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I am trying to solve the following problem:

EXAMPLE 2.10

Determine the reactions on the beam in Fig. 2–30a. Assume $B$ is a pin and the support at $B$ is a roller (smooth surface).

A bent beam is supported on the left at $A$. The beam extends 7 ft rightward and then bends sharply upward. The section to the right of the bend extends 3 ft horizontally and 4 ft vertically to a support at $B$. The horizontal section to the left of the bend is subject to a uniform load of 500 lb/ft in the downward direction.

SOLUTION

Free-Body Diagram. As shown in Fig. 2–30b, the support ("roller") at $B$ exerts a normal force on the beam at its point of contact. The line of action of this force is defined by the 3-4-5 triangle.

Free body diagram of the beam from previous image. The uniform load is replaced by a downward point load of 3500 lb, shown as acting 3.5 ft to the right of $A$. The reactions at $A$ are decomposed into components $\mathbf{A}_x$ and $\mathbf{A}_y$, acting rightward and upward, respectively. A single reaction $\mathbf{N}_B$ is shown acting from below-right at $B$, perpendicular to that section. The horizontal distance between the substituted point load and the reaction at $B$ is shown as 6.5 ft. The sharp bend in the beam is circled in red for emphasis.

Equations of Equilibrium. Resolving $\mathbf{N}_B$ into x and y components and summing moments around $A$ yields a direct solution for $N_B$. Why? Using this result, we can then obtain $A_x$ and $A_y$.

$$\begin {align} \\ \Sigma M_A = 0; \quad & -3500 (3.5) + (\frac{4}{5}) N_B (4) + (\frac{3}{5}) N_B (10) = 0 \\ & \qquad N_B = 1331.5\ \mathrm{lb} = 1.33\ \mathrm{k} \\ \Sigma F_x = 0; \quad & A_x - \frac{4}{5}(1331.5) = 0 & A_x = 1.07\ \mathrm{k}\\ \Sigma F_y = 0; \quad & A_y - 3500 + \frac{3}{5}(1331.5) = 0 & A_x = 2.70\ \mathrm{k}\\ \end {align}$$

For the moment about $A$, why does the reaction at $B$ produce a moment of $(\frac{4}{5})N_B(4)$? I think it's not necessary to include that moment because the moment arm $d = 0.4\ \mathrm{m}$ is not measured directly from point A, but from the bend in the beam, which I circled red in figure (b).

opp

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  • $\begingroup$ I'm not sure why you have a problem with the second term in the sum of moments but not the third term. How would you propose writing that equation instead? $\endgroup$ – Air Jan 11 '17 at 19:42
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The reason is that both the horizontal and the vertical component of $N_B$ generate a moment around $A$. Since the moment due to a force is equal to the product of the force and the perpendicular lever-arm between the force's line of action and the point being considered, this therefore means that the total moment around $A$ is equal to:

$$\begin{align} N_{B,x} &= \dfrac{4}{5}N_B \text{ (horizontal component)} \\ N_{B,y} &= \dfrac{3}{5}N_B \text{ (vertical component)} \\ M_A &= -3500 \cdot 3.5 + 4N_{B,x} + 10N_{B,y} = 0 \end{align}$$

After all, the horizontal distance from $A$ to $B$ is 10 ft, so that's the lever arm for the vertical component, while the vertical distance is 4 ft, so that's used for the horizontal component.

Hoping to make this as clear as possible, here's a diagram of the forces, their lines of action and the respective perpendicular distances. They are color-coded: everything relevant for the horizontal component is green, everything for the vertical component is red.

enter image description here

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    $\begingroup$ I can't see how to answer this question beyond saying that's what it is. The moment is the force times the perpendicular distance, but the question seems to be saying why is the moment the force times the perpendicular distance. $\endgroup$ – achrn Jan 11 '17 at 16:08
  • $\begingroup$ referrring to the last figure uploaded in the original post , we can see that the d (prepedicular distance) for NBy is measured directly from A , but for the NBx ( blue part) , it's measured directly from a point which is 10 m from A . So , for the moment 4NBx , it's not moment about A , am i right ? Why the author consider it in the calculation ? $\endgroup$ – kelvinmacks Jan 12 '17 at 0:21
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    $\begingroup$ @kelvinmacks: See my edit. If this doesn't answer your doubts, I have no idea what will. If you are really having trouble with this stuff, you really need to go back to your very first book on statics. Your doubts don't seem to be about the exercises themselves, but about foundational knowledge you still seem to be struggling with. And I'll repeat a comment I made in another one of your questions: don't ask us about specific exercises, but about the specific conceptual problems you are having. I honestly don't feel like I am helping you here. $\endgroup$ – Wasabi Jan 12 '17 at 1:01
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    $\begingroup$ @kelvinmacks: That would be a more reasonable statement if all of your questions weren't squarely focused on the problem and your attempt to solve it. There is no theoretical discussion or presentation of your premises and arguments. There's no "my understanding is that moment is calculated as [...], and so I did [...]. I see, however, that the book does [...], which contradicts with how I understand [...]." Instead it's "here's the question, I did this, but the book does something different. What's wrong?" The fact that all your questions are 90% images is not a good sign. $\endgroup$ – Wasabi Jan 12 '17 at 1:20
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    $\begingroup$ @kelvinmacks: And to refute your comment, there's nothing stopping you from, instead of showing us the exercise, showing us the section of the book that discusses the subject being applied in the exercise and then going "The book talks about this 'line of action' thing, but I'm having a hard time understanding what it is and how to use it." Unless, of course, you simply have absolutely no clue what the book is doing in this exercise and therefore can't find the relevant section, which is all the more reason for you to go back to the first chapters to really get the foundational stuff down. $\endgroup$ – Wasabi Jan 12 '17 at 1:27

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