© Joseph Jourdain – Josephus Harp Shop – 2019
things that the eyes can not see.
It is a kind of spiritually, but also a soul-revelation”.
I am not an engineer. I worked this out from textbooks using a graphing calculator and spreadsheet. The transfer of force (moment, shear, pressure, tension) are not always easy to conceptualize in complex system which the harp is. Engineering software is available that can simulate my simplified models. If you have access to one, please run it through and let me know what you get or any suggestions to make this better. Thank you.
I believe the values obtained in the summary of calculations are indicative of some of the forces at play in the construction of a harp which a maker should take into consideration. The analysis is nonetheless incomplete and deviates from the real shapes of the neck and pillar. The load distribution, the connecting joints and the choice of materials are all critical in relation to the harp’s acoustic function and aesthetic.
I did not expand the analysis beyond the over simplified methods I used (the centroid load model) with the actual sizes of the neck and pillar by extracting the necessary dimensions from the moment of inertia and section modulus from the formulas provided. (I used a straight beam). I did not expand on the flexure applied to the neck which can be measured before and after the string loading. I did not expand on the effect of the width and depth of the neck and pillar. These would require a much more detailed and cumbersome analysis It takes a structural engineer to really work that out.
The way the joints and support contacts are designed affect the distribution of the internal forces whether they are fixed, free (supported), hinged or mixed. Different formulas have been work out in textbooks for these different models using calculus.
Another factor that is more difficult to characterize is the exactness of a joint or connection adherence in the assembly process. A connection or joint that is 20% NOT fully efficient will increase a stress somewhere else in the design that will need to be compensated for and included in the coefficient of safety if it is not possible to improve the efficiency of the joint or connection. These factors are in the realm of engineering and engineers work them out for a specific design and application. If you have knowledge and experience in that field please contribute to our common knowledge. Once again, thank you.
Stringed musical instruments are dynamic devices with 3 independent physical characteristics that are intrinsically related to each other. They are:
- Potential energy: The load from string’s tension provides the potential energy for the instrument. It is initiated in the string, stored in the instrument, and somewhat released into the air by stimulation.
- Rigidity: The resistance from the internal shear and moment forces gives the instrument its rigidity and structural integrity that must contain the strings’ load and the acoustic resonance.
- Mobility: Acoustic ability to respond to harmonic oscillation coupled with resonance by bending.
These 3 independent relationships are:
Rigidity = 1 ÷Mobility
- Acoustic energy = String Load × Mobility
- Acoustic Power = Acoustic energy × string Load
Rigidity is related to the coefficient of safety and the stiffness of the elements of the structure.
Potential energy is related to energy from the initial force load.
Mobility is related to flexure, the ability to bend and vibrate.
The purpose of this paper is to share my attempts to estimate structural concepts into harp designs without the complex mathematics used in engineering. Give some thoughts to them and see how they apply to your work.
The structural elements of the harp are under significant stress from all the tension of the strings, typically over 1000 pounds for a 36 string harp. This stress is applied to the neck, the pillar, the sound board and the sound box which needs to be balanced with the instrument’s overall mobility for acoustic effectiveness.
Structural engineering is about how stresses from various forces impact a complex system that has a specific function. In the case of a musical instrument, there is a balance that needs to be achieved between rigidity (structural strength) and mobility (ability to vibrate), in addition to ergonomic and artistic considerations.
The methodology offered here will only deal with concepts of structural strength. However, an understanding of that aspect of harp construction will assist you to explore different mobility options for achieving the acoustical qualities you desire.
Important definitions and modulus for Western maple:
Principal axes of wood: as per figure 4-1 from Wood Handbook.
Shear stress: force tending to cause deformation of a material by slippage along a plane or planes parallel to the imposed stress.
Moment force: The turning effect of a force applied to a rotational system at a distance from the axis of rotation.
Moment area: A theorem that provides a way to find slopes and deflections without having to go though a full process of calculus integration in beam theory.
Modulus of Rupture: (tangential) maximum shear forces used in beam evaluation.
Modulus of Elasticity: flexure, bending limit used for beam deflection.
Stress at Extreme Fiber:
- compression parallel to grain (longitudinal): used for post and pillar load
- compression perpendicular to grain (radial): used for flat side of beam
- shear parallel to grain (radial/tangential): used for beam support
For short beam: Modulus of Rupture will not be the limiting factor but compression parallel to grain will most likely be if the supports of the beam are not large enough.
For long beam: Modulus of Rupture will most likely be the limiting factor if the beam supports are sufficient.
For short post: compression parallel to grain will most likely be the limiting factor.
