Understanding Load Paths Part 3; Truss Analysis

Understanding Load Paths Part 3: Truss Analysis and Force Transmission

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So far in this series (part 1 & part 2) we’ve worked out how the wheel loads from a truck make their way into each of the trusses that make up the Fort Atkinson Truss Bridge. In this post we’ll dive into some truss analysis and utilise both the Joint Resolution Method and the Method of Sections. By the end of this article you will have a good understanding truss analysis; in particular how we determine the magnitude of the axial forces in a statically determinate truss.

1 Truss Analysis

First of all we need to nail down the precise geometry of the structure. We’ll do this by moving from a `realistic’ 3D representation of the truss to a simplified 2D visualisation onto which we’ll define the geometry, loading and support conditions. This simplified representation is our structural model used for the truss analysis, Fig. 1 below.

Reducing the realistic representation of the structure down to a working structural model
Figure 1. Reducing the ‘realistic’ representation of the structure (top) down to a working structural model (bottom). Note: members DK and EL are not joined.

For a statically determinate structure like this, the first analysis task is to determine the support reactions. By inspection we can state that the potential horizontal reaction at support A is zero (because all external loads are vertical). Similarly, we can state by inspection that due to symmetry, the vertical reactions at A and H are 40.5 kN each.

At this point we can analyse each joint in turn using the Joint Resolution Method of analysis; we’ll start with joint A. We will be evaluating the sum of the forces meeting at the joint. These forces can be resolved into two orthogonal (mutually perpendicular) directions allowing us to evaluate two equations of force equilibrium. Thus we have two equations from which we can determine two unknowns. So, using the joint resolution method we can only start at a joint that has a maximum of two unknown member forces. This is why joint A (or H) is a good starting point.

We isolate the joint by ‘cutting’ the members meeting at that joint. Making these cuts we ‘reveal’ the internal member forces, for now labelled as TAB and TAN where the T indicates we are assuming (until proven otherwise) that the forces are tension forces. A free body diagram of the joint is then evaluated by taking the sum of the forces in the horizontal and vertical direction and equating them to zero to reveal the unknown member forces. In the following analysis, note that basic geometry gives us the angles \angle NAB = 57.5^\circ and \angle ABN = 32.5^\circ. Now consider joint A, Fig. 2:

Truss analysis: Free body diagram of joint A
Figure 2. Free body diagram of joint A.

As mentioned above, we are initially assuming that the unknown internal member forces are tensile. This is also indicated by the force arrows pointing away from the joint in question. Taking the sum of the forces in the vertical direction and letting it equal to zero (assuming a sign convention with upwards forces positive):

    \begin{align*} \sum F_y &= 0 \:\:(\uparrow +)\\ 40.5+T_{AB}\sin(57.5)&=0\\ T_{AB} &= -48\:\textup{kN} \end{align*}

The negative sign indicates our initial assumption of a tensile force was incorrect and TAB is actual a compression force. Thus the member between nodes A and B is experiencing 48 kN compression. Now considering the sum of the forces in the horizontal direction with forces to the right assumed positive:

    \begin{align*} \sum F_x &= 0 \:\:(\rightarrow +)\\ T_{AB}\cos(57.5)+T_{AN} &=0\\ T_{AN} &=-(-48)\cos(57.5)\\ T_{AN} &= 25.8\: \textup{kN} \end{align*}

In this case the positive number indicates that we were correct to assume a tensile force. Because the structure is both symmetrical and symmetrically loaded, we can conclude that the forces in member HI and HG are the same as those worked out for members AN and AB. Thus we only need to evaluate half of the structure in this example.

Moving on to joint N, Fig. 3, we can conclude by inspection that the force TNB = 0 kN and TNM = 25.8 kN tension.

Truss analysis: Free body diagram of joint N
Figure 3. Free body diagram of joint N.

Now that we have solved members AB and NB, the only other joint with only two unknowns is joint B, Fig. 4. Evaluating the force equilibrium equations yields:

    \begin{align*} \sum F_y &= 0 \:\:(\uparrow +)\\ 40.5-T_{BM}\cos(32.5) &= 0\\ T_{BM} &= 48\:\textup{kN} \:(\textup{tension})\\\\ \sum F_x &= 0 \:\:(\rightarrow +)\\ 25.8 + T_{BC} + T_{BM}\sin(32.5) &= 0\\ T_{BC} &=-25.8-48\sin(32.5)\\ T_{BC} &= -51.6\:\textup{kN} \:(\textup{compression}) \end{align*}

Truss analysis: Free body diagram of joint B
Figure 4. Free body diagram of joint B.

Moving on to joint M, Fig. 5:

    \begin{align*} \sum F_y &= 0 \:\:(\uparrow +)\\ T_{MC}+40.5 &=0\\ T_{MC} &= -40.5\:\textup{kN} \:(\textup{compression})\\\\ \sum F_x &= 0 \:\:(\rightarrow +)\\ -25.8-25.8+T_{MC} &= 0\\ T_{MC} &= 51.6\:\textup{kN} \:(\textup{tension}) \end{align*}

Truss analysis: Free body diagram of joint M
Figure 5. Free body diagram of joint M.

The only joint open for evaluation is joint C, Fig. 6. Evaluating the force equilibrium equations yields:

    \begin{align*} \sum F_y &= 0 \:\:(\uparrow +)\\ 40.5-T_{CL}\cos(32.5)&=0\\ T_{CL}&=48\:\textup{kN} \:(\textup{tension})\\\\ \sum F_x &= 0 \:\:(\rightarrow +)\\ 51.6+T_{CD}+T_{CL}\sin(32.5)&=0\\ 51.6+T_{CD}+48\sin(32.5)&=0\\ T_{CD} &= -77.4\:\textup{kN} \:(\textup{compression}) \end{align*}

Truss analysis: Free body diagram of joint C
Figure 6. Free body diagram of joint C.

