Question 12.2: The structure shown consists of a W10 × 112 rolled-steel bea...
The structure shown consists of a W10 × 112 rolled-steel beam AB and two short members welded together and to the beam. (a) Draw the shear and bending-moment diagrams for the beam and the given loading. (b) Determine the maximum normal stress in sections just to the left and just to the right of point D.
STRATEGY: You should first replace the 10-kip load with an equivalent force-couple system at D. You can section the beam within each region of continuous load (including regions of no load) and find equations for the shear and bending moment.
MODELING and ANALYSIS:
Equivalent Loading of Beam. The 10-kip load is replaced by an equivalent force-couple system at D. The reaction at B is determined by considering the beam to be free body (Fig. 1).
a. Shear and Bending-Moment Diagrams
From A to C. Determine the internal forces at a distance x from point A by considering the portion of beam to the left of section 1. That part of the distributed load acting on the free body is replaced by its resultant, and
\begin{matrix} +\uparrow \sum{F_y=0:} && -3x-V&=&0 && V&=&-3x \text{ kips} \\ +\circlearrowleft \sum{M_1=0:} && 3x({\frac{1}{2}x})+M&=&0 && M&=&-1.5x^2 \text{ kip.ft} \end{matrix}
Since the free-body diagram shown in Fig. 1 can be used for all values of x smaller than 8 ft, the expressions obtained for V and M are valid in the region 0 < x < 8 ft.
From C to D. Considering the portion of beam to the left of section 2 and again replacing the distributed load by its resultant,
These expressions are valid in the region 8 ft < x < 11 ft.
From D to B. Using the position of beam to the left of section 3, the region 11 ft < x < 16 ft is
V = -34 kips M = 226-34x kip.ft
The shear and bending-moment diagrams for the entire beam now can be plotted. Note that the couple of moment 20 kip.ft applied at point D introduces a discontinuity into the bending-moment diagram.
b. Maximum Normal Stress to the Left and Right of Point D.
From Appendix B for the W10 × 112 rolled-steel shape, S = 126 in³ about the X-X axis.
To the left of D: |M| = 168 kip.ft 5 2016 kip.in. Substituting for |M| and S into Eq. (12.3), write
\sigma_m=\frac{|M|}{S}=\frac{2016 \text{ kip.in}}{126 \text{ in}^3} =16.00 \text{ ksi} \quad \quad \quad \sigma_m=16.00 \text{ ksi}
To the right of D: |M| = 148 kip.ft = 1776 kip.in. Substituting for |M| and S into Eq. (12.3), write
\sigma_m=\frac{|M|}{S}=\frac{1776 \text{ kip.in}}{126 \text { in}^3}=14.10 \text{ ksi} \quad \quad \quad \sigma_m=14.10 \text{ ksi}
REFLECT and THINK: It was not necessary to determine the reactions at the right end to draw the shear and bending-moment diagrams. How–ever, having determined these at the start of the solution, they can be used as checks of the values at the right end of the shear and bending-moment diagrams.