For long post: Column Critical Load (to be calculated) and Modulus of Rupture will most likely be the limiting factor.
our example we will use Western maple (Big leaf Maple) with the
Specific gravity = 0.48
Modulus of rupture = 10,700 Lb/in²
Modulus of Elasticity = 1,450,000 Lb/in²
Compression parallel to grain = 5950 Lb/in²
Shear parallel to grain = 1,730 Lb/in²
needed Moment of Inertia and Section Modulus were calculated using
the formulas in the table and dimensions below.
Neck dimensions: as a straight beam
Length: 22.5”, smallest depth: 3” (d), thickness: 1.75” (b)
I = 3.94; y = 1.5”; Z = 2.63; A = 5.25; k = 0.867
dimensions: as a straight beam
Length: 36”, smallest depth: 2” (b), thickness: 1.75 (d)
I = 0.89; y = 0.875”; Z = 1.02; A = 3.5; k = 0.51
Formulas I used for the longitudinal stress on the neck: From The Machinist Handbook
Here we have two options, the first one is the one I used because of the load offset, imagining that all the strings’ pull is at the centriod load (see text). The other option, a uniform load, is also valid although the load offset is not taken into consideration. A more realistic option would be to have a uniform spaced load with a heavier progressive load on one side. I could not find such reference from textbooks so that method needs to be worked out with calculus. My calculus skills are too basic to work that out.
Deflection method I used for the pillar using the dual cantilever system: From The Machinist Handbook
The first thing to establish is to quantify the forces applied on the harp. That is the strings’ tension. You need to calculate that. In 1990’s I wrote a series of 10 articles called “Stringband Evaluation” that were published in the Folk Harp Journal and provided explicit mathematical methods for doing this yourself. Now you can download spreadsheets or online apps that will do this for you; you can also download from my website a freeware application for calculating any single string of mono filament and wrap string made of various materials. Therefore, this aspect will not be part of this discussion.
Let’s assume you have established the total tension of the strings that your harp will have. You also need to know which string will be at the centroid point of that total tension. That location on the neck is the point where if you placed a fulcrum there the forces of the strings will be even on each side of the neck. That is the load applied on the neck/pillar notch and the shoulder that is attached to the sound box. It will look like the drawing above. This drawing will be used as a reference for all simulations. So if all the tension of the strings is 1,000 Lb there will be 500 Lb load on the left of the centroid point toward the pillar end and 500Lb load on the right of the centroid point toward the shoulder end. Those two loads will need to be resisted at the top and bottom of the sound box via the pillar. Because the plane of resistance is not aligned with the strings, the loads will stress the connections in various ways. There will be a significant torque on the neck. This is the engineering challenge the harp maker has to work out.
Step 1: find the longitudinal stresses of the neck
- Step 2: find the torque stress of the neck
- Step 3: find the critical bulking load of the pillar
- Step 4: find the resulting bending load of the torque applied to the pillar from the neck
- Step 5: find the pillar deflection due to the torque load from the neck
If you are good at trigonometry you can do all the load transfer by mathematics. However graphical solutions can be used to work out resultants and resolutions of forces using the vector parallelogram method. Many tutorials are available online regarding those methods.
The drawings below were done using a drafting/drawing software but can easily done on a graph paper with pencil, protractor, square and ruler. The reference harp design is the one used in the “The Challenge” section.
The thick line is the total load from the strings at each end of the neck and is 500 Lb (at the shoulder and at the pillar/neck notch connection). The short side of the triangle at the bottom is at 55° degree from the vertical strings pull. That is the angle at which the neck connects to the sound box from the harp. You find the long line by tracing a perpendicular line from the short line that connect to the top of the thick line. The angle from the top thick line and the long line is 35° (90-55). Where the short and long lines intersect the angle should be 90° degree. The dimensions you get from those lines are the resolution forces from the thick line, the 500 Lb load. So here about 401 Lb load is applied down to the sound box and about 290 Lb in shear load. Because in this design the angle of the neck/pillar notch is similar to the neck and sound box connection then that notch on neck/pillar has to resist 401 Lb in compression and 290 Lb in shear force.
On the drawing below we have the resolution of half the pull of the strings (500 Lb) on each side the sound board. So the effect of all the strings on the sound board will be twice as much (1,000 Lb). The drawing is similar but it is an upside down mirrored image of the above diagram. The string angle to the sound board is 35°. The total pull perpendicular to the center of the sound board will be about 580 Lb (2×290) and a shear load of 802 Lb (2×401). Each side of the sound board and the bottom connection of the pillar to the sound box will have to withstand half of those values. The above internal pull and shear forces are longitudinal in the axis of the strings, neck and sound-box.
Here it is easy to see that by changing the angle of the strings and the joint connections that the resolutions of those loads will be distributed differently.
