Problem 658 | Beam Deflection by Conjugate Beam Method

 

 

From the conjugate beam
$\Sigma M_A = 0$

$LF_2 + \frac{1}{2}(L - a)[M(L - a)/L][a + \frac{1}{3}(L - a)] = \frac{1}{2}a(Ma/L)[\frac{2}{3}a]$

$LF_2 + \dfrac{M(L - a)^2}{2L}[\frac{1}{3}L + \frac{2}{3}a] = \dfrac{Ma^3}{3L}$

$LF_2 + \dfrac{M(L - a)^2(L + 2a)}{6L} = \dfrac{Ma^3}{3L}$

$LF_2 = \dfrac{Ma^3}{3L} - \dfrac{M(L - a)^2(L + 2a)}{6L}$

$F_2 = \dfrac{Ma^3}{3L^2} - \dfrac{M(L - a)^2(L + 2a)}{6L^2}$
 

 

$F_2 = \dfrac{M}{6L^2}[2a^3 - (L - a)^2(L + 2a)]$

$F_2 = \dfrac{M}{6L^2}[2a^3 - (L^2 - 2aL + a^2)(L + 2a)]$

$F_2 = \dfrac{M}{6L^2}[2a^3 - (L^3 - 2aL^2 + a^2L + 2aL^2 - 4a^2L + 2a^3)]$

$F_2 = \dfrac{M}{6L^2}[-L^3 + 3a^2L]$

$F_2 = \dfrac{M}{6L}(3a^2 - L^2)$
 

$M_B = -(L - a)F_2 - \frac{1}{2}(L - a)[M(L - a)/L][\frac{1}{3}(L - a)]$

$M_B = -(L - a)\dfrac{M}{6L}(3a^2 - L^2) - \frac{1}{2}(L - a)[M(L - a)/L][\frac{1}{3}(L - a)]$

$M_B = -\dfrac{M(L - a)}{6L}[(3a^2 - L^2) + (L - a)^2]$

$M_B = -\dfrac{M(L - a)}{6L}[3a^2 - L^2 + L^2 - 2aL + a^2]$

$M_B = -\dfrac{M(L - a)}{6L}(4a^2 - 2aL)$

$M_B = -\dfrac{M(L - a)}{6L}[-2a(L - 2a)]$

$M_B = \dfrac{Ma}{3L}(L - a)(L - 2a)$

$M_B = \dfrac{Ma}{3L}(L^2 - 3aL + 2a^2)$
 

Thus,
$EI \, \delta_B = \dfrac{Ma}{3L}(L^2 - 3aL + 2a^2)$           answer