Solving Stresses in Beams: Reactions at B & F

  • #1
luciriv
5
1
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.
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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.
 
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  • #2
That L shape will impose a punctual moment and a shearing force at point D of the main beam.
 
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  • #3
luciriv said:
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?
 
  • #4
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.
 
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