Page 100 - Shigley's Mechanical Engineering Design

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Load and Stress Analysis 75
an equal but opposite torsional moment. The center of rotation relative to the bolts lies at
the centroid of the bolt cross-sectional areas. Thus if the bolt areas are equal: the center

2
2
of rotation is at the center of the four bolts, a distance of (4/2) + (5/2) = 3.202 in
from each bolt; the bolt forces are equal (R E = R F = R H = R I = R), and each bolt force
is perpendicular to the line from the bolt to the center of rotation. This gives a net torque
from the four bolts of 4R(3.202) = 720. Thus, R E = R F = R H = R I = 56.22 lbf.

3–2 Shear Force and Bending Moments in Beams

Figure 3–2a shows a beam supported by reactions R 1 and R 2 and loaded by the con-
centrated forces F 1 , F 2 , and F 3 . If the beam is cut at some section located at x = x 1 and
the left-hand portion is removed as a free body, an internal shear force V and bending
moment M must act on the cut surface to ensure equilibrium (see Fig. 3–2b). The shear
force is obtained by summing the forces on the isolated section. The bending moment is
the sum of the moments of the forces to the left of the section taken about an axis through
the isolated section. The sign conventions used for bending moment and shear force in this
book are shown in Fig. 3–3. Shear force and bending moment are related by the equation

dM
V = (3–3)
dx
Sometimes the bending is caused by a distributed load q(x), as shown in Fig. 3–4;
q(x) is called the load intensity with units of force per unit length and is positive in the

Figure 3–2 y y

Free-body diagram of simply- F 1 F 2 F 3 F 1
supported beam with V and M V
x x
shown in positive directions. M
x 1 x 1
R 1 R 2 R 1
(a) (b)

Figure 3–3

Sign conventions for bending
Positive bending Negative bending
and shear.

Positive shear Negative shear

Figure 3–4 y q(x)
Distributed load on beam.
x