# Shear force expression for singly supported triangular load

From equilibrium: $$C_y = \frac{w_0L}{2}$$

And the load is given by: $$w(x) = \frac{w_0}{L}x$$

Thus the expressions for shear and moment are given by: $$V(x) =\int{-w(x)dx}=-\frac{w_0}{2L}x^2 + C_1\\M(x) = \int{V(x)} = -\frac{w_0}{6L}x^3 + C_1x + C_2$$

How are $C_1$ and $C_2$ determined?.

Cheers

The original analysis is incorrect, because you failed to account for the sudden load at point C.

$$V(x) = -\frac{w_0}{2L}x^2 + C_1 + C_y*H(x-\frac{2L}{3})$$

Where $$H(x) = \lim_{k\to\infty} \frac{1}{1+e^{-2kx}}$$ is the heaviside step function, As such, M(x) is also different:

$$M(x) = -\frac{w_0}{6L}x^3 + C_1x + C_2 + C_y(x-\frac{2L}{3})H(x-\frac{2L}{3})$$

Now we can determine the constants. In V(x), at x=0, V(x) is 0, so $C_1$ is 0. In M(x), at x=0, M(x) is 0, so $C_2$ is 0. However, integrating further for deflection and rotation, it will be seen that the remaining constants will not be 0, because we will need to balance out the evolving step function with the growing polynomial. We will evaluate those at x = 2L/3 where the rotation must be 0 and the deflection must be 0.

JohnHoltz's answer is correct in that you determine $C_1$ and $C_2$ by looking at the boundary conditions. Looking at the left-hand side of the beam, we know that the cantilever has zero shear and bending moment, which let's us determine (note the use of the auxiliary variable $w$ which describes the load's variation per unit length):

\begin{align} w &= \dfrac{w_0}{L} \\ Q &= \int q\text{ d}x \\ &= \int wx\text{ d}x\\ &=\dfrac{1}{2}w x^2 + C_1 \\ Q(0) &= C_1 = 0 \\ \therefore Q &=\dfrac{1}{2}w x^2 \\ M &= \int Q\text{ d}x \\ &= \int \dfrac{1}{2}w x^2\text{ d}x \\ &= \dfrac{1}{6}w x^3 + C_2 \\ M(0) &= C_2 = 0 \\ \therefore M &= \dfrac{1}{6}w x^3 \\ \end{align}

As mentioned by Mark's answer, however, this only describes the beam from the left cantilever until the support. The discontinuity caused by the support means the equations are no longer valid and must be recalculated.

You can use Mark's heaviside function, or you can just repeat the process above for the other side as well.

Let's start by pretending that we only have the right-hand side, so that $x=0$ is at point C, and therefore define a new length $\ell$ which describes this cantilever, and $w_c$ as the value of the distributed load at point C.

\begin{align} \ell &= \dfrac{L}{3} \\ w_c &= \dfrac{2}{3}w_0 \\ Q &= \int q\text{ d}x \\ &= \int \left(wx + w_c\right)\text{ d}x\\ &=\dfrac{1}{2}wx^2 + w_c x + C_1 \\ Q(\ell) &= \dfrac{1}{2}w\ell^2 + w_c \ell + C_1 = 0 \\ \therefore C_1 &= -\dfrac{1}{2}w\ell^2 - w_c \ell \\ &= -\dfrac{1}{18}w_0 L - \dfrac{2}{9}w_0 L \\ &= -\dfrac{5}{18}w_0 L \\ \therefore Q &=\dfrac{1}{2}wx^2 + \dfrac{2}{3}w_0 x -\dfrac{5}{18}w_0 L \\ M &= \int Q\text{ d}x \\ &= \int \left(\dfrac{1}{2}wx^2 + \dfrac{2}{3}w_0 x -\dfrac{5}{18}w_0 L\right)\text{ d}x \\ &= \dfrac{1}{6}wx^3 + \dfrac{1}{3}w_0 x^2 -\dfrac{5}{18}w_0 Lx + C_2 \\ M(\ell) &= \dfrac{1}{6}w\ell^3 + \dfrac{1}{3}w_0 \ell^2 - \dfrac{5}{18}w_0 L\ell + C_2 = 0 \\ \therefore C_2 &= -\dfrac{1}{6}w\ell^3 - \dfrac{1}{3}w_0 \ell^2 + \dfrac{5}{18}w_0 L\ell \\ &= \dfrac{1}{162}w_0 L^2 + \dfrac{1}{27}w_0 L^2 -\dfrac{5}{54}w_0 L^2 \\ &= -\dfrac{4}{81}w_0 L^2 \\ \therefore M &= \dfrac{1}{6}wx^3 + \dfrac{1}{3}w_0 x^2 -\dfrac{5}{18}w_0 Lx - \dfrac{4}{81}w_0 L^2 \\ \end{align}

Now that we have these values, we just need to shift the origin back to $x=0$ at point A to keep these results compatible with what we got for the left cantilever, so we get:

\begin{align} Q &=\dfrac{1}{2}w\left(x-\dfrac{2}{3}L\right)^2 + \dfrac{2}{3}w_0\left(x-\dfrac{2}{3}L\right) -\dfrac{5}{18}w_0 L \\ M &= \dfrac{1}{6}w\left(x-\dfrac{2}{3}L\right)^3 + \dfrac{1}{3}w_0 \left(x-\dfrac{2}{3}L\right)^2 -\dfrac{5}{18}w_0 L\left(x-\dfrac{2}{3}L\right) - \dfrac{4}{81}w_0 L^2 \end{align}

• Me and my love for discontinuous functions. I can never go backwards. +1 for being able to translate back again for those who need to be able to do so. – Mark Jun 6 '18 at 17:52

The constants of integration are determined from the known conditions. In this case, there is no load at A, so the shear and moment at A are 0.

From your equation for w(x), it appears that x=0 at A, so the equations for V(x) and M(x) result in C1=0, C2=0.