- #1
goldfish9776
- 310
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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?
Instead of just writing down a random equation, you should: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
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 ?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.
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??
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 ?
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.
You can calculate moments about an axis which passes thru a point, but you can't calculate moments about a single point.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 .
I have ΣFx = -AD(5/surd38) -BD(5/surd29) =0SteamKing said:You can calculate moments about an axis which passes thru a point, but you can't calculate moments about a single point.
In any event, have you progressed any further with your analysis of this problem?
You might want to check the x, y and z-components of AD and the length of AD.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
i couldn't see what's wrong with itSteamKing said:You might want to check the x, y and z-components of AD and the length of AD.
Look again at the diagram carefully. The length you have calculated for line AD is incorrect.goldfish9776 said:i couldn't see what's wrong with it
The z coordinate for B is 2, for A is 4?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?
And the length of line AD is?goldfish9776 said:The z coordinate for B is 2, for A is 4?
Surd50... The diagram is unclearSteamKing said:And the length of line AD is?
Surd50... The diagram is unclearSteamKing said:And the length of line AD is?
Seemed pretty clear to me.goldfish9776 said:Surd50... The diagram is unclear
SteamKing said:Seemed pretty clear to me.
The dashed lines have some significance in the diagram here.goldfish9776 said:The 2 m looks like B to Z
No, moment about a point is valid. It will be a vector, defining its own axis.SteamKing said:I don't know what "moment about a point" means. You can evaluate moments only about an axis.
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.
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."
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.
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.
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.