Bending Test Analysis of Steel Beam

Saif aldin ali madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
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[MECHANICS OF MATERIALS Laboratory II]
University of Baghdad
Name: - Saif Al-din Ali -B-

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TABLE OF CONTENTS
ABSTRACT.........................................................................I
OBJECTIVE........................................................................II
INTRODUCTION..............................................................V
THEORY..........................................................................VI
APPARATUS...................................................................VII
Calculations and results................................................VIII
DISCUSSION ...............................................................VIIII

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Name of Experiment: Bending Test
1. ABSTRACT
 The main purpose of the Bend testing is to determine
the ductility, bend strength, fracture strength and
resistance to fracture of the specimen i.e. the
characteristics used to determine whether a material
will fail under pressure and are especially important in
any construction process involving ductile materials
loaded with bending forces.
 If a material begins to fracture or completely fractures
during a three or four point bend test it is valid to
assume that the material will fail under a similar in any
application, which may lead to catastrophic failure.
2. OBJECTIVE
To find the values of deflections and bending stresses of the
beam (steel) supported and carrying a concentrated load at
the center in the case of simply or fixed supported and at free
end in cantilever supported case
3. INTRODUCTION
Generally a bending test is performed on metals or metallic
materials but can also be applied to any substance that can
experience plastic deformation, such as polymers and
plastics. These materials can take any feasible shape but
when used in a bend test most commonly appear in sheets,
strips, bars, shells, and pipes. Bend test machines are
normally used on materials that have an acceptably high
ductility.
One of the more popular uses of bend testing is in the area
of welds. It is done to make sure that the weld has properly
fused to the parent metal and that the weld itself does not

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contain any defects that may cause it to fail when it
experiences bending stresses.
The sample weld is deformed using a guided bend test so
that it forms a “U” subjecting the material on the outer
surface to a tensile force and the material on the inside to a
compressive force. If the weld holds and shows no sign of
fracture it has passed the test and is deemed an acceptable
weld.
4. THEORY
If a beam is simply or fixed supported the ends and carries
a concentrated load at the center, the beam bends concave
upwards. The distance between the original position of the
beam and its position after bending is different at different
points along the length of the beam, being maxim urn at
the center in this case this difference is called "deflection"
Sample test will be made of steel,
young modules
( 𝐸 = 209 ∗ 109 𝑁
𝑚2
= 209 ∗
1015 𝑁
𝑚𝑚2
. 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 (𝑏 =
25𝑚𝑚. ℎ = 3𝑚𝑚)𝑡𝑜 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑐𝑎𝑠𝑒 :-
1 - Cantilever beam,
2 - Simply Supported Beam.
3. Fixed Beam,
In all cases above concentrated loads are applicable and
calculate the maximum deflection values (theory) in each
case according to the following relationship.
𝛔𝐭𝐡 = 𝐰𝐥 𝟑
/𝐜𝐄𝐋
b
h

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b.m.d -WL
1 - Cantilever beam
The constants A and B are required to be found out by utilizing the boundary conditions as defined
below
i.e at x= L ; y= 0 -------------------- (1)
at x = L ; dy/dx = 0 -------------------- (2)
Utilizing the second condition, the value of constant A is obtained as

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2 - Simply Supported Beam

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Boundary conditions relevant for this case are as follows
(i) at x = 0; dy/dx= 0
hence, A = 0
(ii) at x = l/2; y = 0 (because now l / 2 is on the left end or right end support since we have
taken the origin at the centre)

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By symmetry the fixing moments are equal at both ends
Applying Mohr's second theorem for the deflection at mid-span

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5. APPARATUS
• Several loads are applied to the section to calculate the
bending and as shown in the image type of installation as
explained earlier
• Loads are suspended by a handcuff shown in the picture
that carries the weights and the amount of bending is
measured by a gauge with a pool
• The process is repeated 5 times for each type of
installation for different loads for each type

