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

For this FBD, the calculation of Q is straightforward:
Q = (27.5 mm)(50 mm × 6mm) = 8,250 mm 3
Since we′ve made two vertical cuts through the cross section in defining our FBD, we
might initially conclude that t = 2 × (6 mm) = 12 mm. However, a value of t = 12 mm is
incorrect in this instance. To see why, note that one of our vertical cuts here was made on
the y centroidal axis, which is a plane of symmetry for the shape. We know that the shear
stress must be zero on a plane of symmetry that is parallel to the direction of the shear
force. Accordingly, the shear stress on the cutting plane at the y axis must be zero, and a
surface with zero shear stress is not included in the term t. The value of t needed for this
FBD is t = 6 mm. The proper calculation for the shear stress at a is
3
VQ (65,000 N)(8,250 mm )
τ = = = 35.29 MPa
4
It (2,532,789 mm )(6 mm)
Vertical plane of symmetry
Shear flows
in −x direction
on this face.
b
d
(2)
30.5 mm
15.25 mm
74.5 mm
c
a
6 mm 34 mm y
37.25 mm
(1)
b
c
(3)
a
z
d
6 mm
z
y
V
x
Shear flows in +x
direction on this face.
27.5 mm
V
6 mm
105 mm
Approach 2
The second FBD that we will consider is obtained by making
one vertical cut at point a. We will examine that portion
of the cross section from point a to point b to point d.
The FBD exposed by this cut is shown in the accompanying
figure. Again, the FBD has been separated to facilitate
visualization of the shear stresses on the interior surfaces.
The directions of the shear flow and the shear stresses on the
various surfaces are indicated in the figure.
We begin the calculation at point d, where we know
with certainty that the shear stress is zero. We will calculate
the value of Q associated with point c by calculating the first
moment of area (1) (i.e., the area between points c and d)
about the neutral axis:
Q = (- 37.25 mm)(6 mm × 74.5 mm) = -16,650.75 mm
c
Note that a negative value is used for the distance from the z
centroidal axis to the centroid of area (1) because area (1)
lies below the neutral axis.
Next, we consider area (2) in order to calculate the
value of Q associated with point b. Since area (2) lies above
the neutral axis, the distance from the z axis to the centroid
of area (2) is a positive value, and we have
Q = Q + (15.25 mm)(6 mm × 30.5 mm)
b
c
3 3
=- 16,650.75 mm + 2,790.75 mm
=-13,860 mm 3
3
Finally, we consider area (3) in order to determine Q a :
Q = Q + (27.5 mm)(34 mm × 6mm)
a
b
3
=- 13,860 mm + 5,610 mm
=-8,250 mm 3
Note that Q calculated from the FBD of Approach 2 has the same magnitude as Q determined
with the FBD of Approach 1; however, a negative sign appears in the second
calculation. To calculate the magnitude of the shear stress at a, we need only the magnitude
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