Find Load in AD, BD, and DC: Solving Homework Questions

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In summary, the conversation involves finding the load in AD, BD, and DC using equations of static equilibrium. The conversation also discusses the nature of different materials and how they can sustain different types of loads. The equations of equilibrium for this structure are ΣFx = -AD(5/surd38) -BD(5/surd29) =0, ΣFy = AD(3/surd38)-DC=0, and ΣFz= BD(2/surd29) +AD(2/surd38) -800= 0. It is important to carefully analyze the problem and consider the locations of points in order to correctly calculate the lengths and components of the lines involved.
  • #1
goldfish9776
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Homework Statement


I am asked to find the load in AD, BD, and DC...
I have AD(2/surd38) +BD(2/surd29)=800
AD(3/surd38)=DC
I can only formed two equation... But , so many unknown... How to solve this?

Homework Equations

The Attempt at a Solution

 
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  • #2
Here's the diagram
 

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  • #3
goldfish9776 said:

Homework Statement


I am asked to find the load in AD, BD, and DC...
I have AD(2/surd38) +BD(2/surd29)=800
AD(3/surd38)=DC
I can only formed two equation... But , so many unknown... How to solve this?

Homework Equations

The Attempt at a Solution

Instead of just writing down a random equation, you should:

1. Write down the known information in a logical order
2. Write the general equations of static equilibrium (ΣF = 0; ΣM = 0)
3. Analyze the problem to determine if there are simplifications or special conditions which may apply.
4. Write the equations of equilibrium for this problem.
5. Solve these equations to determine the unknown reactions / internal loads.

For this problem, you could first calculate and write down the lengths of the members involved: the bar BD and the lines AD and CD.
Step 2 is pretty much taken care of.
Step 3: The bar BD can take either tension or compression. The lines AD and CD look like they can take only tension. There is only one applied load of 800 kN.

Here's the critical step:

Write the equations of equilibrium for this structure after completing steps 1-3.
 
  • #4
SteamKing said:
Instead of just writing down a random equation, you should:

1. Write down the known information in a logical order
2. Write the general equations of static equilibrium (ΣF = 0; ΣM = 0)
3. Analyze the problem to determine if there are simplifications or special conditions which may apply.
4. Write the equations of equilibrium for this problem.
5. Solve these equations to determine the unknown reactions / internal loads.

For this problem, you could first calculate and write down the lengths of the members involved: the bar BD and the lines AD and CD.
Step 2 is pretty much taken care of.
Step 3: The bar BD can take either tension or compression. The lines AD and CD look like they can take only tension. There is only one applied load of 800 kN.

Here's the critical step:

Write the equations of equilibrium for this structure after completing steps 1-3.
how do u knw The bar BD can take either tension or compression. The lines AD and CD look like they can take only tension??since this is an exercise that involves 3D , then we can only consider moment about x , y and z axis , right ? we can't consider moment about a specific point ?
 
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  • #5
goldfish9776 said:
how do u knw The bar BD can take either tension or compression. The lines AD and CD look like they can take only tension??

Because that's the nature of stiff members like bars. Bars can be stretched in tension, compressed, even bent by applying moments to the ends.

Ropes and wires can only sustain tension loads. Repeat after me: You can't push on a rope.

I'm curious: haven't you ever tried to build anything from scratch, played with Tinker Toys, hung a picture on a wall with a nail and some wire, heck even picked up a stick?
This is how we learn about the mechanical nature of different things, like bars, strings, rope, and the like. There's more to experience in life than banging keys on a keyboard.

since this is an exercise that involves 3D , then we can only consider moment about x , y and z axis , right ? we can't consider moment about a specific point ?

I don't know what "moment about a point" means. You can evaluate moments only about an axis.
 
  • #6
SteamKing said:
Because that's the nature of stiff members like bars. Bars can be stretched in tension, compressed, even bent by applying moments to the ends.

Ropes and wires can only sustain tension loads. Repeat after me: You can't push on a rope.

I'm curious: haven't you ever tried to build anything from scratch, played with Tinker Toys, hung a picture on a wall with a nail and some wire, heck even picked up a stick?
This is how we learn about the mechanical nature of different things, like bars, strings, rope, and the like. There's more to experience in life than banging keys on a keyboard.
I don't know what "moment about a point" means. You can evaluate moments only about an axis.

moment about a point means we simply pick a point , and claculate the moment at it . For example , we take the point A , and we calculate the moment ( force x distance ) about it .
 
  • #7
goldfish9776 said:
moment about a point means we simply pick a point , and claculate the moment at it . For example , we take the point A , and we calculate the moment ( force x distance ) about it .
You can calculate moments about an axis which passes thru a point, but you can't calculate moments about a single point. :wink:

In any event, have you progressed any further with your analysis of this problem?
 
