Deflection of Beams by Moment- area Method

Basis

For small deflections $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$, so $\dfrac{d\theta}{dx} = \dfrac{M}{EI}$, where $\theta$ is the slope. The quantity $M/EI$ plotted along the beam is the $M/EI$ diagram.

        tangent at A
   A ___.----__           
      \        ---.__ B     theta_AB : change in slope
 tangent at B      t_BA : deviation of B from the
                   tangent drawn at A

Theorem I (change of slope)

The change in slope between two points A and B on the elastic curve equals the area of the $M/EI$ diagram between them.

Proof: from $d\theta = \dfrac{M}{EI}dx$, integrate from A to B:

$$\theta_{B} - \theta_{A} = \int_A^B \frac{M}{EI}\,dx = \text{Area}_{AB}\ \text{of the }M/EI\text{ diagram}$$

Theorem II (tangential deviation)

The vertical deviation of point B on the elastic curve from the tangent drawn at A equals the moment of the area of the $M/EI$ diagram between A and B, taken about B.

Proof: consider a small element $dx$ at distance $x$ from B. The change in slope across the element is $d\theta = \dfrac{M}{EI}dx$. The tangents at the ends of the element intercept a vertical distance $dt = x\,d\theta$ on the vertical through B. Integrating,

$$t_{B/A} = \int_A^B x\,\frac{M}{EI}\,dx = \text{Area}_{AB}\times\bar x_B$$

where $\bar x_B$ is the distance of the centroid of the $M/EI$ area from B.

Use

1. Draw the BM diagram, divide by $EI$.
2. Choose a reference tangent where the slope is known: at the fixed end of a cantilever (zero slope) or at the centre of a symmetrically loaded beam.
3. Slope: use Theorem I from the reference point. Deflection: use Theorem II (deviation from the tangent). For a simply supported beam, find the deviation of the supports from a tangent at one support, get the slope there, then find the deflection at a section from geometry.

Sign: positive area (sagging) gives an anticlockwise rotation from A to B and B lies above the tangent at A.

Set-up

    P            P
    |            |
    v____________v
  A   a   |  L-2a   a  B
 R_A = P   R_B = P

By symmetry $R_A = R_B = P$. Bending moment: rises linearly from 0 at A to $Pa$ at the load, then constant $Pa$ between the loads. The $M/EI$ diagram is a trapezoid, symmetrical about mid-span C.

Slope at support A

By symmetry the slope at mid-span C is zero (the tangent at C is horizontal). By Theorem I, the slope at A equals the area of the $M/EI$ diagram between A and C:

$$\theta_A = \frac1{EI}\left[\frac12\,a\,(Pa) + \left(\frac L2-a\right)(Pa)\right] = \frac{Pa(L-a)}{2EI}$$

Deflection at mid-span

The deflection at C equals the deviation of A from the tangent at C, i.e. the moment of the area between A and C about A (Theorem II):

$$y_C = \frac{1}{EI}\left[\frac{Pa^2}{2}\cdot\frac{2a}{3} + Pa\left(\frac L2-a\right)\left(\frac{L}{4}+\frac a2\right)\right]$$

The first term is the triangle (centroid $2a/3$ from A); the second is the rectangle (centroid at $a + \frac12(\frac L2 - a) = \frac L4+\frac a2$).

$$y_C = \frac{1}{EI}\left[\frac{Pa^3}{3} + \frac{PaL^2}{8} - \frac{Pa^3}{2}\right] = \boxed{\frac{Pa\,(3L^2-4a^2)}{24EI}}$$

Numerical values

$$\theta_A = \frac{30\times2\times(6-2)}{2\times25000} = 4.8\times10^{-3}\ \text{rad}$$ $$y_C = \frac{30\times2\times(3\times36-4\times4)}{24\times25000} = \frac{60\times92}{600000} = 9.2\times10^{-3}\ \text{m}$$

Answer: $\theta_A = \theta_B = 4.8\times10^{-3}$ rad $(0.275^\circ)$; maximum deflection at mid-span $= 9.2$ mm downward.

$EI = 200\times10^9\times1.5\times10^{-4} = 30\times10^6\ \text{N m}^2 = 30\,000\ \text{kN m}^2$.

Bending moment diagram

$$R_A = \frac{40\times3}{5} = 24\ \text{kN}, \qquad R_B = 16\ \text{kN}, \qquad M_C = 24\times2 = 48\ \text{kN m}$$

The $M/EI$ diagram is a triangle of base 5 m and height $48/EI$ with its apex at C. Total area $= \frac12\times5\times48 = 120$ (in kN m$^2$, divided by $EI$).

 A      C(2 m)            B
 /\    48
/  \_____
 triangle area = 120/EI

(a) Slopes at the supports

Deviation of B from the tangent at A (moment of area about B). The centroid of a triangle lies one third of the base from the base end opposite the apex: distance from B $= (5+3)/3 = 2.667$ m.

$$t_{B/A} = 120\times2.667 = 320/EI$$ $$\theta_A = \frac{t_{B/A}}{L} = \frac{320}{5\,EI} = \frac{64}{EI} = \frac{64}{30000} = 2.13\times10^{-3}\ \text{rad}$$

Similarly the deviation of A from the tangent at B: centroid from A $= (5+2)/3 = 2.333$ m:

$$t_{A/B} = 120\times2.333 = 280/EI \;\Rightarrow\; \theta_B = \frac{280}{5EI} = \frac{56}{EI} = 1.87\times10^{-3}\ \text{rad}$$

(b) Deflection under the load C

Height of the tangent at A above C: $\dfrac{2}{5}t_{B/A} = \dfrac{128}{EI}$. Deviation of C from the tangent at A is the moment of the area A to C about C: area $\frac12\times2\times48 = 48$, centroid $2/3$ m from C:

$$t_{C/A} = 48\times\frac23 = \frac{32}{EI}$$ $$y_C = \frac{128 - 32}{EI} = \frac{96}{30000} = 3.2\times10^{-3}\ \text{m} = 3.2\ \text{mm}$$

(c) Maximum deflection

The point of zero slope D lies in the longer segment CB (3 m), at a distance $x$ from B. The slope at B equals the area of $M/EI$ from B to D, with $M = R_Bx = 16x$:

$$\theta_B = \frac12\,x\,\frac{16x}{EI} \;\Rightarrow\; 8x^2 = 56 \;\Rightarrow\; x = \sqrt7 = 2.646\ \text{m}\ (<3\text{ m}, \text{ valid})$$

So D is 2.646 m from B, i.e. 2.354 m from A. Max deflection is the deviation of B from the horizontal tangent at D:

$$y_{max} = 8x^2\times\frac{2x}{3}\Big/EI = \frac{16x^3}{3EI} = \frac{16\times18.52}{3\times30000} = 3.29\times10^{-3}\ \text{m}$$

Answer: $\theta_A = 2.13\times10^{-3}$ rad, $\theta_B = 1.87\times10^{-3}$ rad; $y_C = 3.20$ mm; $y_{max} = 3.29$ mm at 2.35 m from A.