Question 9.SP.13: For the beam and loading shown, determine the magnitude and ......

MODELING: Use the free-body diagram of the entire beam in Fig. 1 to obtain

R _A=16.81  kN \uparrow \quad R _B=38.2  kN \uparrow

ANALYSIS:
M/EI Diagram. Draw the M/EI diagram by parts (Fig. 2), considering the effects of the reaction R _A and of the distributed load separately. The areas of the triangle and of the spandrel are

A_1=\frac{1}{2} \frac{R_A L}{E I} L=\frac{R_A L^2}{2 E I} \quad A_2=\frac{1}{3}\left(-\frac{w b^2}{2 E I}\right) b=-\frac{w b^3}{6 E I}

Reference Tangent. As seen in Fig. 3, the tangent to the beam at support A is chosen as the reference tangent. Using the second moment-area theorem, the tangential deviation t_{B / A} of support B with respect to support A is

t_{B / A}=A_1 \frac{L}{3}+A_2 \frac{b}{4}=\left(\frac{R_A L^2}{2 E I}\right) \frac{L}{3}+\left(-\frac{w b^3}{6 E I}\right) \frac{b}{4}=\frac{R_A L^3}{6 E I}-\frac{w b^4}{24 E I}

Slope at A.

\theta_A=-\frac{t_{B / A}}{L}=-\left(\frac{R_A L^2}{6 E I}-\frac{w b^4}{24 E I L}\right)            (1)

Largest Deflection. As seen in Fig. 4, the largest deflection occurs at point K, where the slope of the beam is zero. Using Fig. 5, write

\theta_K=\theta_A+\theta_{K / A}=0               (2)

But          \theta_{K / A}=A_3+A_4=\frac{R_A x_m^2}{2 E I}-\frac{w}{6 E I}\left(x_m-a\right)^3             (3)

Substitute for \theta_A \text { and } \theta_{K / A} from Eqs. (1) and (3) into Eq. (2):

-\left(\frac{R_A L^2}{6 E I}-\frac{w b^4}{24 E I L}\right)+\left[\frac{R_A x_m^2}{2 E I}-\frac{w}{6 E I}\left(x_m-a\right)^3\right]=0

Substituting the numerical data gives

-29.53 \frac{10^3}{E I}+8.405 x_m^2 \frac{10^3}{E I}-4.167\left(x_m-1.4\right)^3 \frac{10^3}{E I}=0

Solving by trial and error for x_m ,              x_m = 1.890 m

Computing the moments of A_3 and A_4 about a vertical axis through A gives

Using the given data, R_A=16.81  kN \text {, and } I=28.7 \times 10^{-6}  m ^4 \text {, }

y_m=6.44  mm \downarrow