Problems on Macaulay's Method

Problem no.:3

Compute the midspan Deflectin for the beam loaded as shown in Figure.3

Take EI is constant.

Figure -3

Solution:3

$\Sigma M_{R2} = 0$

$6R_1 = 600(3)(3.5)$

$R_1 = 1050 \, \text{N}$

$\Sigma M_{R1} = 0$

$6R_2 = 600(3)(2.5)$
$R_2 = 750 \, \text{N}$

$EI \, y'' = 1050x - \frac{1}{2}(600)\langle \, x - 1 \, \rangle^2 + \frac{1}{2}(600)\langle \, x - 4 \, \rangle^2$

$EI \, y'' = 1050x - 300\langle \, x - 1 \, \rangle^2 + 300\langle \, x - 4 \, \rangle^2$
$EI \, y' = 525x^2 - 100\langle \, x - 1 \, \rangle^3 + 100\langle \, x - 4 \, \rangle^3 + C_1$
$EI \, y = 175x^3 - 25\langle \, x - 1 \, \rangle^4 + 25\langle \, x - 4 \, \rangle^4 + C_1x + C_2$

At x = 0, y = 0,

Therefore C₂ = 0

At x = 6 m, y = 0

$0 = 175(6^3) - 25(6 - 1)^4 + 25(6 - 4)^4 + 6C_1$

$C_1 = -3762.5 \, \text{N}\cdot\text{m}^2$

Therefore,

$EI \, y = 175x^3 - 25\langle \, x - 1 \, \rangle^4 + 25\langle \, x - 4 \, \rangle^4 - 3762.5x$

At midspan, x = 3 m

$EI \, y_{midspan} = 175(3^3) - 25(3 - 1)^4 - 3762.5(3)$
$EI \, y_{midspan} = -6962.5 \, \text{N}\cdot\text{m}^3$

Thus,

$EI \, \delta_{midspan} = 6962.5 \, \text{N}\cdot\text{m}^3 \,\,$

Problem no :4

A simply supported beam carries a couple M applied as shown in figure 4. Determine the equation of the elastic curve and the deflection at the point of application of the couple. Also write the equations of deflection when the couple will act at the supports.

Fig.4

Solution :

$EI \, y'' = \dfrac{M}{L}x - M \, \langle \, x - a \, \rangle^0$

$EI \, y' = \dfrac{M}{2L}x^2 - M \, \langle \, x - a \, \rangle + C_1$

$EI \, y = \dfrac{M}{6L}x^3 - \frac{1}{2}M \, \langle \, x - a \, \rangle^2 + C_1x + C_2$

At x = 0, y = 0, therefore C₂ = 0

At x = L, y = 0
$0 = \frac{1}{6}ML^2 - \frac{1}{2}M(L - a)^2 + C_1L$

$0 = \frac{1}{6}ML^2 - \frac{1}{2}M(L^2 - 2La + a^2) + C_1L$

$0 = \frac{1}{6}ML^2 - \frac{1}{2}ML^2 + MLa - \frac{1}{2}Ma^2 + C_1L$

$0 = -\frac{1}{3}ML^2 + MLa - \frac{1}{2}Ma^2 + C_1L$

$C_1L = \frac{1}{3}ML^2 - MLa + \frac{1}{2}Ma^2$

$C_1 = \frac{1}{3}ML - Ma + \dfrac{Ma^2}{2L}$

Therefore,
$EI \, y = \dfrac{M}{6L}x^3 - \frac{1}{2}M \, \langle \, x - a \, \rangle^2 + \left( \frac{1}{3}ML - Ma + \dfrac{Ma^2}{2L} \right)x \,\,$

At x = a
$EI \, y = \dfrac{Ma^3}{6L} + \left( \frac{1}{3}ML - Ma + \dfrac{Ma^2}{2L} \right)a$

$EI \, y = \dfrac{2Ma^3}{3L} + \frac{1}{3}MLa - Ma^2$

$EI \, y = \dfrac{Ma}{3L} (2a^2 + L^2 - 3La)$

$EI \, y = \dfrac{Ma}{3L} (L^2 - 3La + 2a^2) \,\,$

When a = 0
Moment load is at the left support :
$EI \, y = \dfrac{M}{6L}x^3 - \frac{1}{2}M \, \langle \, x - a \, \rangle^2 + \left( \frac{1}{3}ML - Ma + \dfrac{Ma^2}{2L} \right)x$

$EI \, y = \dfrac{M}{6L}x^3 - \frac{1}{2}Mx^2 + \frac{1}{3}MLx$

$EI \, y = \dfrac{Mx}{6L}(x^2 - 3Lx + 2L^2)$

$EI \, y = \dfrac{Mx}{6L}(2L^2 - 3Lx + x^2)$

$EI \, y = \dfrac{Mx}{6L}(L - x)(2L - x) \,\,$

When a = L
Moment load is at the right support :
$EI \, y = \dfrac{M}{6L}x^3 - \frac{1}{2}M \, \langle \, x - a \, \rangle^2 + \left( \frac{1}{3}ML - Ma + \dfrac{Ma^2}{2L} \right)x$

$EI \, y = \dfrac{M}{6L}x^3 + \left( \frac{1}{3}ML - ML + \frac{1}{2}ML \right)x$

$EI \, y = \dfrac{M}{6L}x^3 - \frac{1}{6}MLx$

$EI \, y = \dfrac{Mx^3 - ML^2x}{6L}$

$EI \, y = \dfrac{-Mx (-x^2 + L^2)}{6L}$

$EI \, y = \dfrac{-MLx (L^2 - x^2)}{6L^2}$

$EI \, y = -\dfrac{MLx}{6L^2}(L^2 - x^2)$

$EI \, y = -\dfrac{MLx}{6}\left( 1 - \dfrac{x^2}{L^2} \right) \,\,$

Prof.Kodali Srinivas

Search This Blog