Question 12.S-P.3: Draw the shear and bending-moment diagrams for the beam and ...
Draw the shear and bending-moment diagrams for the beam and loading A shown.
Reactions. Considering the entire beam as a free body, we write
+\curvearrowleft \sum{M_{A}} = 0:
D(24 ft) – (20 kips) (6 ft) – (12 kips) (14 ft) – (12 kips) (28 ft) = 0
D = +26 kips D = 26 kips ↑
+↑\sum{F_{y}} = 0: A_{y} – 20 kips – 12 kips + 26 kips – 12 kips = 0A_{y} = +18 kips A_{y} = 18 kips ↑
\underrightarrow{+} \sum{F_{x}} = 0: A_{x} =0 A_{x} = 0Shear Diagram. Since dV/dx = -ω, we find that between concentrated loads and reactions the slope of the shear diagram is zero (i.e., the shear is constant). The shear at any point is determined by dividing the beam into two parts and considering either part as a free body. For example, using the portion of beam to the left of section 1, we obtain the shear between B and C:
+↑\sum{F_{y}} = 0: +18 kips – 20 kips – V = 0 V = -2 kipsWe also find that the shear is +12 kips just to the right of D and zero at end E. Since the slope dV/dx = -ω is constant between D and E, the shear diagram between these two points is a straight line.
Bending-Moment Diagram. We recall that the area under the shear curve between two points is equal to the change in bending moment between the same two points. For convenience, the area of each portion of the shear diagram is computed and is indicated in parentheses on the diagram. Since the bending moment M_{A} at the left end is known to be zero, we write
M_{B} – M_{A} = +108 M_{B} = +108 kip \cdot ft
M_{C} – M_{B} = -16 M_{C} = +92 kip \cdot ft
M_{D} – M_{C} = -140 M_{D} = -48 kip \cdot ft
M_{E} – M_{D} = +48 M_{E} = 0
Since M_{E} is known to be zero, a check of the computations is obtained.
Between the concentrated loads and reactions, the shear is constant; thus, the slope dM/dx is constant, and the bending-moment diagram is drawn by connecting the known points with straight lines. Between D and E where the shear diagram is an oblique straight line, the bending-moment diagram is a parabola.
From the V and M diagrams we note that V_{max} = 18 kips and M_{max} = 108 kip \cdot ft.