At this point it’s worth taking a look our progress so far, mapped onto the structure, Fig. 7. We can see that joints D and L both have three unknown member forces. Therefore isolating either joint would result in too many unknowns given the available equilibrium equations. The joint resolution method has taken us as far as it can. At this stage we introduce an alternative method of truss analysis known as the Method of Sections.

 Internal member forces evaluated using the joint resolution method.
Figure 7. Internal member forces evaluated using the joint resolution method.

Instead of isolating a single joint, the method of sections involves us making an imaginary cut through the entire structure. In doing so, we reveal the internal member forces in the members our plane cuts through. We can then evaluate equilibrium of the sub-structure created by the cut. This method of truss analysis brings into play a third equilibrium equation; because all of the forces acting on the sub-structure are no longer concurrent (they don’t all pass through the same point), we can take the sum of the moments about any point. Since the structure is in a state of static equilibrium, the sum of the moments (just like the forces) must equal zero. Thus we have three equations of statics at our disposal. The key thing is that our plane cannot cut through more than three unknown members. Making an inclined cut as shown in Fig. 8 will allow us to determine the internal forces in members LD, LE and LK.

Truss analysis: Sub-structure produced by the cutting plane
Figure 8. Sub-structure produced by the cutting plane, revealing three unknown member forces.

Evaluating the sum of the moments about point A and assuming clockwise moments are positive yields:

    \begin{align*} \sum M_A&= 0 \:\:(\textup{clockwise}\: +)\\ -(77.4\times 9) + 40.5\times(3\times5.74)-T_{LD}\times(3\times5.74) &= 0\\ T_{LD}&\approx 0 \end{align*}

Equating the sum of the vertical and horizontal forces to zero yields:

    \begin{align*} \sum F_y &= 0 \:\:(\uparrow +)\\ 40.5-40.5+T_{LE}\sin(57.5)&=0\\ T_{LE}&=0\\\\ \sum F_x &= 0 \:\:(\rightarrow +)\\ -77.4+T_{LK}&=0\\ T_{LK}&=77.4\:\textup{kN} \:(\textup{tension}) \end{align*}

By inspection of joint D we can state that the final unknown member force, TDE = 77.4 kN compression. This completes the load path analysis for the bridge truss. The complete load distribution within the truss is shown in Fig. 9. We’ve successfully traced the truck loading all the way back to the bridge foundation piers…mission accomplished.

Internal member forces for the complete truss.
Figure 9. Internal member forces for the complete truss.

It’s worth saying that before any design could proceed, a much more detailed loading analysis would be undertaken in reality. For example, under different loading conditions (position and magnitude of loading) the members that have zero internal force in this analysis would develop internal forces. Identifying the range of expected internal forces (and therefore stresses) would be the main objective of a robust loading analysis, a precursor to truss analysis.

2 Design Decision following on from Truss Analysis

Now that we have more insight into how this bridge structure works, its worth taking a second look at the actual bridge to see if we can identify the reasoning behind some of design decisions.

Axial forces in the truss
Figure 10. Axial forces in the truss with tension, compression and zero force indicated.

With reference to Fig. 10 above, we notice the following:

  • The four inner vertical members (CM, DL, EK, FJ) are stockier than the outer verticals and the diagonals. More specifically we would say they have a higher flexural rigidity or resistance to bending. This is because they resist compression forces and therefore must be capable of resisting not just direct compression stress but also buckling. Buckling is an instability whereby a compression member can fail (buckle) at a load below its ultimate axial load capacity. Buckling resistance is dependent on both the flexural rigidity of the member and a parameter known as effective length which is a function of both the length of the member and the degree of restraint at its ends. I will talk more about buckling in a future blogtorial, but suffice to say the vertical members identified are stockier in order to resist failure due to buckling.
  • The outer vertical members are much more slender than their vertical neighbours. We can see that for this load arrangement, there is no force in these members. However, for another load arrangement, say if a truck wheel was positioned in line with node N; we could expect to see an axial force develop. However, provided the loading is due to gravity, compression will never develop in these members, only tension. Therefore there is no need to resist buckling, only direct tension. As a result, the members can be much slimmer.
  • The same reasoning applies to the sizing of the diagonal members. They will always be in tension, and can therefore be much slimmer that any member that must also resist buckling.
  • The top chord is in compression while the bottom chord is in tension. The magnitudes of the axial loads increase as we move towards the centre of the truss. This is typical and exactly as we would expect for a simply supported truss. These members will usually be the largest, and sure enough that is also the case with the Fort Atkinson truss; remember, the top chord members must not buckle, while the bottom chord members must resist the added normal stresses due to bending induced by the loading on the deck (discussed earlier).

So hopefully you now have a clearer understanding of truss analysis and how external forces make their way from the point of application on the deck, all the way back into the bridge pier foundations. Once you have this insight, you can start to get inside the minds of the engineers who designed this bridge over 120 years ago!

In the fourth and final instalment of this blogtorial series on load paths, we will take a look at how lateral stability is provided in this bridge structure. We will identify the various structural mechanism at play and show how lateral loading is transmitted through the structure and back down to the foundations.

Remember to sign up to receive your typeset PDF copy of this and future posts. If you would like to see some more examples of truss analysis, you can enroll in my free Fundamentals of Statics course.

Cheers,
Seán


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