# why the constants of the shear and bending are zero

In order to get the bending and the shear function using the singularity functions we integrate loading. could you please tell me why the two constants are zero.

We know that the load is related by the bending M(x) and the shear equation using these equations: q(x)=dV(x)/dx and V=dM/dx. In order to find the M(x) and V(x) Using the singularity method I found that the constants of integration are null: $$V(x) = \int q(x)dx\,.$$ and $$M(x) = \int V(x)dx\,.$$ without constants. Could you please clarify why these constants are always null?

• I added the information from your comment into the question. You can edit the question at anytime to add information.
– hazzey
Mar 28, 2019 at 12:35
• Are you asking why the constant term in the integration is zero? If you, please show exactly what you are questioning. Please don't leave out information that you think is obvious.
– hazzey
Mar 28, 2019 at 12:37
• Thanks Hazzey, Exactely I ask why these two constants are zero here Mar 28, 2019 at 12:45
• So, as per the answer, what are the boundary conditions - they may well be 0... How can we guess... Mar 28, 2019 at 15:44
• To study the defelxion using the singularity methods they use these constants zero I don't know why? Mar 28, 2019 at 15:59

$$\newcommand{\a}[1]{\langle#1\rangle}$$The reason is simple: the singularity-function form of the shear and bending moment equations already perfectly describes the results without the need for integration constants.

This is because the constants serve to reflect the impact of the beam's support reactions on the shear and bending moment, but the singularity-function form already includes these reactions in its basic form.

For example, imagine we have a free-floating beam (so no supports) with a uniformly distributed load $$q$$ throughout its span. This would lead to a rigid-body movement in the direction of the load (since the beam has no supports to resist that external force) and no internal stresses, but just pretend that's not the case: the beam magically resists that motion, generating shear and bending moment.

Using simple integration, we can find these shear and bending moment equations:

\begin{align} q(x) &= q \\ V(x) &= \int q(x)\text{d}x = qx + C_1 \\ M(x) &= \int V(x)\text{d}x = \dfrac{qx^2}{2} + C_1 x + C_2 \end{align}

As to the specific values of $$C_1$$ and $$C_2$$, those are defined by the boundary conditions.

• If the beam is simply-supported, then the constants are ($$R$$ is the support reaction) \begin{align} V(0) &= C_1 = R \\ M(0) &= C_2 = 0 \\ V(x) &= qx + R \\ M(x) &= \dfrac{qx^2}{2} + Rx \end{align}

• If the beam is a cantilever fixed at $$x=L$$ \begin{align} V(0) &= C_1 = 0 \\ M(0) &= C_2 = 0 \\ V(x) &= qx \\ M(x) &= \dfrac{qx^2}{2} \end{align}

• If the beam is a cantilever fixed at $$x=0$$ ($$R_y$$ is the vertical reaction, and $$R_M$$ is the moment reaction) \begin{align} V(0) &= C_1 = R_y \\ M(0) &= C_2 = R_M \\ V(x) &= qx + R_y \\ M(x) &= \dfrac{qx^2}{2} + R_y x + R_M \end{align}

Even if the supports are "within" the span (as in, not at the extremities), the same basic equations will remain valid. The only difference will be that the reactions will lead to discontinuities in the beam results, so you'll need to break the beam down into its component parts and use the basic equations for each part.

As you can see, the basic equation (i.e. $$V = qx + C_1$$) is defined purely by the loading, since that's where it comes from. The support reactions then set the value of the integration constants $$C_1$$ and $$C_2$$ to make that basic equation represent the specific case being studied.

With singularity functions, this happens naturally.

That's because singularity functions aren't described only by the loading, but also by the supports. After all, the support reactions are placed directly into the "loading" function itself.

For example, while the loading equation in the simple-integration case was just $$q(x) = q$$, it now depends on the boundary conditions:

• If the beam is simply-supported, then we get \begin{align} q(x) &= q + R\a{x-0}^{-1} \\ V(x) &= \int q(x)\text{d}x = qx + R\a{x-0}^0 \\ &= qx + R\\ M(x) &= \int V(x)\text{d}x = \dfrac{qx^2}{2} + R\a{x-0}^1 \\ &= \dfrac{qx^2}{2} + Rx \end{align}

• If the beam is a cantilever fixed at $$x=L$$ \begin{align} q(x) &= q \\ V(x) &= \int q(x)\text{d}x = qx \\ M(x) &= \int V(x)\text{d}x = \dfrac{qx^2}{2} \end{align}

• If the beam is a cantilever fixed at $$x=0$$ ($$R_y$$ is the vertical reaction, and $$R_M$$ is the moment reaction) \begin{align} q(x) &= q + R_y\a{x-0}^{-1} + R_M\a{x-0}^{-2} \\ V(x) &= \int q(x)\text{d}x = qx + R_y\a{x-0}^0 + R_M\a{x-0}^{-1} \\ &= qx + R_y \\ M(x) &= \int V(x)\text{d}x = \dfrac{qx^2}{2} + R_y\a{x-0}^1 + R_M\a{x-0}^0 \\ &= \dfrac{qx^2}{2} + R_y x + R_M \end{align}

Comparing these results to those found by traditional integration, they are the same. And that's precisely because of the difference in how the support reactions are treated in each case: traditional integration ignores them and then later adds their effect via the integration constants. Meanwhile, singularity functions include the boundary conditions at the very start in how they define $$q(x)$$.

However, if you continue integrating past the bending moment into the slope and deflection equations, you then have another set of conditions which haven't been taken into account yet: the physical constraints such as $$\theta=\delta=0$$ at fixed supports. Since this effect isn't incorporated into the singularity equations, you'll need to add integration constants for each of these additional integrations.

So, to take the example of the cantilever with fixed end at $$x=0$$ above, the integration up to the deflection is:

\begin{align} q(x) &= q + R_y\a{x-0}^{-1} + R_M\a{x-0}^{-2} \\ V(x) &= \int q(x)\text{d}x = qx + R_y\a{x-0}^0 + R_M\a{x-0}^{-1} \\ M(x) &= \int V(x)\text{d}x = \dfrac{qx^2}{2} + R_y\a{x-0}^1 + R_M\a{x-0}^0 \\ \theta(x) &= \int M(x)\text{d}x = \dfrac{qx^3}{6} + \dfrac{R_y}{2}\a{x-0}^2 + R_M\a{x-0}^1 + C_1\\ \delta(x) &= \int \theta(x)\text{d}x = \dfrac{qx^4}{24} + \dfrac{R_y}{6}\a{x-0}^3 + \dfrac{R_M}{2}\a{x-0}^2 + C_1 x + C_2\\ \end{align}

In this case, $$C_1 = C_2 = 0$$, since the fixed support at $$x=0$$ has $$\theta(0)=\delta(0)=0$$.