# A statics problem containing a distributed triangular load and a linear load

I have came across this problem and I couldn't figure it out

I have used the following equation

\begin{align} y &= 6 - \dfrac{2}{3}x \\ V &= 26 - \dfrac{2}{3}x^2 \\ M_x &= 26x-\dfrac{2}{9}x^3 \end{align}

for the triangular part of it but the answers provided with the question are not matching mine

When you've done an exercise and got the wrong answer, it's always useful to check to see if your result ever passed the "smell test". That is, does your result make much sense.

Now, we can see a few strange things from a quick glance.

The biggest thing which should call our attention is your moment diagram. It starts at 0 at the support and ends at 128 at the free end. This is the exact opposite of what we'd expect from a cantilever: the fixed end should have a bending moment reaction and free ends must, by definition, have zero bending moment. So we know there's something wrong here.

And that takes us to a second question: why was your bending moment zero at the support? Well, because your bending moment equation doesn't have a constant value. We'll see how that happened later, but for now let's also observe that if you had a constant value, it'd obviously be equal to the support's bending moment reaction.

And what is that bending moment reaction? Well, I don't know, because you never calculated it. I can see you correctly calculated the vertical reaction of 32 kN, but you didn't calculate the bending moment.

A pro tip: 99% of the time, you'll need to use at least two of the equilibrium equations (usually $$\sum F_y = 0$$ and $$\sum M = 0$$). If you've only used one, make sure you didn't forget something.

So those are some of the "smells" we must get used to which let us know we might've messed up (happens to the best of us!). Now let's try this again.

Now, I'll solve this problem twice: once using what I think you attempted to do, and another which is slightly easier, but answers what the question asked for and no more. Let's start with your method:

So, the first thing we need to do is calculate the support reactions.

Now, when working with this sort of loading, where there's a triangular load which ends at the same load as a uniform load, I prefer to think of it as a uniform load over the entire beam and a triangular load which goes from (in this case) 4 to 0 kN/m.

\begin{align} \sum F_y&= R_y - 2\cdot10 - \dfrac{1}{2}\cdot4\cdot6 = 0 \\ \therefore R_y &= 32\text{ kN} \\ \sum M_A &= R_M - (2\cdot10)\cdot\dfrac{10}{2} - \left(\dfrac{1}{2}\cdot4\cdot6\right)\cdot\dfrac{1}{3}\cdot6 = 0 \\ \therefore R_M &= 124\text{ kNm} \end{align}

Now that we have the reactions, we can move on to calculate the internal forces. Unfortunately, I didn't understand the method you tried to use to get the shear forces.

The correct way to do it is basically what you did to get the bending moment: we integrate. The shear force is the integral of the loading configuration. So, what is the equation for the loading (which I'll call $$q(x)$$)?

Well, the first thing we notice is that $$q(x)$$ has to be discontinuous: at first it's linear, but then it suddenly becomes a constant. That is impossible to describe with a single equation, so we'll need to use two:

\begin{align} q(x) &= \begin{cases} \dfrac{2}{3}x - 6\text{ kN/m}&\text{ for } x \in [0, 6] \\ -2\text{ kN/m} &\text{otherwise} \end{cases} \end{align}

To get the shear force, we just need to integrate. And here's where you made a mistake when you integrated to get your bending moment equation. When integrating, you must always include a constant $$C$$ to your result, and then figure out what value to use for $$C$$.

\begin{align} V(x) &= \int q(x)\text{d}x \\ &= \int \begin{cases} \dfrac{2}{3}x - 6&\text{ for } x \in [0, 6] \\ -2 &\text{otherwise} \end{cases}\text{d}x \\ &= \begin{cases} \dfrac{1}{3}x^2 - 6x + C_1&\text{ for } x \in [0, 6] \\ -2x + C_2 &\text{otherwise} \end{cases} \end{align}

Now, how can we find the values of $$C_1$$ and $$C_2$$? Well, we know the shear force at the support must be equal to the reaction, so that one's easy:

\begin{align} V(0) &= \dfrac{1}{3}\cdot 0^2 - 6 \cdot 0 + C_1 = 32 \\ \therefore C_1 &= 32 \end{align}

As for $$C_2$$, well, we know the shear force on the left side of the 6 m point where we change from linear to constant load must be equal to the shear force on the right side:

\begin{align} V(6)^- &= \dfrac{1}{3} \cdot 6^2 - 6 \cdot 6 + 32 = 8\\ V(6)^+ &= -2 \cdot 6 + C_2 \\ \therefore C_2 &= 8 + 12 = 20 \end{align}

And then we just repeat to get the bending moment equations:

\begin{align} M(x) &= \int V(x)\text{d}x \\ &= \int \begin{cases} \dfrac{1}{3}x^2 - 6x + 32&\text{ for } x \in [0, 6] \\ -2x + 20 &\text{otherwise} \end{cases}\text{d}x \\ &= \begin{cases} \dfrac{1}{9}x^3 - 3x^2 + 32x + C_3&\text{ for } x \in [0, 6] \\ -x^2 + 20x + C_4 &\text{otherwise} \end{cases} \end{align}

We find $$C_3$$ and $$C_4$$ in the same way as we found $$C_1$$ and $$C_2$$, using the facts that we know the bending moment at the support will be equal to the bending moment reaction and that the bending moment on both sides of the 6 m point will be equal.

