A beam under loading will deflect from the axis os symmetry \(y = 0\) according to the DE (for small deflections):
\(EI \cfrac{d^4y}{dx^4} = w(x)\)
Where \(E\) is the modulas of elasticity, \(I\) is the moment of inertia and \(w\) is the load.
Here, we have boundary conditions which provide information on how to solve for the constants in the general solution to the DE.
See the table below for conditions at \(x = a\):
| Boundary | Conditions |
|---|---|
| Embedded | \(y(a) = 0, \cfrac{dy(a)}{dx} = 0\) |
| Free | \(\cfrac{d^2y(a)}{dx^2} = 0, \cfrac{d^3y(a)}{dx^3} = 0 \) |
| Hinged (Simply Supported) | \(y(a) = 0, \cfrac{d^2y(a)}{dx^2} = 0 \) |
Determine the deflection equation for a beam under constant load that is embedded on the left end and free on the right end.
We start with the general DE and \( w(x) = w_0\) for a constant load:
\(EI \cfrac{d^4y}{dx^4} = w_0 \)
We can just take the integral keeping constant:
\(\cfrac{d^4y}{dx^4} = \cfrac{w_0}{EI}\)
\(\cfrac{d^3y}{dx^3} = \cfrac{w_0}{EI}x + A \)
\(\cfrac{d^2y}{dx^2} = \cfrac{w_0}{2EI}x^2 + Ax + B\)
\(\cfrac{dy}{dx} = \cfrac{w_0}{6EI}x^3 + \cfrac{Ax^2}{2} + Bx + C\)
\(y(x) = \cfrac{w_0}{24EI}x^4 + \cfrac{Ax^3}{6} + \cfrac{Bx^2}{2} + Cx + D\)
Now we can use the boundary conditions to solve for the constants. The left end is embedded thus, \(y(0) = 0, \cfrac{dy(0)}{dx} = 0\).
This means \(C = 0\) and \(D = 0\) and the equation is now:
\(y(x) = \cfrac{w_0}{24EI}x^4 + \cfrac{Ax^3}{6} + \cfrac{Bx^2}{2}\)
The right end is free, thus \(\cfrac{d^2y(L)}{dx^2} = 0, \cfrac{d^3y(L)}{dx^3} = 0\).
We can now determine the value of \(A\):
\(\cfrac{d^3y}{dx^3} = 0 = \cfrac{w_0}{EI}(L) + A\)
\(A = - \cfrac{w_0L}{EI}\)
Next, we can determine the value of \(B\):
\(\cfrac{d^2y}{dx^2} = 0 = \cfrac{w_0}{2EI}L^2 + AL + B \)
\(\cfrac{d^2y}{dx^2} = 0 = \cfrac{w_0}{2EI}L^2 - \cfrac{w_0}{EI}L^2 + B \)
\(B = \cfrac{w_0 L^2}{2EI}\)
Then, we can substitute the known values into the equation:
\(y(x) = \cfrac{w_0}{24EI}x^4 - \cfrac{w_0Lx^3}{6EI} + \cfrac{w_oL^2x^2}{4EI}\)
Finally, we can factor the expression:
\(y(x) = \cfrac{w_0}{24EI} (x^4 - 4Lx^3 + 6L^2x^2)\)
Therefore, we can determine the deflection equation that fits the above criteria is \(\boldsymbol{y(x) = \cfrac{w_0}{24EI} (x^4 - 4Lx^3 + 6L^2x^2)}\).
A thin vertical column under constant load, \( P \), will buckle according to the DE:
\(EI \cfrac{d^2y)}{dx^2} + Py = 0\)
This is a linear equation that can be solved using a characteristic equation:
\(m^2 + \cfrac{EI}{P} = 0\)
\(m = \sqrt{\cfrac{EI}{P}} i\)
\(y(x) = c_1 \cos {(\cfrac{EI}{P} x )} + c_2 \sin {(\cfrac{EI}{P} x)}\)
The boundary conditions are \(y(0) = 0\) and \(y(L) = 0\) since there is no deflection at the fixed ends.