For the neck you need to use beam stress analysis by drawing moment loads (internal stress) and shear loads. It is a little more complex but solution can be done graphically with minimum arithmetic. Many tutorials are available online regarding those methods. To demonstrate that method we will simplify the neck to a straight beam with an off center load of 1,000 Lb. The length of the neck at contact from the pillar to the sound box is 22.5 inches. The centroid point of the total tension is on string 22 of 36 and is 9 inches from the pillar. Two simple formulas are used to determine the off center shear load for each side of the neck.
side (left) shear stress =
length = 600 Lb/in
Neck side (right) shear stress = Load × Left offset ÷ beam length = 400 Lb/in
Above is the graphical solution to the shear stress on the neck as a straight beam.
The total external load of 1,000 Lb is transformed into 1,000 Lb/in internal shear stress in the neck with equal maximum bending moment area of 5400 Lb/in2 on each side (each inch of the beam has to hold the shear stress over the length). That value is useful in predicting the deflection of the beam with known characteristics.
Since the strings are connected to the neck on one side there is a torque stress that needs to be taken into consideration for structural integrity. This is an additional level of complexity.
To model this I will grossly simplify the problem assuming that the total load from all the strings are concentrated at the centriod point. This will be my virtual centroid pin.
Rotational torque on the neck is due to the thickness of the tuning pin which typically is 0.250” on harps. The applied shear force from the pin to the neck is from its outside surface. On the left side the neutral axis of the shear force is on the top of the pin and for the right side the neutral axis of the shear force is at the bottom of the pin. That angular differential creates a leverage that has an axis with a width of 0.125” (½ pin size) for a rotational torque of half the thickness of the board (0.875”). It is small but significant as it creates a torque shear force of about 125 Lb/in (0.875 × 1000 × sin 0.143 rad), for one side for a total of 250 Lb/in torque force. (Each side in the opposite direction, 2 × 125 Lb/in).
On the left is the graphical solution to the torque shear stress on the neck from the above values.
The shaded areas represent the axial maximum bending area moment and is 437 Lb/in2 (500 Lb/in × 0.875inch; each inch of the width of the neck has to hold the shear stress).
So the neck has two axes of forces acting on it. One is axial at the center of the width of the neck. The top side is compressed and the bottom side tension-ed, 500 Lb/in each in the opposite direction for a total shear force of 1,000 Lb/in, and the other rotational and bending of 250 Lb/in (both sides). That rotational load will be resisted by the neck/pillar notch and the neck/sound box connection, each holding 125 Lb/in.
There is no graphical method that I know of to estimate the bulking effect of the pressure put on the pillar without serious engineering methodology because column bulking is not proportional to stress. It has been resolved experimentally with the following method. The first thing you need to know is to find the slenderness ratio. This is the ratio of the pillar's length to the least dimension radius of gyrations of the cross section of the pillar. The radius gyration of a rectangular section is defined for our application as the small side of the cross section divided by the square root of 12. For our harp the length is 36 inches, the least cross section is 1.75 inch and maximum parallel to grain compression for western maple is 5,950 Lb/in2. So radius of gyration is 0.505 and slenderness ratio is 71 (36”÷0.505). That radio defines whether the pillar is classified as a short column or a long column. This matters because the critical load is calculated differently whether it is a short or long column. For short columns the Rankin formula is used when slenderness ratio is between 20 to 120 (130 for oak). For long columns with a higher ratio the Euler formula is used. Also it has been established that when a column with a ratio of length to the smallest section dimension is greater than 11, it will bulk below the maximum compression allowed for that wood specimen. For our example it is 20.5 (36”÷1.75”). So the pillar will experience bulking before the load reaches its maximum structural load allowed. We need to find that limit and we will have to use the above referred formulas for it. When the Euler formula is used where the slenderness ratio is too small a greater critical load will be estimated than the allowed maximum parallel to grain compression load for that wood. If done it will lead to catastrophic failure. Reference from the Machinist handbook says that the critical slenderness ratio for oak with a compression to parallel to grain of 6,500 Lb/in2 at 12% moisture is 130.
Critical slenderness ration for western maple is 119, (5950÷50) - 50 is the ratio of 6500÷130 as per reference so here Rankin formula for short column for critical load should be used.
Doing calculations for both method we have a Critical Pressure of 19,655 Lb/in2 using Euler formula which is much greater than the allowable load for western maple (5,950 Lb/in2). This confirms that the pillar is classified as a short column as predicted by the slenderness ratio calculation. Using the Rankin’s formula we have a Critical Pressure of 4,858 Lb/in2 before bulking happens. That is about 82% of the reference pressure of 5,950 Lb/in2.
The total load on the pillar has been calculated to be 500 Lb plus a torque load of 125 Lb for a total of 625 Lb. This is much below the Critical Pressure for bulking.
