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How can I solve this beam. I was able to find the moments of each section, from there I need help. Please if someone could help me. The method to use is Double Integration. enter image description here

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    $\begingroup$ What do you want to solve for? $\endgroup$
    – NMech
    Oct 1, 2021 at 20:29

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I am going to provide the hints for you to solve this problem.

Hint 1: Since the moment diagram has a singular point (change shapes) at the internal supports B & C, so you need to consider each segment (A-B, B-C & C-D) separately in writing the moment equations.

Hint 2: However, due to symmetry in both beam geometry and loadings, you only need to evaluate over 2 segments (A-B & B-midspan).

  • For segment A-B, write the moment equation $M_1 = \dfrac{w_1x_1^2}{2}$, for $x_1 = 0 - 3$.

  • For segment B-midspan, $M_2 = M_1 - Rx_2 - \dfrac{w_2x_2^2}{2}$, for $x_2 = 0 - 3.5$.

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The first integration of the moment ($EIv"$) will yield $EIv'$ & a constant $C_1$, and the second integration of the moment will yield $EIv$ & a constant $C_2$. After successfully integrations, you can find the constant $C_1$ & $C_2$ by inserting the proper coordinate ($x$) with the boundary conditions - $C_1$ represent rotation, which is "zero" at the support; $C_2$ represent deflection, which is "zero" at the support point as well.

Note, since you have two moment equations ($M_1$ & $M_2$), you will get 4 constants. I suggest naming the constants as $C_{11}$, $C_{12}$, $C_{21}$ & $C_{22}$ to avoid confusion.

The procedure looks somewhat complicated and confusing, but once you start doing it, it will become easy and clear.

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  • $\begingroup$ Thank you so much $\endgroup$
    – RCristianL
    Oct 2, 2021 at 2:07
  • $\begingroup$ You are welcome. Glad that you got it that quick. $\endgroup$
    – r13
    Oct 2, 2021 at 2:14
  • $\begingroup$ In the case of taking the three sections AB, BC and CD, what would be the considerations to be taken? (Continuity equations and others) $\endgroup$
    – RCristianL
    Oct 2, 2021 at 2:18
  • $\begingroup$ I think you are asking what factors that affect how many segments to be taken for any given scenario. Well, 1) bending moment must be a smooth function of "x", 2) change of beam geometry or load types/intensity, 3) the elastic curve must not have gap or kink, the rotation and deflection must be continuous at the junctions where the segments meet. $\endgroup$
    – r13
    Oct 2, 2021 at 2:54
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An easy way is to think of it as two separate beams.

The center beam with L= 7meter will have half of its total 7meters loading as reaction on supports B and C and maximum moment $$M_{center segment}= \frac{\omega, (2236.13kgm) 7^2} {8}$$

Eaxh of the two side legs can be considered a cantilever beam but their end momend, because of symetry will in effect shift the entire moment curve of the center beam up by $$ M_{cantilever}=1936.44*3*3/2=2940.66kgm$$

Then the reactions will be the sum of total load over the full length of the beam divided by two and the positive moment at the center will be $$M_{c}=M_{centersegment}- M_{cantilever}$$ and negative moment at support will be the cantilever moment=2940.66kgm

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  • $\begingroup$ Thank you so much $\endgroup$
    – RCristianL
    Oct 2, 2021 at 2:07

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