Prob. 2.2-4. Consider the free-hanging rod shown in Fig. P2.2-4. The rod has the shape of a conical frustum, with radius Ro at its top and radius R₁ at its bottom, and it is made of material with mass density p. The length of the rod is L. Determine an expression for the normal stress, o(x), at an arbitrary cross section x (0 ≤ x ≤ L), where x is measured downward from the top of the rod. SO 8 L X RL Ro
Prob. 2.2-4. Consider the free-hanging rod shown in Fig. P2.2-4. The rod has the shape of a conical frustum, with radius Ro at its top and radius R₁ at its bottom, and it is made of material with mass density p. The length of the rod is L. Determine an expression for the normal stress, o(x), at an arbitrary cross section x (0 ≤ x ≤ L), where x is measured downward from the top of the rod. SO 8 L X RL Ro
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Prob. 2.2-4. Consider the free-hanging rod shown in
Fig. P2.2-4. The rod has the shape of a conical frustum, with
radius Ro at its top and radius R₁ at its bottom, and it is made
of material with mass density p. The length of the rod is L.
Determine an expression for the normal stress, o(x), at an
arbitrary cross section x (0 ≤ x ≤ L), where x is measured
downward from the top of the rod.
L
X
Ro
RL
P2.2-4 and P2.3-6
![Абкиби 1
Kal
παίρνοντας το
IF
9 |
Ro-RL
W₂ = P.V₂.g
ут
W₂
Έστω Rixi η ακτίνα τυχαίας διατομή του κόνου. Τότε
από ομοιότητα τριγώνων
προφανώς,
L
V = =
1N (R² + R² + RORL). L.
3
xohlize
=
X
буу
=
RL
Ro-R(x)
X
TJ
F
A(x)
Ro
11
T
L.
L
D.REX)
X
Rex) = Ro
Pg. 10. (R²A + RETIRL. R²). (L-X).
2
2 200 Exubiacos V₂ = 1 n (R²x) + R² + R(x)-R₂) (L-X).
Σfy = 0 => F-W₂=0=1₁√| F=W₁₂
+
(RO-R+)
L
= c₁ cf IN
C
X
pg — A² (R²x1 + R²² + RexiR₂). (L-X)
X.R²(x1
(Ro-RL)
L
Ro
Ro-R₂
RL
xRL
Oyy = = = 10.9
+ Pg [ ( "Ro+ Po=B² x ) ² + R₂² + (Ro. 20-Dex) R₂] (L-X)
-p.g
2
[R+ =]]
Rot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F733c2a11-3270-4acb-b93b-1038b7766f6d%2F1d9303cc-85d4-4d06-9676-6eb56241cd76%2Fpf6jkwa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Абкиби 1
Kal
παίρνοντας το
IF
9 |
Ro-RL
W₂ = P.V₂.g
ут
W₂
Έστω Rixi η ακτίνα τυχαίας διατομή του κόνου. Τότε
από ομοιότητα τριγώνων
προφανώς,
L
V = =
1N (R² + R² + RORL). L.
3
xohlize
=
X
буу
=
RL
Ro-R(x)
X
TJ
F
A(x)
Ro
11
T
L.
L
D.REX)
X
Rex) = Ro
Pg. 10. (R²A + RETIRL. R²). (L-X).
2
2 200 Exubiacos V₂ = 1 n (R²x) + R² + R(x)-R₂) (L-X).
Σfy = 0 => F-W₂=0=1₁√| F=W₁₂
+
(RO-R+)
L
= c₁ cf IN
C
X
pg — A² (R²x1 + R²² + RexiR₂). (L-X)
X.R²(x1
(Ro-RL)
L
Ro
Ro-R₂
RL
xRL
Oyy = = = 10.9
+ Pg [ ( "Ro+ Po=B² x ) ² + R₂² + (Ro. 20-Dex) R₂] (L-X)
-p.g
2
[R+ =]]
Rot
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