Page 119 - Shigley's Mechanical Engineering Design

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94 Mechanical Engineering Design
about axis 3–3 and its perpendicular axis (let us call it, say, axis 4–4). Note, for this
cross section, axis 4–4 is an axis of symmetry. Table A–6 is a standard table, and for
brevity, does not directly give all the information needed to use it. The orientation of
the area principal axes and the values of I 2–2 , I 3–3 , and I 4–4 are not given because they
can be determined as follows. Since the legs are equal, the principal axes are oriented
°
±45 from axis 1–1, and I 2–2 I 1–1 . The second-area moment I 3–3 is given by
I 3−3 = A(k 3−3 ) 2 (a)
where k 3–3 is called the radius of gyration. The sum of the second-area moments for a
cross section is invariant, so I 1–1 I 2–2 I 3–3 I 4–4. Thus, I 4–4 is given by
(b)
I 4−4 = 2 I 1−1 − I 3−3
where I 2–2 I 1–1 . For example, consider a 3 3 1 4 angle. Using Table A–6 and Eqs.
4
2
4
(a) and (b), I 3–3 1.44 (0.592) 0.505 in , and I 4–4 2 (1.24) 0.505 1.98 in .
3–11 Shear Stresses for Beams in Bending
Most beams have both shear forces and bending moments present. It is only occasion-
ally that we encounter beams subjected to pure bending, that is to say, beams having
zero shear force. The flexure formula is developed on the assumption of pure bending.
This is done, however, to eliminate the complicating effects of shear force in the devel-
opment. For engineering purposes, the flexure formula is valid no matter whether a
shear force is present or not. For this reason, we shall utilize the same normal bending-
stress distribution [Eqs. (3–24) and (3–26)] when shear forces are also present.
In Fig. 3–17a we show a beam segment of constant cross section subjected to a
shear force V and a bending moment M at x. Because of external loading and V, the
shear force and bending moment change with respect to x. At x + dx the shear force
and bending moment are V + dV and M + dM, respectively. Considering forces in the
x direction only, Fig. 3–17b shows the stress distribution σ x due to the bending
moments. If dM is positive, with the bending moment increasing, the stresses on the
right face, for a given value of y, are larger in magnitude than the stresses on the left
face. If we further isolate the element by making a slice at y = y 1 (see Fig. 3–17b), the
net force in the x direction will be directed to the left with a value of
c
(dM)y

dA
I
y 1
as shown in the rotated view of Fig. 3–17c. For equilibrium, a shear force on the bottom
face, directed to the right, is required. This shear force gives rise to a shear stress τ,
where, if assumed uniform, the force is τbdx. Thus
c
(dM)y

τbdx = dA (a)
I
y 1
The term dM/I can be removed from within the integral and b dx placed on the right
side of the equation; then, from Eq. (3–3) with V = dM/dx, Eq. (a) becomes
V c
τ = yd A (3–29)
Ib
y 1
In this equation, the integral is the first moment of the area A with respect to the neu-

tral axis (see Fig. 3–17c). This integral is usually designated as Q. Thus
c

Q = yd A =¯y A (3–30)
y 1