Strength of Material

1. TENSILE TEST ON MILD STEEL ROD 1.1

AIM

To determine Yield Stress, Ultimate Stress, Breaking stress, percentage reduction in area, percentage elongation over a gauge length and modulus of elasticity of the given specimen.

1.2

APPARATUS

Universal testing machine, extensometer, scale, vernier caliper, etc. I.3

PREPARATION

1.3.1 REFERENCE

IS 1608 Method for tensiie testing of Steel Products. I.3.2 INTRODUCTION

Robert Hooke, an English scientist, in 1678 formulated Hooke'slaw and it states that "within elastic limit, the stress developed is directly proportional to the strain."

P

Stress = il

Strain =

a1

-

I

Stress a Strain

P I Modules of elasticity = E = - x A

dl

When a steel bar of uniform cross section is subjected to tension test, the stress strain graph will be as shown in the figure. The test piece is subjected to uniformly increasing load and the strains at various stresses are measured with an extensometer. In case of mild steel in the initial stages, strain is proportional to stress till the limit of proportionality is reached. I n this range the material obeys Hooke's law. If the stress is increased beyond this limit, the material behaves in an elastic manner. Beyond this limit the rate of increase in strain will be more till the yield point is reached and the stress is known as yield stress. Actually at this point there is a drop in stress and yielding commences. Therefore, there are two yield points, the upper and lower. After yielding and further increase in stress will cause considerable increase in strain and the curve reaches .the maximum point of ultimate load and the stress is known as ultimate stress. At this stress, 1

the bar will develop neck and the stress will decrease and the bar will break (rupture) and the stress is known as breaking stress or rupture strength. As concrete is weak in tension, steel bars (Mild steel and High yield strength steel) are used in reinforced concrete to strengthen the concrete as steel is very strong in tension. Structural elements like beams and slabs are subjected to tensile stresses due to flexure in these elements. It is therefore essential that steel bars should be tested in tension. 1.3.3 GRAPH -

-

Stress Strain Diagram Mild Steel CIA

FARTIAL.1.Y PLASTIC _

,

r

_

_

-

-

"

-

, ; 7 - +.., ; True s:ress-

-

4

Yield

s~ress

( i t is :++e stress a l

-

fai!Ll:e/

0

T Linear range

P E Y U B

strain

f

Limit of Proportionality Elastic Limit Yield Point Ultimate Point Breaking Point

1.3.4 FIXING OF THE LOAD RANGE

Assume ultimate stress of mild steel as 500 ~ l m r n ~ . Ultimate load = area X assumed ultimate stress Choose the load range within the ultimate load. 1.3.5 USE OF EXTENSOMETER

For the use of extensometer, use the extensometer up to say 0.7 times of yield load. Assume yield stress at 250 N/mm2 for MS bars and 400 N/mm2for tor steel Yield load = yield stress x area of cross section Use of extensometer = 0:7 x 250 x cross sectional area

Use the extensometer up to ................N Gauge length for determining percentage elongation = 5.05

mm

Where 'a' is the area of cross section in mm2.

PRE-LAB QUESTIONS Hook's Law. Define Poisson's ratio. Differentiate Ductile and Brittle materials. 4. Derive the relationship between the Young's modulus, Poisson's ratio shear modulus and Bulk modulus. 5. Draw the Stress-Strain diagram for ductile material.

1.3.6 1. 2. 3.

1.4

PROCEDURE

1.

Measure the diameter of the rod and calculate the area of cross section.

2.

Mark the gauge length by means of fine mark to determine elongation.

3.

Choose the load range in the universal testing machine, appropriate to the given diameter of the rod, assuming ultimate stress.

4.

Fix the specimen in between the grips, care should be taken to ensure that test pieces are held in such a way that the load is applied as axially as possible.

5.

Note the gauge length of the extensometer and fix the extensometer to the specimen.

6.

Apply the load to the specimen and note the extensometer readings at uniform increment of loads.

7.

Remove the extensometer before the yield point.

8.

Note down the yield point load shown by the backward movement of the live pointer and stationery for a short duration of time.

9.

Increase the load further and note down the ultimate load.

10.

Note down the breaking load at which the specimen breaks.

11.

Release the specimen from the grips and measure the neck diameter and elongated length of gauge length.

j2.

