Statics: Crane Boom - Why Different Results?

In summary: Your formula is correct for the moment at point A, but you need to calculate the components of the force F in order to find the moment at point E. It is also important to note that the angle t may change at point E, so the value for sin t may be different.
  • #1
newbphysic
39
0
rbd6X51.jpg

1. Homework Statement

W engine = 500 kg * 9.81 = 4905 N
sin t = 3/5

Homework Equations

The Attempt at a Solution


1. use section AE
ΣMa = 0

[itex] 3/5 * Fcd *2 - 4905 * 3 = 0 [/itex]

[itex] F = 12262.5 N [/itex]

2. use section EB

ΣMe = 0

[itex] 3/5 * Fcd *1 - 4905 * 2 = 0 [/itex]

[itex] F = 16350 N [/itex]

why 1 and 2 doesn't result same answer ?
 
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  • #2
newbphysic said:
rbd6X51.jpg

1. Homework Statement

W engine = 500 kg * 9.81 = 4905 N
sin t = 3/5

Homework Equations

The Attempt at a Solution


1. use section AE
ΣMa = 0

[itex] 3/5 * Fcd *2 - 4905 * 3 = 0 [/itex]

[itex] F = 12262.5 N [/itex]

2. use section EB

ΣMe = 0

[itex] 3/5 * Fcd *1 - 4905 * 2 = 0 [/itex]

[itex] F = 16350 N [/itex]

why 1 and 2 doesn't result same answer ?
Beats me.

You should write the equations of static equilibrium for this crane rather than try to jump into the middle of a calculation.

Try drawing a free body diagram for the beam AEB and putting that member in equilibrium.
 
  • #3
SteamKing said:
Beats me.

You should write the equations of static equilibrium for this crane rather than try to jump into the middle of a calculation.

Try drawing a free body diagram for the beam AEB and putting that member in equilibrium.

There is only 3 equations right ?
qDWn1Vk.png

ΣMoment =0
[itex] Fy * distance - weight * distance = 0 [/itex]

Σshear = 0
Σnormal = 0
 
  • #4
newbphysic said:
There is only 3 equations right ?
qDWn1Vk.png

ΣMoment =0
[itex] Fy * distance - weight * distance = 0 [/itex]

Σshear = 0
Σnormal = 0
There are two equations of static equilibrium: ∑F = 0 and ∑M = 0.

The force equation can be applied in the horizontal or the vertical direction, but generally only one moment equation can be written.
 
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  • #5
SteamKing said:
There are two equations of static equilibrium: ∑F = 0 and ∑M = 0.

The force equation can be applied in the horizontal or the vertical direction, but generally only one moment equation can be written.
ok, so moment is caused by force in the vertical direction so formula for ΣM = Fy * d - W * d

For moment at point A
ΣMa = 0
Fy *d - W * d = 0
[itex] sin( t) * Fcd - W *d [/itex]
[itex](3/5 * Fcd *2) - (4905 * 3) = 0 [/itex]

F=12262.5N
is this correct ?

If i cut to segment EB
rxZcgoL.png

sin t will be the same since angle t is not change

ΣM = 0
newbphysic said:
ΣMe = 0

[itex] 3/5 * Fcd *1 - 4905 * 2 = 0 [/itex]

F=16350N

is my formula correct ?
 
  • #6
newbphysic said:
ok, so moment is caused by force in the vertical direction so formula for ΣM = Fy * d - W * d

For moment at point A
ΣMa = 0
Fy *d - W * d = 0
[itex] sin( t) * Fcd - W *d [/itex]
[itex](3/5 * Fcd *2) - (4905 * 3) = 0 [/itex]

F=12262.5N
is this correct ?
Your calculation for F is correct. However, you need to calculate the components of F in order to find the internal forces acting at point E.

I would not use the same letter "d" for both moment arms as this suggests each arm is the same distance.
If i cut to segment EB
rxZcgoL.png

sin t will be the same since angle t is not change

ΣM = 0is my formula correct ?
 

1. Why do different methods of calculating statics for a crane boom result in different answers?

There are several factors that can contribute to different results when calculating statics for a crane boom. One reason is the assumption of ideal conditions, which may not accurately reflect real-world scenarios. Additionally, different methods may use different assumptions and simplifications, leading to variations in the final result. Lastly, inaccuracies in measurements or data input can also affect the outcome.

2. How do I know which method of calculating statics for a crane boom is the most accurate?

The most accurate method will depend on the specific situation and conditions of the crane boom. It is important to carefully consider all assumptions and factors involved in each method, and choose the one that best aligns with the real-world scenario. It may also be beneficial to compare results from multiple methods to ensure consistency and accuracy.

3. Are there any safety concerns when using different methods of calculating statics for a crane boom?

Yes, safety should always be a top priority when using a crane boom. Different methods may yield significantly different results, which can impact the stability and safety of the crane. It is important to carefully analyze and verify all calculations and assumptions to ensure safe operation of the crane.

4. How can I minimize discrepancies in results when using different methods of calculating statics for a crane boom?

There are a few steps that can help minimize discrepancies in results when using different methods. First, double-check all measurements and data input to ensure accuracy. It may also be helpful to use multiple methods and compare results to identify any potential errors. Additionally, consulting with an experienced engineer can provide valuable insights and recommendations for ensuring accurate statics calculations.

5. Can different results from statics calculations for a crane boom impact the overall design and construction process?

Yes, different results can certainly impact the design and construction process of a crane boom. Variations in calculations can affect the structural integrity and stability of the crane, leading to potential safety hazards. It is important to carefully consider and verify all statics calculations to ensure the design and construction process is accurate and safe.

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