A detail to remember is that the sign convention for bending moments on the left side of the cross-section is clockwise positive, so the counter-clockwise reaction created by the support appears as negative for the beam.

\begin{align} M(0) &= \dfrac{1}{9} \cdot 0^3 - 3 \cdot 0^2 + 32 \cdot 0 + C_3 = -124 \\ \therefore C_3 &= -124 \\ M(6)^- &= \dfrac{1}{9} \cdot 6^3 - 3 \cdot 6^2 + 32 \cdot 6 - 124 = -16\\ M(6)^+ &= -6^2 + 20 \cdot 6 + C_4 = -16 \therefore C_4 = -16 + 6^2 - 20 \cdot 6 = -100 \end{align}

And so we have the full equations and can plot the diagrams.

Now, for a slightly faster method. The question doesn't ask for the shear and bending moment equations, just the diagrams. So really we just need to find the shear and bending moment results for a few key points. Really, all we care about is the support and the 6 m point; we know the free end will have zero shear and bending moment.

For the support, we know the shear will be equal to the vertical reaction and the bending moment will be equal to the negative of the bending moment reaction, so we can just copy the work done in the previous method to get those results. Great.

All that's left is the shear and bending moment at the 6 m point.

The shear will be equal to the total force applied to the left or right of the section:

\begin{align} V(6) &= R_y - 2\cdot6 - \dfrac{1}{2} \cdot 4 \cdot 6 \\ &= 32 - 12 - 12 = 8\text{ kN} \end{align}

Calculating from the right would be easier, but would seem to give a negative result ($$-2\cdot4 = -8\text{ kN}$$). But we must remember that the sign convention for shear on the right side of the section is positive downwards, so that negative is actually positive.

And the bending moment will be equal to the moment calculated to either side of the beam. Here I'll calculate from the right side:

\begin{align} M(6) &= (2\cdot4)\cdot\dfrac{1}{2}\cdot4 = 16\text{ kNm (clockwise)} \end{align}

Looking at the sign convention, a clockwise bending moment on the right-side of the section is negative, so we get $$M(6) = -16\text{ kNm}$$.

With these three points (support equal to reactions, 6 m as calculated, free end all zeroes), we can easily plot our graph. We know that shear and bending are the first and second integrals of the loading, respectively. Therefore, if the loading is linear, the shear and bending will be quadratic and cubic, respectively; and if the loading is constant, the shear and bending will be linear and quadratic, respectively. Knowing this we can sketch out the diagrams by hand.

Or, you know, use a program to do it for you (I use Ftool):

• thank you for your answer i will explain the way i used to reach the equation of the shear forces as you can see i we have a line with a slope in the first part [0,6] so i used the equation of a line like this 6-(4/6)x and integrated it until i reached the momentum equation – 123321 May 5 '20 at 15:29
• @YyyUuu, yes, but you seem to have gotten some things confused in that first integration. You start it with a $\sum F_y = V - 6 - \cdots$, which doesn't make sense. $\sum F_y = 0$ is used to find the reactions. You then simply integrate the loading ($6 - (4/6)x$, as you correctly found it) to get the shear. And in your attempt, you changed $2/3x$ to $2/3x^2$, but forgot you need to add a $1/2$ factor, so it should've been $1/3x^2$ instead. And you left the $6$ as $6$, instead of integrating it as well (should've become $6x$). – Wasabi May 5 '20 at 19:10
• yeah to be honest i have missed it up there. thank you again for your hard work – 123321 May 5 '20 at 19:27

One easy way of drawing the shear and moment diagram is to separate the loading, draw the diagrams, and then superpose them.

Let's call the uniformly distributed load W1 and the triangular load W2.

In the diagram, this load and its shear and moment are shown in blue color.

The top part is the W1 loading and its shear is shown as a blue rectangular on the second row. and its moment which is a triangle on the third row in blue.

The fourth row is the triangular loading that stars from 0 at 6 meters and ends at 4 kN at the support. In the diagram, this load and its shear and moment are shown in tan.

• The max shear of the W1 is simply

$$V_1= W_1L=2*10 =20kN \ \text{at support}$$

• The moment of W1 is

$$\ M= -\frac{W_1L^2}{2}=2*10^2/2=-100kNm$$

• The max shear of W2 is the area of the shear diagram

$$\ V_2=4*6/2 =12kN$$

• The max moment of W2 is at the support and is the area of the shear

$$M= -W_2L^2/6= -4*36/6=-24kNm$$

Therefore we have the max shear at the support

$$V_{max}= V_1+V_2= 20+12=32kN$$

And max moment again at the support

$$M_{max}=-24+(-100)=-124kNm$$

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