Parabolic Arch only carries Axial Load

Preface: Paul Zarda of Orlando shows that a parabolic arch, subjected to a uniform load, with pin support conditions, will be shown to carry only a low-stress-level compressive load and not the high-stress-level bending moment.

 Figure 1

Paul Zarda notes that Figure 1 shows a simple arch-like structure that is carrying a uniform load Wy. The structure is assumed to be symmetric, with pin support conditions as shown. A x-sectional cut is shown in the inset in Figure 1. The cut exposes an axial load N, a bending moment M, and a shear force V. In the discussion that follows, it will be shown that if the shape of the arch is parabolic ( y(x) = a parabola), then the bending moment M and the shear force V will be zero. This is, of course, a very effective way to carry this uniform load Wy. The fact that a parabolic shaped arch can be shown to carry only a compressive load is well known in structural mechanics. The proof is presented here as a review and as a foundation to a natural extension of this principle.

Figure 2

Figure 2 shows a free body diagram (FBD) of the overall structure. The support forces at each of the pin supports, as shown, are symmetric based on the symmetry of the structure as presented in Figure 1. Then 

 Sum of forces in the Y-direction = 0:   

                                                                                                               ...Equation-1

                                                                                                               ...Equation 2

 Figure 3

Figure 3 shows a FBD of exactly ½ of the structure exposing the axial force Nc at the plane of symmetry. Then

Sum of forces in the X-direction = 0: 

                                                                                                             ...Equation 3
 Hence                                                            

                                                                                                             ...Equation 4
 Next, from Figure 3, Sum of Moments (counterclockwise) about point A = 0, gives

                                                                                                               ...Equation 5
                                   
Combining and simplifying equations 4 and 5 gives

                                                                                                                ...Equation 6

 Figure 4

 From Figure 4, assuming M and V are 0, then the sum of forces in the x-direction = 0 gives

                                                                                                                  ...Equation 7
From Figure 4, sum of forces in the y-direction = 0 gives

                              

                                                                                                                   ...Equation 8
From Figure 4, sum of moments (counterclockwise) about point A = 0 gives

                                                                                                                    ...Equation 9
Substituting N*cosθ  from Equation 7 and N*sinθ from Equation 8 into Equation 9, one gets

                                                                                                                     ...Equation 10
Substituting Fy from Equation 2 and Fx from Equation 6 into Equation 9 yields  =  0

                                                                                                                      ...Equation 11 

Dividing by Wy and simplifying

                                                                                                                     ...Equation 12 

Adding h terms to both sides

                                                                                                                      ...Equation 13
And noting the right hand side is a perfect square, this gives

                                                                                                                      ...Equation 14

Rewriting

                                                                                                                       ...Equation 15

Paul Zarda who also resides in Sanford notes that Equation 15 is the equation of a parabola that is centered at (x, y) = ( L/2, h) and opening downward as shown in Figure 1. This parabolic arch only carries a compressive load and no bending load.

Paul Zarda notes that the procedure shown here will be applied, in a bi-loading sense, to determine the loads in a tennis racquet under string tensions (future article). It will be shown, for the right combination of loading for the main string tensions and cross string tensions, the head of a racquet can be put in pure axial compression if the shape of the head is an ellipse. While it is well known in the history of mechanical engineering that a parabolic arch under uniform load can carry that load in pure axial compression, the author is not aware that anyone has ever presented the somewhat remarkable finding that if the head of a tennis racquet is elliptical in shape, with the right tensions chosen for the main strings and cross strings, the frame will carry a purely compressive load (with no bending). This will be shown in another article.