0)

I hope this is right by crossing the unit vectors and not just the coordinates...

ΣF_x= -0.5727 - 0.8811 + 0
ΣF_y= 0.1790 + 0.2753 + 0
ΣF_z= 0.2386 - 0.3671 + 0

 

[SteamKing]

Ok, so
OA x AB = (-0.5727, 0.1790, 0.2386)
OA x AC = (-0.8811, 0.2753, -0.3671)
and we already know OA x OA is going to be (0, 0, 0)

I hope this is right by crossing the unit vectors and not just the coordinates...

ΣF_x= -0.5727 - 0.8811 + 0
ΣF_y= 0.1790 + 0.2753 + 0
ΣF_z= 0.2386 - 0.3671 + 0

Let's try this again by way of an example:

Suppose we want to calculate the moment of the force vector F_AB about the point O.
In other words, M_AB = r_OA × F_AB:

r_OA = (-15, -48, 0) in.
F_AB = (0.0, -0.8000, 0.6000) * F_AB lbs.

We can set up the cross product determinant like so:

rOA x FAB =

|  i      j       k     |
|-15     -48      0     |
| 0.0000 -0.8000  0.6000|

rOA x FAB = [(-28.8), (0+9), (12)] * FAB = FAB * (-28.8, 9, 12) in-lbs

where FAB is the unknown magnitude of the force vector F_AB.

To solve this problem, additional moments need to be set up and computed: r_OA × F_AC and r_OA × W
The moment r_OA × F_OA = 0 by definition.

 

[Alison A.]

F_AB = (-28.8, 9, 12)
F_AC = (-44.31, 13.85, -18.46)
W = (48, -15, 0)

Do these look right?

 

[Alison A.]

ΣF_x = -F_AB(28.8) - F_AC(44.31) + W(48)
ΣF_y = F_AB(9) + F_AC(13.85) - W(15)
ΣF_z = F_AB(12) - F_AC(18.46)

 

[SteamKing]

F_AB = (-28.8, 9, 12)
F_AC = (-44.31, 13.85, -18.46)
W = (48, -15, 0)

Do these look right?

These are not forces. They are moments, so they should use different notation. There is also a minor omission in that each moment vector should be multiplied by the magnitude of the associated force:

M_AB = F_AB * (-28.8, 9, 12) in-lbs.
M_AC = F_AC * (-44.31, 13.85, -18.46) in-lbs.
M_W = W * (48, -15, 0) in-lbs.

ΣF_x = -F_AB(28.8) - F_AC(44.31) + W(48)
ΣF_y = F_AB(9) + F_AC(13.85) - W(15)
ΣF_z = F_AB(12) - F_AC(18.46)

These equations should also have similar corrections made to indicate that you are dealing with moments and not forces.
Each of the equations below should be set equal to 0, since at equilibrium, ∑ M = 0.

ΣM_x = -F_AB(28.8) - F_AC(44.31) + W(48) = 0
ΣM_y = F_AB(9) + F_AC(13.85) - W(15) = 0
ΣM_z = F_AB(12) - F_AC(18.46) = 0

Note that the equation ΣM_z = F_AB(12) - F_AC(18.46) = 0 can be used to find F_AB in terms of F_AC

Once you have done this, then the force equations:

∑ Fx = -0.2983 * F_OA - 0.3846 * F_AC = 0
∑ Fy = -0.9545 * F_OA - 0.8000 * F_AB = 0
∑ Fz = 0.6000 * F_AB + 0.9231 * F_AC - W = 0

can be used to find the magnitudes F_AB, F_AC, and F_OA in terms of the load W.

You've done good work up to this point. Keep working. You're only a couple of steps away from finding the maximum value of W.

 

[Alison A.]

These are not forces. They are moments, so they should use different notation. There is also a minor omission in that each moment vector should be multiplied by the magnitude of the associated force:

M_AB = F_AB * (-28.8, 9, 12) in-lbs.
M_AC = F_AC * (-44.31, 13.85, -18.46) in-lbs.
M_W = W * (48, -15, 0) in-lbs.
These equations should also have similar corrections made to indicate that you are dealing with moments and not forces.
Each of the equations below should be set equal to 0, since at equilibrium, ∑ M = 0.

ΣM_x = -F_AB(28.8) - F_AC(44.31) + W(48) = 0
ΣM_y = F_AB(9) + F_AC(13.85) - W(15) = 0
ΣM_z = F_AB(12) - F_AC(18.46) = 0

Note that the equation ΣM_z = F_AB(12) - F_AC(18.46) = 0 can be used to find F_AB in terms of F_AC

Once you have done this, then the force equations:
can be used to find the magnitudes F_AB, F_AC, and F_OA in terms of the load W.

You've done good work up to this point. Keep working. You're only a couple of steps away from finding the maximum value of W.

So I am finding all the forces from the moments first then applying those to the force equations?

 

[SteamKing]

So I am finding all the forces from the moments first then applying those to the force equations?

No, you use the one moment equation I indicated,

ΣMz = F_AB(12) - F_AC(18.46) = 0 to find F_AB in terms of F_AC

Then, you can substitute F_AB back into the force equations to find the other forces in terms of W.

 

[Alison A.]

No, you use the one moment equation I indicated,

ΣMz = F_AB(12) - F_AC(18.46) = 0 to find F_AB in terms of F_AC

Then, you can substitute F_AB back into the force equations to find the other forces in terms of W.

Alright I found F_AB to be (4615/5539)W and F_AC to be (3000/5539)W

 

[Alison A.]

Alright I found F_AB to be (4615/5539)W and F_AC to be (3000/5539)W

And from there force equations I found F_OA to be -0.6983W

 

[Alison A.]

WOOOOOOOOOOOOOO I got the right answer. Wow that was a lot of work, I think I've gone through about 10 pages of paper. Thank you so much for sticking with me even when it seemed like I couldn't grasp the most simple concepts . Could you help me find the last part to my other problem you've been answering? That is my last problem... then I'm finally done.