§AERE 321 Homework 2

Sep 25, 2025

§1.

Given:
ϵz=γxz=γyz=0 \epsilon_z = \gamma_{xz} = \gamma_{yz} = 0
Target:
σz=ν(σx+σy) \sigma_z = \nu (\sigma_x + \sigma_y)
ϵx=1E[(1−ν2)σx−ν(1+ν)σy] \epsilon_x = \frac{1}{E} [(1 - \nu^2) \sigma_x - \nu (1 + \nu) \sigma_y]
ϵy=1E[(1−ν2)σy−ν(1+ν)σx] \epsilon_y = \frac{1}{E} [(1 - \nu^2) \sigma_y - \nu (1 + \nu) \sigma_x]
Now for the derivation. For isotropic materials, the following is true:
ϵz=1E(σz−ν(σx+σy)) \epsilon_z = \frac{1}{E} (\sigma_z - \nu (\sigma_x + \sigma_y))
Substitution makes things clearer:
0=1E(σz−ν(σx+σy)) 0 = \frac{1}{E} (\sigma_z - \nu (\sigma_x + \sigma_y))
0=σz−ν(σx+σy) 0 = \sigma_z - \nu (\sigma_x + \sigma_y)
σz=ν(σx+σy) \boxed{\sigma_z = \nu (\sigma_x + \sigma_y)}
To reach the second target equation, I will be leveraging the same isotropic equation, just on the xx axis:
ϵx=1E(σx−ν(σy+σz)) \epsilon_x = \frac{1}{E} (\sigma_x - \nu (\sigma_y + \sigma_z))
I can now substitute in σz\sigma_z:
ϵx=1E(σx−ν(σy+ν(σx+σy))) \epsilon_x = \frac{1}{E} (\sigma_x - \nu (\sigma_y + \nu (\sigma_x + \sigma_y)))
ϵx=1E[σx−νσy−ν2(σx+σy)] \epsilon_x = \frac{1}{E} [\sigma_x - \nu \sigma_y - \nu^2 (\sigma_x + \sigma_y)]
ϵx=1E[σx−νσy−ν2σx−ν2σy] \epsilon_x = \frac{1}{E} [\sigma_x - \nu \sigma_y - \nu^2 \sigma_x - \nu^2 \sigma_y]
Factoring out the σ\sigma's gives us the target equation:
ϵx=1E[(1−ν2)σx−ν(1+ν)σy] \boxed{\epsilon_x = \frac{1}{E} [(1 - \nu^2) \sigma_x - \nu (1 + \nu) \sigma_y]}
A similar process can be done for the yy axis:
ϵy=1E(σy−ν(σx+σz)) \epsilon_y = \frac{1}{E} (\sigma_y - \nu (\sigma_x + \sigma_z))
ϵy=1E(σy−ν(σx+ν(σx+σy))) \epsilon_y = \frac{1}{E} (\sigma_y - \nu (\sigma_x + \nu (\sigma_x + \sigma_y)))
ϵy=1E[(1−ν2)σy−ν(1+ν)σx] \boxed{\epsilon_y = \frac{1}{E} [(1 - \nu^2) \sigma_y - \nu (1 + \nu) \sigma_x]}
Given:
∂2γxy∂x∂y=∂2ϵy∂x2+∂2ϵx∂y2 \frac{\partial^2 \gamma_{xy}}{\partial x \partial y} = \frac{\partial^2 \epsilon_y}{\partial x^2} + \frac{\partial^2 \epsilon_x}{\partial y^2}
Target:
(∂2∂x2+∂2∂y2)(σx+σy)=−11−ν(∂X∂x+∂Y∂y) \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) (\sigma_x + \sigma_y) = - \frac{1}{1 - \nu} \left( \frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} \right)
Because of how involved I think this problem will get, I will start working term by term. The left hand side has a clean substitution:
γxy=1Gτxy \gamma_{xy} = \frac{1}{G} \tau_{xy}
G=E2(1+ν) G = \frac{E}{2 (1 + \nu)}
γxy=2E(1+ν)τxy \gamma_{xy} = \frac{2}{E} (1 + \nu) \tau_{xy}
∂2γxy∂x∂y=∂2∂x∂y2E(1+ν)τxy \frac{\partial^2 \gamma_{xy}}{\partial x \partial y} = \frac{\partial^2}{\partial x \partial y} \frac{2}{E} (1 + \nu) \tau_{xy}
And since everything here is a constant coefficient but tautau, the partial can be moved:
∂2γxy∂x∂y=2E(1+ν)∂2τxy∂x∂y \frac{\partial^2 \gamma_{xy}}{\partial x \partial y} = \frac{2}{E} (1 + \nu) \frac{\partial^2 \tau_{xy}}{\partial x \partial y}
