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©Lifting Equipment Engineers Association 2012 - training\1-5e5e0000
1
LEEA Correspondence Courses
ASSIGNMENT 1.5
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Name: SONU VARGHESE GEORGE Student number: 17717
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© Lifting Equipment Engineers Association 2012 - training\1-5e5e0000
1
LEEA Correspondence Courses
ASSIGNMENT 1.5
Please note: Use the up and down cursor keys to move between fields in this form. Enter your name and student number in the spaces below. Name: SONU VARGHESE GEORGE
Student number: 17717
Each question has several answers, only one of which is correct.
Select your answer by typing # in the box.
When complete, save the file using the same name. Then use the upload facility to return it for marking.
2. Stress equals: Load divided by strain Strain multiplied by load
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Load divided by cross sectional area Cross sectional area multiplied by load 3. Strain equals:
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Change in length divided by srcinal length Original length divided by change in length Change in diameter divided by srcinal diameter Original diameter divided by change in diameter 4. A test that measures the load over the extension of a test piece is called: A stress test
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A tensile test A strain test A compression test 6. Within the elastic limit the material will return to its srcinal dimensions but if this is exceeded: The diameter will increase It will only return by half the amount of the extension
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It will have become plastic and retain its new dimensions It will become brittle and snap 1. The illustration is a diagrammatic representation of: A tensile load Single shear A compressive load
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Double shear 5. Point A on the graph shown is known as: The yield point The elastic limit A
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The limit of proportionality The ultimate stress
© Lifting Equipment Engineers Association 2012 - training\1-5e5e0000
2 7. Maximum load divided by srcinal cross sectional area equals: Ultimate breaking stress
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Tensile strength Point of plastic deformation Limit of proportionality 8. Hooke’s law relates to:
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Elasticity Plasticity Ductility Fatigue 9. Hooke’s law states, within the limit of proportionality, stress divided by strain: Is a variable Is equal to the reduction in area Is equal to the ultimate breaking point
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Is a constant 10. Young’s Modulus of Elasticity (E):
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Equals stress divided by strain Equals stress multiplied by strain Equals strain divided by stress Equals strain multiplied by stress 11. For mild steel E has a value of: 200,000 MN/mm² 200,000 kN/mm² 200,000 NM/m²
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200,000 MN/m² 13. If a load is suspended at the centre of a simply supported beam the top flange will be: Subject to tensile stress Subject to torsional stress
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Subject to compressive stress Free of stress 14. The deeper a beam in section: The more liable it is to bending The less the load it can carry
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The greater the load it can carry It has no effect on the load it can carry 12. The beam shown in the illustration is said to be: Freely suspended Simply supported
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Encastre Cantilever 15. The point of maximum stress in the lifting beam illustrated will occur at point: A B C
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D
© Lifting Equipment Engineers Association 2012 - training\1-5e5e0000
3 16. The point of maximum stress in the ring shown will occur at point:
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A B C
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D 17. The point of maximum stress in the collar eyebolt illustrated will occur at point:
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A
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B C D 18. A steel bar 20mm x 20mm is subject to a load of 10kN. What is the stress in the bar?
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25MN/m² 250MN/m² 50MN/m² 500MN/m² 19. A bolt in tension supports a load of 6 tons and has a csa of .75 ins². What is the stress in the bolt? 4.5 tons/ins² 6 tons/ins²
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8 tons/ins² 10 tons/ins² 20. A sheave pin in double shear is subject to a force of 5kN. If the pin has a cross sectional area of 100mm², what is the shear stress in the pin? 250N/mm²
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25N/mm² 500N/mm²
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50N/mm²
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Answers still shown in black were correct Answers now shown in red were incorrect – the correct answers are shown in blue
Result 85%

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