Saturday, December 25, 2010

Module 5: Compression Test

Dear Students

The figure below illustrates the compression test for an aluminium test specimen. If the diameter of the specimen is 7 mm :caluclate the following

  1. Maximum Force (Load) in KN

  2. Original Area, mm2

  3. Ultimate compressive Strength(σu), (N/mm2)

  4. Modulus of elasticity

  5. Yield stress
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5 comments:

  1. Mohammad AliJanuary 2, 2011 at 9:14 PM

    1- 12 KN




    2- πr^2 = π ( 7/2)2 = 38.5 mm




    3- σ F/A = 12x1000/38.5=311.6 N/mm2





    4- Ey=0.05mm/44mm=1.1363N/mm2
    σy/Ey =311.6N/mm2/1.1363=274.24N/mm2






    5- σy = Fy/A = 10x1000/38.5=10000/38.5=259.74N/mm2


    Mohammad Ali Al-Jaberi 11-06

    ReplyDelete
  2. h.yJanuary 2, 2011 at 9:23 PM

    1- 12 KN

    2- πr^2 = π ( 7/2)2 = 38.5 mm


    3- σ F/A = 12x1000/38.5=311.6 N/mm2


    4- Ey=0.05mm/44mm=1.1363N/mm2
    σy/Ey =311.6N/mm2/1.1363=274.24N/mm2


    5- σy = Fy/A = 10x1000/38.5=10000/38.5=259.74N/mm2


    hazza yeslam
    11-06

    ReplyDelete
  3. hamdan obaidJanuary 2, 2011 at 11:35 PM

    (1) 12 KN



    (2) πr^2 = π ( 7/2)2 = 38.5 mm




    (3) σ F/A = 12x1000/38.5=311.6 N/mm2





    (4) Ey=0.05mm/44mm=1.1363N/mm2
    σy/Ey =311.6N/mm2/1.1363=274.24N/mm2






    (5) σy = Fy/A = 10x1000/38.5=10000/38.5=259.74N/mm2


    hamdan obaid

    11-6

    ReplyDelete
  4. 90011January 2, 2011 at 11:36 PM

    1- 12 KN




    2- πr^2 = π ( 7/2)2 = 38.5 mm




    3- σ F/A = 12x1000/38.5=311.6 N/mm2





    4- Ey=0.05mm/44mm=1.1363N/mm2
    σy/Ey =311.6N/mm2/1.1363=274.24N/mm2






    5- σy = Fy/A = 10x1000/38.5=10000/38.5=259.74N/mm2


    hussain zaid
    11-06
    90011

    ReplyDelete
  5. Mohammed Al AmoudiJanuary 2, 2011 at 11:52 PM

    1- 12 KN




    2- πr^2 = π ( 7/2)2 = 38.5 mm




    3- σ F/A = 12x1000/38.5=311.6 N/mm2





    4- Ey=0.05mm/44mm=1.1363N/mm2
    σy/Ey =311.6N/mm2/1.1363=274.24N/mm2






    5- σy = Fy/A = 10x1000/38.5=10000/38.5=259.74N/mm2


    Mohammed Al Amoudi 11-06

    ReplyDelete

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