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The data obtained from a rectangular strain gage rosette attached to a stressed steel member are ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = +220 × 10−6.Given E = 2 × 105 N⁄mm2 and Poisson’s ratio = 0.3. The principal stress acting at the point is ___________ MPa
(A) 37.5
(B) 39.5
    Correct answer is option ''. Can you explain this answer?
    Verified Answer
    The data obtained from a rectangular strain gage rosette attached to ...
    Εxx = ε0 = −220 μs
    εyy = ε90o = +220 μs
    Principal strain,
    ε1 , ε2 =
    ε1, ε2 = 0 ± √(−220)2 + (120)2
    = ±250.6 μs
    p1 =
    =
    = 38.55 N/mm2
    p2 = −38.55 N/mm2
    Question_Type: 5
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    The data obtained from a rectangular strain gage rosette attached to ...
    Solution:

    Given,

    ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = 220 × 10−6

    E = 2 × 105 N⁄mm2, Poisson’s ratio = 0.3

    To determine the principal stress acting at the point,

    We know that,

    ε0 = (σ1 - σ2) / E

    ε45° = (σ1 - σ2) / 2E

    ε90° = (σ1 - σ2) / E

    where, σ1 and σ2 are the principal stresses.

    On solving the above equations, we get:

    σ1 + σ2 = E(ε0 + ε90°)

    σ1 - σ2 = 2Eε45°

    σ1 = [E/2(1+ν)] [(ε0 + ε90°) + √(ε0 - ε90°)2 + 4ε2 45°]

    σ2 = [E/2(1+ν)] [(ε0 + ε90°) - √(ε0 - ε90°)2 + 4ε2 45°]

    where, ν is the Poisson's ratio.

    Substituting the given values, we get:

    σ1 = [2 × 105/2(1+0.3)] [(-220 × 10^-6) + (220 × 10^-6) + √((-220 × 10^-6) - (220 × 10^-6))^2 + 4(120 × 10^-6)^2] = 39.5 MPa

    σ2 = [2 × 105/2(1+0.3)] [(-220 × 10^-6) + (220 × 10^-6) - √((-220 × 10^-6) - (220 × 10^-6))^2 + 4(120 × 10^-6)^2] = -2.92 MPa

    The principal stress acting at the point is 39.5 MPa (Option B).
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    The data obtained from a rectangular strain gage rosette attached to a stressed steel member are ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = +220 × 10−6.Given E = 2 × 105 N⁄mm2 and Poisson’s ratio = 0.3. The principal stress acting at the point is ___________ MPa(A) 37.5(B) 39.5Correct answer is option ''. Can you explain this answer?
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    The data obtained from a rectangular strain gage rosette attached to a stressed steel member are ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = +220 × 10−6.Given E = 2 × 105 N⁄mm2 and Poisson’s ratio = 0.3. The principal stress acting at the point is ___________ MPa(A) 37.5(B) 39.5Correct answer is option ''. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The data obtained from a rectangular strain gage rosette attached to a stressed steel member are ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = +220 × 10−6.Given E = 2 × 105 N⁄mm2 and Poisson’s ratio = 0.3. The principal stress acting at the point is ___________ MPa(A) 37.5(B) 39.5Correct answer is option ''. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The data obtained from a rectangular strain gage rosette attached to a stressed steel member are ε0 = −220 × 10−6, ε45° = 120 × 10−6, ε90° = +220 × 10−6.Given E = 2 × 105 N⁄mm2 and Poisson’s ratio = 0.3. The principal stress acting at the point is ___________ MPa(A) 37.5(B) 39.5Correct answer is option ''. Can you explain this answer?.
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