Structures 1

Laboratory Report 1

Student Particulars

Title of Experiment : Buckling Test

Student Name : Sri Kartikeayan S/O Raja Gopal

Student ID : SCM014634

Subject / Subject code : Structures 1 / ECS 3213

Lecturer : Ir. Pan

Date Submitted : 22th August 2014

Abstract:The main purpose of this experiment is to determine the yield stress, ultimate strength, Young’s modulus (elastic Modulus), and behavior of given material when subjected to unaxial loading and plot the graph of stress versus strain. In this experiment, a tensile test machine, steel and aluminum specimen and some measurement apparatus is used. The load is applied to the specimen by the tensile test machine. The load and dial gauge reading is recorded during this experiment. A graph with Stress vs. Strain curve for the specimen and its elastic modulus, yield strength and ultimate tensile strength is pointed out. The reduction in area, percentage of elongation, yield strength and ultimate strength is determined after the experiment has been done. The relationship between stress and strain is established and discussion and conclusion has been made.

Introduction/Theory:Axial loading is produced by two or more collinear forces acting along the axis of a long slender member. This type of loading occurs in many engineering elements that make up machine, bridge and building trusses. When a structure member or machine component is subjected to the external forces (applied loads and support reactions), internal resisting forces will develop within the member or components to balance the external forces.

Stress is the intensity of the internal force on the cross-sectional area of a body. In this test, the internal distribution of internal forces has a resultant force that is normal to exposed cross-sectional area of a simple bar. Thus,

σ= FA (Equation 1)

, where σ = stress (Pa), F = Force (N) and A = cross sectional area (m2)

When the axial loading is applied to the body, individual points of the body generally move with the direction of the external force. This movement of points is generally known as displacement (vector quantity) that will involve a translation and rotation of the body as a whole and neither the size nor the shape of the body is changed. The change in any dimension associated with these displacements is known as deformation.

A normal strain is the quantity used to measure the change in size (elongation) during deformation. The strain may be result of a stress, of a change in temperature, or of other physical phenomena (grain growth or shrinkage). In this test, only strain resulting from changes in stress is considered. The change in length of a simple bar under an axial loading is:

ε= δL0

(Equation 2)

, where ε = Strain, δ = Elongation (m) and L0 = Initial length (m)

The modulus of elasticity is a material property, that describes its stiffness and is therefore one of the most important properties of solid materials. Mechanical deformation puts energy into a material. The energy is stored elastically or dissipated plastically. The way a material stores this energy is summarized in stress-strain curves. Stress is defined as force per unit area and strain as elongation or contraction per unit length. So the formula for elastic modulus is:

E=σε (Equation 3)

, where E = Elastic modulus (MPa), σ = Stress (MPa) and ϵ = Strain

This relationship between loads and deformation in a structure or machine components can be obtained by plotting a stress-strain diagram. It also depends on the dimension of the members as well as type of material they are made.

By analyzing the stress-strain curve of a body, a number of mechanical properties of the material can be determined. Yield strength is defined as the stress required to produce a specified amount of plastic deformation or permanent set in the body. The yield strength is always a practical measure of the limit of elastic action of a material:

Yield Strength=Load at the limit proportionalityArea (Equation 4)

The maximum stress on the original cross-sectional area that develops in the body before fracture is called ultimate strength or ultimate tensile and the term can be modified as compressive or shearing strength:

UltimateStrength= Maximum loadArea (Equation 5)

When the ultimate strength of the body is reached, the cross-sectional area of the body will start to decrease or neck down and the resultant force that can be carried by the specimen decreases until rupture.

Components:

A. SpecimenB. Dial gaugeC. Digital load indicatorD. Power switch

Experimental method :1. The diameter of the specimen at various places is measured and the average is recorded

down.2. The specimen is marked at 10mm interval along its length.3. Two marks are punched on the specimen using the provided puncher.4. The specimen is fixed to the machine by pushing the top grip upwards and the specimen

is inserted into the bottom grip. 5. The extensometer is fixed to the specimen. The screw pins is make sure so that they are

resting in the 2 marks.6. The dial gauge readings are recorded into Table 1.7. The tensile machine is powered on using the power switch.8. The motor switch is pressed to position A.9. The machine is stopped when the reading is about 0.5 – 1kN. This is to preload and take

up all the loose fitting in the machine.10. The zero readings of the dial gauge and load indicator is created.11. The motor switch is turned to position A.12. The reading of the load cell, extensometer and the dial gauge is recorded.13. The extensometer is removed and continues the experiment using the dial gauge only

when the extensometer reading starts to increase at a fast rate.14. The motor is switched off by pressing the motor switch position “O”.15. The specimen is removed from the grips and joined back the specimen.16. The final length is measured and the diameter at the broken section.17. The motor switch is pressed to position M to return the crosshead to the original position.

