Limit the Deflection of Cantilever Beam by Applying Force at the Free End

$EI = 200,000(60.8 \times 10^6)$

$EI = 1.216 \times 10^{13} ~ \text{N}\cdot \text{mm}^2$
 

Part (1)

$\delta = \dfrac{PL^3}{3EI}$

$\delta = \dfrac{200(1.75^3)(1000^4)}{3(1.216 \times 10^{13})}$

$\delta = 29.38 ~ \text{mm}$   ←   answer

 

Part (2)
 

 

$EI \, t_{C/A} = (\text{Area}_{AC}) \cdot \bar{X}_C = 0$

$\frac{1}{2}(3.5)(3.5R)\left[ \frac{2}{3}(3.5) \right] - \frac{1}{2}(1.75)(350)\left[ 1.75 + \frac{2}{3}(1.75) \right] = 0$

$14.291R = 893.229$

$R = 62.5 ~ \text{kN}$   ←   answer
 

Another Solution for Part (2)
Using Superposition Method
$\dfrac{R(3.5^3)}{3EI} = \dfrac{200(1.75^3)}{3EI} + 1.75 \left[ \dfrac{200(1.75^2)}{2EI} \right]$

$\dfrac{343}{24}R = \dfrac{42,875}{48}$

$R = 62.5 ~ \text{kN}$   ←   (okay!)

 

Part (3)
 

 

$EI \, t_{B/A} = (\text{Area}_{AB}) \cdot \bar{X}_B$

$EI \, t_{B/A} = \frac{1}{2}(1.75)(3.5R)\left[ \frac{2}{3}(1.75) \right] + \frac{1}{2}(1.75)(1.75R)\left[ \frac{1}{3}(1.75) \right] - \frac{1}{2}(1.75)(350)\left[ \frac{2}{3}(1.75) \right]$

$EI \, t_{B/A} = 3.573R + 0.893R - 357.292$

$EI \, t_{B/A} = 4.466R - 357.292$

$(1.216 \times 10^{13})(-9.5) = (4.466R - 357.292)(1000^4)$

$R = 54.14 ~ \text{kN}$   ←   answer