  • #8
SteamKing said:
You can calculate moments about an axis which passes thru a point, but you can't calculate moments about a single point. :wink:

In any event, have you progressed any further with your analysis of this problem?
I have ΣFx = -AD(5/surd38) -BD(5/surd29) =0
ΣFy = AD(3/surd38)-DC=0
ΣFz= BD(2/surd29) +AD(2/surd38) -800= 0
 
  • #9
goldfish9776 said:
I have ΣFx = -AD(5/surd38) -BD(5/surd29) =0
ΣFy = AD(3/surd38)-DC=0
ΣFz= BD(2/surd29) +AD(2/surd38) -800= 0
You might want to check the x, y and z-components of AD and the length of AD.
 
  • #10
SteamKing said:
You might want to check the x, y and z-components of AD and the length of AD.
i couldn't see what's wrong with it
 
  • #11
goldfish9776 said:
i couldn't see what's wrong with it
Look again at the diagram carefully. The length you have calculated for line AD is incorrect.

What's the location of point A?
 
  • #12
SteamKing said:
Look again at the diagram carefully. The length you have calculated for line AD is incorrect.

What's the location of point A?
The z coordinate for B is 2, for A is 4?
 
  • #13
goldfish9776 said:
The z coordinate for B is 2, for A is 4?
And the length of line AD is?
 
  • #14
SteamKing said:
And the length of line AD is?
Surd50... The diagram is unclear
 
  • #15
SteamKing said:
And the length of line AD is?
Surd50... The diagram is unclear
 
  • #16
goldfish9776 said:
Surd50... The diagram is unclear
Seemed pretty clear to me. :wink:
 
  • #17
The 2 m looks like B to Z
SteamKing said:
Seemed pretty clear to me. :wink:
 
  • #18
goldfish9776 said:
The 2 m looks like B to Z
The dashed lines have some significance in the diagram here.

There is no purpose in locating the position of the letter Z; there is a purpose in locating the position of point A.
 
  • #19
SteamKing said:
I don't know what "moment about a point" means. You can evaluate moments only about an axis.

here's a case . it give the moment about the x , y , x' and y' axis ... but not moment about a specific point
 

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  • #20
so when we are asked to find the moemnt for a question involving 3D model , we can only use the moemnt about a specific axis to find the moment ? but not a specific point ? (for example. in the above case , we cannot find the moment about the O ( centroid) ) ?
 
  • #21
SteamKing said:
I don't know what "moment about a point" means. You can evaluate moments only about an axis.
No, moment about a point is valid. It will be a vector, defining its own axis.
When you take moments about a selected axis, you get the component of the full moment vector parallel to that axis.
 
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  • #22
Re =post #20: That's right. In 3D there are a lot of rotation axes you can think of for O, so you have to specify which one.

Haru is right in post #21 in the sense that one can add moment vectors. The result is a sum moment with a sum axis. But for your exercise it's more effective to look at the magnitudes of the moments about separate axes one by one. No rotation means they all need to be 0.

Are you in the same class as @werson tan ?
 

1. How do I find the load in AD, BD, and DC?

To find the load in AD, BD, and DC, you will need to use the principles of equilibrium. This means that the forces acting on each joint must be balanced, with the net force and net moment equal to zero. You will also need to use the equations of static equilibrium, which include the sum of forces in the x-direction, the sum of forces in the y-direction, and the sum of moments about a point. By setting up and solving these equations, you can determine the load in each member.

2. What is the difference between a load and a force?

A load refers to the external forces applied to a structure, while a force is a type of load that is a vector quantity with both magnitude and direction. Loads can include forces, moments, and even thermal or environmental effects. In the context of solving homework questions, the term "load" may be used interchangeably with "force."

3. Can I use a computer program to find the load in AD, BD, and DC?

Yes, there are many computer programs and software that can assist with solving homework questions related to finding loads in structures. However, it is important to have a strong understanding of the underlying principles and equations in order to properly use these programs and interpret the results.

4. Are there any assumptions or simplifications made when solving for loads in structures?

Yes, in most cases, there are assumptions and simplifications made in order to solve for loads in structures. These can include assuming certain members are rigid, neglecting the weight of the structure itself, or assuming the structure is in a state of static equilibrium. It is important to clearly state any assumptions or simplifications made when presenting your solution.

5. Can I use the same method to find the load in any type of structure?

The principles and equations used to find loads in structures are applicable to a wide range of structures, including trusses, frames, and beams. However, the specific approach and methods may vary depending on the type of structure and the types of loads being applied. It is important to carefully consider the specific problem and choose the appropriate method for solving it.

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