Question 12.S-P.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 of 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.
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 as a free body.
a. Shear and Bending-Moment Diagrams
From A to C. We 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 we write
+↑\sum{F_{y}} = 0 : -3 x – V = 0 V = -3 x kips+\curvearrowleft \sum{M_{1}} = 0 : 3 x(\frac{1}{2} x) + M = 0 M = -1.5 x^{2} kip \cdot ft
Since the free-body diagram shown 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, we obtain
+↑\sum{F_{y}} = 0 : -24 – V = 0 V = -24 kips+\curvearrowleft \sum{M_{2}} = 0 : 24(x – 4) + M = 0 M = 96 – 24 x kip \cdot ft
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, we obtain for the region 11 ft < x < 16 ft
V = -34 kips M = 226 – 34 x kip \cdot ft
The shear and bending-moment diagrams for the entire beam can now be plotted. We note that the couple of moment 20 kip \cdot 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 App. B we find that for the W10 × 112 rolled-steel shape, S = 126 in^{3} about the X-X axis.
To the left of D: We have |M| = 168 kip \cdot ft = 2016 kip \cdot in. Substituting for |M| and S into Eq. (12.3), we write
σ_{m} = \frac{|M|}{S} (12,3)
σ_{m} = \frac{|M|}{S} = \frac{2016 kip \cdot in.}{126 in^{3}} = 16.00 ksi σ_{m} = 16.00 ksi
To the right of D: We have |M| = 148 kip \cdot ft = 1776 kip \cdot in. Substituting for |M| and S into Eq. (12.3), we write
σ_{m} = \frac{|M|}{S} = \frac{1776 kip \cdot in.}{126 in^{3}} = 14.10 ksi σ_{m} =14.10 ksi