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6. Calculations and results
Calculations
a-Determine the values and locations of maximum deflection maximum
moment and maximum bending stresses for different loads in each case
𝐸𝑒𝑥 =
𝑔L3
3I
∗
𝑚
y 𝑏
𝑬 𝒆𝒙 =
(𝟗. 𝟖𝟏 ∗ 𝟎. 𝟑 𝟑)
𝟑( 𝟓. 𝟔𝟐𝟓 ∗ 𝟏𝟎−𝟏𝟏
)
∗ 𝟏𝟑𝟗 = 𝟐𝟏𝟖. 𝟏 ∗ 𝟏𝟎
𝟗
𝑵/𝒎
𝟐
Error%=(𝐸𝑡ℎ − 𝐸𝑒𝑥)/𝐸𝑡ℎ ∗ 100
Error%=( 𝟐𝟎𝟖 − 𝟐𝟏𝟖. 𝟏)/𝟐𝟎𝟖 ∗ 𝟏𝟎𝟎 = 𝟒. 𝟖%
m=100 g
𝐼 =
𝑏ℎ3
12
=
(25∗10−3)∗(3∗10−3)3
12
= 5.625 ∗ 10−11
EI=11.7 , L=30cm , y=1.5 mm
𝑤 = 𝑚𝑔 → 𝑤 =
100
1000
∗ 9.81 = 0.981 𝑁
B.M = w ⋅ L = 0.981 ∗ 0.3 → B.M = 0.2943 Nm
y 𝑏 =
wL3
3EI
=
0.981∗0.33
3∗11.7
= 7.546 ∗ 10−4 m
σ =
My
I
=
0.2943 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 7.848 𝑀𝑁/𝑚2
100
200
300
400
500
0
100
200
300
400
500
600
0 50 100 150 200 250 300 350 400
m(g)
σ(exp) *10^-2
𝑠𝑙𝑜𝑝𝑒 = (339 − 200) (250 − 150)⁄
= 1.39 ∗ 105 103 = 139⁄

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m=200 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.2 ∗ 9.81 = 1.962 𝑁
B.M = w ⋅ L = 1.962 ∗ 0.3 → B.M = 0.5886 Nm
y 𝑏 =
wL3
3EI
=
1.962 ∗ 0.33
3 ∗ 11.7
= 1.509 ∗ 10−3
𝑚
σ =
My
I
=
0.5886 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 15.69 𝑀𝑁/𝑚2
m=300 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.3 ∗ 9.81 = 2.943 𝑁
B.M = w ⋅ L = 2.943 ∗ 0.3 → B.M = 0.8829 Nm
y 𝑏 =
wL3
3EI
=
2.943 ∗ 0.33
3 ∗ 11.7
= 2.263 ∗ 10−3 𝑚
σ =
My
I
=
0.8829 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 23.544 𝑀𝑁/𝑚2
m=400 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.4 ∗ 9.81 = 3.924𝑁
B.M = w ⋅ L = 3.924 ∗ 0.3 → B.M = 1.1772 Nm
y 𝑏 =
wL3
3EI
=
3.924 ∗ 0.33
3 ∗ 11.7
= 3.018 ∗ 10−3 𝑚
σ =
My
I
=
1.1772 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 31.39 𝑀𝑁/𝑚2
m=500 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.5 ∗ 9.81 = 4.905𝑁
B.M = w ⋅ L = 4.905 ∗ 0.3 → B.M = 1.4715 Nm
y 𝑏 =
wL3
3EI
=
4.905 ∗ 0.33
3 ∗ 11.7
= 3.77 ∗ 10−3
𝑚
σ =
My
I
=
1.4715 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 39.24 𝑀𝑁/𝑚2

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EI=11.7 , L=61cm , y=1.5 mm
m=200 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.2 ∗ 9.81 = 1.962𝑁
B.M =
w ⋅ L
4
= 1.962 ∗ 0.61/4 → B.M = 0.2992 Nm
y 𝑏 =
wL3
48EI
=
1.962 ∗ 0. 613
48 ∗ 11.7
= 7.9298 ∗ 10−4 𝑚
σ =
My
I
=
0.2992 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 7.978 𝑀𝑁/𝑚2
m=400 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.4 ∗ 9.81 = 3.9240 𝑁
B.M =
w ⋅ L
4
= 3.9240 ∗ 0.61/4 → B.M = 0.5984 Nm
y 𝑏 =
wL3
48EI
=
3.9240 ∗ 0. 613
48 ∗ 11.7
= 0.0016 𝑚
𝜎 =
𝑀𝑦
𝐼
=
0.5984 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 15.95 MN/m2
m=600 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.6 ∗ 9.81 = 5.8860 𝑁
B.M =
w ⋅ L
4
= 5.8860 ∗ 0.61/4 → B.M = 0.8976 Nm
y 𝑏 =
wL3
48EI
=
5.8860 ∗ 0. 613
48 ∗ 11.7
= 0.0024 𝑚
𝜎 =
𝑀𝑦
𝐼
=
0.8976 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 23.93664 MN/m2
m=800 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.8 ∗ 9.81 = 7.8480 𝑁
B.M =
w ⋅ L
4
= 7.8480 ∗ 0.61/4 → B.M = 1.1968 Nm