The first term of the right-hand-side can be dealt with similarly:
ϵy=1E(σy−ν(σx+σz)) \epsilon_y = \frac{1}{E} (\sigma_y - \nu (\sigma_x + \sigma_z))
Since everything is in-plane:
ϵy=1E(σy−ν(σx+σz)) \epsilon_y = \frac{1}{E} (\sigma_y - \nu (\sigma_x + \cancel{\sigma_z}))
ϵy=1E(σy−νσx) \epsilon_y = \frac{1}{E} (\sigma_y - \nu \sigma_x)
This can be substituted:
∂2ϵy∂x2=∂2∂x21E(σy−νσx) \frac{\partial^2 \epsilon_y}{\partial x^2} = \frac{\partial^2}{\partial x^2} \frac{1}{E} (\sigma_y - \nu \sigma_x)
∂2ϵy∂x2=1E∂2∂x2(σy−νσx) \frac{\partial^2 \epsilon_y}{\partial x^2} = \frac{1}{E} \frac{\partial^2}{\partial x^2} (\sigma_y - \nu \sigma_x)
Through a similar process, you'd get the second term of the right-hand-side and it's easy to just fork the previous term, saving some work:
∂2ϵx∂y2=1E∂2∂y2(σx−νσy) \frac{\partial^2 \epsilon_x}{\partial y^2} = \frac{1}{E} \frac{\partial^2}{\partial y^2} (\sigma_x - \nu \sigma_y)
Now would be appropriate to combine all the terms:
2E(1+ν)∂2τxy∂x∂y=1E∂2∂x2(σy−νσx)+1E∂2∂y2(σx−νσy) \frac{2}{E} (1 + \nu) \frac{\partial^2 \tau_{xy}}{\partial x \partial y} = \frac{1}{E} \frac{\partial^2}{\partial x^2} (\sigma_y - \nu \sigma_x) + \frac{1}{E} \frac{\partial^2}{\partial y^2} (\sigma_x - \nu \sigma_y)
Things cancel out a little:
2(1+ν)∂2τxy∂x∂y=∂2∂x2(σy−νσx)+∂2∂y2(σx−νσy) 2 (1 + \nu) \frac{\partial^2 \tau_{xy}}{\partial x \partial y} = \frac{\partial^2}{\partial x^2} (\sigma_y - \nu \sigma_x) + \frac{\partial^2}{\partial y^2} (\sigma_x - \nu \sigma_y)
This vaguely resembles the target equation. It would help to expand and rearrange the right-hand-side though:
∂2∂x2(σy−νσx)+∂2∂y2(σx−νσy)=∂2σy∂x2−ν∂2σx∂x2+∂2σx∂y2−ν∂2σy∂y2 \frac{\partial^2}{\partial x^2} (\sigma_y - \nu \sigma_x) + \frac{\partial^2}{\partial y^2} (\sigma_x - \nu \sigma_y) = \frac{\partial^2 \sigma_y}{\partial x^2} - \nu \frac {\partial^2 \sigma_x}{\partial x^2} + \frac{\partial^2 \sigma_x}{\partial y^2} - \nu \frac {\partial^2 \sigma_y}{\partial y^2}
∂2σy∂x2−ν∂2σx∂x2+∂2σx∂y2−ν∂2σy∂y2=∂2σy∂x2+∂2σx∂y2−ν(∂2σx∂x2+∂2σy∂y2) \frac{\partial^2 \sigma_y}{\partial x^2} - \nu \frac {\partial^2 \sigma_x}{\partial x^2} + \frac{\partial^2 \sigma_x}{\partial y^2} - \nu \frac {\partial^2 \sigma_y}{\partial y^2} = \frac{\partial^2 \sigma_y}{\partial x^2} + \frac{\partial^2 \sigma_x}{\partial y^2} - \nu \left( \frac {\partial^2 \sigma_x}{\partial x^2} + \frac {\partial^2 \sigma_y}{\partial y^2} \right)
Now, everything assembled back into the original equation:
2(1+ν)∂2τxy∂x∂y=∂2σy∂x2+∂2σx∂y2−ν(∂2σx∂x2+∂2σy∂y2) 2 (1 + \nu) \frac{\partial^2 \tau_{xy}}{\partial x \partial y} = \frac{\partial^2 \sigma_y}{\partial x^2} + \frac{\partial^2 \sigma_x}{\partial y^2} - \nu \left( \frac {\partial^2 \sigma_x}{\partial x^2} + \frac {\partial^2 \sigma_y}{\partial y^2} \right)
