Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

F
(1)
dx
357
SHEAR STRESS ANd SHEAR
FLOw IN THIN-wALLEd
MEMbERS
B′
y
(1)
B′
t dF
B
dA
(b) FBD of flange element (1)
F + dF
B
M
V
z
V + dV
M + dM
x
t
y
τ zx
B
τ xz
t
y
τ yx
B
τ xy
dx
(a) Thin-walled beam section
FIGURE 9.16 Shear stresses in a thin-walled wide-flange beam.
z
x
(c) Horizontal shear
stress at B in
flange element (1)
z
x
(d) Vertical shear
stress at B in
flange element (1)
member. Next, consider the FBD of a portion of the upper flange, element (1), shown in
Figure 9.16b. On the back side of the beam segment, the bending moment M creates compressive
normal stresses that act on the -x face of flange element (1). The resultant of these
normal stresses is the horizontal force F. Similarly, the bending moment M + dM acting on
the front side of the beam segment produces compressive normal stresses that act on the
+x face of flange element (1), and the resultant of these stresses is the horizontal force F + dF.
Since the resultant force acting on the front side of element (1) is greater than the resultant
force acting on the back side, an additional force dF must act on element (1) to satisfy
equilibrium. This force can act only on the exposed surface BB′ (because all other surfaces
are free of stress). By a derivation similar to that used in obtaining Equation (9.11), the
force dF can be expressed in terms of differentials as
VQ
dF = dx
(9.15)
I
where Q is the first moment of the cross-sectional area of element (1) about the neutral axis
of the beam section. The area of surface BB′ is dA = t dx, and thus, the average shear stress
acting on the longitudinal section BB′ is
z
dF VQ
τ = = (9.16)
dA It z
Note that τ in this instance represents the average value of the shear stress acting on a z
plane [i.e., the vertical surface BB′ of element (1)] in the horizontal direction x—in other
words, τ zx . Since the flange is thin, the average shear stress τ zx will not vary much over its
thickness t. Consequently, τ zx can be assumed to be constant. Because shear stresses acting
on perpendicular planes must be equal (see Section 1.6), the shear stress τ xz acting on an x
face in the z direction must equal τ zx at any point on the flange (Figure 9.16c). Accordingly,
the horizontal shear stress τ xz at any point on a transverse section of the flange can be
obtained from Equation (9.16).