# Indeterminate beam with a guide at each end

I've been battling with a statics question (as attached) and can't for the life of me figure it out.

I've attempted removing the left side guide joint so then there are only 3 reactions and it should be statically determined.

However, I don't know how to reapply the loading conditions for the removed guide as it is axial not transverse to which the boundary condition applies. (X direction in Cartesian coordinates).

Can someone point me in the right direction on this matter?

• You haven't told us what the question is. You have just drawn a picture of a see-saw with a load on one end and said you can't figure something out. What you have drawn is not a statics situation. – alephzero Mar 5 '19 at 17:34
• Apologies alephzero, I was looking to draw a moment diagram of the situation above. And Kamran, does the FEM apply when rollers are used, as the beam is guided not fixed? – HelpNath Mar 6 '19 at 2:06
• @HelpNath, Just to be clear, technically these are support settlement moment, which in many cases are the same as FEM. Consider the rotated beam position, from B to C we have a forced settlement of theta/ b. this causes a moment at B. Same for the A node. – kamran Mar 6 '19 at 5:11

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I've attempted removing the left side guide joint so then there are only 3 reactions and it should be statically determined.

However, I don't know how to reapply the loading conditions for the removed guide as it is axial not transverse to which the boundary condition applies. (X direction in Cartesian coordinates).

I think this is what you are suggesting. By removing the left guide, you get the beam shown in the left-half of the following figure. If you know the correct formula, you can calculate the slope of the beam at end A. Then what you need to do is add a moment to the ends of the beam (like in the right-half of the following figure) and calculate what moment $$M_A$$ is required to bring the slope of point A to zero. The sum of those two are your beam. (Note that you can visualize your original diagram as a half symmetry model about the left guide. My figure below shows a "whole" beam because finding the beam formulas for such an arrangement may be easier. Half of this beam is identical to your original beam.) That is a clever approach if you have the two formulas for the slope at end A!

I do not have the two formulas for those arrangements. Instead of deriving two equations from the two diagrams, I find it easier to derive one equation from the diagram that includes all of the loads. I will use singularity functions which makes it easy to write one equation for the entire half-beam. (A singularity function such as $$\a{x-a}^n$$ is zero when x is less than a, and $$(x-a)^n$$ for x greater than a.) The load diagram for your beam with all of the symbols is as follows:

From $$\sum F_{vertical} = 0$$, we know that the reaction at the pin constraint is $$R_B=P$$. The shear $$V$$, moment $$M$$, and slope $$\theta$$ equations as a function of position (x, measured from end A) can be written as follows: \begin{align} V(x) &= P\a{x-a}^0-P\a{x-L}^0 \tag 1 \\ M(x) &= -M_A\a{x}^0+P\a{x-a}^1-M_C\a{x-L}^0-P\a{x-L}^1 \tag 2 \\ EI \cdot \theta (x) &= -M_A\a{x}^1+ \frac12P\a{x-a}^2-M_C\a{x-L}^1-\frac12P\a{x-L}^2+c_1\tag 3 \end{align}

where L is the total length (L=a+b). For example, equation 2 indicates that the moment is equal to $$-M_A$$ between x=0 and x=a, then it begins to increase by $$P(x-a)$$ when x>a, and decreases by $$M_C$$ and $$P(x-L)$$ when x>L. If you were to plot this, you would see the expected moment diagram for the beam. The constant of integration $$c_1$$ in equation 3 can be found from knowing that the slope is 0 at x=0. From equation 3 (and remembering that <0-a> is 0), \begin{align} EI \cdot 0 &= -M_A \cdot 0+\frac12*P(0)^2-M_C \cdot 0 -\frac12P(0)^2+c_1 \tag 4\\ \therefore c_1 &=0 \tag 5 \\ \end{align}

From knowing that the slope is 0 at x=L, the moment can be found as follows using equation 3 again: \begin{align} EI \cdot 0 &= -M_A\a{L}^1+ \frac12P\a{L-a}^2-M_C\a{L-L}^1-\frac12P\a{L-L}^2 \tag 6 \\ 0 &= -M_A \cdot L+ \frac12P(b)^2-M_C(0)-\frac12P(0)^2 \tag 7 \\ \therefore M_A &= \frac12 \frac PL b^2 \tag 8 \\ \end{align}

The moment at end C can be found by writing the sum of the moments about point B, as follows: \begin{align} \sum M_{about\; B} &= M_A - P\cdot b + M_C=0 \tag 9 \\ \therefore M_C &= P\cdot b -M_A \tag {10} \\ &= P\cdot b - \frac12 \frac PL b^2 \tag {11} \\ &= P \cdot b\left(1-\frac{b}{2L}\right) \tag {12} \\ \end{align}

Let's call the supports A and B from left to right and the mid support C.

The load P creates a moment in the beam, $$M= P \cdot b$$ which will cause the beam to rotate clockwise by an angle $$\theta$$.

And the moments generated at the two supports counter this moment to settle at equilibrium.

We can write \begin{align} \sum M &= 0 \\ P \cdot b &= M_{A} +M_{B } \tag 1 \\ M_{A} &= \left(\frac{ \theta}{AC}\right) \frac {6EI}{L_{AC}} \tag 2 \\ M_B &= \left(\frac{ \theta}{BC}\right)\frac {6EI}{L_{BC}} \tag 3 \end{align} These are the fixed end moments created by support settlement.

Let's say $$BC= K \cdot AC$$, so we substitute in the above equation (1) and we get,

$$P \cdot b = M_{A} = \left(\frac{ \theta}{AC}\right) \frac {6EI}{L_{AC}} + M_{B}=\left(\frac{ \theta}{K \cdot AC}\right) \frac {6EI}{K \cdot L_{AC}} =\frac{\theta(1+K^2)6EI}{K^2 \cdot L_{AC}^2}$$

And from here we calculated the $$\theta$$.

$$\theta = \frac{(P \cdot b)K^2L_{AC}^2}{(1+K^2)6EI}$$

Edit

I added a sketch of the beam, after deflection, and of course, highly exaggerated and inaccurate. $$\theta$$ is the angle the line AB makes with the level original beam.

Note: I have used $$AC, L_{AC}$$ and similar notations for the same lengths just to emphasize where it has been used to delineate an angle and where to show the length.

• I'd suggest explaining what $AC$ and $BC$ are (as opposed to $L_{AC}$ and $L_{BC}$). Also, I think your final equation for $\theta$ isn't meant to have another $\theta$ in the denominator. – Wasabi Mar 6 '19 at 4:06
• @Wasabi, you are right, thank you. I corrected the final equation. I will later add an explanation about your comments and add a bit more detail on my solution. – kamran Mar 6 '19 at 4:43
• @Kamran. Can you clarify what $\theta$ is? Since the slope at A and B are 0, it is confusing to say that the beam rotates an angle $\theta$. (I think $\theta$ is the slope of the beam at the support C.) – JohnHoltz Mar 6 '19 at 16:54
• @JohnHoltz, I added a sketch to my answer, check it. theta would be the angle at C if C was at the center of the beam. – kamran Mar 6 '19 at 19:30
• @Kamran Shouldn't $M_{a}$ be a negative moment? Intuitively it produces a 'hogging' deflection on the beam. Does this mean that the $\theta$ angle would be negative in this situation? – HelpNath Mar 7 '19 at 15:28