# Double Integration Method

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.

• What do you want to solve for?
– NMech
Oct 1, 2021 at 20:29

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 $$x_2 = 0 - 3.5$$.

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.

• Thank you so much Oct 2, 2021 at 2:07
• You are welcome. Glad that you got it that quick.
– r13
Oct 2, 2021 at 2:14
• In the case of taking the three sections AB, BC and CD, what would be the considerations to be taken? (Continuity equations and others) Oct 2, 2021 at 2:18
• 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.
– r13
Oct 2, 2021 at 2:54

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

• Thank you so much Oct 2, 2021 at 2:07