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    CE302 MECHANICS OF MATERIALS Chapter 7 - Tutorial Problems

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    Ahmet Tüken
    CE302 MECHANICS OF MATERIALS Chapter 7 - Tutorial Problems

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    CE302 MECHANICS OF MATERIALS Chapter 7 - Tutorial Problems

    Uploaded by

    Ahmet Tüken
    328 views6 pages
    CE302 MECHANICS OF MATERIALS Chapter 7 - Tutorial Problems

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    CE302 MECHANICS OF MATERIALS Chapter 7 - Tutorial Problems

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    328 views6 pages

    CE302 MECHANICS OF MATERIALS Chapter 7 - Tutorial Problems

    Uploaded by

    Ahmet Tüken
    CE302 MECHANICS OF MATERIALS Chapter 7 - Tutorial Problems

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 6

    Fifth Edition

    CHAPTER MECHANICS OF
    MATERIALS
    7 Ferdinand P. Beer
    Instructors:
    E. Russell Johnston, Jr.
    Prof. M. J. Al-Shannag
    John T. DeWolf Transformation of Stress
    Dr. Ahmet
    David Tuken
    F. Mazurek
    Dr. Hussam Alghamdi
    (Tutorial Problems)
    Lecture Notes:
    Dr. Saleh Aldeghaither
    J. Walt Oler
    Texas Tech University

    These lecture notes are based largely on the slides provided


    by the publisher of the text-book (i.e. McGraw Hill)

    © 2009 The McGraw-Hill Companies, Inc. All rights reserved.


    MECHANICS OF MATERIALS
    Edition
    Fifth Beer • Johnston • DeWolf • Mazurek

    TUTORIAL PROBLEMS

    For the state of stress shown,


    determine (a) the principal planes
    and the principal stresses, (b) the
    stress components exerted on the
    element obtained by rotating the SOLUTION:
    given element counterclockwise • Construct Mohr’s circle
    through 30 degrees. s x + s y 100 + 60
    s ave = = = 80 MPa
    2 2
    R= (CF )2 + (FX )2 = (20)2 + (48)2 = 52 MPa
    © 2009 The McGraw-Hill Companies, Inc. All rights reserved. 7- 2
    MECHANICS OF MATERIALS
    Edition
    Fifth Beer • Johnston • DeWolf • Mazurek

    TUTORIAL PROBLEMS

    • Principal planes and stresses


    XF 48 s max = OA = OC + CA s min = OB = OC - BC
    tan 2q p = = = 2.4
    CF 20 = 80 + 52 = 80 - 52
    2q p = 67.4°
    s max = +132 MPa s min = +28 MPa
    q p = 33.7° clockwise

    © 2009 The McGraw-Hill Companies, Inc. All rights reserved. 7- 3


    MECHANICS OF MATERIALS
    Edition
    Fifth Beer • Johnston • DeWolf • Mazurek

    TUTORIAL PROBLEMS

    f = 180° - 60° - 67.4° = 52.6°


    • Stress components after rotation by 30o
    s x¢ = OK = OC - KC = 80 - 52 cos 52.6°
    Points X’ and Y’ on Mohr’s circle that s y¢ = OL = OC + CL = 80 + 52 cos 52.6°
    correspond to stress components on the
    t x¢y¢ = KX ¢ = 52 sin 52.6°
    rotated element are obtained by rotating
    XY counterclockwise through 2q = 60° s x¢ = +48.4 MPa
    • Please note that s y¢ = +111.6 MPa
    t x¢y¢ = 41.3 MPa

    © 2009 The McGraw-Hill Companies, Inc. All rights reserved. 7- 4


    MECHANICS OF MATERIALS
    Edition
    Fifth Beer • Johnston • DeWolf • Mazurek

    TUTORIAL PROBLEMS

    © 2009 The McGraw-Hill Companies, Inc. All rights reserved. 7- 5


    MECHANICS OF MATERIALS
    Edition
    Fifth Beer • Johnston • DeWolf • Mazurek

    TUTORIAL PROBLEMS

    © 2009 The McGraw-Hill Companies, Inc. All rights reserved. 7- 6

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