Steel Reinforcement Tension Test
Introduction:
Steel Reinforcement Tension Test can be conducted in two experiment session:
* Two deformed grade 460 high yield steel bars
* Two plain round grade 250 mild steel bars
Both steel bars conform to the relevant requirements of BS 4449:1988; have a nominal diameter of 20 mm, and a length of approximately 800 mm.
Objectives:
* Practice tension testing of steel bars as reinforcement for concrete according to Hong Kong current standards.
* Determine geometrical and mechanical properties of steel reinforcements including nominal cross-sectional area, yield stress, tensile strength, nominal stress at fracture, and elongation over gauge length, reduction of cross-sectional area at necking section, and modulus of elasticity.
* Check compliancy of the determined properties of the steel reinforcements with relevant Hong Kong standards.
Apparatus:
. Testing machine equipped with an extensometer with a plotter for drawing load-extension curve (fig. 1)
Fig. 1
2. straight edge (fig. 2)
Fig. 2
3. A caliper (fig. 3)
Fig. 3
4. A metal saw for making marks on the specimens (fig. 4)
Fig. 4
5. weighing equipment (fig. 5)
Fig. 5
6. metal ruler (fig. 6)
Fig. 6
Procedure:
i. The mass (M) and length (L) of the bars were measured. The diameter of plain mild steel bar (Dp) was also measured. Dp is used in this experiment to calculate sectional area for comparison with the area derived from mass.
ii. The specimens were slightly scribed at regular intervals using the metal saw.
iii. The specimen was clamped into testing machine.
iv. The extensometer was mounted on the specimen and the gauge length (Lg) was measured.
v. Testing machine was started applying a tensile load at a rate not more than
10 N/mm2 per second.
vi. When the extension is close to the capacity of the extensometer or the capacity of the plotter, the extensometer was demounted and the plotter was stopped.
vii. The specimen was continuously stretched after the extensometer was removed.
viii. The load was carefully observed and the maximum load the specimen sustained was recorded.
ix. The final load was recorded when the specimen is broken.
x. The broken parts of the specimen were reassembled and the distance between gauge length marks, (Lg') was measured.
xi. Measure the diameter at the necked section for the calculation of section reduction.
Data and calculations:
Table 1: The length and mass of the samples
Sample
Mild Steel
High Yield Steel
A
B
C
D
Length (mm)
800
800
800
800
Mass (kg)
.9917
.9917
.9176
.9367
(1) Nominal Area
A=M / 0.00785L (mm2)
where M is the mass of the bar in kg and
L is the bar in meter.
Nominal Area of bar A= 1.9917/0.00785x0.8
= 317.15mm2
Nominal Area of bar B= 1.9917/0.00785x0.8
= 317.15mm2
Nominal Area of bar C= 1.9176/0.00785x0.8
= 305.35mm2
Nominal Area of bar D= 1.9367/0.00785x0.8
= 308.39mm2
(2) Yield Stress (nominal)
Yield stress = Fy/A
where Fy is the load (in Newton) on the bar when the strain of the bar equals the pre-specified strain for the type of steel, and
A is the nominal area of the bar.
Yield stress of bar A= 89/317.15
=280.62
Yield stress of bar B= 88/317.15
=277.47
Yield stress of bar C= 157/305.35
=514.16
Yield stress of bar D= 160/308.39
=518.82
(3) Tensile strength
Table 2: The maximum loads of samples
Sample
Mild steel
High Yield Steel
A
B
C
D
Maximum load (kN)
50.9
42.2
89.8
91
Tensile strength=Fu/A
where Fu is the maximum load (ultimate load) sustained by the bar.
Tensile strength of bar A =150.9/317.15
=475.8
Tensile strength of bar B =142.2/317.5
=448.37
Tensile strength of bar C =189.8/305.35
=621.58
Tensile strength of bar D =191/308.39
=619.35
(4) Stress at fracture
Table 3: The load which immediately before the fracture of samples
Sample
Mild Steel
...
This is a preview of the whole essay
High Yield Steel
A
B
C
D
Maximum load (kN)
50.9
42.2
89.8
91
Tensile strength=Fu/A
where Fu is the maximum load (ultimate load) sustained by the bar.
