Variable Diameter Tower Under Distributed Axial and Wind Loading
Problem Statement & Figure
A vertical tapered tower of total height 𝐿 = 12 m has a circular cross section that
decreases linearly with height:
𝑑(𝑥) = 𝑑0 − 𝑘𝑥, 𝑑0 = 4.0 m, 𝑘 = 0.25 m per m.
The material has constant Youngs modulus:
𝐸 = 210 × 109 Pa.
A distributed vertical load acts along the tower and increases linearly with height:
𝑥
𝑝(𝑥) = 𝑝0 , 𝑝0 = 100 × 103 N per m.
𝐿
The tower is also subjected to a horizontal wind point load 𝐹 = 6 × 103 Nat the top. The
base is fixed and the top is free.
Determine:
𝐿
1. Axial displacement 𝑢(𝑥) at 𝑥 = 2 and at 𝑥 = 𝐿.
2. Maximum axial stress anywhere along the tower.
3. Lateral displacement 𝑣(𝑥) at 𝑥 = 𝐿/2 and at 𝑥 = 𝐿.
4. Produce and include two plots: axial displacement 𝑢(𝑥) for 0 ≤ 𝑥 ≤ 𝐿and lateral
𝐿
deflection 𝑣(𝑥) for 0 ≤ 𝑥 ≤ 𝐿. Mark the values at 2 and 𝐿 on the plots.
Part A — Axial deformation under distributed load
Internal axial force
Equilibrium of a vertical element gives
𝑑𝑁(𝑥)
= −𝑝(𝑥).
𝑑𝑥
Integrating from 𝑥 to 𝐿,
𝐿
𝐿
𝑁(𝑥) = ∫ 𝑝(𝑠) 𝑑𝑠 = ∫ 𝑝0
𝑥
𝑥
𝑠
𝑝0 2
𝑑𝑠 =
(𝐿 − 𝑥 2 ).
𝐿
2𝐿
Strain displacement relation
𝜀𝑥𝑥 =
𝑑𝑢
𝑁(𝑥)
𝜋[𝑑(𝑥)]2 𝜋(𝑑0 − 𝑘𝑥)2
=
, 𝐴(𝑥) =
=
.
𝑑𝑥 𝐸 𝐴(𝑥)
4
4
Integration approach for displacement
𝑥𝑐
𝑢(𝑥𝑐 ) − 𝑢(0) = ∫
0
𝑁(𝑥)
𝑑𝑥, 𝑢(0) = 0,
𝐸 𝐴(𝑥)
hence
𝑥𝑐
𝑥𝑐
𝑁(𝑥)
2𝑝0
𝐿2 − 𝑥 2
∫
𝑢(𝑥𝑐 ) = ∫
𝑑𝑥 =
𝑑𝑥 .
𝜋𝐸𝐿
(𝑑0 − 𝑘𝑥)2
0 𝐸 𝐴(𝑥)
0
Values of interest:
𝐿/2
𝐿
2𝑝0
∫
𝑢( ) =
2
𝜋𝐸𝐿
0
𝐿
𝐿2 − 𝑥 2
2𝑝0
𝐿2 − 𝑥 2
∫
𝑑𝑥
,
𝑢(𝐿)
=
𝑑𝑥 .
(𝑑0 − 𝑘𝑥)2
𝜋𝐸𝐿
(𝑑0 − 𝑘𝑥)2
0
Axial stress and maximum value
𝜎(𝑥) =
𝑁(𝑥) 2𝑝0 (𝐿2 − 𝑥 2 )
=
.
𝐴(𝑥) 𝜋𝐿(𝑑0 − 𝑘𝑥)2
The maximum occurs at the base 𝑥 = 0:
𝜎max =
2𝑝0 𝐿
.
𝜋𝑑02
Part b — Lateral deflection under wind point load at the top
Moment of inertia and moment along the height:
𝐼(𝑥) =
𝜋[𝑑(𝑥)]4
, 𝑀(𝑥) = 𝐹(𝐿 − 𝑥).
64
Small deflection relations:
𝑑𝜃
𝑀(𝑥) 𝑑𝑣
=
,
= 𝜃(𝑥).
𝑑𝑥 𝐸 𝐼(𝑥) 𝑑𝑥
Form using differences from base values
𝑥
𝑥
𝑀(𝑠)
𝜃(𝑥) − 𝜃(0) = ∫
𝑑𝑠 , 𝑣(𝑥) − 𝑣(0) = ∫ 𝜃(𝑠) 𝑑𝑠 .
0 𝐸 𝐼(𝑠)
0
For the cantilever base 𝑥 = 0, 𝜃(0) = 0 and 𝑣(0) = 0, so
𝑥
𝑥
𝑀(𝑠)
𝑑𝑠 , 𝑣(𝑥) = ∫ 𝜃(𝑠) 𝑑𝑠 .
0 𝐸 𝐼(𝑠)
0
𝜃(𝑥) = ∫
General integral form with constants
𝜃(𝑥) = ∫
𝑀(𝑥)
𝑑𝑥 + 𝐶1 , 𝑣(𝑥) = ∫ 𝜃(𝑥) 𝑑𝑥 + 𝐶2 , 𝐶1 = 0, 𝐶2 = 0.
𝐸 𝐼(𝑥)
Tip and mid height deflections
𝐿
𝑠
𝐿/2
𝑠
𝐹(𝐿 − 𝑟)
𝐿
𝐹(𝐿 − 𝑟)
𝑑𝑟 𝑑𝑠 , 𝑣( ) = ∫
∫
𝑑𝑟 𝑑𝑠 .
2
0 𝐸 𝐼(𝑟)
0
0 𝐸 𝐼(𝑟)
𝑣(𝐿) = ∫ ∫
0
Part c — Visualization and plots
Produce the following:
1. Plot of axial displacement 𝑢(𝑥)for 0 ≤ 𝑥 ≤ 𝐿 under 𝑝(𝑥) = 𝑝0 𝑥/𝐿. Mark values at
𝑥 = 𝐿/2 and 𝑥 = 𝐿.
2. Plot of lateral deflection 𝑣(𝑥)for 0 ≤ 𝑥 ≤ 𝐿 under the tip load 𝐹. Mark values at 𝑥 =
𝐿/2 and 𝑥 = 𝐿.
Results
Matlab Results:
Quantity
Symbol / Expression Result
Axial displacement at mid-height u(L/2)
0.001956 mm
Axial displacement at tip
u(L)
0.004592 mm
Maximum axial stress
σmax
47.746 kPa
Lateral deflection at mid-height v(L/2)
0.000729 mm
Lateral deflection at tip
0.005238 mm
Figures:
v(L)
Figure 1: Axial Displacement
Figure 2: Lateral Deflection