This allows me to almost factor out a 1−ν1 - \nu out but the partials are swapped in the parenthesis. For the home stretch, I'll have to apply the boundary conditions:
∂σx∂x+∂τyx∂y+X=0 \frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + X = 0
∂σy∂y+∂τxy∂x+Y=0 \frac{\partial \sigma_y}{\partial y} + \frac{\partial \tau_{xy}}{\partial x} + Y = 0
Taking a single partial of the boundary conditions:
∂2σx∂x2+∂2τyx∂y∂x+∂X∂x=0 \frac{\partial^2 \sigma_x}{\partial x^2} + \frac{\partial^2 \tau_{yx}}{\partial y \partial x} + \frac{\partial X}{\partial x} = 0
∂2σy∂y2+∂2τxy∂x∂y+∂Y∂y=0 \frac{\partial^2 \sigma_y}{\partial y^2} + \frac{\partial^2 \tau_{xy}}{\partial x \partial y} + \frac{\partial Y}{\partial y} = 0
Solving for XX and YY partials:
∂X∂x=−∂2σx∂x2−∂2τyx∂y∂x \frac{\partial X}{\partial x} = -\frac{\partial^2 \sigma_x}{\partial x^2} - \frac{\partial^2 \tau_{yx}}{\partial y \partial x}
∂Y∂y=−∂2σy∂y2−∂2τxy∂x∂y \frac{\partial Y}{\partial y} = -\frac{\partial^2 \sigma_y}{\partial y^2} - \frac{\partial^2 \tau_{xy}}{\partial x \partial y}
Note that the following is true due to a theorem from Differential Equations that I don't remember anymore because I took it a long time ago:
∂2τxy∂y∂x=∂2τxy∂x∂y \frac{\partial^2 \tau_{xy}}{\partial y \partial x} = \frac{\partial^2 \tau_{xy}}{\partial x \partial y}
This lets us rewrite the XX and YY by adding them together:
∂X∂x+∂Y∂y=−∂2σx∂x2−∂2τyx∂y∂x−∂2σy∂y2−∂2τxy∂x∂y \frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} = -\frac{\partial^2 \sigma_x}{\partial x^2} - \frac{\partial^2 \tau_{yx}}{\partial y \partial x} - \frac{\partial^2 \sigma_y}{\partial y^2} - \frac{\partial^2 \tau_{xy}}{\partial x \partial y}
∂X∂x+∂Y∂y=−∂2σx∂x2−∂2σy∂y2−2∂2τxy∂x∂y \frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} = -\frac{\partial^2 \sigma_x}{\partial x^2} - \frac{\partial^2 \sigma_y}{\partial y^2} - 2\frac{\partial^2 \tau_{xy}}{\partial x \partial y}
Hey, that's familiar!
−∂X∂x−∂Y∂y+∂2σx∂x2+∂2σy∂y2=2∂2τxy∂x∂y - \frac{\partial X}{\partial x} - \frac{\partial Y}{\partial y} + \frac{\partial^2 \sigma_x}{\partial x^2} + \frac{\partial^2 \sigma_y}{\partial y^2} = 2\frac{\partial^2 \tau_{xy}}{\partial x \partial y}
This makes the right-hand-side:
−(1+ν)(∂2σx∂x2+∂2σy∂y2+∂X∂x+∂Y∂y) -(1 + \nu) \left( \frac {\partial^2 \sigma_x}{\partial x^2} + \frac {\partial^2 \sigma_y}{\partial y^2} + \frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} \right)
Finally, after factoring out the partials because you can factor out operators like that, you get the target equation:
(∂2∂x2+∂2∂y2)(σx+σy)=−11−ν(∂X∂x+∂Y∂y) \boxed{\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) (\sigma_x + \sigma_y) = - \frac{1}{1 - \nu} \left( \frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} \right)}

§2.

