Question 8.SP.2: The overhanging beam AB supports a uniformly distributed loa......

MODELING and ANALYSIS:
Reactions at A and D. Draw the free-body diagram (Fig. 1) of the beam.
From the equilibrium equations \Sigma M_D=0 \text { and } \Sigma M_A=0 , the values of R_A and R_D are as shown.

Shear and Bending-Moment Diagrams. Using the methods discussed in Secs. 5.1 and 5.2, draw the diagrams (Fig. 1) and observe that

|M|_{\max }=239.4  kip \cdot ft =2873  kip \cdot in . \quad|V|_{\max }=43  kips

Section Modulus. For |M|_{\max }=2873 \text {  kip } \cdot \text {  in. and } \sigma_{\text {all }}=24 ksi, the minimum acceptable section modulus of the rolled-steel shape is

S_{\min }=\frac{|M|_{\max }}{\sigma_{\text {all }}}=\frac{2873  kip \cdot in .}{24  ksi }=119.7  in ^3

Selection of Wide-Flange Shape. Choose from the table of Properties of Rolled-Steel Shapes in Appendix E the lightest shapes of a given depth that have a section modulus larger than S_{\min } .

The lightest shape available is W21 × 62

Shearing Stress. For the beam design, assume that the maximum shear is uniformly distributed over the web area of a W21 × 62 (Fig. 2). Write

\tau_m=\frac{V_{\max }}{A_{ web }}=\frac{43  kips }{8.40  in ^2}=5.12  ksi <14.5  ksi

Principal Stress at Point b. The maximum principal stress at point b in the critical section where M is maximum should not exceed \sigma_{\text {all }}=24 ksi.
Referring to Fig. 3, we write

\begin{aligned} & \sigma_a=\frac{M_{\max }}{S}=\frac{2873  kip \cdot in .}{127  in ^3}=22.6  ksi \\ & \sigma_b=\sigma_a \frac{y_b}{c}=(22.6  ksi ) \frac{9.88  in .}{10.50  in .}=21.3  ksi \end{aligned}

Conservatively, \tau_b=\frac{V}{A_{\text {web }}}=\frac{12.2  kips }{8.40  in ^2}=1.45  ksi

Draw Mohr’s circle (Fig. 4) and find

\sigma_{\max }=\frac{1}{2} \sigma_b+R=\frac{21.3  ksi }{2}+\sqrt{\left(\frac{21.3  ksi }{2}\right)^2+(1.45  ksi )^2}

\sigma_{\max }=21.4  ksi \leq 24  ksi