Pure bending; deflection of SS beam under UDL

Pure bending

Pure bending is the bending of a beam under a constant bending moment with zero shear force over the length considered. Only normal (bending) stresses act on the cross-section; there is no shear stress.

Example: a simply supported beam carrying two equal loads $P$ placed symmetrically at distance $a$ from each support. Between the loads the shear force is zero and $M = Pa$ is constant, so that middle portion is in pure bending.

   P                P
   |<--a-->|        |
   v                v
 A ^----------------------^ B
   SFD:  +P  | 0 |  -P
   BMD:  rises to Pa, flat, falls

Maximum deflection of SS beam under UDL

Take A as origin, span $L$, load $w$ per unit length. Reactions $R_A = R_B = wL/2$.

Bending moment at distance $x$:

$$EI\frac{d^2y}{dx^2} = M = \frac{wL}{2}x - \frac{w x^2}{2}$$

Integrate once (slope):

$$EI\frac{dy}{dx} = \frac{wL x^2}{4} - \frac{w x^3}{6} + C_1$$

Integrate again (deflection):

$$EI\,y = \frac{wL x^3}{12} - \frac{w x^4}{24} + C_1 x + C_2$$

Boundary conditions: at $x=0$, $y=0$ gives $C_2 = 0$. At $x=L$, $y=0$:

$$0 = \frac{wL^4}{12} - \frac{wL^4}{24} + C_1 L \;\Rightarrow\; C_1 = -\frac{wL^3}{24}$$

So

$$y = \frac{w}{24EI}\left(2Lx^3 - x^4 - L^3 x\right)$$

By symmetry the slope is zero and deflection is maximum at mid-span $x = L/2$:

$$y_{max} = \frac{w}{24EI}\left(\frac{2L^4}{8} - \frac{L^4}{16} - \frac{L^4}{2}\right) = -\frac{5wL^4}{384EI}$$

Answer: $y_{max} = \dfrac{5wL^4}{384EI}$ downward, at mid-span.