There is a measured bulking of about 0.125” on the actual pillar. This is due solely by the rotational torque exerted by the neck. To solve that we need to work out a solution where the pillar is now considered as a dual cantilever system with uniform load and where the fixed ends are the bottom and top section of the pillar and the center free but attached (guided) by a hinge. The bottom section will be the mirror of the top section where it is attached to the neck as per the diagram. Using that model we can see how the forces are distributed on the entire structure.
In our previous section we established that the rotational force on the neck from the virtual bridge pin is equals to 250 Lb/in. That load will be shared between the neck/pillar notch and the neck/sound box connection. That torque force will be the load 125 Lb/in (250 Lb/in ÷ 2) at the fixed end of the cantilever, that is at the pillar where the notch is. The length of the pillar for that harp is 36 inches. Because it is a cantilever system the shear force is 125 Lb/in and the moment area is 125 Lb/in × 36” (pivoting arm is the length of the pillar) and gives a moment area to the pillar of 4,500 Lb/in2.
The estimation of the deflection at the center of the pillar is obtained by calculating the imaginary uniform (virtual) load that would be required for the deflection. We are doing reversed calculations for the moment force at the torque of the neck (4,500 Lb/in2 internal forces) to an external virtual force on the side of the pillar.
For that dual system the moment force of the cantilever at the neck/pillar connection is defined as:
For the “center of the pillar” ends of cantilever the moment load is defined as:
This is the moment force on one side of the virtual hinge that connects the 2 cantilevers system. So we need to add the other side to balance the system for a total moment force of 4,500 Lb/in2 (2,250×2) at the fixed end of the other side and that gives a total virtual load of 1,500 Lb (750×2) to the pillar. This is matching our initial calculation because the stress on that theoretical hinge connection is continuous and similar at both sides of the center of the pillar. It is like there is no hinge. That virtual hinge becomes an integral element of the entire pillar as one complete system as it really is. Our moment force in the pillar is 4,500LB/in2.
Now the last step we have to do is to compute the deflection from the virtual load we have found from our calculation. For our dual cantilever system it is defined as:
W = 750 and gives a result of 0.141”.
The measured deflection was 0.125”. Here the correcting factor is not important but it demonstrates that the dual cantilever model is a workable solution for this application. Indeed, you get exactly the same results using the single beam model fixed at both ends with uniform load. (See notes)
The multiple steps used for this evaluation demonstrate that the over-simplified models applied give some indication of the stresses on the harp from the initial string’s tension of 1,000 Lb on the virtual pin that gives a torque force to the neck of 250 Lb/in that resulted to the bulking of the pillar with a deflection of of 0.125”.
These are not true acceptable values as explained in my foreword note and they need to be established by an engineer.
1: Longitudinal stress on the neck
Shear force on the pillar side 600Lb
Shear force on the neck side 400Lb
Maximum bending moment 5400Lb
2: Torque stress of the neck
Total torque stress 250Lb
Torque stress at the pillar notch and shoulder 125Lb each
3: Critical bulking load of the pillar
Slenderness ratio of the pillar is 119
Short column critical load is 4,858Lb
4 :Resulting bending load of the torque applied to the pillar from
Maximum load on pillar 625Lb
Torque load at the notch 125Lb
Bending moment on the pillar 4,500Lb
Resulting virtual load from bending moment at the center of the pillar 1,500Lb
5: Pillar deflection resulting from the bending load made by the
Deflection is 0.125 inch
Deflection angle 0.4 degree
Force load on sound board 1000Lb
Stress pull perpendicular to sound board 580Lb
Stress shear load longitudinal to sound board 802Lb
Force load on each side 500Lb
Stress pull perpendicular on each side 290Lb
Stress shear load longitudinal on each side 401Lb
Resultant stresses on neck-pillar notch and neck shoulder connection
For this design the angle of the notch and shoulder in relation to the string angle is the same, so they will have similar stress resultants. If they were different so would be the stress resultants. Note they are the reversed stress forces from the sound box sides except for the torque for this design.
stress 125Lb each (notch and shoulder)
Compression stress 401Lb each (notch and shoulder)
Shear stress 290Lb each (notch and shoulder)
method for deflection of the pillar using virtual load.
Instead of using the dual cantilever system for the pillar you can use the continuous load, uniformly distributed, beam fixed at both ends modeling. The graphic below makes it easy to see the difference between M and M1 and the visualization is more accurate but harder to explain with words. However it validates my reasoning that describes my method since the results are identical.
Checking the results: W=1500; M1=4500; M=2250; L=36; D=0.141; I=0.89, V=750; E=1450,000
The advantage of this diagram (Civil Engineering Formulas by Tyler Mc Graw-Hill) is that it shows what the moments and shears forces are in the pillar because of the torque. Here V = shear force.