# Resolving forces in a simple truss

I'm trying to re-remember how to resolve forces in pin joint frames. I seem to have gotten stuck rather quickly and the examples I've been through don't replicate the same issue.

Here is my truss I'm working on (terrible image, sorry) When trying to calculate the reaction force at B1, B2 and B3 I get different answers depending on where I take moments about from.

Example:

F(B2) = 2 x F(B3)

6 x F(B3) + 3 x F(B2) = -10 x 3

Therefore:

F(B3) = +2.5kN

F(B2) = +5kN

or

Take Moment about B2, F(B1) = F(B3)

3 x F(B3) + 3 x F(B1) = -10 x 3

Therefore

F(B1) = +5kN

F(B3) = +5kN

or

F(B2) = 2 x F(B1)

6 x F(B1) + 3 x F(B2) = -10 x 3

Therefore:

F(B1) = +2.5kN

F(B2) = +5kN

Different answers for the each restraints. Can someone please advise where I am going wrong? Thanks Rich

• Can you please edit your question showing your work which got you those results? – Wasabi Aug 24 '16 at 14:41
• This is a statically indeterminate truss. You cannot solve for the reactions simply by summing moments. For a truss to be solvable from equilibrium alone it must satisfy the following equation: $m + r \leq 2j$. Where m is the number of members (7), r is the number of reactions (6 - 2 per support how you have drawn it), and j is the number of joints (5). Therefore we have: $7 + 6 \nless 2(5)$. You will need to use another method, such as the Force Method to solve this. – atom44 Aug 24 '16 at 15:42
• Thanks mg4w. A brief look into the Force Methods suggest I have a lot f reading yet to do. Or maybe some cheating by simplifying the problem. – Rich Aug 24 '16 at 16:15
• @Rich, I would recommend starting with determinate trusses. Once you are confident with those move on to indeterminate. – atom44 Aug 24 '16 at 17:23
• This will probably help a lot. – F.Bek Aug 25 '16 at 14:18

As @mg4w already mentioned in a comment, this is a statically indeterminate structure. This means that the structure can't be trivially solved as you have tried (unless you use a simplifying assumption, as I describe towards the end).

To show the problem, these are global equilibrium equations we have at our disposal:

\begin{align} \sum F_x &= H_{top} + H_{mid} + H_{bot} = 0 \\ \sum F_y &= V_{top} + V_{mid} + V_{bot} - 10 = 0 \\ \sum M_{mid} &= -3.00H_{top} + 3.00H_{bot} - 3.00\times10 = 0 \\ &= -H_{top} + H_{bot} + 10 = 0 \\ \therefore H_{top} &= H_{bot} + 10 \end{align}

but this is as far as we can go. I could substitute $$H_{bot}$$ in the $$\sum F_x = 0$$, but that wouldn't get me anywhere. We're stuck. That's the trouble with hyperstatic (statically indeterminate) structures. Global equilibrium won't get you anywhere.

Now, it can be solved by hand, but it takes a bit of work. Here's your structure: Let's start by simplifying your structure as far as we can to keep things clean once we reach the point where trivial methods will get stuck. For a start, observe that bars 4 and 7 will never take on any load. These are truss bars (hinged on both ends) and therefore only take axial load. For them to take an axial load, they would have to deform along their axes, but since both extremities are constrained, they can't deform and therefore will never take a load. So we can remove them from our model.

Also, the region where the load is applied can be trivially solved. The load is vertical and only bar 2 has a vertical component, so it will have to absorb the entirety of the external force. So the vertical component of the axial force in bar 2 is equal to -10 kN. The horizontal component will have to be such that the resultant force is parallel to the bar, so we can find that

$$\begin{gather} \frac{-10}{h} = \tan\theta_2 = \frac{3}{-2} \\ \therefore h = 6.67\ \text{kN} \\ N_2 = \sqrt{10^2 + 6.67^2} = 12.02\ \text{kN (compression)} \end{gather}$$

Now, if bar 2 is generating a horizontal force of 6.67 kN on that node, bar 1 must balance that out, generating a horizontal force of -6.67 kN, which puts the bar in tension.

We can now remove bars 1 and 2 from the model as well, replacing them with their internal forces. We therefore end up with the following model (the supports are hinged, so no need for the hinge "ball"): Now, the horizontal load applied by bar 1 is simply absorbed by the support and we can mostly forget about it.

