# Reduction of a simple distributed loading (correct question)

I am not yet at the chapter of equations of equilibrium, plus it says the couple moment is not 0, so I assume it just about the loadings and not including the reactive forces and reactive moment.

So if I call the top one F1 and the bottom loading F2 I got:

$$F_1+F_2=0= -4b\dfrac{1}{2}+2.5(b+a)\dfrac{1}{2}$$

solving this i got $a=0.6b$

Moment around the free end of the bar (not $A$, but opposite side) due to the loadings is (counter clockwise positive):

$$M= -F_1\cdot\text{center of triangle} + F_2\cdot\text{center of triangle}$$

Because you can replace the loading with force $F_1$ and force $F_2$ and its line of action is through the center of the triangle area ($\dfrac{1}{3}\text{base}$):

$$M= -8= -4b\dfrac{1}{2}\cdot\dfrac{1}{3}b+2.5(b+a)\dfrac{1}{2}\cdot\dfrac{1}{3}(b+a)$$

So what am I doing wrong? Because M can never be negative with me, plus the answer should be $b= 5.625$ and $a= 1.539$. But this to me makes no sense, because then $F_1+F_2\neq0$. And if I should take the reactive forces into account at $A$ then you can never have still a moment, because then it is not static anymore.

• I am really sorry, but the other question that I uploaded about this, had the wrong image. Sorry for the inconvenience Feb 25 '16 at 15:18
• Did the answer to your previous question not help you with this one as well?
– hazzey
Feb 25 '16 at 15:24

## 1 Answer

The answer can be found with the same process used in the previous question.

Defining $F_1$ as the downwards load, we have

$$F_1 = -\dfrac{4b}{2} = -2b$$

Defining $F_2$ as the upwards load, we have

$$F_2 = \dfrac{2.5(a+b)}{2} = 1.25(a+b)$$

As you stated, $F_1+F_2 = 0 \therefore b = \dfrac{5}{3}a$.

Now, the moment due to a force couple is $M = F \times D$, where $D$ is the distance between the forces in the couple. Now, the centers of action of $F_1$ and $F_2$ are (from the free end):

\begin{align} D_{F_1} &= \dfrac{b}{3} \\ D_{F_2} &= \dfrac{a+b}{3} \end{align} Therefore $D = \dfrac{a+b}{3} - \dfrac{b}{3} = \dfrac{a}{3}$. Thus, $M = 2b \times \dfrac{a}{3} = \dfrac{10}{3}a \times \dfrac{a}{3} = 8 \therefore a = \sqrt{7.2} \therefore b = \dfrac{5}{3}\sqrt{7.2}$.

Checking our work: \begin{align} F_1 &= -\dfrac{10}{3}\sqrt{7.2} \\ F_2 &= 1.25(\sqrt{7.2}+\dfrac{5}{3}\sqrt{7.2}) = \dfrac{10}{3}\sqrt{7.2} \\ &\therefore F_1 + F_2 = 0 \text{ OK!}\\ M &= \dfrac{10}{3}\sqrt{7.2}\cdot\dfrac{\sqrt{7.2}}{3} = \dfrac{72}{9} = 8\text{ OK!} \end{align}

• But so then the book is wrong in saying that a= 1.539 an b=5.625? Feb 25 '16 at 15:44
• And with your couple you get a counterclockwise moment of 8 kN*m right? While the text states a clockwise moment. Feb 25 '16 at 16:02
• Yes, I believe your book is wrong. As you yourself noticed, with those values for $a$ and $b$, you get $F_1\neq F_2$. And yes, the moment should be counter-clockwise. You have two equal forces and obviously the downwards force will be to the left of the upwards force since the latter has a longer base. This creates a counter-clockwise moment.
– Wasabi
Feb 25 '16 at 17:13