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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.

enter image description here

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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:

enter image description here

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:

enter image description here

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