$\Sigma M_{R2} = 0$]

$9R_1 + 4(60) = \frac{1}{2}(6)(100)(2)$

$R_1 = 40 \, \text{ lb}$

$\Sigma M_{R1} = 0$

$9R_2 = \frac{1}{2}(6)(100)(7) + 13(60)$

$R_2 = 320 \, \text{ lb}$

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

$EI \, t_{A/B} = \frac{1}{2}(9)(360)(6) - \frac{1}{4}(6)(600)(\frac{39}{5})$

$EI \, t_{A/B} = 2700 \, \text{ lb}\cdot\text{ft}^3$

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

$EI \, t_{C/B} = -\frac{1}{2}(4)(240)(\frac{8}{3})$

$EI \, t_{C/B} = -1280 \, \text{ lb}\cdot\text{ft}^3$

The negative sign indicates that the elastic curve is below the tangent line. It is shown in the figure indicated as *t*_{C/B}. See Rules of Sign for Area-Moment Method.

$\dfrac{y_C}{4} = \dfrac{t_{A/B}}{9}$

$y_C = \frac{4}{9}t_{A/B}$

$EI \, y_C = \frac{4}{9}EI \, t_{A/B}$

$EI \, y_C = \frac{4}{9}(2700)$

$EI \, y_C = 1200 \, \text{ lb}\cdot\text{ft}^3$

Since the absolute value of *EI t*_{C/B} is greater than the absolute value of *EI y*_{C}, the elastic curve is below the undeformed neutral axis (NA) of the beam.

Therefore,

$EI \, \delta_C = 1280 - 1200$

$EI \, \delta_C = 80 \, \text{ lb}\cdot\text{ft}^3$ below *C* (deflection is down) *answer*

## Comments

## The problem is in English

The problem is in English system but the solution is in SI

## I can't quite believe on the

I can't quite believe on the answer or maybe I don't understand it, 1280 should be add to 1200 because the tangent is above the NA of the beam. My answer is 2480 lb ft³