Question 8.S-P.2: The overhanging beam AB supports a uniformly distributed loa...

Reactions at A and D.     We draw the free-body diagram of the beam.

From the equilibrium equations \Sigma M_{D}=0 and \Sigma M_{A}=0 we find the values of R_{A} and R_{D} shown in the diagram.

Shear and Bending-Moment Diagrams.     Using the methods of Secs. 5.2 and 5.3, we draw the diagrams and observe that

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

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

Selection of Wide-Flange Shape.     From the table of Properties of Rolled-Steel Shapes in Appendix C, we compile a list of the lightest shapes of a given depth that have a section modulus larger than S_{\min }.

We now select the lightest shape available, namely                      W21 × 62

Shearing Stress.     Since we are designing the beam, we will conservatively assume that the maximum shear is uniformly distributed over the web area of a W21 × 62. We write

t _{m}=\frac{V_{\max }}{A_{\text {web }}}=\frac{43 kips }{8.40 in ^{2}}=5.12 ksi <14.5 ksi                        (OK)

Principal Stress at Point b.     We check that the maximum principal stress at point b in the critical section where M is maximum does not exceed s _{\text {all }}=24 ksi. We write

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

We draw Mohr’s circle and find

s _{\max }=21.4 ksi \leq 24 ksi            (OK)