σx=Ay \sigma_x = A y
y=±h  ⟹  σy=τxy=0 y = \pm h \implies \sigma_y = \tau_{xy} = 0
This is a pure bending moment applied to this section. Because there is no shear forces and normal forces (at y=0y = 0), this is likely caused by equal and opposite couple-moments on either sides of the beam.The boundary conditions would include positive external moment on the left and vice versa. In other words, the external loads would be:
M−L=−M0 M_{-L} = - M_0
M+L=M0 M_{+L} = M_0
Additional boundary conditions:
y=0, x=0  ⟹  σx=u=v=0 y = 0, ~ x = 0 \implies \sigma_x = u = v = 0
y=±h  ⟹  σy=τxy=0 y = \pm h \implies \sigma_y = \tau_{xy} = 0
x=±L  ⟹  v=0 x = \pm L \implies v = 0
I will be solving the Airy functions by integration, starting with σx\sigma_x:
σx=∂2Φ∂y2=Ay \sigma_x = \frac{\partial^2 \Phi}{\partial y^2} = Ay
∂Φ∂y=12Ay2+f1(x) \frac{\partial \Phi}{\partial y} = \frac{1}{2} A y^2 + f_1(x)
Φ=16Ay3+f1(x)y+f2(x) \Phi = \frac{1}{6} A y^3 + f_1(x)y + f_2(x)
Now I can differentiate it with xx:
∂Φ∂x=f1′(x)y+f2′(x) \frac{\partial \Phi}{\partial x} = f_1'(x)y + f_2'(x)
∂2Φ∂x2=σy=0=f1′′(x)y+f2′′(x) \frac{\partial^2 \Phi}{\partial x^2} = \sigma_y = 0 = f_1''(x)y + f_2''(x)
Since f1′′(x)f_1''(x) has a coefficient of yy, the other term doesn't, and their sum must equal 00 regardless of the value of yy, both terms must be 00 independently:
f1′′(x)=0  ⟹  f1(x)=Bx+C f_1''(x) = 0 \implies f_1(x) = B x + C
f2′′(x)=0  ⟹  f2(x)=Dx+E f_2''(x) = 0 \implies f_2(x) = D x + E
Updating Φ\Phi:
Φ=16Ay3+(Bx+C)y+Dx+E \Phi = \frac{1}{6} A y^3 + (B x + C) y + D x + E
Φ=16Ay3+Bxy+Cy+Dx+E \Phi = \frac{1}{6} A y^3 + B x y + C y + D x + E
This can be simplified thanks to the shear:
τxy=0=−∂2Φ∂x∂y=∂2Φ∂x∂y \tau_{xy} = 0 = - \frac{\partial^2 \Phi}{\partial x \partial y} = \frac{\partial^2 \Phi}{\partial x \partial y}
0=∂Φ∂x(12Ay2+Bx+C) 0 = \frac{\partial \Phi}{\partial x} \left( \frac{1}{2} A y^2 + B x + C \right)
0=B 0 = B
This simplifies the expression for Φ\Phi:
Φ=16Ay3+Cy+Dx+E \Phi = \frac{1}{6} A y^3 + C y + D x + E
Confirming σy=0\sigma_y = 0:
σy=0=∂2Φ∂x2 \sigma_y = 0 = \frac{\partial^2 \Phi}{\partial x^2}
0=∂Φ∂x(D)=0 0 = \frac{\partial \Phi}{\partial x} \left( D \right) = 0
Now, a few more terms disappear because they play no role after two or more derivatives. I looked this case up and apparently they're called "null functions":
∂2∂y2Cy=∂2∂x2Cy=∂2∂x∂yCy=0 \frac{\partial^2}{\partial y^2} C y = \frac{\partial^2}{\partial x^2} C y = \frac{\partial^2}{\partial x \partial y} C y = 0
∂2∂y2Dx=∂2∂x2Dx=∂2∂x∂yDx=0 \frac{\partial^2}{\partial y^2} D x = \frac{\partial^2}{\partial x^2} D x = \frac{\partial^2}{\partial x \partial y} D x = 0
∂2∂y2E=∂2∂x2E=∂2∂x∂yE=0 \frac{\partial^2}{\partial y^2} E = \frac{\partial^2}{\partial x^2} E = \frac{\partial^2}{\partial x \partial y} E = 0
So this leaves:
Φ=16Ay3 \Phi = \frac{1}{6} A y^3
Checking the biharmonic equation:
∇4Φ=0 \nabla^4 \Phi = 0
(∇⋅∇)(∇⋅∇)Φ=0 (\nabla \cdot \nabla) (\nabla \cdot \nabla) \Phi = 0
