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 loaded as shown in Fig. 2.3. Show that the stress function
\phi=A x^{2}+B x^{2} y+C y^{3}+D\left(5 x^{2} y^{3}-y^{5}\right)
is valid for the beam and evaluate the constants A, B, C, and D.
Repeat using the Symbolic Math Toolbox in MATLAB.
The stress function must satisfy Eq. (2.9). From the expression for \phi,
\frac{\partial^{4} \phi}{\partial x^{4}}+2 \frac{\partial^{4} \phi}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} \phi}{\partial y^{4}}=0 (2.9)
\frac{\partial \phi}{\partial x}=2 A x+2 B x y+10 D x y^{3}
\frac{\partial^{2} \phi}{\partial x^{2}}=2 A+2 B y+10 D y^{3}=\sigma_{y} (i)
Also,
\frac{\partial \phi}{\partial y}=B x^{2}+3 C y^{2}+15 D x^{2} y^{2}-5 D y^{4}
\frac{\partial^{2} \phi}{\partial y^{2}}=6 C y+30 D x^{2} y-20 D y^{3}=\sigma_{x} (ii)
and
\frac{\partial^{2} \phi}{\partial x \partial y}=2 B x+30 D x y^{2}=-\tau_{x y} (iii)
Further,
\frac{\partial^{4} \phi}{\partial x^{4}}=0, \quad \frac{\partial^{4} \phi}{\partial y^{4}}=-120 D y, \quad \frac{\partial^{4} \phi}{\partial x^{2} \partial y^{2}}=60 D y
Substituting in Eq. (2.9) gives
\frac{\partial^{4} \phi}{\partial x^{4}}+2 \frac{\partial^{4} \phi}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} \phi}{\partial y^{4}}=2 \times 60 D y-120 D y=0
The biharmonic equation is therefore satisfied and the stress function is valid. From Fig. 2.3, \sigma_{y}=0 at y = h, so that, from Eq. (i),
2 A+2 B H+10 D h^{3}=0 (iv)
Also, from Fig. 2.3, \sigma_{y}=-q \text { at } y=-h, so that, from Eq. (i),
2 A-2 B H-10 D h^{3}=-q (v)
Again, from Fig. 2.3, \tau_{x y}=0 \text { at } y=\pm h, giving, from Eq. (iii),
2 B x+30 D x h^{2}=0
so that
2 B+30 D h^{2}=0 (vi)
At x = 0, no resultant moment is applied to the beam; that is,
M_{x=0}=\int_{-h}^{h} \sigma_{x} y d y=\int_{-h}^{h}\left(6 C y^{2}-20 D y^{4}\right) d y=0
so that
M_{x=0}=\left[2 C y^{3}-4 D y^{5}\right]_{-h}^{h}=0
or
C-2 D h^{2}=0 (vii)
Subtracting Eq. (v) from (iv),
4 B h+20 D h^{3}=q
or
B+5 D h^{2}=\frac{q}{4 h} (viii)
From Eq. (vi),
B+15 D h^{2}=0 (ix)
so that, subtracting Eq. (viii) from Eq. (ix),
D=-\frac{q}{40 h^{3}}
Then,
B=\frac{3 q}{8 h}, \quad A=-\frac{q}{4}, \quad C=-\frac{q}{20 h}
and
\phi=\frac{q}{40 h^{3}}\left[-10 h^{3} x^{2}+15 h^{2} x^{2} y-2 h^{2} y^{3}-\left(5 x^{2} y^{3}-y^{5}\right)\right]
Using Fig. 2.3, derivations of the constants A, B, C, and D, along with validation of the stress function \phi obtained through the following MATLAB file:
% Declare any needed symbolic variables
syms A B C D x y L h q sig_y sig_x tau_xy
% Define the given stress function
phi = A*x^2 + B*x^2*y + C*y^3 + D*(5*x^2*y^3 – y^5);
% Check Eq. (2.9)
check = diff(phi,x,4) + 2*diff(diff(phi,x,2),y,2) þ diff(phi,y,4);
if double(check) == 0
disp(‘The stress function is valid’)
disp(‘ ’)
% Calculate expressions for sig_y, sig_x, and tau_xy
eqI = diff(phi,x,2) – sig_y;
eqII = diff(phi,y,2) – sig_x;
eqIII = diff(diff(phi,x),y) + tau_xy;
% From Fig. 2.3, sig_y=0 at y=h so that from eqI
eqIV = subs(subs(eqI,y,h),sig_y,0);
% From Fig. 2.3, sig_y=-q at y¼-h so that from eqI
eqV = subs(subs(eqI,y,-h),sig_y,-q);
% From Fig. 2.3, tau_xy=0 at y=+/-h so that from eqIII
eqVI = subs(subs(eqIII,y,h),tau_xy,0);
% Calculate the expression of the applied moment
M = int(solve(eqII,sig_x)*y, y, -h, h);
% From Fig. 2.3, M=0 at x=0
eqVII = subs(M,x,0);
% Solve eqIV, eqV, eqVI, and eqVII for A, B, C, and D
D_expr = solve(subs(eqIV-eqV,B,solve(eqVI,B)), D);
C_expr = solve(subs(eqVII,D,D_expr), C);
B_expr = solve(subs(eqVI,D,D_expr), B);
A_expr = solve(subs(subs(eqIV,D,D_expr),B,B_expr), A);
% Substitute the expressions for A, B, C, and D into phi
phi = subs(subs(subs(subs(phi,A,A_expr),B,B_expr),C,C_expr),D,D_expr);
% Output the expression for phi to the Command Window
phi = factor(phi)
else
disp(‘The stress function does not satisfy the biharmonic equation (Eq. (2.9))’)
disp(‘ ’)
end
The Command Window outputs resulting from this MATLAB file are as follows. The stress function is valid.
phi = -(q*(10*h^3*x^2 – 15*h^2*x^2*y + 2*h^2*y^3 + 5*x^2*y^3 – y^5))/(40*h^3)