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y 𝑏 =
wL3
48EI
=
7.8480 ∗ 0.613
48 ∗ 11.7
= 0.0032 𝑚
𝜎 =
𝑀𝑦
𝐼
=
1.1968 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 31.9152 MN/m2
m=1000 g
𝑤 = 𝑚𝑔 → 𝑤 = 1 ∗ 9.81 = 9.81 𝑁
B.M =
w ⋅ L
4
= 9.81 ∗ 0.61/4 → B.M = 1.4960 Nm
y 𝑏 =
wL3
48EI
=
9.81 ∗ 0.613
48 ∗ 11.7
= 0.0040 𝑚
𝜎 =
𝑀𝑦
𝐼
=
1.1968 ∗ 1.5 ∗ 10−3
5.625 ∗ 10−11
= 39.894 MN/m2
3. Fixed Beam,
EI=11.7 , L=60cm , y=1.5 mm
m=300 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.3 ∗ 9.81 = 2.9430 𝑁
B.M =
w ⋅ L
8
= 2.9430 ∗ 0.61/8 → B.M = 0.2207 Nm
y 𝑏 =
wL3
192EI
=
2.9430 ∗ 0.613
192 ∗ 11.7
= 2.8298 ∗ 10−4
𝑚
m=600 g
𝑤 = 𝑚𝑔 → 𝑤 = 0.6 ∗ 9.81 = 5.8860 𝑁
B.M =
w ⋅ L
8
= 5.8860 ∗ 0.61/8 → B.M = 0.4415 Nm
y 𝑏 =
wL3
192EI
=
5.8860 ∗ 0.613
192 ∗ 11.7
= 5.947 ∗ 10−4
𝑚
m=900 g

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𝑤 = 𝑚𝑔 → 𝑤 = 0.9 ∗ 9.81 = 8.8290𝑁
B.M =
w ⋅ L
8
= 8.8290 ∗ 0.61/8 → B.M = 0.6622 Nm
y 𝑏 =
wL3
192EI
=
8.8290 ∗ 0.613
192 ∗ 11.7
= 8.4894 ∗ 10−4 𝑚
m=1200 g
𝑤 = 𝑚𝑔 → 𝑤 = 1.2 ∗ 9.81 = 11.7720𝑁
B.M =
w ⋅ L
8
= 8.8290 ∗ 0.61/8 → B.M = 0.8829 Nm
y 𝑏 =
wL3
192EI
=
8.8290 ∗ 0.613
192 ∗ 11.7
= 0.0011 𝑚
m=1500 g
𝑤 = 𝑚𝑔 → 𝑤 = 1.5 ∗ 9.81 = 14.7150𝑁
B.M =
w ⋅ L
8
= 8.8290 ∗ 0.61/8 → B.M = 1.1036Nm
y 𝑏 =
wL3
192EI
=
8.8290 ∗ 0.613
192 ∗ 11.7
= 0.0014 𝑚

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B-Determine the percentage error of the deflection in each case
Error %=(𝑦𝑡ℎ − 𝑦𝑒𝑥)/𝑦𝑡ℎ ∗ 100
 yex = 7 ∗ 10−4
yth = 7.546 ∗ 10−4
Error%= 7.2356%
 yex = 1.54 ∗ 10−3
yth = 1.509 ∗ 10−3
Error%= 2.0543%
 yex = 2.3 ∗ 10−3
yth = 2.263 ∗ 10−3
Error%= 1.6350%
 yex = 2.87 ∗ 10−3
yth = 3.018 ∗ 10−3
Error%= 4.9039%
 yex = 3.75 ∗ 10−3
yth = 3.77 ∗ 10−3
Error%= 0.5305%

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 yex = 6.7 ∗ 10−4
yth = 7.9298 ∗ 10−4
Error%= 15.5086%
 yex = 1.42 ∗ 10−3
yth = 0.0016
Error%= 11.2500%
 yex = 2.2 ∗ 10−3
yth = 0.0024
Error%= 8.3333%
 yex = 2.93 ∗ 10−3
yth = 0.0032
Error%= 8.4375%
 yex = 3.62 ∗ 10−4
yth = 0.004
Error%= 9.5%