(∂2∂x2+∂2∂y2+∂2∂z2)(∂2∂x2+∂2∂y2+∂2∂z2)Φ=0 \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \Phi = 0
(∂2∂x2+∂2∂y2+∂2∂z2)(∂2∂x2+∂2∂y2+∂2∂z2)16Ay3=0 \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \frac{1}{6} A y^3 = 0
(∂2∂y2)(∂2∂y2)y3=0 \left( \frac{\partial^2}{\partial y^2} \right) \left( \frac{\partial^2}{\partial y^2} \right) y^3 = 0
0=0 0 = 0
So, confirmed, this is the defacto solution for Φ\Phi for this loading:
Φ=16Ay3 \boxed{\Phi = \frac{1}{6} A y^3}
Onto the displacement equations:
∂u∂x=ϵx=1E[σx−ν(σy+σz)] \frac{\partial u}{\partial x} = \epsilon_x = \frac{1}{E} [ \sigma_x - \cancel{\nu(\sigma_y + \sigma_z)} ]
∂u∂x=1Eσx \frac{\partial u}{\partial x} = \frac{1}{E} \sigma_x
∂u∂x=AEy \frac{\partial u}{\partial x} = \frac{A}{E} y
u=AEyx+f(y) u = \frac{A}{E} y x + f(y)
Boundary condition:
x=0  ⟹  u=0=AEyx+f(y) x = 0 \implies u = 0 = \cancel{\frac{A}{E} y x} + f(y)
f(y)=0 f(y) = 0
u=AEyx u = \frac{A}{E} y x
Rewriting it to the final form:
u=AxyE \boxed{u = \frac{A x y}{E}}
As for the other axis:
∂v∂y=ϵy=1E[σy−ν(σx+σz)] \frac{\partial v}{\partial y} = \epsilon_y = \frac{1}{E} [ \cancel{\sigma_y} - \nu( \sigma_x + \cancel{\sigma_z} ) ]
∂v∂y=−νEσx \frac{\partial v}{\partial y} = - \frac{\nu}{E} \sigma_x
∂v∂y=−νEAy \frac{\partial v}{\partial y} = - \frac{\nu}{E} A y
v=−νA2Ey2+f(x) v = - \frac{\nu A}{2 E} y^2 + f(x)
This f(x)f(x) can be solved for using the shear strain condition:
∂u∂y+∂v∂x=0 \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} = 0
AxE+f′(x)=0 \frac{A x}{E} + f'(x) = 0
f′(x)=−AxE f'(x) = -\frac{A x}{E}
f(x)=−A2Ex2+C1 f(x) = -\frac{A}{2 E} x^2 + C_1
This makes the equation for vv:
v=−νA2Ey2−A2Ex2+C1 v = - \frac{\nu A}{2 E} y^2 - \frac{A}{2 E} x^2 + C_1
The last boundary condition:
x=±L  ⟹  v=0 x = \pm L \implies v = 0
0=−A2EL2+C1 0 = - \frac{A}{2 E} L^2 + C_1
A2EL2=C1 \frac{A}{2 E} L^2 = C_1
Finally, vv ends up being:
v=−νA2Ey2−A2Ex2+A2EL2 v = - \frac{\nu A}{2 E} y^2 - \frac{A}{2 E} x^2 + \frac{A}{2 E} L^2
And this can be rewritten to the final form:
v=A(L2−x2−νy2)2E \boxed{v = \frac{A(L^2 - x^2 - \nu y^2)}{2 E}}
Onto the comparison of my derivation with what we learned in Mechanics of Materials. Since in Mechanics of Materials we did not concern ourselves with the displacement in the xx axis and only bothered with calculating the displacement of the neutral axis in the yy axis where y=0y = 0, I, thereby declare:
v321=A(L2−x2−νy2)2E v_{321} = \frac{A(L^2 - x^2 - \cancel{\nu y^2})}{2 E}
v321(x)=A(L2−x2)2E v_{321}(x) = \frac{A(L^2 - x^2)}{2 E}
Note that I am using the course numbers for ME 324 and AERE 321 as the subscripts. As for the equation for the displacement according to Mechanics of Materials:
Superposition is dead simple:
M324(x)=−M0⟨x+L⟩0 M_{324}(x) = -M_0 \lang x + L \rangle^0
θ324(x)=−M0EI⟨x+L⟩1+C2 \theta_{324}(x) = -\frac{M_0}{E I} \lang x + L \rangle^1 + C_2
v324(x)=−2M0EI⟨x+L⟩2+C2x+C1 v_{324}(x) = -\frac{2 M_0}{E I} \lang x + L \rangle^2 + C_2 x + C_1
This would be an appropriate time to address II:
I=112wh3 I = \frac{1}{12} w h^3
Here, ww is some unit value.
