From

Case No. 7 of

Summary of Beam Loadings, the deflection equations are

$EI \, y = \dfrac{Pbx}{6L}(L^2 - x^2 - b^2) \text{ for } 0 \lt x \lt a$

$EI \, y = \dfrac{Pb}{6L} \left[ \dfrac{L}{b}(x - a)^3 + (L^2 - b^2)x - x^3 \right] \text{ for } a \lt x \lt L$

The point under the load $P$ is generally located at $x = a$ and at this point, both equations above will become

$EI \, y = \dfrac{Pab}{6L}(L^2 - a^2 - b^2)$

**Deflection under the 500 N load**

EIδ = EIδ due to 500 N load + EIδ due to 800 N load

$EI \, \delta = \dfrac{500(2)(3)}{6(5)}(5^2 - 2^2 - 3^2) + \dfrac{800(1)(2)}{6(5)}(5^2 - 2^2 - 1^2)$

$EI \, \delta = 1200 + 1066.67$

$EI \, \delta = 2266.67 \, \text{ N}\cdot\text{m}^3$ *answer*

**Deflection under the 800 N load**

EIδ = EIδ due to 500 N load + EIδ due to 800 N load

$EI \, \delta = \dfrac{500(3)}{6(5)} \left[ \dfrac{5}{3}(4 - 2)^3 + (5^2 - 3^2)(4) - 4^3 \right] + \dfrac{800(1)(4)}{6(5)}(5^2 - 4^2 - 1^2)$

$EI \, \delta = 666.67 + 853.33$

$EI \, \delta = 1520 \, \text{ N}\cdot\text{m}^3$ *answer*

## How come the value of x in

How come the value of x in finding the deflection under the 500N load is the same? shouldn't x=4 for the EIy due to 800N load?

## $x$ is the location where…

In reply to How come the value of x in by Hi (not verified)

$x$ is the location where your are solving for the deflection. You are referring to the location of the load. The location of the load is not necessarily the location of the deflection you wish to calculate.

For the 500 N load:$x = 2$ is the location of the 500 N force to cause a deflection at $x = 2$ also.

For the 800 N load:$x = 4$ is the location of the 800 N force, while $x = 2$ is the location of the deflection caused by the force. Notice that the location of the force is not the same as the location of the deflection.