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

τ
y
dA
ϕ
dϕ
y
Shear Stress
The variation of the shear stress τ can now be expressed in terms of the angle φ:
2
VQ Vrt ( sin φ)
2V
τ = = =
3 sin φ
It ⎛ πrt
πrt
t
⎝ ⎜ ⎞
2 ⎠
⎟
z N.A.
r
C
Moments about C
The resultant force dF acting on the element of area dA is expressed as dF = τdA =
τ (r t dφ), or
t
dF
2rtV
= sinφ
dφ
=
πrt
2V
sin φdφ
π
z
P
O
y
dF = τ ( r t dϕ)
ϕ
dϕ
r
C
e
The moment of dF about point C is
dM
C
= rdF =
2rV
sinφ
dφ
π
Integrating this expression between φ = 0 and φ = π gives the moment produced by the
shear stresses:
M
C
π 2rV
= ∫ dMC
= ∫ sinφdφ
=
0 π
4rV
π
t
To satisfy moment equilibrium, the moment M C of the shear stress τ about the center C of
the thin-walled cross section must equal the moment of the load P about that same point:
M
C =
Pe
The resultant of the shear stress is the shear force V, which must equal the applied load P
to satisfy vertical equilibrium. It follows that the distance e to the shear center is
MC
MC
4r
e = = = ≅ 1.27r
Ans.
P V π
This result shows that the shear center O is located outside of the semicircular cross section.
Sections consisting of Two Intersecting Thin Rectangles
Next, we will consider thin-walled open sections made up of two intersecting rectangles.
Consider an equal-leg angle section, such as that shown in Figure 9.30. When a vertical
V
q
F 1
F 1
45°
45°
O
F 2
F 2
(a) Shear flow
(b) Resultant shear forces
FIGURE 9.30 Shear center of equal-leg angle shapes.
(c) Shear center
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