Tensile strength of bar A =150.9/317.15
=475.8
Tensile strength of bar B =142.2/317.5
=448.37
Tensile strength of bar C =189.8/305.35
=621.58
Tensile strength of bar D =191/308.39
=619.35
(4) Stress at fracture
Table 3: The load which immediately before the fracture of samples
Sample
Mild Steel
High Yield Steel
A
B
C
D
Load immediately before the fracture (kN)
28
24
25
39
Stress at fracture= Ff/A
Stress at fracture of bar A=128/317.15
=403.59
Stress at fracture of bar B=124/317.15
=390.98
Stress at fracture of bar C=125/305.35
=409.37
Stress at fracture of bar D=139/308.39
=450.73
(5) Young's modulus
Table 4: The?F, ?L and the nominal cross sectional area of samples
Sample
Mild Steel
High Yield Steel
A
B
C
D
?F (kN)
40
37
47
29
?L (mm)
0.066
0.056
0.086
0.05
?F/?L
606060.61
660714.29
546511.63
580000
Nominal cross sectional area (mm2)
314.16
314.16
314.16
314.16
Young's modulus=stress increment/strain increment
=?F.Lg/A.?L
where ?F is the load increment
?L is the extension increment
Lg is gauge length
A is the nominal cross sectional area of the bar
Young's modulus of bar A =606060.61.100/314.16
=192915.08
Young's modulus of bar B =660714.29.100/314.16
=210311.89
Young's modulus of bar C =546511.63.100/314.16
=173960.05
Young's modulus of bar D =580000.100/314.16
=184619.73
(6) Percentage elongation over gauge length
Table 5:
The Gauge length after the test and the formula results (Lg'-Lg) of samples
Sample
Mild Steel
High Yield Steel
A
B
C
D
Gauge length after the test (mm)
35
33
22
23
Lg'-Lg (mm)
35
33
22
23
Percentage elongation over gauge length=Lg'-Lg/Lg.100
where Lg is the original gauge over gauge length, and
Lg' is gauge length after the test.
Percentage elongation over gauge length of bar A=35/100.100
=35%
Percentage elongation over gauge length of bar A=33/100.100
=33%
Percentage elongation over gauge length of bar A=22/100.100
=22%
Percentage elongation over gauge length of bar A=23/100.100
=23%
(7) Percentage reduction of area
Table 6:
The Diameter after stretching, Cross sectional area after stretching and the results of A-A' of samples
Sample
Mild Steel
High Yield Steel
A
B
C
D
Diameter after stretching (mm)
3.95
4.62
2.97
3.19
Cross sectional area after stretching, A' (mm2)
52.84
67.87
32.12
36.64
A-A'
64.31
49.28
73.23
71.75
Percentage section area reduction = A-A'/A.100
where A' is the area at neck derived from the diameter at that section
A is the nominal area of the section.
Percentage section area reduction of bar A=164.31/317.15.100
=51.81%
Percentage section area reduction of bar B=140.28/317.15.100
=47.07%
Percentage section area reduction of bar C=173.23/305.35.100
=56.73%
Percentage section area reduction of bar D=171.75/308.39.100
=55.69%
Discussion:
. The difference between steel and iron.
The major part of steel is iron nearly up to 97%. The remaining part is carbon or some alloying elements are used such as manganese, chromium, vanadium, and tungsten. In our case, the alloying element is manganese. By adding this elements, it can prevent dislocations in the iron atom crystal lattice from sliding past one another.
By varying the alloying element, the variation of hard ductility and tensile strength of the steel can be achieved. For example, increasing carbon content can made steel harder and stronger than iron, but is also more brittle. The type of steel with high carbon content is called cast iron.
2. What are the differences between high yield steel and mild steel bars?
Steel is metal alloy and its major components are iron and carbon. Steel can not be found in natural sources.
Mild steel has a relatively low tensile strength, but it is cheap and malleable. High yield steel bars provide greater strength and tensile stress compared with the mild steel bars. For mild steel, it is low in carbon content (below 0.3% by weight). But high yield steel bars have higher carbon content (above 0.3 - 1.7% by weight). The difference in carbon content will affect the strength and the ductility of the steel. That means the high yield steel bars have higher strength and less ductile than mild steel.
The differences of the mild steel and high yield steel bars are shown below:
Mild Steel
High Yield Steel
Appearance
* Plain
* Deformed
Composition
* Carbon 0.1-0.25%
* Mn 0.5%
* Carbon 0.6-0.99%
* Mn 1.5%
Properties
* Good machinability
* Ductility
* Weldability
* Toughness
* Greater yield strength
* Lower ductility
Recognition
* High density
* High stiffness
* become rusty easily in damp atmosphere
* High density
* High stiffness
* More springy
* More friction
Cost
* cheaper
* comparatively expensive
Products
* Window frames
* Pipes
* Rivets
* High strength wires
* Bolts
* Springs
3. Why their loadings are different?
There are two factors affect the loadings.
Firstly, it is because of the difference on the content of manganese(Mn). Manganese can stiffening the steels and improve the loading capacity. So, high yield steel bars (1.5%) have better loading than Mild steel (0.5%) because of the higher manganese content.
Secondly, is because of the appearance. Deformed steel have greater surface area and the uneven surface, it can have more friction than plain steel. The higher friction can be a good support for loading. So, the high yield steel bars take the higher loading.
4. Comparison of the Young's modulus (E) between high yield steel and mild steel
The elastic modulus of high yield steel bar is lower than that of mild steel bar. However, it is not reflect the strength of the steel.
The modulus of elasticity is to measure the stiffness. It is not related to the strength. As Young's modulus (E) equal to stress increment divided by strain increment. Where strain is a dimensionless number, the unit of the modulus of elasticity is the same as that of stress. Besides, the modulus of elasticity is a measure of the deformability of a material. The higher modulus of elasticity the material has the less deformation the material exists. As the high yield steel bars are deformed, they have lower modulus' value.
5. Calculation for yield stress
When we calculate the yield stress, we need to decide the pre-specified strains of both mild steel and high yield steel.
Under table is showing the pre-specified strain for the determination of yield stress.
Mild Steel
High Yield Steel
Grade of steel bars(BS)
250
460
Pre-specified strain at which the yield stress is determined
0.33%
0.43%
Then, we need to use the above Formula.
Yield Stress
A = the nominal area of bar and
Fy = the load on the bar in Newton. This loading is due to the strain of the bar equals the pre-specified strain for the type of steel
The other information needed in the formula was found in the graph paper given. Our graph paper will be placed in attachments for reference.
6. Definition of the ductility
The ductility of a metal can be defined as the amount of plastic strain. It can accept before fracture when subjected to a tensile load. Difficult property to quantify and various tests are currently used for assessing and quantifying it. Amongst these is the tensile test.
In the tensile test, the Percentage Elongation is one of the two methods used to measure ductility. It is calculated as follows:
Percentage Elongation = L1 - L / L x 100%
Where L1 = gauge length after fracture in mm; and
L = original gauge length in mm (100 mm for this experiment)
7. Define the quality of the sample with accordance BS
Here are the requirements of the cross-sectional area and mass of each steels according to BS 4449 : 1988
Nominal size (mm)
Cross-sectional area (mm2 )
Mass per metre run (kg)
6
28.3
0.222
8
50.3
0.395
0
78.5
0.616
2
13.1
0.888
6
201.1
.579
20
314.2
2.466
25
490.9
3.854
32
804.2
6.313
40
256.6
9.864
50
963.5
5.413
The tolerance on mass of each steels according to BS 4449 : 1988
Nominal size (mm)
Tolerance on mass per metre run(%)
6
± 9
8 and 10
± 6.5
2 and over
± 4.5
(I) By comparing the cross-sectional area and mass per meter run
According to BS 4449: 1988, the required cross-sectional area and mass per meter run with 20 mm nominal size are 314.2 mm2 and 2.466 kg respectively.
The cross-sectional area of sample of the mile steel and high yield bar are 317.15mm2 and the average nominal size of high yield bar is 306.87mm2. Owing to the use of uniform effective cross-sectional area formula to calculate the non-uniform effective cross-sectional area of High Yield Steel , there must be some difference in data , so it was acceptable to BS requirement
The mass per meter run required by British standard is 2.466kg/m. The tolerance on mass per meter run is ± 4.5. It showed that the accepted range of the mass was between 2.36 kg and 2.58 kg Compare with the samples, the mass per meter run of mile steel and high yield bar are 2.4896 and 2.409kg respectively. Hence, the sample for the test fulfil the requirement of BS 4449 : 1988
(II) By comparing the percentage elongation
According to BS 4449 :1988, the required percentage elongation of Mild Steel and High yield steel are at least 22 % and 12% respectively. In our sample, the average percentage of elongation of mile steel is 34% and that of high yield steel is 22.5.
Hence, these steel bars can fulfil the requirement of BS.
(III) By comparing the specified characteristic strength
According to BS 4449:1988 (Table 8), the required specified characteristic strength of Mild Steel and High Yield Steel is at least 232.5 N/mm2and 427.8 N/mm2 respectively. In our sample, the average of specified characteristic strength for mild steel and high yield steel is 279.05 N/mm2and 516.49 N/mm2.
Hence, it can fulfil the requirement of British standard.
There is a relationship between tensile strength and reduction of area.
Formula of Young's modulus=stress increment/strain increment
=?F.Lg /A.?L
where ?F is the load increment
?L is the extension increment
Lg is gauge length
A is the nominal cross sectional area of the bar
As young modulus, cross sectional area of the bar is a constant.
Constant=?F .?L/ Lg
So, we know that ?F disproportional to ?L/ Lg, it means that the elongation over gauge length will be decreased if the load is increased. The high yield bar have a higher strength than mild steel bar, so that the high yield bar can afford a higher tension force than mild steel bar. Hence, the extension of high yield bar is less than mild steel bar. This is testified by the experiment result.
Precautions:
. The marks on the steel bar should be made clear and precisely within the distance of 50 mm. When we mark the steel bar, we should not saw the steel bar too deep. Otherwise the breaking point may be occuring at the position of the markings.
2. The plotter pen should be placed at the original of the graph paper when the FORNEY 2700 KN universal testing machine starts to apply the force on the steel bar. Also, the loading rate should not exceed 10N/mm2 per second.
3. A double check should be carried out in order to make sure that both sides of the steel were locked tightly when we put in in the testing machine. And the protective cage should be locked when the machine is switched on. Otherwise, the broken steel bars may explode out.
4. After the load-extension recorder was disconnected, it would make sure to wear the headphone to protect the ears.
5. There may be some errors when we are reading the results. We should look carefully on reading of the meter of the machine. It is because the steel bar would drop very quickly when the steel bar has broken and the machine cannot apply the loading of the steel bar.It Will break suddenly on a particular meter and the reading of the meter would not stop when the sample fails.
Conclusion:
The table has shown the calculated results on nominal Area, yield Stress (nominal), tensile strength, stress at fracture, Young's modulus, percentage elongation over gauge length, percentage reduction of area of samples respectively.
Sample
Mild Steel
High Yield Steel
A
B
C
D
Length (mm)
800
800
800
800
Mass (kg)
.9917
.9917
.9176
.9367
Nominal Area (mm2)
317.15
317.15
305.35
308.39
Maximum load (kN)
50.9
42.2
89.8
91
Load immediately before the fracture (kN)
28
24
25
39
Extension (mm)
35
33
22
23
Diameter after stretching (mm)
3.95
4.62
2.97
3.19
Cross sectional area after stretching (mm2)
52.84
67.87
32.12
36.64
Yield stress
280.62
277.47
514.16
518.82
Tensile strength
475.8
448.37
621.58
619.35
Young's modulus
92915.08
210311.89
73960.05
84619.73
Stress at fracture
403.59
390.98
409.37
450.73
% elongation over gauge length (%)
35
33
22
23
% reduction of area (%)
51.81
47.07
56.73
55.69
After the steel reinforcement tension test, we found that high yield steel has higher loading capacity compared with mild steel bars. It is shown in the part of yield stress, tensile strength, modulus of elasticity, percentage elongation, reduction of cross-sectional area and the part of nominal stress at fracture. The uneven surface of the high yield steel bars provided more friction and hardness.
