The English mathematician Airy proposed a stress functionφdefined by the equations σx = ∂2φ
∂y2 σy = ∂2φ
∂x2 τxy= − ∂2φ
∂x∂y (2.8)
Clearly, substitution of Eqs (2.8) into Eqs (2.6) verifies that the equations of equilibrium are satisfied by this particular stress–stress function relationship. Further substitution into Eq. (2.7) restricts the possible forms of the stress function to those satisfying the biharmonic equation
∂4φ
∂x4 +2 ∂4φ
∂x2∂y2 +∂4φ
∂y4 =0 (2.9)
The final form of the stress function is then determined by the boundary conditions relating to the actual problem. Therefore, a two-dimensional problem in elasticity with zero body forces reduces to the determination of a functionφof x and y, which satisfies Eq. (2.9) at all points in the body and Eqs (1.7) reduced to two dimensions at all points on the boundary of the body.
2.3 Inverse and semi-inverse methods
The task of finding a stress function satisfying the above conditions is extremely difficult in the majority of elasticity problems although some important classical solutions have been obtained in this way. An alternative approach, known as the inverse method, is to specify a form of the functionφsatisfying Eq. (2.9), assume an arbitrary boundary and then determine the loading conditions which fit the assumed stress function and chosen boundary. Obvious solutions arise in whichφ is expressed as a polynomial.
Timoshenko and Goodier1consider a variety of polynomials forφand determine the associated loading conditions for a variety of rectangular sheets. Some of these cases are quoted here.
Example 2.1
Consider the stress function
φ=Ax2+Bxy+Cy2
where A, B and C are constants. Equation (2.9) is identically satisfied since each term becomes zero on substituting forφ. The stresses follow from
σx = ∂2φ
∂y2 =2C σy = ∂2φ
∂x2 =2A τxy= − ∂2φ
∂x∂y = −B
Fig. 2.1Required loading conditions on rectangular sheet in Example 2.1.
To produce these stresses at any point in a rectangular sheet we require loading conditions providing the boundary stresses shown in Fig. 2.1.
Example 2.2
A more complex polynomial for the stress function is φ= Ax3
6 + Bx2y
2 + Cxy2 2 + Dy3
6 As before
∂4φ
∂x4 = ∂4φ
∂x2∂y2 = ∂4φ
∂y4 =0
so that the compatibility equation (2.9) is identically satisfied. The stresses are given by
σx = ∂2φ
∂y2 =Cx+Dy σy = ∂2φ
∂x2 =Ax+By τxy= −∂2φ
∂x∂y = −Bx−Cy
We may choose any number of values of the coefficients A, B, C and D to produce a variety of loading conditions on a rectangular plate. For example, if we assume A=B=C=0 thenσx=Dy,σy=0 andτxy=0, so that for axes referred to an origin at
2.3 Inverse and semi-inverse methods 51
Fig. 2.2(a) Required loading conditions on rectangular sheet in Example 2.2 forA=B=C=0; (b) as in (a) but A=C=D=0.
the mid-point of a vertical side of the plate we obtain the state of pure bending shown in Fig. 2.2(a). Alternatively, Fig. 2.2(b) shows the loading conditions corresponding to A=C=D=0 in whichσx=0,σy=By andτxy= −Bx.
By assuming polynomials of the second or third degree for the stress function we ensure that the compatibility equation is identically satisfied whatever the values of the coefficients. For polynomials of higher degrees, compatibility is satisfied only if the coefficients are related in a certain way. For example, for a stress function in the form of a polynomial of the fourth degree
φ= Ax4
12 + Bx3y
6 + Cx2y2
2 + Dxy3 6 + Ey4
12 and
∂4φ
∂x4 =2A 2 ∂4φ
∂x2∂y2 =4C ∂4φ
∂y4 =2E Substituting these values in Eq. (2.9) we have
E= −(2C+A) The stress components are then
σx = ∂2φ
∂y2 =Cx2+Dxy−(2C+A)y2 σy = ∂2φ
∂x2 =Ax2+Bxy+Cy2 τxy= − ∂2φ
∂x∂y = −Bx2
2 −2Cxy− Dy2 2
The coefficients A, B, C and D are arbitrary and may be chosen to produce various loading conditions as in the previous examples.
Example 2.3
A cantilever of length L and depth 2h is in a state of plane stress. The cantilever is of unit thickness, is rigidly supported at the end x=L and is loaded as shown in Fig. 2.3.
Show that the stress function
φ=Ax2+Bx2y+Cy3+D(5x2y3−y5) is valid for the beam and evaluate the constants A, B, C and D.
The stress function must satisfy Eq. (2.9). From the expression forφ
∂φ
∂x =2Ax+2Bxy+10Dxy3
∂2φ
∂x2 =2A+2By+10Dy3 =σy (i) Also
∂φ
∂y =Bx2+3Cy2+15Dx2y2−5Dy4
∂2φ
∂y2 =6Cy+30Dx2y−20Dy3=σx (ii) and
∂2φ
∂x∂y =2Bx+30Dxy2= −τxy (iii)
Further
∂4φ
∂x4 =0 ∂4φ
∂y4 = −120Dy ∂4φ
∂x2∂y2 =60 Dy
h
q/unit area
h
L
y
x
Fig. 2.3Beam of Example 2.3.
2.3 Inverse and semi-inverse methods 53
Substituting in Eq. (2.9) gives
∂4φ
∂x4 +2 ∂4φ
∂x2∂y2 +∂4φ
∂y4 =2×60Dy−120Dy=0
The biharmonic equation is therefore satisfied and the stress function is valid.
From Fig. 2.3,σy= 0 at y=h so that, from Eq. (i)
2A+2BH+10Dh3=0 (iv)
Also from Fig. 2.3,σy= −q at y= −h so that, from Eq. (i)
2A−2BH−10Dh3= −q (v)
Again from Fig. 2.3,τxy=0 at y= ±h giving, from Eq. (iii) 2Bx+30Dxh2 =0 so that
2B+30Dh2=0 (vi)
At x=0 there is no resultant moment applied to the beam, i.e.
Mx=0= h
−h
σxy dy= h
−h
(6Cy2−20Dy4) dy=0 i.e.
Mx=0 =[2Cy3−4Dy5]h−h =0 or
C−2Dh2=0 (vii)
Subtracting Eq. (v) from (iv)
4Bh+20Dh3=q or
B+5Dh2= q
4h (viii)
From Eq. (vi)
B+15Dh2=0 (ix)
so that, subtracting Eq. (viii) from Eq. (ix) D= − q
40h3
Then
B= 3q
8h A= −q
4 C = − q 20h and
φ= q
40h3[−10h3x2+15h2x2y−2h2y3−(5x2y3−y5)]
The obvious disadvantage of the inverse method is that we are determining prob- lems to fit assumed solutions, whereas in structural analysis the reverse is the case.
However, in some problems the shape of the body and the applied loading allow sim- plifying assumptions to be made, thereby enabling a solution to be obtained. St. Venant suggested a semi-inverse method for the solution of this type of problem in which assumptions are made as to stress or displacement components. These assumptions may be based on experimental evidence or intuition. St. Venant first applied the method to the torsion of solid sections (Chapter 3) and to the problem of a beam supporting shear loads (Section 2.6).