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Determine the load

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1. 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???

2. Homework Equations


3. The Attempt at a Solution
 

SteamKing

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1. 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???

2. Homework Equations


3. 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.
 
310
1
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 cant consider moment about a specific point ?
 
Last edited:

SteamKing

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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 cant consider moment about a specific point ?
I don't know what "moment about a point" means. You can evaluate moments only about an axis.
 
310
1
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 .
 

SteamKing

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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?
 
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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
 

SteamKing

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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.
 

SteamKing

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i couldnt see whats 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?
 
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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?
 

SteamKing

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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.
 
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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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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) ) ?
 

haruspex

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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.
 
Last edited:

BvU

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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 ?
 

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