# Reduction of a simple distributed loading

This is the question, but with me 6 kN/m is 4 kN/m and 2 kN/m is 2.5 kN/m. And the length is not 4 but 9 metres.

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.

• Sorry I forgot to mention that the length is not 4 but 9 meter in my question. Feb 25 '16 at 11:25

Defining $F_1$ as the triangular load, we have

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

Defining $F_2$ as the uniform load, we have

$$F_2 = 2.5(a+b)$$

As you stated, $F_1+F_2 = 0 \therefore b = -5a$. This result may seem odd, given that it's negative, but it's correct. After all, the resultant force of a triangular load is half of what it would be if the load were uniform and equal to the maximum value. So, for a triangular load to have the same resultant force as a uniform load while both occupy the same length ($a=0$), the triangular load must have a maximum value which is double that of the uniform load. If the triangular load's maximum value is greater than double, then it can occupy a shorter length while having the same resultant force ($a>0$, that's the case displayed in the original question with 6 and 2 kN/m). However, if the triangular load's maximum value is less than double the uniform load (as is your case), then the triangular load must occupy a greater length to have the same resultant force ($a<0$).

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}{2} \end{align} Therefore $D = \left|\dfrac{a+b}{2} - \dfrac{b}{3}\right| = \left|\dfrac{3a+b}{6}\right| = \left|\dfrac{-2a}{6}\right| = \dfrac{|a|}{3}$. Thus, $M = -2b \times \dfrac{|a|}{3} = 10a \times \dfrac{|a|}{3} = -8 \therefore a = -\sqrt{2.4} \therefore b = 5\sqrt{2.4}$.

Checking our work: \begin{align} F_1 &= -10\sqrt{2.4} \\ F_2 &= 2.5(-\sqrt{2.4}+5\sqrt{2.4}) = 10\sqrt{2.4} \\ &\therefore F_1 + F_2 = 0 \text{ OK!}\\ M_A &= -10\sqrt{2.4}\left(4-\dfrac{5\sqrt{2.4}}{3}\right) + 10\sqrt{2.4}\left(4-\dfrac{4\sqrt{2.4}}{2}\right) = 10\sqrt{2.4}\left(\dfrac{5\sqrt{2.4}}{3}-\dfrac{4\sqrt{2.4}}{2}\right) \\ &= 24\left(\dfrac{5}{3}-2\right) = -\dfrac{24}{3} = -8\text{ OK!} \end{align}

• I am so sorry, I worked on it all night so I was very tired. This morning I tried again and was so frustrated that i searched for image and uploaded it. But now I realize it is not the same question (aside from the incorrect data). I have uploaded a new question with the correct question. I apologize that you spend time on this, I hope you could check out the correct question. Feb 25 '16 at 15:15
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