Draw a graph stress (Y-axis) versus strain (X-axis) for ductile and brittle materials.

1.5

OBSERVATION Specimen Diameter of the rod Area of the spe,cimen Least count of the extensometer The gauge length 1, The setting length l2 Yield point load Ultimate load Breaking load Final neck diameter Final gauge length Neck area

i) ii) iii)

iv) v) vi ) vii) viii) xi) x) xi) xii)

: Mild Steel 1 Tor Steel.

-

-

= Idiv = 0.01 mm 1,OOmm 300mm ................ N

-

-

................N .................N ............... .mm

............... .mm2 ................ mm

1.5.1 TABULATION:

Straine= Deformation(?I)

Load

[T)

SI.

No.

\

Based on Kg

N

Extenso meter(mm)

Scale (mm)

Extensomete r reading

Stress

'P' Scale readin g

(N /mm2)

1.5.2 CALCULATIONS ?.

Yield stress

-

Yield po int load ~/mm* Cross Sectional Area

Young's modulus 'E' E = ple (N/,m2)

2.

Ultimate stress

3.

Actual Breaking stress

=

4.

Nominal Breaking stress

=

Rreliking lotrri -Cross Section~~l ,I seu

5.

Percentage elongation

=

Change in length xlO0 % Originctl Ietzgt h

6.

Percentage reduction in cross sectional area=

HI-ecr king -loc10 iYeck : I r e ~

~ / m m ~

C'hujigr in ur-erl ~ 1 0 0O/O Origirzitl ureu

I.5.3 FROM GRAPH

in which E P -A -I ?I = I.6

'

Young's modulus ~ l r n m ' Load in N Area of cross section in mm2 Gauge length o f specimen in mm Change in length of the specimen in mm

POST-LAB QUESTIONS 1. In what region of a stress vs. strain graph do you find Young's Modulus?

2. What is the difference between the engineering and true stress - strain? 3. In stress-strain diagram, why the curve lowering and then raising after reaching yield point? 4. In machining process, how to differentiate ductile and brittle materials from visual?

1.7

RESULT

1.

Yield stress

-

2.

Ultimate stress

- ...................... Nlmm2

3.

Actual Breaking stress

-

..................... Nlmm2

4.

Nominal Breaking stress

-

.................... Nlmm2

5.

Percentage elongation

-

.....................%

6.

Percentage reduction in cross sectional area

=

.................... %

7.a.

Modulus of elasticity (from graph)

-

....................Nlrnm2

7.b.

Modulus of elasticity (mean value) from tabulation

-

.........!..........Nlmm2

..................... N h m2

SIGNIFICANCE OF THE TEST

The stress characteristics of mild steel and torsteel can be found with the help of this test. The working stress of steel is important in design purposes. As per IS the permissible stresses for mild steel is 140 ~ l m m 'upto 20 mm and that for High yield strength bars is 230

~ / m2. m

2. TORSION TEST

AIM

2.1

To find the modulus of rigidity of the given material and to plot a graph between torque and twist.

2.2

APPARATUS I.Torsion testing machine 2. Scale 3. Vernier Caliper 4. Thread

2.3.1 THEORY The purpose of torsion testing usually parallels that of uniaxial tension tests. From the experiment, the shear elastic modulus (G), shear proportional stress (tp), shear yield stress (t y), and the stress-strain behavior in general, can be obtained. For a circular shaft, when subjecting to twisting couples or torques, there will be torsional deformation occurring. The distribution of stress in the cross section of a circular shaft is statically indeterminate. However, every cross section remains plane and undistorted. I.Strain :

The shearinq strain in a small element with sides parallel and perpendicular to the axis of the shaft and at a distance pfrorn that axis can be expressed as:

where 4 is the angle of the twist for a length L of the shaft. The above equation shows that the shearing strain in a circular shaft varies linearly with the distance from the axis of the shaft. Therefore, the strain is maximum at the surface.

