How Do Beam Forces and Support Reactions Relate in Equilibrium Analysis?

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The discussion focuses on calculating the reaction forces at points A and B for a uniform beam supported by a roller and a pin, subjected to specific forces. Using equilibrium equations, the upward reaction force at point B is determined to be approximately 24.186 lb, while the downward reaction force at point A is calculated to be 108.47 lb, though this is noted to be incorrectly labeled as downward. The participants emphasize the importance of correctly accounting for moments and forces in the equilibrium equations, particularly the beam's weight. Additionally, they discuss the relationship between the x components of the reaction forces at A and B, suggesting a geometric approach to find these components. Accurate calculations and corrections are crucial for determining the correct support reactions in this equilibrium analysis.
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Homework Statement


As shown, a roller at point A and a pin at point B support a uniform beam that weighs 42.0 lb. The beam is subjected to the forces F1 = 21.0 lb and F2 = 22.0 lb. The dimensions are L1 = 1.30 ft and L2 = 8.00 ft. What are the magnitudes FA and FB of the reaction forces FA and FB at points A and B, respectively? The beam's height and width are negligible.

Homework Equations


\SigmaFx = 0
\SigmaFy = 0
\SigmaMA = 0

FA = sqrt{Ax^2 + Ay^2}
FB = sqrt{Bx^2 + By^2}

The Attempt at a Solution


\SigmaMA = 0
F1L1 + F2cos(15)*(L1+L2) = By(L1+L2)
21(1.3) + 22cos(15)(9.3) = 9.3By
By = 24.186 lb (upward)

\SigmaFy = 0
Ay + By = -F1 -F2cos(15) - 42 = 0
Ay + 24.186 = -21 - 22cos(15) - 42
Ay = 108.47 lb (downward)

\SigmaFx = 0
Ax + Bx = F2sin(15)= 0
Ax + Bx = 5.694 lb
How do I find the x components for A and B?

\theta = tan-1(3/4) = 36.87\circ

I'm not sure about my A and B components.
 

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sami23 said:

Homework Statement


As shown, a roller at point A and a pin at point B support a uniform beam that weighs 42.0 lb. The beam is subjected to the forces F1 = 21.0 lb and F2 = 22.0 lb. The dimensions are L1 = 1.30 ft and L2 = 8.00 ft. What are the magnitudes FA and FB of the reaction forces FA and FB at points A and B, respectively? The beam's height and width are negligible.


Homework Equations


\SigmaFx = 0
\SigmaFy = 0
\SigmaMA = 0

FA = sqrt{Ax^2 + Ay^2}
FB = sqrt{Bx^2 + By^2}

The Attempt at a Solution


\SigmaMA = 0
F1L1 + F2cos(15)*(L1+L2) = By(L1+L2)
21(1.3) + 22cos(15)(9.3) = 9.3By
By = 24.186 lb (upward)
You forgot to include the moment from the beams weight
\SigmaFy = 0
Ay + By = -F1 -F2cos(15) - 42 = 0
Ay + 24.186 = -21 - 22cos(15) - 42
Ay = 108.47 lb (downward)
you mean upward. The terms on the right should all be plus terms ( that is , if Q - P = 0, then +Q = +P) But first correct the error in the moment equation
\SigmaFx = 0
Ax + Bx = F2sin(15)= 0
Ax + Bx = 5.694 lb
How do I find the x components for A and B?

\theta = tan-1(3/4) = 36.87\circ

I'm not sure about my A and B components.
Once you correctly solve for Ay, then Ax is related to Ay by the properties of the 3-4-5 triangle (the reaction at A must be perprendicular to the diagonal, since it is a roller support)
 

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