SAIF ALDIN ALI MADIN سيف الدين علي ماضي S96aif@gmail.com Experiment Name:- Deflection of curved beam 2. Introduction The deflection of a beam or bars must be often be limited in order to provide integrity and stability of structure or machine. Plus, code restrictions often require these members not vibrate or deflect severely in order to safely support their intended loading. This experiment helps us to show some kind of deflection and how to calculate the deflection value by using Castigliano’s Theorem and make a comparison between result of the experiment and the theory.

1. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 1 | P a g e
[Strength of Material Laboratory II]
University of Baghdad
Name: - Saif Al-din Ali -B-

2. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 2 | P a g e
TABLE OF CONTENTS
Name..................................................................................I
INTRODUCTION.............................................................II
Objective.........................................................................III
THEORY...........................................................................V
APPARATUS...................................................................VI
Calculations and results................................................VII
DISCUSSION ...............................................................VIII

3. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 3 | P a g e
1. Experiment Name:- Deflection of curved beam
2. Introduction
The deflection of a beam or bars must be often be limited in order to provide
integrity and stability of structure or machine. Plus, code restrictions often require
these members not vibrate or deflect severely in order to safely support their
intended loading.
This experiment helps us to show some kind of deflection and how to calculate the
deflection value by using Castigliano’s Theorem and make a comparison between
result of the experiment and the theory.
3. OBJECTIVE
The object of this experiment was to determine the deflections in the
horizontal and vertical directions under loading of a guardant, half Circle
and full Circle beam by means of experiment and compare the
experimental values of deflection to calculated, theoretical values
4. Theory
Castiglione’s theorem can be employed to determine the deflection
incurred by a force of loading in a curved beam. Castiglione's theorem
states that the component in a given direction of the deflection caused
by an external force on an clastic body is equivalent to the partial
derivative of the work of deformation with respect to the component of
the force in the given direction. The work of deformation in this case is a
moment "induced by a loading force on the beam. The general
expression of Castiglione’s theorem is as follows:
δv = ∫ (
𝑴
𝑬𝑰
) (
𝒅𝑴
𝒅𝑾
) 𝒅𝒔
𝒔
𝟎
δH = ∫ (
𝑴
𝑬𝑰
) (
𝒅𝑴
𝒅𝑷
) 𝒅𝒔
𝒔
𝟎
The work of deformation, or the moment, can be expressed as the
product of the loading force, W, the radius from the center of curvature
of the beam R and the sine of the angle of curvature. The moment can
be expressed by the following equation
M=WR sinθ + PR (1-cosθ)

4. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 4 | P a g e
The integrating factor ds of the general Castigliano equation can be
expressed as follows:
ds = R dθ
The partial derivative of the work of deformation with respect to the
component of the force is expressed as a function of the radius of the
beam and angle of the deflected beam:
For the vertical deflection, the partial derivative is written as
(dM/dW)v = Rsinθ
and for the horizontal deflection of a curved beam, the partial derivative
is written as
(dM/dW)H = R (1-cosθ)
Sample test will be made of steel, young modules ( E = 30 * 10 "
ib / in2 ), cross section dimensions ( B=1 in . h = 0.125 in), and
radius of the curvature (R=6) in to the following cases:-
1-Quadrant beam.
2- Half Circle Beam.
3- Full Circle Beam.
In all above cases a concentrated loads are applied and calculate the
horizontal and vertical deflections values (theory) in each case according to
the following general relationships
δv = C1*W
𝑹 𝟑
𝑬𝑰
δH = C2*W
𝑹 𝟑
𝑬𝑰
Where
W = vertical applied load (Ib)
δv , δH vertical and horizontal deflection (in)
R = radius of the curvature (in)
E = young's modulus of material of the beam. N/mm2
I = second moment of area of the cross section (moment of inertia) of the
beam, about the neutral axis mm4
C1, C2: constants

5. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 5 | P a g e
The Cases used in testing:
1- Quadrant beam
δv=
𝝅
𝟒
× 𝑭 ×
𝑹 𝟑
𝑬𝑰
δH=
𝟏
𝟐
× 𝑭 ×
𝑹 𝟑
𝑬𝑰
2- Half circle beam.
δv=
𝟑𝝅
𝟐
× 𝑭 ×
𝑹 𝟑
𝑬𝑰
δH=𝟐 × 𝑭 ×
𝑹 𝟑
𝑬𝑰
3. Full circle Beam.
δv=(
𝝅
𝟒
−
𝟐
𝝅
) × 𝑭 ×
𝑹 𝟑
𝑬𝑰
δH=(
𝟐
𝝅
−
𝟏
𝟐
) × 𝑭 ×
𝑹 𝟑
𝑬𝑰

6. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 6 | P a g e
5. APPARATUS
1) Position the empty load hangers on the
beam of the testing apparatus
2) Zero the dial indicators.
3) Place a 0.25 lb weight on the hanger
4) Record both the horizontal and vertical
deflection of the beam.
5) Add 0.25 Ib additional weight to the
weight hanger. Again, record the horizontal
and vertical deflection of the beam
6) Repeat Step 5 until a total weight applied
to the weight hanger is equal to 1.25 Ibs.

7. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 7 | P a g e
6. Calculations and results
Theoretical Analysis:-
I =
𝑏ℎ3
12
=
1× (0.125)3
12
 I=1.627 × 10−4
𝑖𝑛4
For quadrant beam
1. δv =
𝝅
𝟒
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝝅
𝟒
× 𝟎. 𝟐𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv = 8.68× 𝟏𝟎−𝟑
in =0.2 mm
2. δv =
𝝅
𝟒
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝝅
𝟒
× 𝟎. 𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv = 0.0173 in =0.43 mm
3. δv =
𝝅
𝟒
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝝅
𝟒
× 𝟎. 𝟕𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv = 0.026 in =0.66 mm
4. δv =
𝝅
𝟒
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝝅
𝟒
× 𝟏 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv = 0.034 in = 0.86 mm
5. δv =
𝝅
𝟒
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝝅
𝟒
× 𝟏. 𝟐𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv = 0.043 in = 1.09 mm
1. δH =
𝟏
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟏
𝟐
× 𝟎. 𝟐𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH=5.53× 𝟏𝟎−𝟑
𝒊𝒏=0.14 mm
2. δH =
𝟏
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟏
𝟐
× 𝟎. 𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH=0.011 𝒊𝒏=0.279 mm
3. δH =
𝟏
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟏
𝟐
× 𝟎. 𝟕𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH=0.0165 𝒊𝒏=0.419 mm
4. δH =
𝟏
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟏
𝟐
× 𝟏 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH=0.022 𝒊𝒏 =0.5588 mm
5. δH =
𝟏
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟏
𝟐
× 𝟏. 𝟐𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH=0.0276 𝒊𝒏 = 0.701 mm

8. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 8 | P a g e
For Half circle beam
1. δv =
𝟑𝝅
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟑𝝅
𝟐
× 𝟎. 𝟐𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv =0.052 in = 1.3208 mm
2. δv =
𝟑𝝅
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟑𝝅
𝟐
× 𝟎. 𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv =0.104 in = 2.6416 mm
3. δv =
𝟑𝝅
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟑𝝅
𝟐
× 𝟎. 𝟕𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv =0.156 in =3.9624 mm
4. δv =
𝟑𝝅
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟑𝝅
𝟐
× 𝟏 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv =0.2 in = 5.08 mm
5. δv =
𝟑𝝅
𝟐
× 𝒘 ×
𝑹 𝟑
𝑬𝑰
=
𝟑𝝅
𝟐
× 𝟏. 𝟐𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δv =0.26 in = 6.604 mm
1. δH=𝟐 × 𝒘 ×
𝑹 𝟑
𝑬𝑰
= 𝟐 × 𝟎. 𝟐𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH =0.022 in =0.558 mm
2. δH=𝟐 × 𝒘 ×
𝑹 𝟑
𝑬𝑰
= 𝟐 × 𝟎. 𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH = 0.044 in =1.117 mm
3. δH=𝟐 × 𝒘 ×
𝑹 𝟑
𝑬𝑰
= 𝟐 × 𝟎. 𝟕𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH = 0.066 in =1.676 mm
4. δH=𝟐 × 𝒘 ×
𝑹 𝟑
𝑬𝑰
= 𝟐 × 𝟏 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH = 0.0885 in =2.248 mm
5. δH=𝟐 × 𝒘 ×
𝑹 𝟑
𝑬𝑰
= 𝟐 × 𝟏. 𝟐𝟓 ×
𝟔 𝟑
𝟑𝟎 ×𝟏𝟎 𝟔 ×𝟏.𝟔𝟐𝟕 ×𝟏𝟎−𝟒
 δH = 0.11 in = 2.794 mm
Results:-

9. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 9 | P a g e
Slope experimental=1.2
Slope theoretical =1.7
Error%=
𝑡𝑎𝑚 𝜃𝑡ℎ−tan 𝜃 𝑒𝑥𝑝.
tan 𝜃 𝑡ℎ
∗ 100%
Error%=
1.7−1.2
1.7
∗ 100%= 29.41%
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
W(ib)
δH(mm)
w&δH
Linear (Experimental) Linear (Theoretical)
Quadrant beam
W(ib) Experimental Theoretical
δH(mm) δv(mm) δH(mm) δv(mm)
0.25 0.1 0.19 0.14 0.2
0.50 0.3 0.51 0.279 0.43
0.75 0.425 0.73 0.419 0.66
1.00 0.66 1.04 0.558 0.86
1.25 0.83 1.31 0.7 1.09

10. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 10 | P a g e
Slope experimental=1.2
Slope theoretical =1.7
Error%=
𝑡𝑎𝑚 𝜃𝑡ℎ−tan 𝜃 𝑒𝑥𝑝.
tan 𝜃 𝑡ℎ
∗ 100%
Error% =
1.05−1
1.05
∗ 100%= 4.7%
Half circle beam
W(ib)
Experimental Theoretical
δH(mm) δv(mm) δH(mm) δv(mm)
0.25 0.22 0.84 0.55 1.32
0.50 0.85 2.22 1.11 2.64
0.75 1.36 3.73 1.67 3.96
1.00 1.74 4.5 2.24 5.08
1.25 2.24 5.59 2.79 6.6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.2 0.4 0.6 0.8 1 1.2 1.4
W(ib)
δv(mm)
W&δv
Linear (Experimental) Linear (Theoretical)

11. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 11 | P a g e
Slope experimental=0.48
Slope theoretical =0.5
Error%=
𝑡𝑎𝑚 𝜃𝑡ℎ−tan 𝜃 𝑒𝑥𝑝.
tan 𝜃 𝑡ℎ
∗ 100%
Error% =
0.5−0.48
0.5
∗ 100%= 4%
Slope experimental=0.15
Slope theoretical =0.17
Error%=
𝑡𝑎𝑚 𝜃𝑡ℎ−tan 𝜃 𝑒𝑥𝑝.
tan 𝜃 𝑡ℎ
∗ 100%
Error% =
0.17−0.15
0.17
∗ 100%= 11
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.5 1 1.5 2 2.5 3
W(ib)
δH(mm)
W&δH
Linear (Experimental) Linear (Theoretical)
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 1 2 3 4 5 6 7
W(ib)
δv(mm)
w&δv
Linear (Experimental) Linear (Theoretical)

12. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 12 | P a g e
7. DISCUSSION
1-Drive the horizontal and vertical deflections in each case
As the value of the horizontal displacement is
required to be determined at the free end,
and as there is no load acting in this direction, it
will be necessary to assume an imaginary
load H, as shown by the dashed line. Later in the
calculation, it will be necessary to note
that H=O.
Consider an element dx at any given angle (J
from the base. At this point the bending
moment is
M= -WRcos (J- HR(1-sin (J)
NB The sign convention for bending moment is not important when using Castigliano's
first theorem, providing that moments increasing curvature are of opposite sign to
moments decreasing curvature.

13. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 13 | P a g e
𝑀 = 𝑤𝑅(1 − cos 𝜃) + 𝑝𝑅 sin 𝜃
𝜎𝑣 =
1
𝐸𝐼
∫ 𝑀
ⅆ𝑀
ⅆ𝑊
ⅆ𝑆
𝜋
0
𝜎𝑣 =
1
𝐸𝐼
∫ (𝑤𝑅(1 − cos 𝜃) + 𝑝𝑅 sin 𝜃) 𝑅(1 − cos 𝜃) ⅆ𝜃
𝜋
0
𝜎𝑣 =
1
𝐸𝐼
∫(𝑤𝑅(1 − cos 𝜃) + 𝑝𝑅 sin 𝜃) 𝑅(1 − cos 𝜃) ⅆ𝜃
𝜋
0
𝜎𝑣 =
𝑤𝑅3
𝐸𝐼
∫(1 − cos 𝜃)2
ⅆ𝜃
𝜋
0
𝜎𝑣 =
𝑤𝑅3
𝐸𝐼
∫(1 − 2 cos 𝜃 + cos2
𝜃) ⅆ𝜃
𝜋
0
𝜎𝑣 =
𝑤𝑅3
𝐸𝐼
∫(1 − 2 cos 𝜃 + cos2
𝜃) ⅆ𝜃
𝜋
0
𝜎𝑣 =
𝑤𝑅3
𝐸𝐼
[0 − 2𝑆𝑖𝑛𝜃 +
1
2
𝜃 +
1
4
𝑆𝑖𝑛𝜃 ]0
𝜋
𝝈 𝒗 =
𝒘𝑹 𝟑
𝑬𝑰
∗
𝟑
𝟐
𝝅
dӨ
p

14. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 14 | P a g e
𝜎 𝐻 = ∫
𝑀
𝐸𝐼
ⅆ𝑀
ⅆ𝑃
𝑅 ⅆ𝜃
𝜋
0
𝜎 𝐻 =
1
𝐸𝐼
∫(𝑤𝑅 sin 𝜃 ∗ 𝑅(1 − cos 𝜃) 𝑅 ⅆ𝜃 + ∫(𝑤𝑅 sin 𝜃 ∗ 𝑅(1 − cos 𝜃) 𝑅 ⅆ𝜃
𝜋
𝜋/2
𝜋
0
𝜎 𝐻 =
𝑤𝑅3
𝐸𝐼
∫(sin 𝜃 (1 − cos 𝜃) ⅆ𝜃 + ∫(sin 𝜃 (1 − cos 𝜃) ⅆ𝜃
𝜋
𝜋/2
𝜋
0
𝜎 𝐻 =
𝑤𝑅3
𝐸𝐼
([− cos 𝜃 −
1
2
sin 𝜃 ]0
𝜋
+ [− cos 𝜃 −
1
2
sin 𝜃 ] 𝜋/2
𝜋
)
𝜎 𝐻 =
𝑤𝑅3
𝐸𝐼
(0 −
1
2
+ 1 + 0 + 1 − 0 +
1
2
)
𝝈 𝑯 =
𝒘𝑹 𝟑
𝑬𝑰
∗ 𝟐
2.
 Discuss the difference between experimental reading and theoretical
calculations.
A. Non-calibration of devices _ measuring device
B. Human errors when taking readings
C. Crystal defects of the metal test
D. Non-attention to the internal cycles of the measuring device and the rounding
process when taking readings
 Discuss the factors that affecting on the amount of deflection
A. Beam Section, dominantly depth (Moment of Inertia and also selfweight of
beam)
B. Span of beam , The nature of the end supports whether they are fixed or
simply supported.
C. Creep and shrinkage in flexural members
D. Loading (Dead and Live Load)

15. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
26/11/2018 15 | P a g e
3. What are the applications of the experiment ?