- #1
Zaza669
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How to find forces in bars BD CD and CE?
Homework Equations
Static[/B]
The Attempt at a Solution
Bd: -2.5 KN
Cd: 0.866 KN
Ce: 1.73 KN
Find Forces in Truss Bars BD, CD and CE
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SUMMARY
The forces in truss bars BD, CD, and CE have been calculated as follows: BD experiences a force of -2.5 kN, CD has a force of 0.866 kN, and CE has a force of 1.73 kN. The analysis assumes that torque stresses at the joints can be ignored, and it is concluded that the force in bar CE is zero due to symmetry and the nature of the supports at joint G. This interpretation relies on the understanding of how horizontal forces affect the movement of the joints in the truss structure.
PREREQUISITES- Understanding of static equilibrium in truss structures
- Knowledge of force analysis in structural engineering
- Familiarity with joint support types and their effects on force distribution
- Basic principles of symmetry in mechanical systems
- Study the principles of static equilibrium in truss analysis
- Learn about different types of truss joints and their implications on force calculations
- Explore methods for calculating forces in truss structures using the method of joints
- Investigate the effects of support types on force distribution in structural systems
Structural engineers, civil engineering students, and anyone involved in analyzing truss systems and force distributions in mechanical structures.
How to find forces in bars BD CD and CE?
Static[/B]
Bd: -2.5 KN
Cd: 0.866 KN
Ce: 1.73 KN
- #3
Buzz Bloom
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Hi Zaza:
Since no one else has posted an answer, I will offer a thought which I hope will be helpful, although I have never previously attempted to analyze this kind of problem.
Since no mention has been made about the nature of the joints between the beams, I assume that torque stresses on the joints can be ignored. You have calculated that the stress on CE is 1.73 KN. I notice that the joint G has a rolling support, and joints C and E are unsupported. I interpret this to mean that any stresses at G due to horizontal forces on AC, CE, and EG will result in G moving left or right, and this motion will relieve any horizontal stresses in AC, CE, and EG. Since any force in CE must be entirely horizontal due to symmetry, the force in bar CE is zero.
Regards,
Buzz
Since no one else has posted an answer, I will offer a thought which I hope will be helpful, although I have never previously attempted to analyze this kind of problem.
Since no mention has been made about the nature of the joints between the beams, I assume that torque stresses on the joints can be ignored. You have calculated that the stress on CE is 1.73 KN. I notice that the joint G has a rolling support, and joints C and E are unsupported. I interpret this to mean that any stresses at G due to horizontal forces on AC, CE, and EG will result in G moving left or right, and this motion will relieve any horizontal stresses in AC, CE, and EG. Since any force in CE must be entirely horizontal due to symmetry, the force in bar CE is zero.
Regards,
Buzz
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