1. the distributed load $w_0$ at the end of D of the beam for which the reaction at B is zero
  2. the corresponding reaction at C.

enter image description here

Please show me how to get the right answer to this question.: $$\begin{gather} R_1 = 1/2 \cdot w_0 \cdot 9 = 0/2 w_0 \\ R_2 = 1/2 \cdot 3.5 \cdot 9 = 63/4\text{ kN} \\ \text{For }B = 0 \\ \sum M_c = 0 \\ 9/2 w_0 \cdot 8 - 63/4 * something \\ \end{gather}$$ Thats where I stopped I don't know what to do May someone please show me the solution and explain to me his/her steps?

  • 1
    $\begingroup$ while homework questions are allowed, you need to show your work and have a specific question about the homework $\endgroup$
    – user16
    Apr 3, 2019 at 21:41
  • $\begingroup$ I'm sorry I didn't know the rule; I added my attempt but I am really stuck if you can please show me the solution with an explanation of the steps. So sorry again for breaking the rule and thanks for the help $\endgroup$ Apr 3, 2019 at 21:57

2 Answers 2


In summary the steps are:

  1. Write equilibrium equations in terms of unknown load magnitude Wo.
  2. Set the vertical reaction RB = 0.
  3. Solve equation for Wo.

It helps me to break the trapezoidal distribution into a rectangular distribution magnitude Wo, and a triangular distribution with peak magnitude = (3.5 - Wo).

From the steps you have shown. There is already a problem when you calculate the resultant of the triangular distribution. Look closely at the diagram at the top of the solution below to see the problem.

enter image description here


The distance from the nearest support is 1m.

The distributed load is 3.5kN/m.

As the load is not uniform, you need to calculate the rectangular area load and halve it to obtain triangular load which is half base by height. $$W_0 = 0.5×3.5=1.75kN$$

This is effectively a point load half-way between D and the nearest support. While I've not calculated the reaction moment, I think you might be able to figure it out.

  • $\begingroup$ You don't seem to have included the concentrated bending moment at the left end. $\endgroup$
    – Wasabi
    Apr 18, 2019 at 2:45

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