2. Stress : Within the elastic range, by using the Hooke's law, the stress z is

'

where G is the modulus of rigidity of the material. It shows that the shearing stress in a circular shaft also varies linearty with the distance from the axis of the shaft. Therefore, the stress is also maximum at the surface. 3. Relation between torque and the angle of twist

Within the elastic range, the angle of the twist torque T applied to it with the relationship of

4

of a circular shaft is proportional to the

The value of modulus of rigidity can be found out through observations made during the experiment by using the torsion equation:

I

I B Where T=torque applied, I,= polar moment of inertia, C=modulus of rigidity, = Angle of twist (radians). and

1,

I= gauge length. 2.3.2 PRE-LAB QUESTIONS 1 . What are the assumptions made in the theory of torsion. 2. What is resisting Torque? 3. Define polar moment of ipertia. 4. Write the torsion formula. 5. Write the relation between the shear modulus, Young's modulus, and Poisson ratio for brass, aluminum, and steel.

2.4

PROCEDURE

I.

The diameter and length of the specimen is measured using vernier caliper and scale.

2.

The specimen is kept in its position by making the angle of twist at zero position. This is done by making the torque 'zero' by rotating the lever clockwise or anticlockwise and tropometer made to read zero.

3.

A torque was then applied by rotating the lever and corresponding relative angle of

twist is noted.

4.

Similarly the torque is applied by consecutive increase step by step b y rotating the lever and the corresponding twist is noted.

5.

Then after some maximum torque, the torque was decreased step by step and relative angle of twist is noted.

6.

The mean angle of twist

I?'

in degrees and '?' inradians are calculated

.

7.

Modulus of Rigidity & Shear stress for the material are worked out

8.

A graph is plotted between the torque versus relative angle of twist and Modulus of rigidity is calculated.

FORMULA Torsion Equation :

T -----

J

where, T

=

Torque N rnm

J

--

Polar Moment if Inertia mm4

C

=

Modulus of rigidity ~ / r n m-~

? Q

= -

Angle of twist in radians X

I

'

Length of specimen in mrn

d

--

Diameter of specimen in mm

R

=

Radius of specimen in mm

F,

=

Shear stress of specimen in N / mm2

TORQUE DIAL Least count 1 div

= 400kg cm = 40000 N rnm

Tropometer, 1 division = 1"

2.5

OBSERVATION

Diameter of specimen = Length o f specimen =

(mm)

(rnm) Polar moment of Inertia = J = $d4 132 (rnm4)

4

=

Fs

--------R

--

C? ------I

2.5.1

TABULATION:

2.5.3 GRAPH

FROM GRAPH

'f

C = ?H

2.6

POST-LAB QUESTIONS I.What is torsional rigidity? 2 . What are the effects of torques? 3. Why hollow circular shafts are preferred when compared to solid circular shafts?

2.7

RESULT Modulus of rigidity of the given specimen a)

FromTable

-

b)

From Graph

-

3. BENDING TEST

AIM

3.1

To determine the young's modulus of the material of the beam by conducting a deflection test o n simply supported beam. 3.2

APPARATUS

Knife edge supports, weights with hanger, Deflectometer, Calipers, Scale etc. PREPARATION

3.3

3.3.1 INTRODUCTION All structural and machine elements whether, cantilever, simply supported, fixed or continuous undergo deflection when subject to external loads. The deflection o f a member should always be within the specified limits. W e can determine the deflection of beams subject to any type of loading by using standard deflection formulae. The actual deflection of the member is directly proportional to the load and span and is inversely proportional to flexural rigidity (El). Actual deflection so calculated should be less fhan the permissible deflection. In the laboratory, however the deflection is directlydetermined and the young's modulus is calculated using the deflection formulae. Material

Steel

~(~lrnrn*) 2.2 1o5

Aluminium 1.195 x

lo5

3.3.2 PRE-LAB QUESTIONS 1. List the different types of beams and supports. 2. Define pure bending. 3. What are the assumptions made in the theory in simple beams?

4. 5. 6. 7. 8.

Write the flexure formula. Define Moment of Inertia. Define neutral axis of beam. Define Section Modulus. What is point of contra-flexure?

3.4

PROCEDURE

1.

Measure the breadth and depth of the beam

2.

Fix the position of the span, load and deflectometer a s per the sketch.

3.

Note the initial reading of the dial gauge.

4.

Apply the loads at suitable increments, each time noting the corresponding deflection.

Decrease the load uniformity at the same rate and note the corresponding deflection

5.

each time. 6.

Determine the Modulus of Elasticity of the material

7.

Then draw the load ( in Y - axis) Versus Deflection ( X - axis) graph.

8.

Conduct the experiment for the given materials Mild steel, Aluminium and Timber.

9.

Choose y and W from the graph and calculate the Young's Modulus for the material.

3.5

OBSERVA'TIONS Specimen Breadth of cross section Depth of cross section Least count of deflectometer

................. .................mm. .................mm.

-

.................

3.5.1 TABUtA'rION

(a) For central loading (Mild steel, Aluminium and Timber) SI. No.

Load in

Typeof material

Deflection (mm)

N.

Kg.

Loading

Unloading

Mean deflection I,,( mm

Young's Modulus (El ~lmrn~

1.

2.

3. 4.

5. (b) SI. No.

.

For non-central loading (Mild Steel, Aluminium & Timber) (Case I & Case II) Type of material

Kg

1.

2.

3. 4. 5.

Deflection (mm)

Load N

Loading

Unloading

Mean deflection (,,I mm

Young's Modulus (E) ~ l m m ~

3.5.2 CALCULATIONS 1

Moment of Inertia

I

For Central Loading

E = ----48431

= -- x

hl/'

12

Hf/

For Non-Central Loading Case I (x < a)

Case It (x =. a) (x < L) bVbX W ( s - a)' E = ----- (L* - a 2 -x2)+-----6ylL ~ Y I

Y =

Deflection in m m

W =

Load in N

E =

Young's modulus in ~ l r n r n '

I

3.6

=

Moment of inertia in mm4

L =

Span of the beam

a =

Distance between L.H. Support and Load

b =

Distance between R. H. support and load

x =

Distance between L.H. Support and deflectrometer in mm.

POST-LAB QUESTIONS 1. Mention the limitations of double integration method in simply supported beam subjected to multi point loads.

2. How the bending stresses varies from top to bottom layer of the beam? 3. How the shear stresses varies from top to bottom layer of the beam? 4. What are the methods for finding out the slope and deflection at a section?

3.7

RESULT The Young's Modulus of Elasticity the given material = 1. Mild steel For central loading

For Non-central loading Case4 = Case II = 2. Aluminium For central loading For Non-central loading Case I = Case II = 3. Timber For central loading For Non-central loading Case l = Case II =

-

--

SIGNIFICANCE OF THE TEST

If the Young's modulus of the material of the specimen is equal to the standard value specified for ,the material, the deflection is found to be valid.

Formulas Moment of Inertia = I

'

bd 12

=-

.

tvi

(1)

For central loading y =48EI

(2)

For non-central loading (Case I) a > b and x < a (x = Position where deflection is measured)

(3)

For non-central loading (case II) A>banda 5. (b) Shearing, when U D s 2.5. (c) Double barreling, when U D > 2.0 and friction is present at the contact surfaces. (d) Barreling, when U D < 2.0 and friction is present at the contact surfaces. (e) Homogenous compression, when U D < 2.0 and no friction is present at the contact surfaces. (f) Compressive instability due to work-softening material.

Post-lab Questions

1.

Why Perform a Compression Test? Ans: Axial compression testing is a useful procedure for measuring the plastic flow behavior and ductile fracture limits of a material.

2.

What happens to ductile materials such as mild steel when placed in compression? Ans; It starts buckling.

10. SPRING 'TEST

Pre-lab Questions 1.

What is strain energy? Ans: Whenever a body is strained, the energy is absorbed in the body. The energy, which is absorbed in the body due to straining effect, is known as strain energy.

2

What is the principle of conservation of energy? Ans: The principle of conservation of energy states that energy cannot be created or destroyed, although it can be changed from one form to another.

3.

What is a spring? Ans: A spring is an elastic member, which deflects, or distorts under the action of load and regains its original shape after the load is removed.

4.

What are the types of springs? Ans: Helical Springs, Leaf springs and Torsion springs

5.

Define Stiffness of spring. Ans: It is defined as the load per unit deflection.

K= wl6

6.

Define pitch o f the spring. Ans: The distance between the center to center of the wires in adjacent active coils

7.

Define spring index. Ans: It is the ratio of the mean coil diameter to the diameter of the spring wire. C = Dld

10

. ;...7,.'..