SECTION 12.3 Centroids of
Centroids of Composite
Gomposite Areas
Areas
solved bv using
Section 12.3 are tobe
for
problems
The
theformulas for
centroid Cof atrapezoid
the distance F to the
Determine
dividing the
12.3-1
Problem
8, Appendix D) by
composite areas.
altitude h (see
having baSes aand b and
trapezoid into two
Case
triangles.
Solution 12.3-1 Centroid of a
trapezoid
ah
bh
2
3
al
2
bh
A=XA,=+
la + b)
2
2
e.-SA-9-au
e. h(2a + b)
j= A 3(a + b)
(see
Problem 12.3-2 One quarter of a square of side ais removed
figure).
a
remaining area?
What are the coordinates and ù of the centroid Cof the
Fiti4
PROBS. 12.3-2 and 12.5-2
Solution 12.3-2 Centroid of a composite area
3a
u
4
A, =
a
A
A=
3a
A, = 4
e.- E54--5)
A .12
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+ b)
3
4
CHAPTER 12 Review of Centroids and Moments of Inertia
Problem 12.3-3 Calculate the distance vto the centroid Cof the channel
Sehon shown in the feure ifa = 6in., b = lin., and c= 2in.
B
PROBS. 12.3-3, 12.3-4, and 12.5-3
Solution 12.3-3 Centroid of a channel section
A, = be =2 in.
b
A, =ab =6 in.?
A=A,= 2A, +A, = 10 in.?
Q,- y, A, = 2y;A, +VzA, = 11.0 in.
2
a=6in
h= l in.
y,=btc/2 = 2in.
V, ==
2 0.5 in.
= 1.10 in.
A
c=2 in.
Problem 12.3-4 What must be the relationship between the
dimensions a,
b, and c of the channel section shown
Cwill lie on line BB?
the figure in order that the centroid
Solution 12.3-4 Dimensions of channel section
b
B
b
2
A, = b
V, =b+c/2
b
A, = ab
V, =b/2
(A)
A=
A,= 2A,t A, = b(2: t u)
,=Sy,A, = 2y,A, +V,A, = b/2(4bc + 2¢' +ab)
4bu +2+ ab
y=
2(2c + a)
Set y = b and solve: 2¢' = ab
A
Problem 12.3-5 The cross section ofa beam constructed of aVW 24 X
162 wide-flange section with an 8 in. X 3/4 in. cover plate welded to the
top flange is shown in the figure.
Determine the distance y from the base of the beam to the centroid C
of the cross-sectional area.
PROBS, 12.3-5 and 12.5-5
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-Plate 8in. X in
.W24 X 162
Problem 12.3-7 Determine the coordinales Tand v of the centroid Cof the
L-shaped arca shown in the figure.
0.5 in.
6 in.
C
PROBS. 12.3-7, 12.4-7, 12.5-7 and 12.7-7
0.5 in.
4 in.
Solution 12.3-7 Centroid of Lshaped area
A, = (3.5)(0.5) = 1.75 in.2
y, =0.25 in. X= 2.25 in.
A, = (6)(0.5) = 3.0 in,2
y = 3.0 in. X,=0.25 in.
A;
A=
A, =A,+ A, =4.75 in.?
e=,A, = A,t+ IyA, = 4.688 in.
6 in.
A
4 in.
= 0.99 in.
e.A
A, = yA, + A, =9.438 in.
= 1.99 in.
Problem 12.3-8 Determine the coordinatesIand vof the centroid C of the
area shown in the figure.
170mm
280 mm
50 mm
50 mm
150 mm
80
80
mm
mm
80
80
IMm
300 imm