Deflection of Beams by Integration Method

Derivation

When a beam bends, the neutral surface takes the shape of a curve, the elastic curve. From the flexure formula

$$\frac{M}{I} = \frac{E}{R} \;\Rightarrow\; \frac1R = \frac{M}{EI}$$

From calculus the curvature of the curve $y(x)$ is

$$\frac1R = \frac{d^2y/dx^2}{\left[1+(dy/dx)^2\right]^{3/2}}$$

The slope $dy/dx = \theta$ is very small for stiff beams, so $(dy/dx)^2 \ll 1$:

$$\boxed{EI\frac{d^2y}{dx^2} = M}$$

Integrating once gives the slope and twice gives the deflection:

$$EI\frac{dy}{dx} = \int M\,dx + C_1, \qquad EI\,y = \iint M\,dx\,dx + C_1x + C_2$$

Sign convention

$x$ to the right, $y$ positive upward, sagging bending moment positive. With this convention the downward deflection comes out negative. (Some books take $y$ downward and use $EI\,y'' = -M$.) The magnitude is the same.

Boundary conditions

Continuity (Macaulay's method)

If the loading changes along the span, $M$ has a different expression in each segment, giving extra constants. At the junction between two segments both the slope and the deflection must be equal on either side. Macaulay's method writes $M$ for the whole beam in one expression using brackets $\langle x-a\rangle^n$ that are zero for $x<a$. The bracket is integrated as a whole, so there are only two constants $C_1,C_2$ for the entire beam. A uniformly distributed load that stops before the end must be extended to the end and balanced by an equal and opposite load.

(a) Integration

Take the origin at the fixed end A and $x$ towards the free end B. The bending moment at a section is hogging, from the load on the part to its right:

$$M = -\frac{w(L-x)^2}{2}$$ $$EI\frac{d^2y}{dx^2} = -\frac{w(L-x)^2}{2}$$ $$EI\frac{dy}{dx} = \frac{w(L-x)^3}{6} + C_1, \qquad EI\,y = -\frac{w(L-x)^4}{24} + C_1x + C_2$$

At the fixed end $x = 0$: slope $= 0$ and $y = 0$.

$$C_1 = -\frac{wL^3}{6}, \qquad C_2 = \frac{wL^4}{24}$$

At the free end $x = L$:

$$EI\,\theta_B = -\frac{wL^3}{6} \;\Rightarrow\; |\theta_B| = \frac{wL^3}{6EI}, \qquad EI\,y_B = -\frac{wL^4}{6}+\frac{wL^4}{24} = -\frac{wL^4}{8} \;\Rightarrow\; |y_B| = \frac{wL^4}{8EI}$$

The negative signs mean the free end slopes and deflects downward.

(b) Superposition

Total deflection is the sum of the deflections due to each load acting alone (valid for linear elastic behaviour and small deflections).

$EI = 200\times10^9\times8\times10^{-5} = 16\times10^6\ \text{N m}^2 = 16\,000\ \text{kN m}^2$.

Answer: slope $= 7.03\times10^{-3}$ rad $(0.40^\circ)$, deflection $= 14.8$ mm downward at the free end.

Reactions

$$R_A = \frac{30\times4}{6} = 20\ \text{kN}, \qquad R_B = 10\ \text{kN}$$

Macaulay equations (origin at A, $y$ upward, sagging $M$ positive)

$$EI\frac{d^2y}{dx^2} = 20x - 30\langle x-2\rangle$$ $$EI\frac{dy}{dx} = 10x^2 - 15\langle x-2\rangle^2 + C_1$$ $$EI\,y = \frac{10x^3}{3} - 5\langle x-2\rangle^3 + C_1x + C_2$$

Boundary conditions:

• $x = 0$, $y = 0$ gives $C_2 = 0$.
• $x = 6$, $y = 0$: $\dfrac{10\times216}{3} - 5\times4^3 + 6C_1 = 0$, so $720 - 320 + 6C_1 = 0$, so $C_1 = -66.67$.

