Question 12.4: The W360 × 79 rolled-steel beam AC is simply supported and c...
The W360 × 79 rolled-steel beam AC is simply supported and carries the uniformly distributed load shown. Draw the shear and bending-moment diagrams for the beam, and determine the location and magnitude of the maximum normal stress due to bending.
STRATEGY: A load is distributed over part of the beam. You can use the equations in this section in two parts: for the load and for the no-load regions. From the discussion in this section, you can expect the shear diagram will show an oblique line under the load, followed by a horizontal line. The bending-moment diagram should show a parabola under the load and an oblique line under the rest of the beam.
MODELING and ANALYSIS:
Reactions. Considering the entire beam as a free body (Fig. 1)
\pmb{\text{R}}_A=80 \text{ kN}\uparrow \quad \quad \quad \pmb{\text{R}}_C=40 \text{ kN}\uparrow
Shear Diagram. The shear just to the right of A is V_A = +80 \text{ kN}.
Since the change in shear between two points is equal to minus the area under the load curve between the same two points, V_B is
\begin{matrix} V_B-V_A&=&-(20 \text{ kN/m})(6 \text{ m})=-120 \text{ kN} \\ V_B&=&-120+V_A=-120+80=-40 \text{ kN} \end{matrix}
The slope dV/dx = –w is constant between A and B, and the shear diagram between these two points is represented by a straight line. Between B and C, the area under the load curve is zero; therefore,
V_C-V_B=0 \quad \quad V_C=V_B=-40 \text{ kN}
and the shear is constant between B and C.
Bending-Moment Diagram. Note that the bending moment at each end is zero. In order to determine the maximum bending moment, locate the section D of the beam where V = 0.
V_D-V_A=-\omega x \\ 0-80 \text{ kN}=-(20 \text{ kN/m})x
Solving for x, x = 4 m
The maximum bending moment occurs at point D, where dM/dx = V = 0.
The areas of various portions of the shear diagram are computed and given (in parentheses). The area of the shear diagram between two points is equal to the change in bending moment between the same two points, giving
M_D-M_A=+160 \text{ N}\cdot \text{m} \quad \quad \quad M_D=+160 \text{ N}\cdot \text{m}\\ M_B-M_D=-40\text{ N}\cdot \text{m} \quad \quad \quad M_B=+120 \text{ N}\cdot \text{m} \\ M_C-M_B=-120\text{ N}\cdot \text{m}\quad \quad \quad M_C=0
The bending-moment diagram consists of an arc of parabola followed by a segment of straight line. The slope of the parabola at A is equal to the value of V at that point.
Maximum Normal Stress. This occurs at D, where |M| is largest.
From Appendix B, for a W360 × 79 rolled-steel shape, S = 1270 mm³ about a horizontal axis. Substituting this and |M| = |M_D| = 160 × 10³ N·m into Eq. (12.3),
\sigma_m=\frac{|M_D|}{S}=\frac{160 \times 10^3 \text{ N}\cdot \text{m}}{1270 \times 10^{-6} \text{ m}^3} =126.0 \times 10^6 \text{ Pa}
Maximum normal stress in the beam = 126.0 MPa
