Report strength of materials

CONTENT Lab 1,2: TENSION EXPERIMENT OF DUCTILE AND BRITTLE MATERIALS Lab 3,4: COMPRESSION OF DUCTILE AND BRITTLE MATERIALS STEEL AND CAST IRON Lab 5: ELASTIC MODULUS E WHEN PULLED OR COM

Ho Chi Minh City University of Technology

Civil Engineering

REPORT Strength of Materials

Lecturer: Mr LÊ ĐỨC THANH

Nguyễn Trung Kiên 1852489

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CONTENT

Lab 1,2: TENSION EXPERIMENT OF DUCTILE AND BRITTLE MATERIALS

Lab 3,4: COMPRESSION OF DUCTILE AND BRITTLE MATERIALS (STEEL AND CAST IRON)

Lab 5: ELASTIC MODULUS E WHEN PULLED OR COMPRESSED

Lab 6: ELASTIC MODULUS G WHEN TWISTED

Lab 7: DISPLACEMENT AND ROTATION OF CANTILIVER BEAMS SUBJECTED TO ASYMMETRIC BENDING

Lab 8: DISPLACEMENT OF CANTILIVER BEAMS SUBJECTED TO COMBINED BENDING MOMENTS

Lab 9: CRITICAL AXIAL FORCE

TENSION EXPERIMENT OF DUCTILE AND BRITTLE MATERIALS 1.1 The purpose of this experiment

-Understand the relationship between force and strain when pulling steel and cast iron samples, and identify mechanical characteristics of steel and cast iron: +About steel: Find - Yield stress  y

-In the chapter about tension and compression, we can know the graph of the

relationship between tensile force and deformation The length of the tensile specimen

is as follow:

Figure 1 Figure 2 The relationship between P-∆L

Stress limits Steel Cast iron

Ultimate stress   u P u /A0   u Pu /A0

Ductility properties for steel:

Relative elongation (%) ô  100%

0 0

1 

L L L

Necking ratio (%)   100%

0 1

0 

A A A

In which L1 is the length at failure;

L0 is the original length;

A1 is the cross-sectional area at failure;

A0 is the original cross-sectional area

+ The cross sectional area A = a x b 0 0 0

+ The length L =11.30 :ýā for long sample; L = 5.65 0 :ýā for short sample;

L0 = 4:ýā for brittle sample

+ The length L = L + b /2 0 0

The radius R and the head size of specimen depend on the testing machine

1.4 Experimental tools

- A caliper with precision 1/50mm

- A technical scale with precision 0.01g

- A marking tool to mark spacing on the sample

- Tension testing machine M.A.N

1.5 Preparing for the experiment

- Measure the diameter do, L for rectangular samples or a and b for circular samples 0 0 0

- Marking N equal 1-cm intervals on the length Lo of specimens These marks are then used to determine the length L1 at failure

- Predict the materials’ stress limits and tensile strength of the specimen to determine the appropriate load levels

- Select the tensioning clamp and load level of the machine appropriately to the specimen

- Place the specimen in the tensioning clamp, control the force and pen on the plotter

Determine the length L of the specimen at failure: L =17.098 cm 1 1

Relative elongation (%): ô 100%

0 0 1





L L L

1414098.17

0 

A A A

With d = 1.421 cm => A = (1.4210 0 2�㔋)/4 = 1.586 (cm2 )

d = 0.920 cm => A = (0.9201 1 2�㔋)/4= 0.665 (cm ) 2

With d , d0 1 are the initial and failure diameter of sample

A , A are the initial and failure area of cross section 0 1

586.1665.0586.1

For cast iron

Ultimate stress  u = Pu/A0 = 38500/1.586 = 24274.9 (N/cm2)

Graph

1.8 Comments on the results

- After clamping the sample, start increasing the force step by step In the first stage, we see that the force increases and the strain also increases P and ∆L are related to each other

in a first-order straight line This stage is called the elastic phase

-Then continue to increase the load, but the load meter increases insignificantly while the strain gauge increases rapidly, the graph has an almost horizontal curve This stage is called the yield phase We can determine Py = 49000(N), the stress of yield strength is  y = 30895.3 (N/cm ) 2

