Civil Engineering Reference
In-Depth Information
D
B
h
O
C
A
Figure 2.23
Tetrahedron ABCD.
15
.
∂
∂
γ
h
72 3
∂
∂
v
h
∂
∂
s
h
723
∂
∂
v
h
sv
s
h
∂
∂
72 3
∂
∂
v
h
∂
∂
s
h
15
.
15
.
05
.
=
s
−
v
=
s
−
15
.
=
s
−
15
.
v
3
3
25
.
s
s
s
∂
∂
γ
h
72 3
∂
∂
v
h
∂
∂
s
h
∂
∂
v
h
∂
∂
s
h
1
3
1
3
=
0
s
−
15
.
v
=
0
or s
−
1 5
. v
=
0
As
=
15
.
Ah
(
6
h
)
2.
s
or
2
2
2
2
2
2
2
2
9
hsAB
== +++
BC
CA
OA
+
OB
+
OC
+
3
h
2
2
2
2
2
2
2
6
h
=
B CC
+
+ AAOAOBOC
+
+
+
However,
1
3
2
2
2
2
2
2
OA
+
OB
+
OC
=
(
AB
+
BC
+
CA
)
Hence,
4
3
2
9
2
2
2
2
2
2
2
2
6
h
=
(
B CCAorh
+
+
)
= ++
(
AB
BC
CA
)
(2.4)
1
3
3
Ah
v
72 3
v
83
9
A
43
A
γ=
72 3
=
=
83
=
=
++
=α
32
/
32
/
3
2
2
2
2
s
h
h
AB
BC
CA
9
2
19
32
h
2
+
h
2
+
3
h
2
Thus, γ attains the maximum value of α at the centroid O of triangle ABC with a height
h given by Equation 2.4. Exercise: γ attains the highest value at the centroid of triangle
ABC.
