Application of Moment - Area Method for Simply supported Beams

The deflection δ at some point B of a simply supported beam can be obtained by the following steps:

1. Compute $t_{C/A} = \dfrac{1}{EI}(Area_{AC}) \, \bar{X}_C$ = ZCA

2. Compute $t_{B/A} = \dfrac{1}{EI}(Area_{AB}) \, \bar{X}_B$ = ZBA

3. Slope at support A = O A = t CA /L = ZCA / L

4. Slope at support C = O C = tCA/ L = ZCA / L

5. Solve δ by ratio and proportion (see figure above).

$\dfrac{\delta + t_{B/A}}{x} = \dfrac{t_{C/A}}{L}$

Problem:1

The middle half of the beam shown in Fig. has a moment of inertia 1.5 times that of the rest of the beam. Find the midspan deflection.

(Hint: Convert the M diagram into an M/EI diagram.)

Simple beam with different moment of inertia over the span

Solution :

M/EI diagram of a simple beam with changing moment of inertia

$t_{A/C} = \dfrac{1}{EI}(Area_{AC}) \, \bar{X}_A$

$t_{A/C} = \frac{1}{2}a \left( \dfrac{Pa}{2EI} \right)(\frac{2}{3}a) + a \left( \dfrac{Pa}{3EI} \right)(\frac{3}{2}a) + \frac{1}{2}a \left( \dfrac{2Pa}{3EI} - \dfrac{Pa}{3EI} \right)(\frac{5}{3}a)$

$t_{A/C} = \dfrac{Pa^3}{6EI} + \dfrac{Pa^3}{2EI} + \dfrac{5Pa^3}{18EI}$

$t_{A/C} = \dfrac{17Pa^3}{18EI}$

Therefore,
$\delta_{midspan} = \dfrac{17Pa^3}{18EI} \,\,$ → answer

Problem : 2

Determine the value of EIδ at the right end of the overhanging beam shown in Fig.

Solution :

$\Sigma M_B = 0$
$aR_A = w_ob(\frac{1}{2}b)$
$R_A = \dfrac{w_ob^2}{2a}$

$EI \, t_{A/B} = (Area_{AB}) \, \bar{X}_A$

$EI \, t_{A/B} = \frac{1}{2}a(\frac{1}{2}w_ob^2)(\frac{2}{3}a)$

$EI \, t_{A/B} = \frac{1}{6}w_oa^2b^2$

$EI \, t_{C/B} = (Area_{BC}) \, \bar{X}_C$

$EI \, t_{C/B} = \frac{1}{3}b(\frac{1}{2}w_ob^2)(\frac{3}{4}b)$

$EI \, t_{C/B} = \frac{1}{8}w_ob^4$

$\dfrac{y_C}{b} = \dfrac{t_{A/B}}{a}$

$y_C = \dfrac{b}{a}t_{A/B}$

$EI \, y_C = \dfrac{b}{a}EI \, t_{A/B}$

$EI \, y_C = \dfrac{b}{a}(\frac{1}{6}w_oa^2b^2)$

$EI \, y_C = \frac{1}{6}w_oab^3$

$\delta_C = y_C + t_{C/B}$

$EI \, \delta_C = EI \, y_C + EI \, t_{C/B}$

$EI \, \delta_C = \frac{1}{6}w_oab^3 + \frac{1}{8}w_ob^4$

$EI \, \delta_C = \frac{1}{24}w_ob^3(4a + 3b) \,\,$ → answer

Comments

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