- Continue to load the load meter increases and the strain meter also increases, so the material begins to cope with the force, the graph has a curve that does not follow a definite function This stage is called the re-stable stage, Pu =56000 (N) is determined, the stress of the ultimate strength is  u = 35308.9 (N/cm ) and at this position the sample begins to 2

p

fluctuates and after a while, there is an explosion and the sample is broken at the knot position

- Then take out the sample to check, we see that the diameter at the waist is only 0.920 (mm), 0.501 (mm) down compared to the original, the relative constriction is 58.07% The length after breaking is 17.098 (cm), increasing by 3.098 (mm), we determine the relative elongation is 22.13%

- When starting to increase the load for P, the strain increases, after a period of time, until

P is large enough to make the ribbed steel bar break Observing the severed steel section, the knot formation is not significant

Testing specimens

Cast iron

Steel

COMPRESSION OF DUCTILE AND BRITTLE MATERIALS (STEEL AND

CAST IRON) 2.1 Objectives

Find yield compression stress for steel and ultimate compression stress for cast iron

2.2 Theoretical background

In this chapter, the relationship between longitudinal elongation L and the tension force 

P as shown in the Figure 8

Figure 8 Figure the relationship between P-∆L when compress cast iron Yield stress:y =P /Ay 0

2.3 Test specimen

Cylinder specimen with the height h, the depth d0 and 1h/d0 3 is shown in figure 9

p

A caliper with precision 1/50mm

2.5 Preparing for the experiment

-Measured d and h 0

-Calculate A , the predicted largest force for determining the load level of the machine 0

- Place the specimen between the two compressing plates to be ready for axial compression

- Adjust the number 0 (if required), control the diagram plotter

2.6 Experiment procedure

- Turn on the machine to increase the force slowly

- Observe the force-deformation diagram and read force Py in the yield stage, continue to increase

the force to 70-80% of the load level and then stop

- Release the force and take the specimen

2.7 Calculation

For steel

Figure of steel before Figure of steel

being compressed after being compressed

For cast iron

Ultimate stress when compression:u =Pu/A0 =123000/(�㔋1.6422/4) = 58085.7 (N/cm2)

P :u Force when the sample is damaged

A: Area of cross section, A = (�㔋d2/4)

Initial

2.8 Comments on the results

-Mechanical properties of steel in compression

Steel is a material that has good compressive strength and does not break apart like it does when pulled

In the compression process, it is very difficult to determine Pch

Two compression and drag processes for σchk≈ σchn P increases much, ∆L increases little under compression

Explain the type of steel sample and failure mode after testing

The sample has the shape of a drum after compression because under the action of force P and the friction of the table, it prevents the expansion of the two ends, leading to the sample having a blank shape (different from the hypothesis of a round prismatic shape)

- For a flowing (horizontal) material there is no plastic deformation, other than showing elastic deformation One feature of crevice failure is that the two broken faces can be joined together to restore the original material form The stress-strain curve for flowing materials has a linear form Mechanical testing on many identical samples will have different stress-strain results The tensile strength is very small compared to the compressive strength and

it is generally assumed to be zero for many applications It can be explained by the Stress

ELASTIC MODULUS G WHEN TWISTED

6.1 Objectives

- Determine the shear modulus G of the steel and test Hooke's law

6.2 THEORETICAL BASIS

- In pure twisting of a circular cross-section bar, the relative torsion angle between the

two cross sections A and B at a distance LAB equal to:

ýþ=Āÿ ÿ ýþ

�㔺�㔼 �㕃 => G =Āÿ ÿ ýþ

�㔼 � 㕃  ýþ Where: M - torque (constant over Lz AB length)

I - the polar moment of inertia of the cross-section If we determine get P

Mz , LAB, I and measuring P AB, G can be inferred

6.3 SAMPLE FOR EXPERITMENT

Measuring instruments slip modulus G

The test specimen (1) is a bar with a circular cross-section, one end is clamped to the

jaw, the other end put in ball bearings (bearings) (2) can be rotated freely, outside there is

1 spare end to attach a crossbar (3) for hanging weights producing torque M Between z

mount and ball bearing fitted with 2 crossbars (4) at A and at B, at the end of each

crossbar, place a successive displacement (see figure)

When placing the weight, the bar is subjected to pure torsion, at A, B there are torsion

angles A, B (angle absolute twist between A, B with the jaws) causes the two 

horizontal bars (4) to rotate and the ends of the two rods horizontal displacement With

the displacement meter we can measure the displacements A, B and have

ý~�㕡āýÿ= �㔑&ý þ≈ �㕡ā�㔑ÿ � 㕩=&þ

From there we get: �㔑ýþ= �㔑ý2 �㔑þ

6.4 TOOLS EXPERIENCE EQUIPMENT –

- Calipers; Weighing hangers and weights

- Two displacement meters accurate to 0.01mm

6.5 TEST PREPARATION

- Measure the inferred diameter of the sample �㔼 � 㕃4=�㔋Ă0

32

- Measure distance b to deduce the maximum load class (weight) placed on the system

- Measure distance LAB and a

- Attach the weight hanger to the system

- Set the displacement next to the horizontal bar

- Create a table to record the results as follows:

displacement meter

weight

No Load Readings on displacement (x10 mm)

6.6 CONDUCTING EXPERIMENT

- See the weight of the hanger and crossbar (3) as the initial load P , read A , B over 2 0 0 0

displacement meters (reads can be adjusted to 0) - Put a 1kG weight on the hanger (i.e P 1

= P + P= P +1kG) read the readings A , B corresponding 0  0 1 1

- Place one more 1kg weight in turn on the hanger and read the corresponding readings

6.7 CALCULATION

- a=100mm, b=380mm, d =20mm, L0 AB=100mm

- Calculate the torsion moment (torque): Mz = P.b = 1.380 = 380 kg.mm 

- Calculate the average of the two reading numbers on displacement meters:

&ý ÿ�㕣ă =&ýÿ

ÿ =17.8x103 −2= 5.93 10 ý 22 ÿÿ; &þ ÿ�㕣ă =&þÿ

- Calculate the average relative angle of twist between A and B:

&�㔑 ÿ�㕣ă = �㔑 ý, ÿ�㕣ă 2 �㔑 þ, ÿ�㕣ă = 5.93 10 ý 24 2 2.93 10 ý 24 = 3ý10 24

- Calculate the elastic modulus G when twisted:

�㔺 =&Āÿ ÿ ýþ

&�㔑 �㕎�㕣�㕒 �㔼 � 㕃= 380ý100

3ý10 −4 ý�㔋ý20432 = 8063 kg/mm2

i i

6.8 COMMENT ON THE RESULTS

- Comment on the linearity of the readings on the displacement (Hooke's law test): The graph represents the relationship P and the readings Ai and B are straight lines i

because the values of &A and &B fluctuate slightly around the mean value Combined with the graph, we see that Hooke's law is satisfied

- Compare the G result found in the experiment with G calculated by the formula: �㔺ā/ăĀÿþ2(=1+�㔇ā) or Gtheory=8100kg/mm2

- For E, μ obtained from the experiment we can calculate Gtheory (E = 200.000N / mm2, 

= 0.3)

%&= |�㔺ā/ăĀÿþ 2 �㔺 ÿăÿýÿāþ

�㔺 ÿăÿýÿāþ | ý 100% = |81008100 2 8063| ý100% = 0 46%

- With a small error %&=0.46%,so the results are relative accurate

- Accuracy of measuring instruments, measurements and calculation methods are quite accurate, not too complicated for experimenters Pay close attention to the process of

0

5.8

11.8 17.8

0 1 2 3 4 5 6 7 8 9 10

DISPLACEMENT AND ROTATION OF CANTILIVER BEAMS SUBJECTED

TO ASYMMETRIC BENDING

7.1 Objectives

Measure the displacement (deflection) and the angle of twist (at a number of cross sections)

of a rectangular beam under asymmetric bending Compare the measured results with the value calculated according to the formula in order to check the formula

6ā�㔼 ý (3�㔿 ÿ 2 �㔿 þ )

�㔑ÿ2ā�㔼=2�㕃�㔿ÿ

ý

It is possible to use displacement meters to directly measure the displacements at points B,

C and D on beams, and then compare with the displacement calculated using the theoretical formula above In addition, the beam's elastic line in the CD segment is the first order, so

it is possible to calculate the angle of twist at C based on the displacements:

7.5 Preparing for the experiment

Measure the dimension b and h of the specimen cross section and distance L , L , L B C D

Install the displacement meters at positions B, C, D

Place the weight to create the force at C

Record of the results:

-Consider the weight of the hanger is P , adjust the displacement meter to 0 0

- Place the weights to create the force P , P , P , with the constant loading increment of 1 2 3

7.8 Comments on the results

The read numbers are almost linear Based on the graph, we see that the values of

△ þÿ, △ ÿÿ, △ Āÿ are almost constant The graph of the relationship between forces and displacements B, C, D is a linear function This means that when the force P increases by

a constant amount of △ �㕃, �㕡/ÿ △ þ,△ ÿ,△ Ā increase by a constant amount Observing,

we see that the recorded data has errors between the practical values and the theoretical values The cause may be due to inaccurate operator manipulation, inaccurate reading and

2.02 3.62

DISPLACEMENT OF CANTILIVER BEAMS SUBJECTED TO COMBINED

BENDING MOMENTS 8.1 Objectives

Determine the direction and total displacement f of cantilever beams (console beams) subjected to combined bending moments

Compare with the theoretical results

In which, l is the distance from the load P to the fixed end; �㔼ý12=; �㔼Ā/3 þ12=Ā/3The total displacement of the center O: Ā = Ā√ý2+ Āþ2

The angle between the displacement f and the symmetry axis y is :  �㕡�㕎Ā �㗼 =Ąý

+ Record the results:

- Consider the hanger weight and bar weight as the initial load P , read the numbers on 0

two displacement meters x , y 0 0.

- Place the 1kG weights (i.e add P = 1kG) to create the load P , P Record the  1 2

displacements x , y1 1, x , y2 2 corresponding to each load

- The increment of the displacement should be a constant If not, it is necessary to review the layout of the experiment and redo

- Draw graph representing the relationship between the load P and Āý,Āþ

- Calculate Āý, Āþ based on theory

=>Āý =�㕃ý ý 3

3ā�㔼 þ=� 㕃 Āÿÿ �㔑ý 3 3ā�㔼 þ =3⋅2000⋅25921⋅Āÿÿ 300⋅500= 0.3 402 ÿÿ

=>Āý =�㕃þ ý 3

3ā�㔼 ý=3ā�㔼�㕃 Āÿÿ �㔑ý3

ý =3⋅2000⋅58321⋅Āÿÿ 300⋅500= 0.309ÿÿ 3

8.8 Comments on the results

- The graph shows that the relationship between P, f , f is a linear first-order x y

0 50 100 150 200

The critical axial force of a bar is determined by Euler's formula as follows

- Two pin connections μ = 1 (case 1)

- One pin and one fix μ = 0.7 (case 2)

- Two fixed ends μ = 0.5 (case 3)

- One fixed end and one free end μ = 2 (case 4)

We need to gradually increase the value of the compression force P and note the force value when the bar is bent without returning to the original form and the theory has shown that

the force P is the critical force

Steel bars (2) are made of spring steel of cross section b x h and the length L with elastic modulus E

The load is applied using a set of 5N and 1N scales