# Finding the bending moment diagram for an indeterminate continuous beam

In a recent lecture, we had the following example of an indeterminate continuous beam where we must find the redundant support reaction at $$A$$:

This was found by removing the support at $$A$$, and then finding the individual deflections from the UDL load ($$\Delta_A^q$$), deflection from the point load ($$\Delta_A^P$$) and then finally, the deflection that would have been contributed by the redundant support ($$\Delta_A^{R_A}$$) as a function of $$R_A$$.

These deflections are all found using the force (virtual) method, which are then superimposed together to solve for $$R_A$$ using the following continuity equation:

$$\Delta_A=\Delta_A^P+\Delta_A^q+\Delta_A^{R_A}=0$$

Upon finding the reaction support $$R_A$$, we can then solve for the remaining support reactions ($$R_B, R_C$$) as well.

My issue is that I don't understand how to draw the bending moment diagram of the real system without being given a numerical value for $$R_B$$ first, which can only be determined after finding the redundant reaction support $$R_A$$ (which is the purpose of the problem).

The method I use to find the bending moment equations is to take cuts before each discontinuity, from left to right, so there would be a cut between $$AB$$ and $$BC$$.

For the UDL deflection, I was able to find the bending moment equation for cut $$AB$$ since this does not rely on $$R_B$$ and is: $$M(x)=\frac{qx^2}{2}$$

For the cut $$BC$$, without being given a numerical value for $$R_B$$, the bending moment equation would be: $$M(x)=-2Lqx+6L^2q+R_Bx-2LR_B$$

But, knowing conceptually that the bending moment equation for cut $$BC$$ is a linear equation ending with a zero moment at $$C$$, I was able to find the gradient (using rise over run) and some linear translations to find a final bending moment equation for cut $$BC$$ that is:

$$M(x)=-Lqx+4L^2q$$

This assumption of a linear equation for the cut $$BC$$ is supported by the bending moment diagram already given in the lecture slides for the UDL load as follows:

So for the UDL load, I was able to find the correct bending moment diagram relatively straightforward using what I established from the cut $$AB$$ and the fact that the bending moment must be zero at $$C$$ (i.e. $$x=4L$$).

However, when it comes to developing the point load bending moment equations, I am unsure on how to obtain these equations without knowing either the reaction supports $$R_B$$ or $$R_C$$ or by looking at the bending moment diagram for the point load, which was given in the lecture slides as follows:

Note, this bending moment diagram has a cut $$AB$$ and new cuts $$BD$$ and $$DC$$ where $$D$$ is the position of the point load being applied along the continuous beam (i.e. $$x=3L$$). The moment equations for these three cuts (as a function of $$R_B$$) are:

Cut $$AB$$: $$M(x)=0$$

Cut $$BD$$: $$M(x)=-R_Bx+2LR_B$$

Cut $$DC$$: $$M(x)=4Lqx-R_Bx+2LR_B-12L^2q$$

Or by rewriting these equations not as a function of $$R_B$$ (which I observed using the above bending moment diagram in the lecture slides and linearly translating the equations):

Cut $$AB$$: $$M(x)=0$$

Cut $$BD$$: $$M(x)=-2Lqx+4L^2q$$

Cut $$DC$$: $$M(x)=2Lqx-8L^2q$$

So I was wondering how can I obtain the bending moment equations for the cuts $$BD$$ and $$DC$$ without knowing beforehand what the bending moment diagram would look like for the point load, or without writing these bending moment equations as functions of $$R_B$$ or $$R_C$$? How do we know the maximum moment at point $$D$$ (i.e. $$x=3L$$) is $$2L^2q$$?

Other than that, I'm ok to find the virtual bending moment diagrams, and subsequently the deflections for each of these cases using integration on the product of the bending moment equations for each real and virtual system cut. I'm just unsure on how to develop the bending moment diagram for the point load.

Any input would be of much help! Thanks!

• Just remember the BM diagram is the integral of the shear force diagram. Always draw your SF diagram first. That is not a complete solution to this problem, but at least makes constructing the BM diagram obvious. I'd solve it using deflections at the pivots myself. Feb 12, 2023 at 21:14
• @GregLocock Thanks for your suggestion. I know I could use a shear force diagram and integration to find the bending moment equations/diagram, but this would still have functions of R_B or R_C no? We were told to solve for the deflection at point A and no where else unfortunately. Feb 13, 2023 at 5:35
• You could use superposition of the reaction forces for each of the two loads, which are directly obvious. Feb 13, 2023 at 7:23
• Superposition is a linear combination so it should require just linear elasticity and small deformations assumptions, which are most commonly used for problems like these anyway. Static determinism is not an issue. Feb 13, 2023 at 18:05
• for load 1 work out the reaction forces B1 and C1, repeat for load 2 to give B2 and C2, then for both loads the reactions are B1+B2 at B and C1+C2 at C Feb 13, 2023 at 19:26