Question 12.S-P.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.
Reactions. Considering the entire beam as a free body, we find
R_{A} = 80 kN ↑ R_{C} = 40 kN ↑
Shear Diagram. The shear just to the right of A is V_{A} = +80 kN. Since the change in shear between two points is equal to minus the area under the load curve between the same two points, we obtain V_{B} by writing
V_{B} – V_{A} = -(20 kN/m) (6 m) = -120 kN
V_{B} = -120 + V_{A} = -120 + 80 = -40 kN
The slope dV/dx = -w being constant between A and B, 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 V_{C} = V_{B} = -40 kN
and the shear is constant between B and C.
Bending-Moment Diagram. We note that the bending moment at each end of the beam is zero. In order to determine the maximum bending moment, we locate the section D of the beam where V = 0. We write
V_{D} – V_{A} = -wx
0 – 80 kN = -(20 kN/m)x
and, solving for x: x = 4 m
The maximum bending moment occurs at point D, where we have dM/dx = V = 0. The areas of the various portions of the shear diagram are computed and are given (in parentheses) on the diagram. Since the area of the shear diagram between two points is equal to the change in bending moment between the same two points, we write
M_{D} – M_{A} = + 160 kN \cdot m M_{D} = +160 kN \cdot m
M_{B} – M_{D} = – 40 kN \cdot m M_{B} = +120 kN \cdot m
M_{C} – M_{B} = – 120 kN \cdot m 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. It occurs at D, where |M| is largest. From App. B we find that for a W360 × 79 rolled-steel shape, S = 1270 mm^{3} about a horizontal axis. Substituting this value and |M| = |M_{D}| = 160 × 10^{3} N \cdot m into Eq. (12.3), we write
σ_{m} = \frac{|M|}{S} (12,3)
σ_{m} = \frac{|M_{D}|}{S} = \frac{160 × 10^{3} N \cdot m}{1270 × 10^{-6} m^{3}} = 126.0 × 10^{6} Pa
Maximum normal stress in the beam = 126.0 MPa
