UNIVERSITI TENAGA NASIONAL

COLLEGE OF ENGINEERING

DEPARTMENT OF MECHANICAL ENGINEERING

MEMB221 - MECHANICSAND MATERIALS LAB(SEMESTER 1, 2015/16)

EXP. TITLE : BENDING TEST

GROUP MEMBERS: ALVIN CHAN HSIEN YI ME093017

NAZARIAH BINTI ZAINAL ME094151

NUR ADILAH BINTI MOHD JAMALUDDIN ME093319

NOEL TEH HONG GUAN ME094552

SECTION : 06 GROUP: 04

INSTRUCTOR : MISS. NURASLINDA BINTI ANUAR

Performed Date Due Date* Submitted Date5 AUGUST 2015 12 AUGUST 2015 12 AUGUST 2015

TABLE OF CONTENT

CONTENT PAGE

ABSTRACT

OBJECTIVE

THEORY

EQUIPMENT AND DESCRIPTION OF EXPERIMENTAL APPARATUS

PROCEDURE

DATA AND OBSERVATIONS

ANALYSIS AND RESULTS

DISCUSSIONS

CONCLUSIONS

REFERENCES

ABSTRACT

Bending test was conducted to examine the relationship between load, span, width,

height and deflection of a beam. Also, to ascertain the coefficient of elasticity for each

specimens. The specimens used are steel, brass, aluminum and wood. The test then started

with the first task which is to find the relationship between load and deflection. Next, to

determine the relationship between span and deflection followed by to investigate the

relationship between the width and deflection of the test specimen. Also, the relationship

between the height and deflection of the test specimen. For the second task, the main

objective is to ascertain the coefficient of elasticity.

OBJECTIVE

The objectives of this experiment are to investigate the relationship between load,

span, width, height and deflection of a beam, placed on two bear affected by a concentrated

load at the center and to ascertain the coefficient of elasticity for steel, brass, aluminum, and

wood.

THEORY

The stress-strain behavior of brittle materials is not usually ascertained by tensile test.

For more suitable transverse bending test is usually employed, in which a rod specimen either

a circular or rectangular cross section is bent until crack using a three- or four-point loading

technique. The appraisals are led by following to Standard Test Method for Flexural Strength

of Advanced Ceramics at Ambient Temperature (ASTM Standard C 1161). The apparatus

has been outline to enable students to completes experiments on simply supported and

cantilever beams in order to investigate the relationship between the deflections and the

applied loads also the effect of variations in length and cross sectional.

- Simply supported beam with central point load.

It can be shown that the deflection under the load i.e. maximum deflection

- Determination of coefficient of elasticity

Where :

δ = Deflection (mm)

L = Span (mm)

Mb = Moment of Flexure (Nmm)

Wb = Resistance to Flexure (mm3)

σb = Flexural Stress (N/mm2)

E = Coefficient of Elasticity

I = Inertia Factor

F1 = Load occasioned by weight of load device = 2.5N

F2 = Load occasioned by additional weight (N)

EQUIPMENT AND DESCRIPTION OF EXPERIMENTAL APPARATUS

Twist and Bend Test Machine MT 210

PROCEDURE

Task 1: To investigate the relationship between load, span, width, height and deflection of a

beam, placed on two bear affected by a concentrated load at the center.

A: Investigate the relationship between load and deflection

1- A span of 600 mm is obtained when the berries are set. The interval between each

groove on the shafts of the apparatus is 100 mm.

2- The specimen was placed on the bearers with dimensions 6 x 25 mm. then, the centre

of the specimen was mounted by the load device

3- The top of the gauge is centered on the upper plane of the load device when the

testing device is set. Lower the gauge so that its small hand is at about 10 and the

gauge is set to be zero by twisting its outer ring. the weight was loaded and deflection

reading is taken.

4- A graph of deflection versus loading is needed to be drawn.

B: Investigate relationship between span and deflection

1- The test specimen was employed with dimensions of 6x25 mm and was loaded with

weight of 10N. The span was varied as indicated. The deflection reading is taken.

2- A graph of deflection versus span is needed to be drawn.

C: Investigate the relationship between width and deflection of the test specimen

1- The bearers for a span of 500 mm are set. With 5N weight of load, the test specimen

indicated was employed. The deflection reading is taken.

2- A graph of the width of the deflection versus the test specimen is needed to be drawn.

D: Investigate the relationship between the height and deflection of the test

specimen.

1- The bearers for a span of 500 mm are set. With 5N weight of load, the test specimen

indicated was employed. The deflection reading is taken.

2- A graph of the height of the deflection versus the test specimen is needed to be drawn.

Task 2: To ascertain the coefficient of elasticity for steel, brass, aluminum and wood.

1- The span is set at 500 mm.

2- A circular test specimen of steel is employed.

3- The load device is mounted and the testing device is set.

4- Load with weights (5N,10N,15N)

5- The deflection reading is taken.

6- The test is repeated with test specimens of brass, aluminum and wood.

At the point when E is calculated, the initial load brought on by the load device has no

significance since the gauge at zero with the device set up. Be that as it may, when

calculating flexural stress, F1 is incorporate.

DATA & OBSERVATION

Task 1

Part A: The relationship between load and deflection.

Span: 600mm

Test Specimen: 6 x 25 mm

Load (N) Deflection (mm)

5 0.22

10 0.46

15 0.70

20 0.95

Table 1

Part B: The relationship between span and deflection.

Load: 10 N

Test Specimen: 6 x 25 mm

Span (mm) Deflection (mm)

300 0.04

400 0.12

500 0.34

600 0.46

Table 2

Part C: The relationship between width and deflection of the test specimen.

Span: 500 mm

Load: 5 N

Test Specimen Width

(mm)

Deflection (mm)

15 0.74

20 0.54

25 0.45

30 0.32

Table 3

Part D: The relationship between the height and deflection of the test specimen.

Span: 500 mm

Load: 5 N

Test Specimen Height (mm) Deflection (mm)

3 0.57

4 0.44

6 0.10

8 0.03

Table 4

Task 2

To ascertain the Coefficient of Elasticity (E) for steel, brass, aluminum and wood.

Span: 500 mm

Diameter: 8mm (Rod)

Dimension: 5 x 30 mm (Wood)

Material Load, F (N)

Deflection

δ

(mm)

Wood

5 1.47

10 3.10

15 4.23

Aluminum

5 0.97

10 1.82

15 2.79

Brass

5 0.63

10 1.33

15 2.05

Steel

5 0.30

10 0.65

15 0.98

Table 5

ANALYSIS & RESULTS

Task 1:

Part A: Graph of Deflection against Loading

Graph 1

Part B: Graph of Deflection against Span

Graph 2

5 1 0 1 5 2 00

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Defl ecti on against Load

Load (N)

Defle

ction

(mm

)

3 0 0 4 0 0 5 0 0 6 0 00

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Defl ecti on against Span

SPAN (MM)

Defle

ction

(mm

)

Part C: Graph of Width of Deflection against Test Specimen

Graph 3

Part D: Graph of Height of the Deflection against Test Specimen

3 4 6 80

0.1

0.2

0.3

0.4

0.5

0.6

Graph 1.4 : Defl ecti on against Height

Height (mm)

Defle

ction

(mm

)

Graph 4

1 5 2 0 2 5 3 00

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Graph 1.3 : Defl ecti on against Width

Width (mm)

Defle

ction

(mm

)

Calculation:

Material Load,

F (N)

Deflection

δ

(mm)

Moment of

Flexure, Mb

(Nmm)

Flexural

Stress

σb

(N/mm2)

Coefficient Of Elasticity

E

(N/mm2)

Eave

(N/mm2)

Wood

5 1.47 937.5 7.5 28344.6712

28259.073

0

10 3.10 1 562.5 12.5 26881.7204

15 4.23 2 187.5 17.5 29550.8274

Aluminum

5 0.97 937.5 18.6510 66763.2183

69187.720

5

10 1.82 1 562.5 31.0849 71165.1887

15 2.79 2 187.5 43.5189 69634.7546

Brass

5 0.63 937.5 18.6510 102794.161

5 98316.435

510 1.33 1 562.5 31.0849 97383.9425

15 2.05 2 187.5 43.5109 94771.2025

Steel

5 0.30 937.5 18.6510 215867.739

1 204458.71

6810 0.65 1 562.5 31.0849 199262.528

4

15 0.98 2 187.5 43.5189 198245.882

9

Table 6To calculate the coefficient of elasticity of steel, brass, aluminum and wood, the deflection

formula is:

δ= FL3

48 EI E= FL3

48 Iδ

To determine the flexural stress:

σ b=

M b

W b M b=( F+F1 )

L4

When rectangular it is I=bh3

12 and W b=

bh2

6

When circular it is I=πd 4

64 and W b=

πd3

32

δ = Deflection (mm) E = Coefficient of Elasticity

L = Span (mm) = 500 mm I = Inertia Factor

Mb = Moment of Flexures (Nmm) F = Load occasioned by additional weight (N)

Wb = Resistance to Flexure (mm3) σb = Flexural Stress (N/mm2)

F1 = Load occasioned by weight of Load Device (N) = 2.5 N

Moment of flexure is the same for every specimen according to the load weight used.

Moment of Flexure M b=( F+F1 )

L4

5 N M b=(5+2. 5)500

4=937 .5 Nmm

10 N M b=(10+2. 5)500

4=1562.5 Nmm

15 N M b=(15+2. 5)500

4=2187 .5 Nmm

Flexural stress for wood:

Dimension: 5 × 30 mm

W b=bh2

6=30×52

6=125 mm3

5 N σ b=

937 .5125

=7 .5 Nmm2

10 N σ b=

1562. 5125

=12. 5 Nmm2

15 N σ b=

2187 .5125

=17 . 5 Nmm2

Flexural Stress for Steel, Brass and Aluminum are the same:

Dimension: 8 × 650 mm

σ b=M b

W b W b=

πd3

32=π×83

32=50 .2655 mm3

5 N σ b=

937 .550. 2655

=18 .6510 Nmm2

10 N σ b=

1562. 550. 2655

=31. 0849 Nmm2

15 N σ b=

2187 .550. 2655

=43 .5189 Nmm2

Inertia Factor for Wood specimen:-

I=bh3

12=30×53

12=312. 5 mm4

Inertia Factor for Steel, Brass and Aluminum specimens:-

I=πd 4

64= π×84

64=201.0619 mm4

Wood:-

5 N E= FL3

48 Iδ= 5×5003

48×312 .5×1 .47=28344 .6712 N /mm2=28 .345GPa

10 N E= FL3

48 Iδ=10×5003

48×312 .5×3 .10=26881 .72043 N /mm2=26 . 882 GPa

15 N E= FL3

48 Iδ=15×5003

48×312 .5×4 .23=29550 .82742 N /mm2=29 . 551GPa

Hence, Eave=

28 .345+26 .882+29.5513

=28.259 GPa

Theoretical Value = 13.1 GPa (Referring to Appendix B from the text book)

% error =

|13 .1−28 . 259|13 . 1

×100=115 .72 %

Aluminum:-

5 N E= FL3

48 Iδ= 5×5003

48×201 .0619×0 .97=66763 .2183 N /mm2=66 .763 GPa

10 N E= FL3

48 Iδ=10×5003

48×201 .0619×1 .82=71165. 1887 N /mm2=71. 165 GPa

15 N E= FL3

48 Iδ=15×5003

48×201 .0619×2 .79=69634 . 7546 N /mm2=69 .635GPa

Hence, Eave=

66 .763+71.165+69 .6353

=69.188 GPa

Theoretical Value = 70 GPa (Referring to Appendix B from the text book)

% error =

|70−69 .188|70

×100=1 .16 %

Brass:-

5 N E= FL3

48 Iδ= 5×5003

48×201 .0619×0 .63=102794 .1615 N /mm2=102 .794 GPa

10 N E= FL3

48 Iδ=10×5003

48×201 .0619×1 .33=97383 . 9425 N /mm2=97 .384 GPa

15 N E= FL3

48 Iδ=15×5003

48×201 .0619×2 .05=94771. 20254 N /mm2=94 .771 GPa

Hence, Eave=

102 .794+97 . 384+94 .7713

=98 .316 GPa

Theoretical Value =105 GPa (Referring to Appendix B from the text book)

% error =

|105−98 .316|105

×100=6 .37 %

Steel:-

5 N E= FL3

48 Iδ= 5×5003

48×201 .0619×0 .30=215867 .7391 N /mm2=215 .868 GPa

10 N E= FL3

48 Iδ=10×5003

48×201 .0619×0 .65=199262. 5284 N /mm2=199 .263GPa

15 N E= FL3

48 Iδ=15×5003

48×201 .0619×0 .98=198245 .8829 N /mm2=198 . 246 GPa

Hence, Eave=

215 . 868+199 .263+198. 2463

=204 .459 GPa

Theoretical Value = 200 GPa (Referring to Appendix B from the text book)

% error =

|200−204 .459|200

×100=2 .23 %

(Note: all the answers in the above calculations are tabulated in ‘Table 6’)

DISCUSSION

In Task 1, it is to investigate the relationship between loads, span, width, height towards

deflection of a beam. This beam is placed on two bearers and the load is concentrated at the

center. There are 4 parts all together. In part A the relationship between load and deflection is

investigated. From the graph plotted (refer to Table 1 & Graph 1) it can be deduced that as

load increases the deflection also increases, hence a linear relationship between them.

Meanwhile the relationship between span and deflection is investigated in part B. Similarly

(refer to Table 2 & Graph 2) in here deflection increases with span, also a linear relationship

between them.

In part C the width of the specimen is varied while maintaining the same load weight. In this

case as seen from the data (refer to Table 3 & Graph 3) the deflection decreases as width is

increased. The same also applied to investigation in part D where the height the specimen is

the variable here (refer to Table 4 & Graph 4). Again deflection decreases as height

increases. Hence from part C and D, deflection is inversely proportional to width and height

respectively.

In Task 2 the Coefficients of Elasticity, E, of 4 different specimens was determined. Again

the same apparatus as in task 1 is used. After experiments are done and after calculation, E

for wood is obtained as 28.259 GPa. This count to a percentage error of 115.72% from the

theoretical value which is 13.1 GPa. Aluminum has a theoretical E of 70 GPa but from the

experiment analysis it was calculated to be 69.188 GPa, a small error only occurred on it

which is 1.16%. E for brass is found to be 98.316 GPa, with an error of only 6.37% from

theoretical value of 105 GPa. Finally E of steel was found to be 204.459 GPa, an error of

115.72% from the actual value of 200 GPa. The percentage error of the wood is the highest

among the 4 specimens, which is 115.72% and the percentage error of the Aluminum is

lowest in this experiment (task 2) which is 1.16%.

Simple bridge can be one of the application which apply the concept of simply supported

beam with central point load. Bridge as we can see in both of it are been supported so that it

can balanced the bridge. Deflection may occur if the force or load are stand in the center of

the bridge.

The deflection may be decreased if the material of the bridge are selected wisely as if the

wood are been selected, the deflection occur may be high which that the bridge may break as

the wood cannot support too much of load. Steel is the main material in the bridge

construction. As we know, steel is a strong material and it is the most suitable construction

material. Moreover, the deflection occurred in steel also not higher as the deflection of

aluminum, brass and wood. Hence, steel is the most suitable material for the skeleton of the

bridge because it has a high value of the Modulus of Elasticity.

CONCLUSION

REFERENCES