Shear Force Diagram of a Simply Supported Beam with triangular load distribution

Question

I've been attempting this fundamental shear force diagram problem for several days, but can't seem to get the correct result. I'm trying to calculate the shear force diagram in terms of $x$, but I'm unsure about the intensity $w(x)$ of the triangular load distribution between $0m \le x \lt 3m$. I am able to calculate the correct result for the latter section $3m \lt x \le 6m$, so I'm a little confused as to what the correct intensity of the triangular load distribution is and how to calculate the correct shear force using the correct intensity $w(x)$?

Below I've attached the problem and calculated the support reactions, which are $A_y=15kN$ and $B_y=15kN$.

Now, I've attached my free body diagram of the first section between $0m \le x\lt 3m $ and indicated the positive sign convention for this beam.

I then proceeded to find the shear force in terms of $x$ as follows:

$\sum F_y=0:$

$$15-w(x)·x·\frac12 - v_1 = 0 \quad (eq\ 1)$$

Where $w(x)·x·\frac12$ is the area of the triangular load distribution.

This is where I get confused. My understanding of triangular load distribution in terms of the intensity $w(x)$ is that:

$$w(x)=\frac{w_0x}{L}$$

Where $w_0 = 10$ and $L=3$ for this problem.

But substituting these values into the intensity $w(x)$ and back into $(eq\ 1)$ gets me the wrong result of: $$v_1=15-\frac53 x^2$$

After reading multiple textbooks and watching several videos, I finally found out that if the maximum load of a triangular load distribution is at the initial point $x=0$ then the following formula should be applied:

$$w(x)=\frac{w_0x}{L}-w_0$$

I now understand this a bit, but I am wondering where I could get a good explanation as to why?

I'm struggling to find a good explanation as almost every example I've found in textbooks/videos use triangular load distributions that increase from the initial point and not decrease.

However, after utilising this formula, I still get the wrong solution. My working out is as follows:

$\sum F_y=0:$

$$15-\Bigl(w(x)·x·\frac12 \Bigr) - v_1 = 0$$ $$15-\biggl(\Bigl(\frac{10x}{3}-10\Bigr)·x·\frac12 \biggr)- v_1 = 0$$ $$15-\biggl(\Bigl(\frac{10x}{6}-\frac{10}{2}\Bigr)·x\biggr) - v_1 = 0$$ $$15-\biggl(\Bigl(\frac{5x}{3}-5\Bigr)·x \biggr)- v_1 = 0$$ $$15-\frac{5x^2}{3}+5x - v_1 = 0$$ $$\Rightarrow v_1=15-\frac{5x^2}{3}+5x$$

The actual solution is: $$v_1=15+\frac{5x^2}{3}-10x$$

So I'm not sure whether I'm using the correct intensity $w(x)$ and/or whether the triangle area has been correctly calculated using this intensity $w(x)$.

For the second section $3m\le x\lt6m$ I am able to calculate the correct shear force in terms of $x$, this solution is:

$$v_2=-15-\frac{5x^2}{3}+10x$$

Plotting a diagram of the correct shear forces $v_1$ and $v_2$ in terms of $x$ looks the following:

For your reference, this problem (F11.6) can be found in chapter 11 of Statics and Mechanics of Materials (4th Ed. SI edition) by Hibbeler.

I'd appreciate if someone could explain intensity loads for situations similar to above and where I went wrong in my calculations.

Thank you.

Edit:

After reading a few examples, I found that if I calculate the shear force from the left end I am able to get the correct shear force using my initial intensity $w(x)=\frac{w_0x}{L}$ and not the latter intensity $w(x)=\frac{w_0x}{L}-w_0$.

However I'm unsure why I can't calculate this from the right end? Does it have something to do with the left support $A_y=15kN$ creating a discontinuity? If I calculate from the left end am I correct in changing the section's range to $0m \lt x \le 3m$ to not include the left support $A_y$?

My working out is as follows:

$\sum F_y=0:$

$$-\Bigl(w(x)·x·\frac12 \Bigr) + v_1 = 0$$ $$-\biggl(\Bigl(\frac{10}{3}(3-x)\Bigr)·(3-x)·\frac12 \biggr)+ v_1 = 0$$ $$-\biggl(\Bigl(10-\frac{10x}{3}\Bigr)·(3-x)·\frac12 \biggr)+ v_1 = 0$$ $$-\biggl(\bigl(30-10x-10x+\frac{10x^2}{3}\bigr)·\frac12 \biggr)+ v_1 = 0$$ $$-\biggl(\bigl(30-20x+\frac{10x^2}{3}\bigr)·\frac12 \biggr)+ v_1 = 0$$ $$-\bigl(15-10x+\frac{10x^2}{6}\bigr)+ v_1 = 0$$ $$-15+10x-\frac{5x^2}{3}+ v_1 = 0$$

$$\Rightarrow v_1=15+\frac{5x^2}{3}-10x$$

This is the correct solution.

Answer 1

Your procedure is correct, but you have made a mistake with the sign convention. Apparently you are using the same convention as I do, where a vertical load/force is negative, and an upward force is positive.

Consequently, as you've written correctly $$ w(x)=\frac{w_0}{L}x-w_0 $$

Your mistake happens as you formulate the force equilibrium equation. With this definition of $w(x)$ you already comply to the sign convention. If you now formulate the shear force equation and write $$ V_1=15-\int w(x)dx $$ you basically reverse the sign convention again. To formulate the force equilibrium equation you have to sum all forces, not subtract them, thus $$ V_1=15+\int w(x)dx $$ which leads to $$ V_1=15+\frac{5}{3}x^2-10x $$ which is the correct result.

Sign convention [edited]

Take a look at your $w(x)$. It's a force pointing downwards, so it should be negative. You have written it as $$ w(x)=\frac{w_0}{L}x-w_0 \qquad \mbox{for}\qquad x=\{0...3\}$$ Thus $w(0)=-w_0$, which means your load $w(x)$ is already defined in the coordinate system you specified. If you now sum or integrate and add a minus sign in front of $w(x)$ you basically turn the downward load into an upwards facing load.

Consider, e.g. $w(x)=const.=-w_0$, i.e. an evenly distributed downwards load, thus a negative value. It's easy to figure out, that the reaction at the bearing is $A_y=\frac{1}{2} w_0L $. Now to find the shear force distribution, if we use your method, we'd write: $$ V_1=A_y-\int_0^L (-w_0)dx=\frac{1}{2} w_0L - (-w_0)L=\frac{3}{2}w_0L $$ This is the wrong result, because for the shear force distribution calculation we suddenly switch sign convention (subtracting during integration instead of adding).

In the first part you said $w(x)$ was $w(x)=-w_0+\frac{x}{L}w_0$, which goes linearly from $-w_0$ to $0$, which in your example would be $w(x)=\frac{10}{3}(x-3)$. During your second calculation you managed to get the right result because you changed the sign convention of your $w(x)$, as you inserted $w(x)=\frac{10}{3}(3-x)$ (which is an upwards load), but then you turned it downwards again by adding a negative sign in front of the whole term.

I really recommend sticking to one sign convention. Either you say you sum all forces, and defined downward loads as negative, or you subtract downward loads from upward reactions.