Since bars 3 and 6 have the same tangent (in modulus), the vertical force of 10 kN carried by bar 2 will be equally divided between them. Just as with bar 2, this will generate horizontal forces within them. Using the same method as above, we find that $$h = \pm1.67\ \text{kN}$$ (positive for bar 6, negative for bar 3). The resultant axial force in the bars due to this load is therefore equal to 5.27 kN (tension for bar 3, compression for bar 6). Since the horizontal components generated by bars 3 and 6 cancel themselves out, bar 5 is unaffected.

And now we get to the point where we get stuck: the 6.67 kN horizontal force applied by bar 2. Bars 3, 5 and 6 all have horizontal components and can therefore participate in absorbing this force. We therefore have to figure out how the force is parcelled out between them.

Bars can be replaced with springs which resist the displacement of the nodes. The stiffness of the springs for truss bars is equal to $$K = \dfrac{EA}{L}$$. The springs work according to Hooke's Law, which basically states that

$$F = K\delta$$

where $$\delta$$ is the deformation of the bar.

Now, the node will clearly only deform in the horizontal direction. But bars 3 and 6 are inclined, so their stiffness is only partial, so we need to get the horizontal component of that stiffness, which is equal to $$\overline{K}_3 = \overline{K}_6 = \dfrac{EA}{L}\cos\theta$$ (where $$\theta$$ is the angle of bars 3 and 6 with the horizontal axis. The sign doesn't matter for cosine). The deformation felt by bars 3 and 6 is also inclined, so we also have to get just the horizontal component of that deformation, which gives us $$\overline{\delta}_3 = \overline{\delta}_6 = \delta\cos\theta$$.

Therefore, Hooke's Law for bars 3 and 6 for horizontal deformations actually looks like:

$$F = K\delta\cos^2\theta = \frac{EA}{L}\cos^2\theta\delta$$

The total stiffness felt by the node is therefore equal to (assuming equal $$EA$$ for all bars):

\begin{align} K &= K_3 + K_5 + K_6 \\ K_3 = K_6 &= \frac{EA}{L_3}\cos^2\theta \\ K_5 &= \frac{EA}{L_5} \end{align}

So, what fraction of the horizontal force will go to bar 5?

\begin{align} f_5 &= \dfrac{\dfrac{EA}{L_5}}{\dfrac{EA}{L_5} + 2\dfrac{EA}{L_3}\cos^2\theta} \\ f_5 &= \dfrac{\dfrac{1}{L_5}}{\dfrac{1}{L_5} + 2\dfrac{1}{L_3}\cos^2\theta} \\ f_5 &= \dfrac{1}{1 + 2\dfrac{1}{\sqrt{10}}\left(\dfrac{1}{\sqrt{10}}\right)^2} = 0.9405 \\ \therefore f_3 = f_6 &= \frac{1 - 0.9405}{2} = 0.0297 \end{align}

So, bar 5 gets 94.05% of the horizontal load and bars 3 and 6 get 2.97% each. Bar 5 therefore has an axial compression of $$0.9405\times6.67 = 6.27\ \text{kN}$$, while the horizontal component of the axial force in bars 3 and 6 is equal to $$0.0297\times6.67 = 0.20\ \text{kN}$$. Using the same method as used previously for bars 2, 3 and 6, we can also find the vertical components for bars 3 and 6 as equal to $$v = 0.20\times\dfrac{\pm3}{1} = \pm0.60\ \text{kN}$$ (positive for bar 3, negative for bar 6). This generates a resultant axial force on bars 3 and 6 of 0.63 kN (compression in both).

Adding the axial forces in bars 3 and 6 due to both vertical and horizontal components of the forces applied by bar 2, we get:

\begin{alignat}{2} N_3 &= 5.27 - 0.63 = 4.64\ \text{kN}&&\text{ (tension)} \\ N_6 &= -5.27 - 0.63 = -5.90\ \text{kN}&&\text{ (compression)} \\ \end{alignat}

To summarize, the forces in each of the bars is: \begin{alignat}{2} N_1 &= 6.67\ \text{kN}&&\text{ (tension)} \\ N_2 &= 12.02\ \text{kN}&&\text{ (compression)} \\ N_3 &= 4.64\ \text{kN}&&\text{ (tension)} \\ N_4 &= 0.00\ \text{kN}&& \\ N_5 &= 6.27\ \text{kN}&&\text{ (compression)} \\ N_6 &= 5.90\ \text{kN}&&\text{ (compression)} \\ N_7 &= 0.00\ \text{kN}&& \\ \end{alignat}

For the reactions, we can just use these internal components.