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3. Fixed Beam,
 yex = 2.5 ∗ 10−4
yth = 2.8298 ∗ 10−4
Error%= 11.6545%
 yex = 5.2 ∗ 10−4
yth = 5.947 ∗ 10−4
Error%= 12.561%
 yex = 7.8 ∗ 10−4
yth = 8.4894 ∗ 10−4
Error%= 8.1207%
 yex = 1.13 ∗ 10−3
yth = 0.0011
Error%= 2.7273%
 yex = 1.36 ∗ 10−4
yth = 0.0014
Error%= 2.8571%

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Results;-
Cantilever beam
σ(
𝑀𝑁
𝑚2
)
B.M
(Nmm)
y 𝑏
(th)
∗ 10−2
(𝑚𝑚)
y 𝑏
(ex)
∗ 10−2
(𝑚𝑚)
W(N)m(g)
7.848294.375.46700.981100
15.69588.6150.91541.962200
23.544882.9226.32302.943300
31.391177.2301.82873.924400
39.241471.53773574.905500
Simply Supported Beam
7.978299.279.298671.962200
15.95598.41601423.9240400
23.93664897.62402205.8860600
31.91521196.83202937.8480800
39.8941496.04003629.811000
Fixed Beam
5.887220.728.29252.9430300
11.772441.559.47525.8860600
17.658662.284.894788.8290800
23.544882.911011311.7721200
29.431103.614013614.71501500
C.N
Error%
7.2356
2.0543
1.6350
4.9039
0.5305
S.S.B
15.5086
11.25
8.3333
8.4375
9.5
FX.B
11.6545
12.561
8.1207
2.7273
2.8571

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0
200
400
600
800
1000
1200
0 50 100 150 200 250 300 350 400 450
m(g)
y(b)
Simply Supported Beam
y(th) Linear (y(ex))
𝑠𝑙𝑜𝑝𝑒(𝑡ℎ) = (840 − 605) (350 − 249) = 2.3267⁄
𝑠𝑙𝑜𝑝𝑒(𝑒𝑥) = (450 − 235) (180 − 90)⁄ = 2.3888
Error %=(𝑠 𝑡ℎ − 𝑠 𝑒𝑥)/𝑠 𝑡ℎ ∗ 100% =2.669%
𝑠𝑙𝑜𝑝𝑒(𝑡ℎ) = (380 − 300) (290 − 230) = 1.333⁄
𝑠𝑙𝑜𝑝𝑒(𝑒𝑥) = (200 − 100) (154 − 80)⁄ = 1.351
Error %=(𝑠 𝑡ℎ − 𝑠 𝑒𝑥)/𝑠 𝑡ℎ ∗ 100% =1.35%

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7. DISCUSSION
1. Discuss the difference between experimental reading and
theoretical calculations.
The difference in outputs between the two cases is the
result of the inaccuracy of taking the readings and rounding in the
calibration calibration of the device and the measuring device. This
is what is included in the charts above. We have neglected points or
pass between the points to deliver the case to a linear process. Also
the sensitivity of the device to external influences resulting from
vibrations or simple touch
0
200
400
600
800
1000
1200
1400
1600
0 20 40 60 80 100 120 140 160
m(g)
y(b)
Fixed Beam
Linear (y(ex)) Linear (y(th))
𝑠𝑙𝑜𝑝𝑒(𝑡ℎ) = (600 − 399) (60 − 41) = 10.5⁄
𝑠𝑙𝑜𝑝𝑒(𝑒𝑥) = (1200 − 1000) (112 − 92)⁄ = 10
Error %=(𝑠 𝑡ℎ − 𝑠 𝑒𝑥)/𝑠 𝑡ℎ ∗ 100% =4.761%

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2. Discuss the factors that affecting on the amount of
deflection
1. Errors in the deflection computation of flexural members
2. Loading of flexural members
3. Flexural stiffness
4. Factors affecting fixity
5. Construction variations of flexural members
6. Creep and shrinkage in flexural members
7. Beam Material (Elastic Modulus)
8. Beam Section, dominantly depth (Moment of Inertia and also
self-weight of beam)
9. Loading (Dead and Live Load)
10. Span of beam
11. Support Type (Fixed, Hinged)
3. What are the applications of each case?
Cantilever beam Simply Support Fixed Beam

Bending Test Analysis of Steel Beam

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