Determing the Force in Each Member of the Loaded Truss

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SUMMARY

The discussion focuses on determining the forces in each member of a loaded truss, specifically members AB, BC, and AC. The calculations show that the force in member AB is 1753.2 N in tension, while the force in member BC is -1321.1 N, indicating compression. The correct force in member AC is calculated as 795.05 N using the cosine of the angle at joint C. The analysis emphasizes the importance of understanding the direction of forces and the equilibrium equations applied at each joint.

PREREQUISITES
  • Understanding of static equilibrium principles in truss analysis
  • Familiarity with vector components and trigonometric functions
  • Knowledge of tension and compression forces in structural members
  • Ability to apply equilibrium equations at joints in a truss system
NEXT STEPS
  • Study the method of joints in truss analysis
  • Learn about the role of roller supports in structural mechanics
  • Explore advanced truss analysis techniques using software like SAP2000
  • Review the application of equilibrium equations in two-dimensional structures
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Structural engineers, civil engineering students, and anyone involved in analyzing truss systems for construction and design purposes.

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Homework Statement



Determine the force in each member of the loaded truss. The force is positive if in tension, negative if in compression.

I have attached an image of the problem


Homework Equations





The Attempt at a Solution



I defined the positive y-axis to be from B to A and hence the positive x-axis is from B to C.

To find AB I took the sum of the force in the y-direction on point B, my calculations are:

0 = AB - (224kg)(9.8)cos(37)

AB = 1753.2 N

This answer is correct.

Then to find BC I found the forces in the x-direction at point B

0 = -BC + (224)(9.8)sin(37)
BC = 1321.1 N

However it says my answer is wrong. I suspect that it should be -1321.1 N but I have not tried this.

I am stuck on finding AC.

Some of the my classmates tried:

(BC)cos(37) which gives me 1055 N but it says this is wrong.

Any help would be appraciated.
 

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I have gotten the answer now.

BC = -1321.1 N

AC = BCcos(53) = 795.05 N but I don't understand why these are correct.
 
Your method of solving for AB and BC is very good. Note that at joint B, the comp. of the weight along your chosen x-axis points in toward the joint, to the left; thus, the force in BC must point in toward the joint to the right. Forces in members that point in toward the joint are compressive (designated as minus), while forces that point away from the joint are tensile (designated as plus).
For the force in AC, look at joint C and use conventional horiz and vert x and y axes, and apply the equilibrium equations. Note that the roller support at C cannot support vertical forces.
 

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