At the top support, we have bar 1 and bar 3. \begin{alignat}{2} V_{top} &= 5.00 - 0.60 &&= 4.40\ \text{kN} \\ H_{top} &= 6.67+1.67-0.20 &&= 8.14\ \text{kN} \\ \end{alignat}

At the middle support, we only have bar 5, so it only has a horizontal component of -6.27 kN. \begin{align} V_{mid} &= 0.00\ \text{kN} \\ H_{mid} &= -6.27\ \text{kN} \\ \end{align}

At the bottom support, we only have bar 6. \begin{alignat}{2} V_{bot} &= 5.00 + 0.60 &&= 5.60\ \text{kN} \\ H_{bot} &= -1.67-0.20 &&= -1.87\ \text{kN} \\ \end{alignat}

You'll notice that these reactions satisfy the global equilibrium equations. And here's the computer model to check our work (allowing for errors of rounding): Now, there is a way to trivially solve this problem by adopting a simplifying assumption. All you have to do is assume that bar 5 is axially rigid, meaning it will not deform at all. The validity of such an assumption is questionable if the bar all have the same $$EA$$. However it is less than a third of the length of bars 3 and 6, meaning it is more than three times as stiff. If that's enough for you to consider it rigid (I personally believe that's only acceptable for things at least one order of magnitude stiffer than its neighbors, but I don't know of any code "suggestions" on the matter), the solution becomes trivial.

Since bar 5 is rigid, the central node won't suffer any horizontal displacements from the horizontal force applied by bar 2. Therefore, bars 3 and 6 won't deform and won't be affected by that component, meaning bar 5 will absorb the entirety of that horizontal component. And that's that. Obviously, this will change the results for bars 3 and 6, as well as the reactions. For comparison, here's the result in this case: All figures obtained via Ftool, a free 2D frame analysis tool

Working on the idea that equilibrium exists in the members already, we can suggest that the vertical force is equal to the horizontal force at point A. This means we need only calculate the resultant for the truss diagonal starting at A. Take A as a point moment for 3 forces:

$$M_A=0$$ $$3F_A \sin 270 -B_1 \cos 0 -R_A \sin \beta=0$$ $$3F_A \sin 270 -B_1 \cos 0 =R_A \sin \beta$$ $$\frac{3F_A \sin 270 -B_1 \cos 0}{ \sin \beta} =R_A$$

Since $$R_A$$ exerts an horizontal force on all members $$B_1, B_2, B_3$$ we should be able now to calc the rest of the truss.

Therefore the horizontal reaction at B1 would be 10kN and the diagonal descent force from A would be 58.614kN compressionThis results in a compressive perpendicular force to B2.

$$58.614 \cos (180-\arctan \frac 3{2})=32.513kN$$

That force must therefor be distributed over the 3 vertical pin joints.

In spite of all this, a few examples below explain how the calculations work with statically determinate structures.

• I'm not following. What do you mean by "the vertical force is equal to the horizontal force at point A"? Which vertical force, the one actually applied? But how is that equal to the horizontal force? Which horizontal force? Do you mean the horizontal component of the diagonal member on A? But those components are obviously not equal, since that diagonal isn't at a 45° angle. And what are "members $B_1$, $B_2$, $B_3$"? Those are supports, not members. And what's $R_A$, and why does it "exert a horizontal force" on them? – Wasabi Nov 25 '18 at 16:15
• Please look at the software results in my answer (or run a model yourself) and see how your results match up with what I got. – Wasabi Nov 25 '18 at 16:16
• My comments unfortunately still stand after your edit. And what are you trying to show with those images? Since you changed the boundary conditions to make the structure statically determinate, and since you don't really show the work needed to get to those results, I don't see what they add to your answer. – Wasabi Nov 26 '18 at 21:13
• By turning each portion of truss into a determinate system we can calculate the internal stress in each member. Where there is external forces applied to the entire system, the 3 pins should share the load equally and therefore is 30kN/3. In reality there is going to be shared forces of an indeterminate amount in each but if the truss is designed to be statically determinate then logically it would be overdesigned and therefore safe in all loading conditions. – Rhodie Nov 27 '18 at 22:57
• Not necessarily, since this "turning each portion [...] into a determinate system" is arbitrary. For example, on the first model, why did you put an XY support at the bottom and only a Y support at the actual support? Sure, if you invert the supports (XY at the real support, only Y at the bottom), you get the same result. But why did you choose to put a Y restraint at the bottom? – Wasabi Nov 28 '18 at 15:11