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

What is the significance of Equation (9.1)? if the internal bending moment in a beam
is not constant (i.e., ∆M ≠ 0), then an internal horizontal shear force F H must exist at y =
y 1 in order to satisfy equilibrium. Furthermore, note that the term Q pertains expressly
to area A′. (See Figure 9.5b.) Since the value of Q changes with area A′, so does F H . In
other words, at every value of y possible within a cross section, the internal shear force
F H required for equilibrium is unique.
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THE SHEAR STRESS FORMuLA
Before continuing, it may be helpful to apply Equation (9.1) to the problem discussed in
Section 9.2. In that problem, the internal bending moments on the right and left sides of
the beam segment (which had a length Dx = 150 mm) were M B = 1.350 kN · m and M C =
2.025 kN · m, respectively. From these two moments, DM = 2.025 kN · m - 1.350 kN · m =
0.675 kN · m = 675 kN · mm. The moment of inertia was given as I z = 33,750,000 mm 4 .
The area A′ pertinent to this problem is simply the area of member (1), the 50 mm by
120 mm board at the bottom of the cross section (Figure 9.2, repeated here for convenience).
The first moment of area, Q, is computed from ∫y dA′. Let the width of member (1) be denoted
by b. Since this width is constant, the differential area dA′ can be conveniently expressed as
dA′ = b dy. In this instance, area A′ starts at y = -25 mm and extends away from the neutral axis
in a downward direction, to an outermost boundary at y = -75 mm. With b = 120 mm, we have
-75
b
Q bydy y
2
- 75
= ∫ = [ ] = 300,000 mm
-25
- 25
2
and, from Equation (9.1), the horizontal shear force F H required to keep member (1) in
equilibrium is
3
MQ (675 kN mm)(300,000 mm )
FH
= D ⋅
=
= 6kN
4
I
33,750,000 mm
z
This result agrees with the horizontal force determined in Section 9.2.
3
z
y
120 mm
(2)
(1)
100 mm
50 mm
FIGURE 9.2 (repeated) Beam
cross-sectional dimensions.
Shear Stress in a Beam
Equation (9.1) can be extended to define the shear stress produced in a beam subjected to
nonuniform bending. The surface upon which F H acts has a length Dx. Depending upon the
shape of the beam cross section, the width of area A′ may vary, so the width of area A′ at
y = y 1 will be denoted by the variable t. (See Figure 9.5b.) Since stress is defined as force
divided by area, the average horizontal shear stress acting on horizontal section b–e can be
derived by dividing F H in Equation (9.1) by the area of the surface upon which this force
acts, which is t Dx:
FH
MQ
τ H ,avg = = D t D x t Dx I
z
M Q
= D Dx
It
z
(e)
Implicit in equation (e) is the assumption that the shear stress is constant across the width
of the cross section at any y position. That is, at any specific y position, the shear stress is
constant for any z location. This derivation also assumes that the shear stresses τ are parallel
to the vertical sides of the cross section (i.e., the y axis).
In the limit as Dx → 0, DM/Dx can be expressed in terms of differentials as dM/dx, so
Equation (e) can be enhanced to give the horizontal shear stress acting at any location x
along the beam’s span:
dM Q
τ H = (f)
dx It
z