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Determination Of The Characteristic Strength Properties Of Mild Steel Reinforcement
[A CASE STUDY OF ILORIN METROPOLIS]
CHAPTER FOUR -- [Total Page(s) 8]
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CHAPTER FOUR
4.0. ANALYSIS, RESULTS AND DISCUSSION
This chapter contains the analysis, results and discussion of the 8mm, 10mm, 12mm, 16, 20mm and 25mm specimens if mild steel bar.
4.1 ANALYSIS OF SPECIMENS
All calculations are according to (William, 2004).
4.1.1 ANALYSIS FOR 8mm MILD STEEL SPECIMENS
i. ULTIMATE STRENGTH OR TENSILE STRENGTH
This is the ordinate of the maximum point on the stress strain curve.
Specimen 1 = 440.53N/〖mm〗^2
Specimen 2 = 460.78N/〖mm〗^2
Specimen 3 = 440.5 3N/〖mm〗^2
Average ultimate strength = ((440.53+ 460.78 + 440.53)N/〖mm〗^2)/3
=447 N/〖mm〗^2
ii. YIELD STRENGTH
This is the ordinate of the yield point on the stress strain curve.
Specimen 1=195.8N/〖mm〗^2
Specimen 2 = 146.8N/〖mm〗^2
Specimen 3 = 195.8N/〖mm〗^2
Average yield strength = ((195.8+146.8+195.8)N/mm2)/3
=179 N/〖mm〗^2
iii. BREAKING STREGTII
The ordinate of the point on the stress strain curve at which rupture occurs.
Specimen 1 = 430.74N/〖mm〗^2
Specimen 2=440.53N/〖mm〗^2
Specimen 3=430.74N/〖mm〗^2
Average breakmg strength=((430.74+440.53+430.74)N/〖mm〗^2)/3
=434.0 N/〖mm〗^2
iv. MODULUS OF RESILIENCE
This may be calculated as the area under the stress— strain curve from the origin up to the proportional limit.
Specimen1=1/2 bh=1â„2 ×190N/〖mm〗^2 ×0.0016=0.152
Specimen2=1/2 bh =1/2×O.0012x1400=0.084
Specimen 3 =1/2 bh=1/2×190×0.0016 = 0.152
Average modulus of resilience = ((0.152÷0.084+0.152)N /〖mm〗^2)/3
= 0.129N/mm^2
v. PERCENTAGE ELONGATION
The increase in length (gauge length) after fracture divided by the initial length and multiplied by 100.
Specimen I = (40.6 -37.5 )/37.5×100 = 8.3%
Specimen 2 = (40.5 - 37.5)/37.5 x 100 = 8.0%
Specimen 3 = (40.6 -37.5)/37.5 x 100 = 8.3/o
Average percentage elongation = ((8.3 +8.0 +8.3))/3 = 8.2 %
vi. PERCENTAGE REDUCTION IN AREA
π 〖(5.1)〗^2 π (4.13)^2
Specimen 1 = ( 4 4 )/□(〖π(5.0)〗^2/4) x 100
= 34.4%
π 〖(5.1)〗^2 π (4.48)^2
Specimen 2 = ( 4 4 )/□(〖π(5.1)〗^2/4) x 100
= 22.86%
π 〖(5.1)〗^2 π (4.38)^2
Specimen 3 = ( 4 4 )/□(〖π(5.1)〗^2/4) x 100
= 26.24%
Average =27.8%
CHAPTER FOUR -- [Total Page(s) 8]
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ABSRACT - [ Total Page(s): 1 ]ABSTRACT WILL BE HERE SOON ... Continue reading---
APPENDIX A - [ Total Page(s): 6 ] ... Continue reading---
TABLE OF CONTENTS - [ Total Page(s): 1 ]TABLE OF CONTENTS CHAPTER ONE 1.0 Introduction 1.2 Statement of the Problem 1.3 Aim and Objectives of the Study 1.4 Justification of the Study 1.5 Scope of the Study 1.6 Proposed Methodology CHAPTER TWO2.0 Literature Review 2.1 Nigerian Steel Industry (Historical Development) 2.2 Engineering Materials and Properties 2.2.1 Cement and Concrete 2.2.2 Aggregates and Sand 2.2.3 Timber and Plywood 2.3 Strength of Materials ... Continue reading---
CHAPTER ONE - [ Total Page(s): 2 ]CHAPTER ONE 1.0 INTRODUCTION Steel is a man-made material containing 95% of iron. The remaining constituent are small amount of element derived from the raw-material use in the making of the steel, as well as other element added to improve certain characteristics or properties of the product (Marcus, 1964). Steel reinforcement are used generally in the form of bars of circular cross-section in concrete structure. They are like a skeleton in human body. Plain concrete without s ... Continue reading---
CHAPTER TWO - [ Total Page(s): 10 ]CHAPTER TWO2.0 LITERATURE REVIEW2.1 NIGERIAN STEEL INDUSTRY (HISTORICAL DEVELOPMENT)Planning for the Nigerian steel industry started around 1958. Many international Organizations and consulting firms had been commissioned at various times to study the feasibility of steel plants under the aegis of the Federal Ministry of Industries In 1971 an extra-ministerial agency was established by Decree to focalize efforts required to actualize a steel plant. The agency was called ‘Nigerian Steel Dev ... Continue reading---
CHAPTER THREE - [ Total Page(s): 3 ]3.3.2 Principle of OperationWith every 2 revolutions made on the hand or motor driven gear box of high mechanical advantage, a force of 20kN (2000kgf) is applied to a test piece held in the chuck pins. The force deflects the spring beam and this deflection operates a level acting on a piston in a cylinder containing mercury. It should be noted that the mercury inside the sleeve must be at zero point before the drive is made, and this can be alone using the mercury adjuster. The recording graph i ... Continue reading---
CHAPTER FIVE - [ Total Page(s): 1 ]CHAPTER FIVE5.0 CONCLUSION AND RECOMMENDATION From the test carried out and the results obtained, the average yield strength for specimens diameter of 8mm, 10mm, 12mm, 16mm, 2Omm and 25mm were 79N/mm2, 225 N/mm2, 261 N/mm2, 277 N/mm2, 295 N/mm2 and 297 N/mm2 respectively. It was therefore observed that specimen of 8mm and 10m do not meet the BS8110 specification of 250 N/mm2 for mild steel.However, the analysis shows that the average ultimate strength obtained for the specimens of 8mm, 10mm 12mm ... Continue reading---
REFRENCES - [ Total Page(s): 1 ]REFERENCESAlbert, G.G., (1960), ‘Elements of Physical Metallurgy’, 2 Edition, Addison Wesley Publishing Co. Inc., London, pp337-340Arthur, H.N., et aL, (2004), ‘Design of Concrete Structures’, 13th Edition, Tata McGraw Hill Companies, India, pp38-50Bakare, O.S., (2006), Thesis on Determination of Ultimate Tensile Strength of High Tensile Steel Specimens, Civil Engineering Department, University of Ilorin, Nigeria.Kenneth. L -. Dionisio. B.. (1997), ‘Reinforced con ... Continue reading---
