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Determine:

  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?

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    $\begingroup$ while homework questions are allowed, you need to show your work and have a specific question about the homework $\endgroup$ – user16 Apr 3 '19 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 $\endgroup$ – Andre_van_stone Apr 3 '19 at 21:57
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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

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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.

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  • $\begingroup$ You don't seem to have included the concentrated bending moment at the left end. $\endgroup$ – Wasabi Apr 18 '19 at 2:45

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