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

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

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

1.3 Test specimen

Based on TCVN 197-85 (197-2000), the test sample is shown in the Figure 3 The cross section of samples can be circle or rectangle as shown in Figure 4

For circular cross-sectional samples: + The length L0=10d0 or 5d 0

+ The length L= L + d 00

+ d

+ The cross sectional area A = a x b 000

+ 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 00

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 000

- 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

1.6 Experiment procedure

Slowly increase the tension force, monitor the force meter and the plotter, and read the yield force P (where the force does not increase but the deformation increases), and the y

ultimate force P (maximum force when the specimen is failed) based on the force-u

deformation graph

When the specimen is failed, release the force and take the specimen

1.7 Calculation

Determine the length L of the specimen at failure: L =17.098 cm 11

Relative elongation (%): ô 100%

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

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=Āÿÿýþ

�㔺�㔼�㕃 => G =Āÿÿýþ�㔼�㕃 ýþ Where: M - torque (constant over LzAB 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Ă032

- 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)

- See the weight of the hanger and crossbar (3) as the initial load P , read A , B over 2 000

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  011

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

6.7 CALCULATION

- a=100mm, b=380mm, d =20mm, L0AB=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:

ÿ=17.8x103 −2= 5.93 10ý22ÿÿ; &þÿ�㕣ă=&þÿ

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

&�㔑ÿ�㕣ă= �㔑ý,ÿ�㕣ă2 �㔑þ,ÿ�㕣ă= 5.93 10ý242 2.93 10ý24= 3ý1024

- Calculate the elastic modulus G when twisted: �㔺 =&Āÿ.ÿýþ

&�㔑�㕎�㕣�㕒.�㔼�㕃= 380ý100 3ý10−4ý�㔋ý20432 = 8063 kg/mm2

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: - 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

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

7.2 Theoretical background

A cantilever beams with the stiffness EIx is subjected to the force P as shown The displacements at B, C and D are:

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 BCD

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 123

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

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

8.2 Theoretical background

A cantilever beam is subjected to a force P, the angle between the force P and the symmetry axis y is , the displacement at the free end of the beam consists of: 

 The displacement in y direction: fy=�㕃þý3

In which, l is the distance from the load P to the fixed end; �㔼ý12=; �㔼Ā/3 þ12=Ā/3 The 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 =ĄýĄþ

The displacements f x and fy can be measured by using displacement meters Then, the total displacement and the angle are determined f 

8.3 Test specimen

The steel cross-section bxh and suspension system are arranged as shown in the following

+ Place the specimen, measure the angle , place the displacement meters and the  + Place the specimen, measure the angle , place the displacement meters and the 

weight hanger + 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 00.

- Place the 1kG weights (i.e add P = 1kG) to create the load P , P Record the  12

displacements x , y11, x , y22 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

8.8 Comments on the results

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

- The cause of the error: reading the measurement result cannot be accurate, the device is not stable

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)