(a) Slopes

At A ($x=0$):

$$\theta_A = \frac{C_1}{EI} = \frac{-66.67}{12000} = -5.56\times10^{-3}\ \text{rad (clockwise)}$$

At B ($x=6$):

$$EI\theta_B = 10(36) - 15(16) - 66.67 = 53.33 \;\Rightarrow\; \theta_B = 4.44\times10^{-3}\ \text{rad}$$

(b) Deflection under the load ($x = 2$)

$$EI\,y_C = \frac{10\times8}{3} - 0 - 66.67\times2 = 26.67 - 133.33 = -106.67$$ $$y_C = \frac{-106.67}{12000} = -8.89\times10^{-3}\ \text{m} = 8.89\ \text{mm downward}$$

(c) Maximum deflection

Maximum deflection occurs where the slope is zero. The longer segment is CB, so test $x>2$:

$$10x^2 - 15(x-2)^2 - 66.67 = 0 \;\Rightarrow\; -5x^2 + 60x - 126.67 = 0$$ $$x^2 - 12x + 25.33 = 0 \;\Rightarrow\; x = 6 - \sqrt{36-25.33} = 6 - 3.266 = 2.734\ \text{m}$$

(this lies beyond 2 m, so it is valid.)

$$EI\,y_{max} = \frac{10(2.734)^3}{3} - 5(0.734)^3 - 66.67(2.734) = 68.12 - 1.98 - 182.27 = -116.1$$ $$y_{max} = \frac{-116.1}{12000} = -9.68\times10^{-3}\ \text{m}$$

Answer: $\theta_A = 5.56\times10^{-3}$ rad, $\theta_B = 4.44\times10^{-3}$ rad; $y_{under\ load} = 8.89$ mm; $y_{max} = 9.68$ mm downward at 2.73 m from A.

Statically indeterminate beam

 ||A===================B
 ||  w = 20 kN/m       ^ R_B (prop)
 ||<------ 4 m ------>

Let $R_B$ be the prop reaction. Take the origin at A, $x$ towards B, sagging moment positive, $y$ upward.

Let $R_A = wL - R_B$ be the vertical reaction at the fixed end A and $M_A$ the hogging fixed-end moment. The bending moment at a section $x$ from A is

$$EI\,y'' = R_Ax - M_A - \frac{wx^2}{2}$$ $$EI\,y' = \frac{R_Ax^2}{2} - M_Ax - \frac{wx^3}{6} \quad (C_1=0\text{ since } y'=0\text{ at }x=0)$$ $$EI\,y = \frac{R_Ax^3}{6} - \frac{M_Ax^2}{2} - \frac{wx^4}{24} \quad (C_2=0)$$

(a) Prop reaction

Conditions: $y = 0$ at $x = L$ (prop) and moment $M(L) = 0$ at the free end B (no moment applied):

$$R_AL - M_A - \frac{wL^2}{2} = 0 \;\Rightarrow\; M_A = R_AL - \frac{wL^2}{2}$$ $$EI\,y_B = \frac{R_AL^3}{6} - \frac{M_AL^2}{2} - \frac{wL^4}{24} = 0$$

Substitute $M_A$: $\dfrac{R_AL^3}{6} - \dfrac{R_AL^3}{2} + \dfrac{wL^4}{4} - \dfrac{wL^4}{24} = 0$, hence $\dfrac{R_AL^3}{3} = \dfrac{5wL^4}{24}$, so $R_A = \dfrac{5wL}{8}$.

$$R_A = \frac{5\times20\times4}{8} = 50\ \text{kN}, \qquad R_B = wL - R_A = 80 - 50 = 30\ \text{kN}$$

(b) Fixed-end moment

$$M_A = R_AL - \frac{wL^2}{2} = 50\times4 - 160 = 40\ \text{kN m (hogging)} \;\left(=\frac{wL^2}{8}\right)$$

Check: $\sum M_A$: $30\times4 - 80\times2 + 40 = 0$. ✓

(d) Contraflexure

$M(x) = 50x - 40 - 10x^2 = 0 \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow x = 1\ \text{m}$ or $4\ \text{m}$. So the point of contraflexure is at $x = 1$ m from A. Maximum sagging moment is where $V = 0$, $x = 50/20 = 2.5$ m: $M = 50(2.5) - 40 - 10(6.25) = 22.5$ kN m.

(c) Maximum deflection ($y'=0$)

$$EI\,y' = 25x^2 - 40x - \frac{20x^3}{6} = 0 \;\Rightarrow\; 3.333x^2 - 25x + 40 = 0$$ $$x = \frac{25 - \sqrt{625 - 533.3}}{6.667} = \frac{25 - 9.57}{6.667} = 2.314\ \text{m}\ (=0.5785L\text{ from A, i.e. }0.4215L\text{ from the prop})$$ $$EI\,y = \frac{50(2.314)^3}{6} - \frac{40(2.314)^2}{2} - \frac{20(2.314)^4}{24} = 103.2 - 107.1 - 23.9 = -27.7$$ $$y_{max} = \frac{-27.7}{16000} = -1.73\times10^{-3}\ \text{m} = 1.73\ \text{mm downward}$$

Answer: $R_B = 30$ kN, $R_A = 50$ kN, $M_A = 40$ kN m; contraflexure at 1 m from A; $y_{max} = 1.73$ mm at 2.31 m from A.