Results and Analysis:

Aluminium:Aluminium

Load, F (N)

Cross-sectional area

(mm2)

Calculated Stress, σ = F/A

(N/mm2)

Calculated Length ∆L (mm)

Calculated Strain, ε = ΔL/L0

(mm/mm)

Young's Modulus, E (N/mm2)

0 0.00 0.00 0.0000 0301 10.58 0.05 0.0009 11485.805603 21.19 0.11 0.0020 10458.985906 31.83 0.16 0.0029 10803.716

1201 42.20 0.23 0.0042 9962.771507 52.95 0.29 0.0053 9914.7141801 63.28 0.34 0.0063 10106.4762099 73.75 0.39 0.0072 10268.6362399 84.29 0.44 0.0081 10402.6152708 95.15 0.49 0.0090 10544.2943003 27.34 105.52 0.52 0.0096 11018.3593304 116.09 0.58 0.0107 10868.6843603 126.60 0.62 0.0114 11087.5993910 137.39 0.67 0.0123 11134.4024200 147.58 0.71 0.0131 11286.4114509 158.43 0.77 0.0142 11172.6054807 168.90 0.82 0.0150 11253.345109 179.52 0.86 0.0158 11334.55408 190.02 0.91 0.0168 11338.6215702 200.35 0.99 0.0182 10988.9696000 210.82 1.12 0.0206 10221.1126300 221.36 1.40 0.0258 8585.7346397 224.77 1.62 0.0298 7534.011

Yield strength=190.02N/mm2Ultimate strength=224.77N/mm2

0.0000 0.0030 0.0060 0.0090 0.0120 0.0150 0.0180 0.0210 0.0240 0.0270 0.03000.00

50.00

100.00

150.00

200.00

250.00

Stress versus Strain

Strain (mm/mm)

Stress, σ (N/mm2)

Young modulus = 11338.62N/mm2

Aluminium:

Specimen diameter = 6.0mm

Cross-sectional area = 28.46mm2

Gauge length = 60mm

Percentageof elongational=Final length−Initial length

Initial length×100 %

¿64mm−60mm

64mm×100 %

¿6.67 %

STEEL:

Steel

Load, F (N)

Cross-sectional area

(mm2)

Calculated Stress, σ = F/A

(N/mm2)

Calculated Length ∆L (mm)

Calculated Strain, ε = ΔL/L0

(mm/mm)

Young's Modulus,

E (N/mm2)

0 0.00 0.00 0.0000 0343 12.05 0.03 0.0006 21814.125525 18.45 0.05 0.0009 20033.38730 25.65 0.07 0.0013 19897.099

1031 36.23 0.13 0.0024 15131.441391 48.88 0.19 0.0035 13968.1361601 56.25 0.23 0.0042 13280.9281997 70.17 0.29 0.0053 13138.4762229 78.32 0.33 0.0061 12887.2742695 94.69 0.39 0.0072 13184.3613009 28.27 105.73 0.43 0.0079 13351.153390 119.11 0.47 0.0087 13761.5323698 129.94 0.50 0.0092 14111.1314014 141.04 0.54 0.0099 14182.3614434 155.80 0.56 0.0103 15106.8044722 165.92 0.58 0.0107 15533.2715105 179.37 0.59 0.0109 16508.545464 191.99 0.61 0.0112 17090.1475708 200.56 0.63 0.0116 17286.5516035 212.05 0.65 0.0120 17714.4986386 224.39 0.67 0.0123 18185.246627 232.85 0.68 0.0125 18594.0076970 244.91 0.71 0.0131 18730.0687354 258.40 0.73 0.0135 19141.887605 267.22 0.75 0.0138 19346.5217981 280.43 0.77 0.0142 19775.6848281 290.97 0.78 0.0144 20255.9738559 300.74 0.79 0.0145 20670.9718916 313.28 0.81 0.0149 21001.4849255 325.19 0.84 0.0155 21021.4219541 335.24 0.85 0.0157 21416.0769942 349.33 0.88 0.0162 21555.397

10269 360.82 0.89 0.0164 22014.20910583 371.86 0.91 0.0168 22188.72410875 382.12 0.93 0.0171 22310.6

11123 390.83 0.95 0.0175 22338.97611520 404.78 0.98 0.0180 22428.04111856 416.58 1.01 0.0186 22396.58212070 424.10 1.03 0.0190 22358.10412320 432.89 1.06 0.0195 22175.31412669 445.15 1.08 0.0199 22381.20812980 456.08 1.11 0.0204 22310.87713348 469.01 1.16 0.0214 21954.47913643 479.37 1.18 0.0217 22059.35513955 490.34 1.23 0.0227 21646.59914286 501.97 1.25 0.0230 21805.47614525 510.37 1.28 0.0236 21650.65814894 523.33 1.31 0.0241 21692.26915101 530.60 1.35 0.0249 21342.08615434 542.30 1.39 0.0256 21185.00815739 553.02 1.44 0.0265 20853.5316033 563.35 1.51 0.0278 20258.2916328 573.72 1.63 0.0300 19112.184

Yield strength:290.70 N/mm2

Ultimate strength=573.72N /mm2

0.0000 0.0030 0.0060 0.0090 0.0120 0.0150 0.0180 0.0210 0.0240 0.0270 0.03000.00

100.00

200.00

300.00

400.00

500.00

600.00

Stress versus Strain

Strain (mm/mm)

Stress, σ (N/mm2)

Young modulus: 14182.36N/mm2

Steel:

Specimen diameter = 6.0mm

Cross-sectional area = 28.46mm2

Gauge length = 60mm

Percentageof elongationsteel=Final length−Initial length

Initial length×100 %

¿63mm−60mm

60mm×100%

¿5%

Reference:1. http://tpm.fsv.cvut.cz/student/documents/files/BUM1/Chapter15.pdf 2. http://esminfo.prenhall.com/engineering/shackelford/closerlook/pdf/Shackelford_Ch6.pdf