I am having difficulties trying to work out one problem. I've attached a picture explaining the situation - the task is to determine internal forces in C but the bending moment M(x) doesn't seem to get me the correct results (I used equivalance and went from LHS). I would greatly appreciate if somebody helped me construct a correct equation or find the mistake, thank you.

picture of the problem


1 Answer 1


Let's start by simplifying this for ourselves.

In such a simple beam, horizontal loads don't have any influence in anything other than the axial loads. You apparently need to solve for those as well, but you're only asking for the bending moment, so I'm going to ignore them, so let's see how the load looks considering just the vertical component.

$$\begin{align} q_0 &= 3\sin{48} = 2.23\text{ kN/m} \\ q_1 &= 9\sin{48} = 6.69\text{ kN/m} \end{align}$$

Let's also already calculate some intermediate values for the load at the left support ($A$) and at $C$, which we'll need later on:

$$\begin{align} q_A &= q_0 + (q_1-q_0)\dfrac{1.2}{10.2} = 2.75\text{ kN/m} \\ q_C &= q_0 + (q_1-q_0)\dfrac{2.2}{10.2} = 3.19\text{ kN/m} \end{align}$$

By moment, we can see that the reaction at the left support ($A$) is:

$$\begin{gather} M_B = -9R_A + \dfrac{q_0 \cdot 10.2^2}{2} + \dfrac{(q_1-q_0) \cdot 10.2^2}{2\cdot3} = 0 \\ \therefore R_A = \dfrac{10.2^2}{9}\left(\dfrac{q_0}{3} + \dfrac{q_1}{6}\right) = 21.48\text{ kN} \end{gather}$$

We can then calculate the bending moment at $C$: $$\begin{align} M_C &= -\dfrac{q_0 \cdot 2.2^2}{2} - \dfrac{(q_C-q_0) \cdot 2.2^2}{2\cdot3} + R_A \cdot 1 \\ &= -2.2^2\left(\dfrac{q_0}{3} + \dfrac{q_C}{6}\right) + R_A = 15.31\text{ kNm} \end{align}$$

If you'd rather do this by finding the bending moment equation, you'll need to use the beam equation. In the main span, the distributed load has the following equation:

$$q(x) = q_A + (q_1-q_A)\dfrac{x}{9}$$

Integrating that, we get the shear equation:

$$\begin{align} Q(x) &= \int q\text{d}x = -q_Ax - (q_1-q_A)\dfrac{x^2}{18} + C_1 \\ Q(0) &= C_1 = \text{shear at }A^+ = -\dfrac{(q_0 + q_A)}{2}\cdot1.2+R_A = 18.492\text{ kN} \\ \therefore Q(x) &= -q_Ax - (q_1-q_A)\dfrac{x^2}{18} + 18.492 \\ \therefore Q_C &= Q(1) = 15.523\text{ kN} \end{align}$$

Integrating again, we get the bending moment:

$$\begin{align} M(x) &= \int Q\text{d}x = -\dfrac{q_Ax^2}{2} - (q_1-q_A)\dfrac{x^3}{54} + 18.492x + C_2 \\ M(0) &= C_2 = \text{moment at }A = -1.2^2\left(\dfrac{q_0}{3} + \dfrac{q_A}{6}\right) = -1.73\text{ kNm} \\ \therefore M(x) &= -\dfrac{q_Ax^2}{2} - (q_1-q_A)\dfrac{x^3}{54} + 18.492x - 1.73 \\ \therefore M_C &= M(1) = 15.31\text{ kNm} \end{align}$$


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