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

ExAmpLE 9.14
Find the shear center O of the semicircular thin-walled cross section shown.
y
Plan the Solution
Shear stresses are created in the wall of the semicircular cross section in response to the
applied load P. The moment produced by these shear stresses about the center C of the
thin-walled cross section must equal the moment of the load P about center C if the section
is to bend without twisting. We will develop an expression for the differential moment
dM acting on an area dA of the wall. Then, we will integrate dM to determine the
total twisting moment produced by the shear stresses and equate that expression to the
moment created by the external load P acting at the shear center O. From this resulting
equation, the location of the shear center O can be derived.
z
O
P
t
e
r
C
SolutioN
Moment of inertia
From the sketch, observe that the distance y from the z axis to a differential area dA of the
wall can be expressed as y = r cos φ. The differential area dA can be expressed as the
product of the differential arclength ds and the thickness t; thus, dA = t ds. Furthermore,
the differential arclength can be expressed as ds = r dφ. As a result, the differential area
can be expressed in polar coordinates r and φ as dA = r t dφ. From these relationships for
y and dA, the moment of inertia of the semicircular thin-walled cross section can be derived
as follows:
z N.A.
ds
dϕ
r
y
dA
ϕ
C
y
2 2 3 2
Iz
= ∫ y dA = ∫ ( r cos φ) rtdφ = rt∫
cos φ dφ
3
1 1
= rt
⎡
φ + sinφ cosφ
⎤
⎣⎢ 2 2 ⎦⎥
3
πrt
=
2
π
0
π
0
π
0
t
First Moment of Area, Q
The value of Q can also be determined by integration in polar coordinates. From the
accompanying sketch, the value of Q for the area of the cross section above an arbitrarily
chosen angle θ is to be determined.
From the definition of Q, the first moment of area dA about the neutral axis (N.A.)
can be expressed as dQ = y dA. Substituting the previous expressions for y and dA into this
equation gives the following equation for dQ in terms of r and φ:
dQ = ydA = ( rcos φ)
rt dφ
z
N.A.
y
dA
ϕ
dϕ
θ
C
r
y
Integrating dQ between φ = 0 and φ = θ gives a general expression for Q:
t
θ
2
Q = ∫ dQ = ∫ r t cosφ dφ
0
2
= rt[sin φ]
2
= rtsinθ
θ
0
θ
0
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