I=112h3 I = \frac{1}{12} h^3
v324(x)=−24M0Eh3⟨x+L⟩2+C2x+C1 v_{324}(x) = -\frac{24 M_0}{E h^3} \lang x + L \rangle^2 + C_2 x + C_1
Boundary condition:
v324(−L)=0 v_{324}(-L) = 0
v324(−L)=0=−C2L+C1 v_{324}(-L) = 0 = - C_2 L + C_1
C2L=C1 C_2 L = C_1
v324(x)=−24M0Eh3⟨x+L⟩2+C2x+C2L v_{324}(x) = -\frac{24 M_0}{E h^3} \lang x + L \rangle^2 + C_2 x + C_2 L
v324(x)=−24M0Eh3⟨x+L⟩2+C2(x+L) v_{324}(x) = -\frac{24 M_0}{E h^3} \lang x + L \rangle^2 + C_2 (x + L)
The other boundary condition:
v324(L)=0 v_{324}(L) = 0
0=−24M0Eh3(2L)2+2C2L 0 = -\frac{24 M_0}{E h^3} (2 L)^2 + 2 C_2 L
0=−24M0Eh32L2L+2C2L 0 = -\frac{24 M_0}{E h^3} 2 L 2 L + 2 C_2 L
0=−24M0Eh32L2L+2LC2 0 = -\frac{24 M_0}{E h^3} 2 L \cancel{2 L} + \cancel{2 L} C_2
0=−24M0Eh32L+C2 0 = -\frac{24 M_0}{E h^3} 2 L + C_2
24M0Eh32L=C2 \frac{24 M_0}{E h^3} 2 L = C_2
C2=48M0LEh3 C_2 = \frac{48 M_0 L}{E h^3}
v324(x)=24M0Eh3⟨x+L⟩2+48M0LEh3(x+L) v_{324}(x) = \frac{24 M_0}{E h^3} \lang x + L \rangle^2 + \frac{48 M_0 L}{E h^3} (x + L)
And since x+Lx + L will never be negative, we can ignore the Macaulay brackets:
v324(x)=24M0Eh3(x+L)2+48M0LEh3(x+L) v_{324}(x) = \frac{24 M_0}{E h^3} (x + L)^2 + \frac{48 M_0 L}{E h^3} (x + L)
And since I plan to plot this just for quick eyeball comparison, I am normalizing vv by removing coefficients:
V324(x)=2L(x+L)−(x+L)2 V_{324}(x) = 2 L (x + L) - (x + L)^2
Same for the vv I calculate using this course's methods:
V321(x)=L2−x2 V_{321}(x) = L^2 - x^2
This is so obviously different but I shall plot it with L=1L = 1 (unitless). The Python code:
import numpy as np import matplotlib.pyplot as plt L = 1 def v_324(x): return 2 * L * (x + L) - (x + L) ** 2 def v_321(x): return L**2 - x**2 x = np.linspace(-L, L, 400) fig, (ax1, ax2) = plt.subplots(2, 1, figsize=(6, 8)) ax1.plot(x, v_324(x), color="tab:red") ax1.set_title("v_324(x)") ax1.set_ylabel("v_324(x)") ax1.grid(True) ax2.plot(x, v_321(x), color="tab:blue") ax2.set_title("v_321(x)") ax2.set_xlabel("x") ax2.set_ylabel("v_321(x)") ax2.grid(True) plt.tight_layout() plt.show()
Python • 506B
The plot:
That is undoubtedly the same shape. But it is for certain that including the values for M0M_0, EE, hh (or II directly) will result in subtle differences. Moreover, the simpler Mechanics of Materials method doesn't predict deformation in the xx axis and anything on the yy axis but the neutral surface.