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I have the following structure, in which I am failing to calculate the two reactions: Cz (apparently vertical) and Dz, in the roller support of the beam. The beam length is 7.5 m, the reactions Az, Cx, and the momentum in C are calculated correctly. The distributed load replaced with concentrated on the first beam. Thank you in advance for advice and help.
The reasoning here is very similar to that in your previous question.
Start by noting that the structure can be separated into three different isostatic structures:
Where $F$ is obviously the sum of the reactions of the imaginary supports on the beams.
The left-hand side you've already solved. All that's left is the right hand side, and that's just as easy: $R = \frac{1}{2} \cdot 5.1 \sqrt{4.5^2+6^2} = 19.13\text{ kN}$.
And then with that you obviously know that the vertical reaction on the central column is equal to $19.13 + 14.28 = 33.43\text{ kN}$. The horizontal and moment reactions on that column you've already solved.
To check our work:
