Emm 3490 Lab Sheet Strength of Material e

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1

EQ1. TENSILE TEST

Objective

The objectives of this experiment are:1. To develop an understanding of stress–strain curves.

2. To determine the various mechanical properties of engineering material.

Theory

Stresses may be tensile, compressive or shear in nature. Figure 1 shows a metal bar in tension, i.e.

the force F is stretching force which thus increases the length of the bar and reduces its cross-

section. The area used in calculations of stress is generally the original area A0 that existed before

the application of the forces, not the area after the force has been applied. This stress is thus

referred to as the engineering stress σ:

0/ AF =σ [N/m2 or Pa] (1)

The dimensional change caused by a stress is called strain. In tension (or compression), the strain is

the ratio of the change in length to the original length. The term strain is defined as:

00t / )( 100(%) l l l e −×= (2)

Where l – l 0 = Δl , the change in length. Since strain is a ratio of two lengths it has no units. Strain is

frequently expressed as a percentage.

Results of such a tensile test can be represented in the form of engineering stress–strain curve.

Figure 2 is typical of ductile metals such as copper tested at room temperature. The tensile

strength, also known as ultimate tensile strength (UTS), is defined as the maximum stress which a

material can withstand. It is obtained by dividing maximum load by original cross-sectional area of

tensile specimen.

0max /AF UTS =σ [N/m2 or Pa] (3)

Figure 3 and 4 show the two types of stress–strain curves. The yield stress σy is defined as the

stress at which plastic deformation (elongation) of the tensile specimen takes place at a constant

load (Figure 3). Such behavior is generally observed in carbon steels.

Fig. 1 Metal bar in tension

F

l 0 l

A0

F

Fig. 2 Typical engineering stress–strainbehavior to fracture

Strain

S t r e s s

UTS

F: Fracture

F



2

Some steels, especially non-ferrous alloys, do not show the presence of sharp yield point (Figure 4).

For such steels, proof stress is reported instead of yield stress. Proof stress σ0.2 is that stress at

which some small amount of permanent deformation, say equal to 0.2 percent strain, take place. In

other words, it is that stress which produces a permanent elongation of 0.2 percent in the tensilespecimen on the removal load.

At the beginning of the test, the force increases rapidly and proportionately to strain: the stress–

strain curve obeys Hooke’s law

tEe =σ [N/m2 or Pa] (4)

The proportionality constant (the slope of the curve) is called the elastic modulus or Young’s

modulus E (Figure 5) and straint

e .

If the specimen is unloaded in this range, it will return to its original length, i.e. all deformation is

elastic.

Tensile test is carried out by gripping the ends of a suitably prepared standardized specimen in a

tensile testing machine, and then applying a continually increasing uni-axial load until such time as

failure occurs. Before the test, the gauge length L0, and the cross-sectional area A0 are measured to

enable calculations of percent elongation and percent area reduction to be made. Figure 6 shows

dimensions of a specimen for tensile test.

Ductility is a measure of a material's ability to deform plastically without fracture. The two most

common methods of ductility measurement are:

a. Percent elongation is determined by setting a gauge length on a specimen prior to

loading and after tensile failure measuring the final distance of these gauge marks. Then

a percent elongation value is calculated as equation (2).

b. Percent area reduction is calculated by putting the two ends of the fractured specimentogether and measuring the diameter at the break. Calculate the area at the break at this

point of fracture. This final area is then compared with the original area of the specimen

and a percent reduction in area is then calculated.

Reduction in area (%) = 00 / )( 100 A A A −× (5)

Fig. 4 General stress–strain

Strain

0.002

σ y

PlasticElastic

P S t r e s s

Fig. 3 Typical stress–strain

Strain

Lower yieldpoint

Upper

yield point

S t r e s s

σ y



3

Original diameter d 0

Diameter at failure d Reduction indiameter

Originalgauge length

L 0 Plastic

deformation

Gauge length at failure L

Fig. 7 Schematic of tensile test specimenfor before and after testing

Specimen and Equipments

Fig. 5 Hooke’s law: Stressproportional to strain

Strain

S t r e s s

Slope = modulusof elasticity

Unload

Load

0

Fig. 6 Standard tensile test specimen withcircular cross section

Gaugelength L 0

Parallel length L C

Radius r Diameter d 0

Cross-sectionalarea A0

0065.5 AL =



4

1. Universal testing machine – Instron Series 8500 (10 kN).

2. Vernier caliper.

3. Tensile specimen: steel / brass / copper / aluminium

Fig. 8 Instron Series 8500 (10 kN)

Procedures

1. Refer to Figure 7, use vernier caliper to measure the original diameter of the specimen.

Take measurements in at least three locations and average.

2. Calculate the value of gauge length, and make two marks on the parallel part of the

specimen to register the gauge length.

3. Grip the specimen in the gripping heads of the machine.

4. Set the required parameters on the control panel.

5. Adjust the load recorder on the front panel controller to zero, to read load applied.

6. Press start button to start the tensile

test.

7. Monitor the sample and note when constriction begins. From now on, the force will no

longer increase, but instead, will tend to decrease until fracture occurs.

8. Remove the tested specimen from the gripping heads, and measure dimensions of tested

specimen. Fit the broken parts together and measure reduced diameter and final gauge

length (Figure 7).

Results1. Show all the measurements of specimen.

i. Original diameter d 0 [mm].

ii. Original gauge length L0 [mm].

iii. Final diameter d [mm].

iv. Final gauge length L [mm].



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2. Calculate the following.

i. Original cross-sectional area A0 [mm2].

ii. Final cross-sectional area A [mm2].

iii. Percent elongation.

iv. Percent reduction in area.

v. Young’s modulus E [GPa].

3. Plot the stress–strain graph.

4. On the stress-strain curve show the following points, and verify the value (i) to (iii).

i. Ultimate tensile strength (UTS) [MPa].

ii. Yield stress σy or Proof stress σ0.2 [MPa].

iii. Fracture stress σf [MPa].

iv. Elastic limit.

v. Proportional limit.

vi. Elastic and plastic deformation regions.

vii. Necking region.

Discussions

1. Discuss on the shape of obtained stress–strain curve.

2. Compare and discuss on the experimental results with the theory.

3. Discuss the difference between Engineering Stress and True Stress, and whether there is a

significant difference between these values at failure.

4. Explain the necking process, and discuss how the necking of the specimen relates to the

shape of the stress–strain curve.

5. Discuss on the mechanical properties of the tested specimen.

6. Discuss on the factors that can be affected to the experimental result.

7. Compare and discuss the affect differences in the stress-strain behavior of various

materials.

Conclusions

1. Give an overall conclusion based on the obtained experimental results.

2. Conclude on the applications of the experiment.



6

Fig. 1 Simply-supported beam withtwo symmetric concentrated loads andsupported by pinned and rollersupports

W W

Pinned Roller

Fig. 2 Overhanging beam withconcentrated and distributed loads,and supported by pinned supports

Pinned Pinned

W w

Fig. 3 Cantilever beam with aconcentrated load and supported by fixed-end support

Fixed-end

W

EQ2. REACTION OF BEAM TEST

Objective

The objectives of this experiment are:1. To identify the supports reaction in simply-supported and overhanging beams.

2. To develop an understanding of beam apparatus, and to determine its sensitivity and

accuracy.

Theory

1. General

A beam is a member which has the primary function of resisting transverse loading. Beam is one of

the simplest structures in design but one of the most complexes to analyze in terms of the external

and internal forces acting on it. The complexity of its behavior under load depends on how it is

supported - at one or both ends - and how its ends are attached to the supports. Three basic beam

types are the simply-supported, overhanging, and cantilever beams.

A beam supported by a support at the ends and having one span is called a simply-supported beam

(Figure 1). A support will develop a reaction normal to the beam but will not produce a couple. If

either or both ends of the beam project beyond the supports, it is called overhanging beam (Figure

2). A cantilever beam (Figure 3) is one in which one ends is built into a wall or other support so

that the built-in end can neither move transversely nor rotate.



7

Fig. 4 Supports reaction of the simply-supported beam with concentrated loads

W 1 W 2

a b

l / 2 l / 2 R 1 R 2

2. Types of Load

A beam is normally horizontal, the loads being vertical, other cases which occur being locked upon

as exceptions. The two types of loads for beams are concentrated and distributed loads.i) A concentrated load W [N] is one which is considered to act at a point, although in

practice it must really be distributed over a small area (Figures 1 and 3).

ii) A distributed load w [N/m] is one which is spread in some manner over the length of

the beam. The rate of loading may be uniform, or may vary from point to point along the

beam (Figure 2).

3. Types of Support

The deformations and stresses which result in a beam owing to a particular load (concentrated

load) or group of loads (distributed load) are dependent on the manner in which the beam is

supported. The three basic types of supports for beams are roller, pinned and fixed-end.

i) A roller support is one which exerts a reactive force having a known line of action(Figure 1).

ii) A pinned support in one which allows the beam freedom to rotate but prevents it from

any linear movement (Figures 1 and 2).

iii) A fixed-end support is one which prevents the beam from translating or rotating at the

point of support (Figure 3).

4. Supports Reaction of the Simply-Supported Beam with Concentrated Loads

Referring to the loading in Figure 4, the left-hand support reaction R1 is first required and the

reactions can be found from the equations of force and moment equilibrium.

2121 W W R R +=+

(1)( ) ( ) ( ) b W a W W W l b l W a l W l R 21212

121

221

11 −++=−++= (2)

Therefore

( ) ( ) ( )l b W l a W W W R 212121

1 −++= (3)

Substitute (3) to (1).

( ) ( ) ( )l b W l a W W W R 212121

2 +−+= (4)



8

Specimens and Equipments

1. Beam apparatus – SM104

2. Vernier caliper

3. Load cells

4. Dial gauges

5. Weight hangers

6. Weights: 5 N, 10 N

7. Steel blocks

8. Beams: Steel / Brass / Aluminium

Supports Reaction of the Simply-Supported Beam with Concentrated Loads

Procedures

1. Measure the thickness and width of the beam.

2. Measure the length of the beam and mark it at mid-span and at 1/4-span points.

3. Set up load cells 1/4-span to the left and right of the mid-span reading, and lock the knife

edge.

4. Place the beam in position with 1/4-span overhang at either end.

5. Position two weight hangers equidistant from the mid point of the beam.

6. Place a dial gauge in position on the upper cross-member so that the ball end rests on the

centre-line of the beam immediately above the left-hand support.

7. Check that the stem is vertical and the bottom O-ring has been moved down the stem.

8. Adjust the dial gauge to zero read and then lock the bezel in position.

9. Move the dial gauge to a position above the right-hand support, check that the beam is

parallel to the cross-member, then adjust the height of the knife edge so that the dial gauge

reads zero.

10. Remove the dial gauge and unlock both knife edges. Adjust the load cell indicators to read

zero.

11. Apply loads to the weight hangers in a systematic manner, tap the beam very gently and

take readings of the load cells.

12. Process the results and plot graphs from the experimental results.

Fig.5 Beam apparatus – SM104



9

Results

1. Show all the measurements of beam.

i. Beam length L [mm]

ii. Beam width b [mm]

iii. Beam thickness h [mm]

iv. Beam working length l [mm]

2. Record the values of R1 and R2, and calculate the R1+R2, Δ and % in Table 1.

3. Plot the graphs.

i. R1 and R2 against W 1, when W 2 = 0.

ii. R1 and R2 against W 2, when W 1 = 0.

iii. R1 and R2 against W 1 = W 2 = 5 N, 10 N, … 30 N.

Discussions

1. Discuss on the verification of equations (3) and (4).

2. Discuss on the obtained graphs from the experiment.

3. Calculate the theoretical values of R1 and R2, by using equations (3) and (4). Plot the graph

of theoretical values of R1 and R2. Compare and discuss the theoretical and experimental

graphs.

4. What is the percentage error? Discuss the factor that can be affected to the experiment

results.

W 1

Fig. 6 Experimental set up for supports reaction of the simply-supported beam with concentrated loads

l

Load cell

L / 4

W 2

R 1 R 2

L

4

l

4

l

L / 4 L / 4 L / 4



10

Table 1 Experimental results of simply-supported beam with concentrated loads

Conclusions

Give an overall conclusion based on the obtained experimental results.

W 1 [N] W 2 [N] R 1 [N] R 2 [N] R 1+R 2 [N] Δ [N] %

5 0

10 0

15 020 0

25 0

30 0

0 5

0 10

0 15

0 20

0 25

0 30

5 5

10 10

15 15

20 20

25 25

30 30

* Δ = ( R 1+R 2 ) – ( W 1+W 2 ) % = 100Δ / ( W 1+W 2 )



11

EQ3. COMPRESSION TEST

Objective

The objectives of this experiment are:1. To study and observe the techniques of the compression testing.

2. To determine the mechanical properties on three different sizes of the tested specimens.

Theory

The compression test is simply the opposite of the tension test with respect to the direction of

loading. It is often stated that materials behave the same in tension and compression. That is true

for most ductile materials. However, there are some materials that are very weak in tension and

extremely strong in compression.

Prior to the yield point tension and compression results are similar. The major difference with the

compression test compared to the tensile test is that the specimen compresses or the area increases

after the yield point is reached. For some ductile materials the specimen will compress until a flat slug is reached. However brittle materials will fail suddenly after their ultimate strength is

exceeded. These brittle materials have much greater compression strength than tensile strength.

That is why these materials are mostly tested in compression.

When a force (or load) is applied to a material (Figure 1), it produces a stress in the material. The

stress σ acting on the material is the force F exerted per unit area A0:

0/AF σ = [N/m2 or Pa] (1)

The dimensional change caused by a stress is called strain. In compression (or tension), the strain

is the ratio of the change in length to the original length. The term strain is defined as:

00c/ )(100(%) llle −×= (2)

Where l – l 0 = Δl , the change in length. Since strain is a ratio of two lengths it has no units. Strain is

frequently expressed as a percentage.

F

l 0 l

A0

F

Fig.1 Metal bar in compression



12

In compression testing the sample is squeezed while the load and the displacement are recorded.

Compression tests result in mechanical properties that include the compressive yield stress,

ultimate compressive stress (brittle materials), compressive modulus of elasticity, and proportional

limit.

Compressive yield stress is measured in a manner identical to that done for tensile testing. Ultimate

compressive strength σmax is the stress required to rupture a specimen. It is obtained by dividingmaximum compressive load F max by the original specimen cross sectional area A0.

0maxmax /AF σ = [N/m2 or Pa] (3)

Compression members, such as columns, are mainly subjected to axial forces. The failure of a short

compression member resulting from the compression axial force looks like in Figure 2.

However, when a compression member becomes longer, the role of the geometry and stiffness

(Young's modulus) becomes more and more important. For a long (slender) column, buckling

occurs way before the normal stress reaches the strength of the column material as shown in

Figure 3. For an intermediate length compression member , kneeling occurs when some areas yieldbefore buckling, as shown in the Figure 4.

Brittlematerial

Ductilematerial

Shortcompression

member

F F

F F

Fig. 2 Short member in compression testing

Long compression

member Buckling

F

F

Fig. 3 Long member in compression testing

Intermediatecompression

memberKneeling

(Inelastic buckling)

F

F

Fig. 4 Intermediate member in compression testing



13

The failure of a compression member has to do with the strength and stiffness of the material and

the geometry (slenderness ratio) of the member. Whether a compression member is considered

short, intermediate, or long depends on these factors.

In practice, for a given material, the allowable stress in a compression member depends on the

slenderness ratio L/r and can be divided into three regions: short, intermediate, and long.

Short columns are dominated by the strength limit of the material. Intermediate columns arebounded by the inelastic limit of the member. Finally, long columns are bounded by the elastic limit

(i.e. Euler's formula). These three regions are depicted on the stress/slenderness graph below

(Figure 5).

The short/intermediate/long classification of columns depends on both the geometry (slenderness

ratio) and the material properties (Young's modulus and yield strength).


1. Universal testing machine – Instron Series 8500.

2. Vernier caliper

3. Compression specimen: wood

Fig. 5 Relationship between compressionstrength with slenderness ratio

S h o r t

I n t e r m e d i a t e

L o n g

L / r

F / A

Euler’s formula(Elastic stability limit)

Inelastic stability limit

σU

(Strength limit)

Fig. 6 Instron Series 8500 (100kN)



14

Procedures

1. Use vernier caliper to measure the original size of the specimen.

2. Center the specimen between the compression test plates.



5. Press start button to start the compression test.

6. Observe the specimen, as the load is gradually applied.

7. Record the maximum load and continue loading until complete failure.

8. Stop the machine and remove the specimen.

9. Repeat experiment with other specimens.

10. Observe and describe the type of failure for each specimen.

PRECAUTIONS

I. NEVER OPERATE THE UTM WHEN SOMEONE’S HANDS ARE BETWEEN THE GRIPS.

II. ENSURE ALL LAB PARTICIPANTS ARE CLEAR OF EQUIPMENT BEFORE BEGINNING OR

RESUMING TESTING.

iii. STOP THE UTM AS SOON AS THE SPECIMEN FAILS.

Results

1. Show all the measurements of specimen.

i. Original depth d [mm]

ii.

Original width w [mm]iii. Original length L [mm]

2. Calculate the original cross-sectional area A0 [mm2].

3. Plot the load–displacement graph for the tested specimen.

4. On the load–displacement curve show and verify the value for the following points.

i. Maximum load F max [N]

ii. Load taken by the specimen at the time of failure F f [N]

5. Plot the compressive stress–compressive strain graph.

6. On the stress-strain curve show the following points, and verify the value (i) to (iii).

i. Yield stress σy [MPa]

ii. Ultimate compressive strength σmax [MPa]

iii. Modulus of elasticity E [GPa]

iv. Proportional limit



15

Maximum Load,maxF

[N]

Yield Stress

σ [MPa]

Original length

L [mm]

Total length

∆L [mm]

0

Discussions

1. Discuss on the shape of obtained stress–strain curve.


3. Discuss on the mechanical properties of the tested specimen.

4.

Discuss on the factors that can be affected to the experimental result.Conclusions





16

EQ4. TORSION TEST

Objective

The objectives of this experiment are:1. To develop the relationship between torque T and shear stress τ with angle of twist θ.

2. To determine the modulus of rigidity G of the material.

3. To determine the maximum shear stress at the elastic limit and at failure of the material.

Theory

1. General

The torsion test differs from a tensile test in that the former has a stress gradient across the cross-

section of the specimen. Thus at the limit of the elastic range, yielding will first occur in the

outermost fibers whilst the core is still elastic. In a tensile test, yielding occurs relatively evenly

throughout the bar.As the specimen is twisted further into the plastic region, a greater proportion of the cross-section

yields until there is a plastic zone through to the centre of the bar. A typical torque–twist diagram

(Figure 1) is very similar to a load–extension diagram from a tensile test.

2. Stress, Strain and Angle of Twist

Consider a solid circular shaft of radius r and length L, subjected to a torque T [Nm] at one end, the

other end being fixed (Figure 2). Under the action of this torque, a radial line at the free end of the

shaft twists through an angle θ [rad], point A moves to B, and AB subtends an angle γ at the fixed

end. This is then the angle of distortion of the shaft, i.e. shear strain.

Arc AB = r θ = L γ

Angle of twist

T o r q u e

Yield

Slope = GJ /L (Torsional rigidity)

Fig. 1 Torque–twist diagram



17

L θ r γ /= (1)

But γ = τ /G, where τ [N/m2 ] is the shear stress and G [N/m2 ] is the modulus of rigidity .

By substitution and rearranging

L θ Gr τ // = (2)

The torque can be equated to the sum of the moments of the tangential stresses on the elements

2πrδr .

( )∫= r rdr π τ T 2 ( ) ( )∫= 22/ r rdr π L θ G ( ) J L θ G /= (3)

where J [m4] is called the polar second moment of area (= πr 4/2).

Combining (2) and (3) produces the so-called simple theory of torsion, and gives the fundamental

relationship between shear stress, torque and geometry.

L θ Gr τ J T /// == (4)

Showing that, for a given torque, the shear stress is proportional to the radius.

The shear stresses which are developed in a shaft subjected to pure torsion are indicated as

( )r L θ Gτ /= (5)

Now from the definition of the modulus rigidity G, τ = G γ . It therefore follows that the two

equations may be combined to relate the shear stress and strain in the shaft to the angle of twist per

unit length, thus

( ) γ Gr L θ Gτ == / (6)

This equation indicates that the shear stress and shear strain vary linearly with radius and have

their maximum value at the outside radius (Figure 3).

Fig.3 Maximum shear stress

r

τ max

τ max

Fig. 2 Schematic representationof torsional deformation

L

T

T

θ

r

A

B

γ



18


1. Torsion testing machine – Norwood 50 Nm

2. Vernier caliper

3. Torsion specimens: steel, brass, aluminium

Fig. 5 Norwood 50 Nm

Procedures

1. Measure the initial length and initial gauge length diameter of the specimen.

2. The specimen between the loading device and the torque-measurement unit into the

straining hexagon sockets.

3. Turn the hand-wheel as required aligning the specimen.

4. Slide the tailstock unit so that the specimen is fully into the hexagon sockets.

5. Ensure that there is no preload on the specimen.

6. Zero the pointer on the zero degree point on the protractor scale.

7. Adjust the digital torque meter reads zero.

8. Turn the hand-wheel clockwise slowly to load the specimen. Turn it only for a defined angle

increment.

9. Read the torque value from the digital torque meter and notice it together with the

indicated angle of twist.

10. Continue the process of steps 8 and 9 until fracture occurs.

11. Repeat the experiment for other specimens.



19

Results

1. Specimen Dimensions:

DimensionDetermination no.

Average

1 2 3

Initial gauge length diameterd 0 [mm]

Initial lengthL 0 [mm]

Final gauge length diameterd [mm]

Final lengthL [mm]

2. Angle of Twist and Torque.

Angle of Twistθ [°]

Angle of Twistθ [rad]

TorqueT [Nm]

10

20

30

40

50

60

.

.

.

θ F (Fracture angle)

Table 1 Experimental results

3. Plot the torque T versus angle of twist θ graph for the tested specimen. From the graph,

calculate the modulus of rigidity G, and determine the slope of the elastic part.

4. Calculate the shear stress τ at the proportionality.

5. Plot the shear stress τ against angle of twist θ graph.

6. Plot the shear stress τ against shear strain γ graph. Tabulate the following values and show

them on the τ– γ curves

i. Proportional limit shear stress in torsion

ii. Shear modulus of elasticity

iii. shear stress

Discussions


2. Discuss on the mechanical properties of the tested specimens in shear.



20

3. Discuss on the factors that can be affected to the experimental result.

Conclusions





21

EQ5. FLEXURAL TEST

Objective

To study and examine the flexural properties of materials.

Theory

Figure 1 shows a length of beam under the action of a bending moment M [Nm]. O is the centre of

curvature, and R is the radius of curvature of the neutral surface NN. The beam subtends an angle θ

at O. Let σ [N/m2] be the longitudinal (bending) stress in a filament AB at a distance y from NN.

Then the strain in AB is

( ) NN/NNab/ −=Eσ ( )[ ] θ θ θ R R y R /−+=

R y /= or

R E y // =σ (1)

where E [N/m2 ] is Young’s modulus.

Consider a cross-section of the beam. If δA is an element of cross-sectional area at a distance y from

the neutral axis XX (Figure 2) then for pure bending the net normal for the cross-section must be

zero, i.e.

0=⋅∫ dAσ or ( ) 0/ =⋅∫ dA y R E

This is the condition that XX passes through the centroid of the section.

Fig. 1 Beam subjected to purebending, after the moment M has been applied

θ

O

M M

NNa b

R

y

Fig. 2 Beam cross-section

y

δA

X X

Y

Y



22

The bending moment is balanced by the moment of the normal forces about XX, i.e.

( )∫ ∫ ⋅⋅=⋅= dA y R EdA y M 2/σ R EI /= or

R EI M // = (2)

where I [m4] is a property of the cross section known as the moment of inertia or second moment

of area. For rectangular I = bh3/12, b: width, h: thickness; and for circular I = πd4 / 64, d: diameter.

Equations (1) and (2) now can be combined and written as the convenient form.

R E y I M /// ==σ (3)

From equation (3), the maximum stress obtained in any cross-section is given by

tmaxtmaxt // Z M I My ==σ

cmaxcmaxc // Z M I My ==σ (4)

where the subscripts denote tension and compression. The quantities I / yt max and I / yc max are

functions of geometry only; they are termed the section modulus and are denoted by Zt and Zc [m3].

For rectangular Z = bh2 / 6, b: width, h: thickness; and for circular Z = πd 3 / 32, d: diameter.

In the unsymmetrical sections such as T-sections where the values of ymax will be different on each

side of the neutral axis (NA) (Figure 3), and here two values of section modulus are often quoted,

11 / y I Z = and 22 / y I Z = (5)

each being then used with the appropriate value of allowable stress.

In case of a beam (b: width, h: thickness) with concentrated load W [N] at the centre (Figure 4), the

beam deflection y [mm] can be expressed in the form.

σ t

σ c

+

–

NA y 1

y 2

Fig. 3 Typical bending stress distributions

σ t

σ c

+

–

NA y 1 = y –

y 2 = y –



23

EI Wl y 48/3= (6)

where l is the beam working length, EI [Nm2] is known as the flexural rigidity of the member.

The flexural stress σ [N/m2] caused by bending is calculated by the following formula:22/3/ bh Wl Z M ==σ (7)

where bending moment M = Wl / 4. If the load recorded corresponds to the value at which failure

occurs, then σ corresponds to the flexural strength σf .

The maximum strain εf due to bending (compression at the side contacted by the loading and

tensile at the opposite face) is estimated by using Hooke’s Law,

2

maxf /6/ l hy E == σ ε (8)

Specimen and Equipments

1. Instron Series 8500 (5kN).

2.

Vernier caliper.3. Test jig

4. Loading block

5. Flexural specimen

W

Fig. 4 The beam with concentrated loadW at the centre

l / 2

x

y

l

Fig. 5 Instron Series 8500 (5kN)



24

Procedures

1. Measure the thickness and width of the beam.

2. Grip the loading block and test jig in the upper and lower gripping head, respectively.

3. Locate the specimen (Figure 5) so that the upper surface is to the side and centered in

loading assembly.4. Operate the machine until the loading block is bought into contact with the upper surface of

the specimen. Ensure to secure full contact between the loading (and supporting) surfaces

and the specimen.



7. Press Start button to start the flexural test.

8. Observe the specimen, as the load is gradually applied.

9. Record the maximum load and continue loading until complete failure.

Results

1. Show all the measurements of beams.

Beam lengthL [mm]

Beam widthb [mm]

Beam thicknessh [mm]

Beam working lengthl [mm]

Specimen 1

Specimen 2

Specimen 3

2. Plot the load–deflection graph for the tested specimen.3. Complete the table below.

Experiment Theory

Flexural strength σ f [MPa]

Maximum flexural strain ε f

Flexural modulus Ef [GPa]

Discussions

1. Discuss on the shape of obtained load–deflection graph.

2. What is the percentage error (%) between experiment results with the theory? Why?

3. What is the critical application of the experiment in industry?

Conclusions




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Fig. 3 Cantilever beam with aconcentrated load and supported by

fixed-end support

Fixed-end

W

Fig. 2 Overhanging beam withconcentrated and distributed loads,and supported by pinned supports

Pinned Pinned

W w

Fig. 1 Simply-supported beam withtwo symmetric concentrated loads andsupported by pinned and rollersupports

W W

Pinned Roller

EQ6. CANTILEVER TEST

Objective

1. To identify the supports reaction in simply-supported and overhanging beams.

2. To develop an understanding of beam apparatus, and to determine its sensitivity and

accuracy.

Theory

1. General

A beam is a member which has the primary function of resisting transverse loading. Beam is one of

the simplest structures in design but one of the most complexes to analyze in terms of the external

and internal forces acting on it. The complexity of its behavior under load depends on how it is

supported - at one or both ends - and how its ends are attached to the supports. Three basic beam

types are the simply-supported, overhanging, and cantilever beams.

A beam supported by a support at the ends and having one span is called a simply-supported beam(Figure 1). A support will develop a reaction normal to the beam but will not produce a couple. If

either or both ends of the beam project beyond the supports, it is called overhanging beam (Figure

2). A cantilever beam is one in which one ends is built into a wall or other support so that the built-

in end can neither move transversely nor rotate (Figure 3).

2. Types of Load

A beam is normally horizontal, the loads being vertical, other cases which occur being locked upon

as exceptions. The two types of loads for beams are concentrated and distributed loads.

i) A concentrated load W [N] is one which is considered to act at a point, although

in practice it must really be distributed over a small area (Figures 1, 2 and 3).



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ii) A distributed load w [N/m] is one which is spread in some manner over the

length of the beam. The rate of loading may be uniform, or may vary from point

to point along the beam (Figure 2).

3. Types of Support

The deformations and stresses which result in a beam owing to a particular load (concentrated

load) or group of loads (distributed load) are dependent on the manner in which the beam issupported. The three basic types of supports for beams are roller, pinned and fixed-end.

i) A roller support is one which exerts a reactive force having a known line of

action (Figure 1).

ii) A pinned support in one which allows the beam freedom to rotate but prevents

it from any linear movement (Figures 1 and 2).

iii) A fixed-end support is one which prevents the beam from translating or rotating

at the point of support (Figure 3).

4. Deflection of Cantilever

The deflection under the load for a cantilever loaded at the free end is given by

EI

WL z

3

3

= (1)

If EI and L are maintained constant then:

W k z .1

= (2)

Where1k is constant

Similarly if EI and W are maintained constant:

3

2. Lk z = (3)

Likewise E

k z 3= and

I

k z 4= if E and I respectively are made the variables.


1. Beam apparatus – SM104

2. Cantilever support

3. Vernier caliper

4. Load cells

5. Dial gauges

6. Steel blocks

7. Beams: Steel / Brass / Aluminium

Fig. 4 Beam apparatus – SM104



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W

L

Procedures

1. Set up a load cell at a convenient position near to one side of the frame.

2. Set up the clamp to give a cantilever of convenient length.

3. Pass one end of the beam through the clamp and rest the other end on the load cell. (It is

convenient to lock the knife edge during assembly). Tighten the clamp and tie up thefree end of the beam using a short piece of string.

4. Place the dial gauge near to the clamp and set the zero. Move the dial gauge to the free

end of the cantilever, unlock the knife edge and adjust it so that the dial gauge returns to

zero. Set the pointer of the load cell to zero.

5. Adjust the knife edge upwards to give a convenient reading on the load cell. Record the

load and the dial gauge reading.

6. Adjust the knife edge upwards to give a number of load increments recording loads and

dial gauge readings.

7. Return the knife edge to its initial position; lock the knife edge; slacken the clamp and

move it to a new position (this is more convenient than moving the load cell).8. Repeat the experiment for several lengths of cantilever.

9. Use the Aluminum, Brass and Steel beams 6 mm thick to varying E value.

10. Use the steel beams 3 mm, 4.5 mm and 6 mm thick to vary I.

Load cell

Dial Gauge(z)

Cantileversupport

Fig. 5 Schematic of Experimental Setup

W

L

Load cell

Dial Gauge(z)

Cantileversupport



Results

1. Record the values of z according to the following table.

Load

[N]

Deflection z [mm]

L=200mm L=300mm L=400mm L=500mm L=600mm

2

4

6

8

10

2. Plot graphs and verify the accuracy of the results

i. z against W to verify equation (2) is correct, with k 1 for each length being given

by gradient of the graph.

ii. z against L3 verifies that equation (3) is correct, with k 2 for each load being given

by the respective gradient of the graph.

Conclusions


Emm 3490 Lab Sheet Strength of Material e

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