beam - New

[logistic_guy]

Determine the internal normal force, shear force, and bending moment at point $\displaystyle C$ in the beam.

 

[logistic_guy]

Let $\displaystyle N_{C}$ be the internal normal force, $\displaystyle V_{C}$ be the internal shear force, and $\displaystyle M_{C}$ be the internal bending moment.

Since there are no horizontal forces at point $\displaystyle C$, $\displaystyle N_{C} = 0$.

 

[nessmell]

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[logistic_guy]

Let us calculate the torques around point $\displaystyle A$.

$\displaystyle 3F_B - 200(1.5)(2.25) - 100(1.5)(0.75) = 0$

This gives:

$\displaystyle F_B = 262.5 \ \text{N}$

 

[logistic_guy]

Since I have found $\displaystyle F_B$ and not $\displaystyle F_A$, I have to work on the right-half of the beam.

I will choose upward to be positive. The sum of the vertical forces on the right-half of the beam are:

$\displaystyle F_B - 200(1.5) - V_C = 0$

$\displaystyle 262.5 - 300 - V_C = 0$

$\displaystyle V_C = 262.5 - 300 = -37.5 \ \text{N} \rightarrow$ internal shear force

 

[khansaheb]

Without a FBD, at the section C, the work above will fetch 50 - 70 percent credit.

 

[logistic_guy]

Without a FBD, at the section C, the work above will fetch 50 - 70 percent credit.

Even $\displaystyle 50\%$ is still too much!