Solving Stresses in Beams: Reactions at B & F

[luciriv]

Homework Statement

Figure shows a supported beam, where a=2 and b=3, and the cross section is a rectangle with the sides a=2 in and b=3 in. Determine the support reactions, maximum normal and shearing stresses. Draw the shearing force and bending moment diagrams.

Relevant Equations

In the figure, pulg means inches. The shearing stress is $$\tau = \dfrac{VQ}{Ib},$$ where $V$ represents the shearing force, $Q$ is the first moment of area, $I$ is the moment of inertia of the entire cross section, and $b$ is the width of the beam at the position where $\tau$ acts.

To determine the reactions at B and F, I don't know how to handle that L-shaped beam connected to the horizontal beam. My attempt is:
$$\sum F_{y} \colon\;\;\; -200 + R_{B} - 75 + 32 + R_{F} - 1632 = 0$$
from which it follows that $R_{B} + R_{F} = 1875$.
$$\sum M_{B} \colon\;\;\; -5 \times 200 - 8 \times 75 + 25 \times 32 + 33R_{F} - 90 = 0.$$
So $R_{F} = 26.97\, lb$ and $R_{B} = 1848.03\, lb$.
Is this right? Any hint or help to determine the shearing forces around that L-shaped beam is welcome.

 

[Lnewqban]

That L shape will impose a punctual moment and a shearing force at point D of the main beam.

 

[Chestermiller]

Homework Statement:: Figure shows a supported beam, where a=2 and b=3, and the cross section is a rectangle with the sides a=2 in and b=3 in. Determine the support reactions, maximum normal and shearing stresses. Draw the shearing force and bending moment diagrams.
Relevant Equations:: In the figure, pulg means inches. The shearing stress is $$\tau = \dfrac{VQ}{Ib},$$ where $V$ represents the shearing force, $Q$ is the first moment of area, $I$ is the moment of inertia of the entire cross section, and $b$ is the width of the beam at the position where $\tau$ acts.

View attachment 286775

To determine the reactions at B and F, I don't know how to handle that L-shaped beam connected to the horizontal beam. My attempt is:
$$\sum F_{y} \colon\;\;\; -200 + R_{B} - 75 + 32 + R_{F} - 1632 = 0$$
from which it follows that $R_{B} + R_{F} = 1875$.
$$\sum M_{B} \colon\;\;\; -5 \times 200 - 8 \times 75 + 25 \times 32 + 33R_{F} - 90 = 0.$$
So $R_{F} = 26.97\, lb$ and $R_{B} = 1848.03\, lb$.
Is this right? Any hint or help to determine the shearing forces around that L-shaped beam is welcome.

what is your assessment of the bending moment and force at point D are for the L-shaped beam?

 

[PhanthomJay]

For finding end reactions, your treatment of the force at E in determining its moment about B is good. But you have signage errors in determining the sum of moments and you missed some loads and I don't know what is the 90.