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A member formed by connecting a steel bar to an aluminium bar is shown in figure. Assuming that the bars are prevented from bucking sidewise, calculate the magnitude of the force P that will cause the total length of the member to decrease by 0.25 mm. The values of the elastic modulus for steel and aluminium are 2.1 × 105 N/mm2 and 0.7 × 105 N/mm2 respectively. What is the total work done by the force P?
(A) 27.5
(B) 28.5
    Correct answer is option ''. Can you explain this answer?
    Verified Answer
    A member formed by connecting a steel bar to an aluminium bar is show...
    Area of the steel bar = As = 50 × 50 = 2500 mm2
    A!rea of aluminium bar = Aa = 100 × 100 = 10000 mm2
    Total change in length = δ
    = δ
    = 0.25mm
    P × 0.11143 × 10−5 = 0.25
    ∴ P = 224356 N
    Total work done = 1/2 × load × deformation
    =
    = 28044 Nmm
    = 28.044 Nm = 28.044 Joule
    Question_Type: 5
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    A member formed by connecting a steel bar to an aluminium bar is show...
    Area of the steel bar = As = 50 × 50 = 2500 mm2
    A!rea of aluminium bar = Aa = 100 × 100 = 10000 mm2
    Total change in length = δ
    = δ
    = 0.25mm
    P × 0.11143 × 10−5 = 0.25
    ∴ P = 224356 N
    Total work done = 1/2 × load × deformation
    =
    = 28044 Nmm
    = 28.044 Nm = 28.044 Joule
    Question_Type: 5
    Free Test
    Community Answer
    A member formed by connecting a steel bar to an aluminium bar is show...
    Area of the steel bar = As = 50 × 50 = 2500 mm2
    A!rea of aluminium bar = Aa = 100 × 100 = 10000 mm2
    Total change in length = δ
    = δ
    = 0.25mm
    P × 0.11143 × 10−5 = 0.25
    ∴ P = 224356 N
    Total work done = 1/2 × load × deformation
    =
    = 28044 Nmm
    = 28.044 Nm = 28.044 Joule
    Question_Type: 5
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    A member formed by connecting a steel bar to an aluminium bar is shown in figure. Assuming that the bars are prevented from bucking sidewise, calculate the magnitude of the force P that will cause the total length of the member to decrease by 0.25 mm. The values of the elastic modulus for steel and aluminium are 2.1 × 105 N/mm2 and 0.7 × 105 N/mm2 respectively. What is the total work done by the force P?(A) 27.5(B) 28.5Correct answer is option ''. Can you explain this answer?
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    A member formed by connecting a steel bar to an aluminium bar is shown in figure. Assuming that the bars are prevented from bucking sidewise, calculate the magnitude of the force P that will cause the total length of the member to decrease by 0.25 mm. The values of the elastic modulus for steel and aluminium are 2.1 × 105 N/mm2 and 0.7 × 105 N/mm2 respectively. What is the total work done by the force P?(A) 27.5(B) 28.5Correct 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 A member formed by connecting a steel bar to an aluminium bar is shown in figure. Assuming that the bars are prevented from bucking sidewise, calculate the magnitude of the force P that will cause the total length of the member to decrease by 0.25 mm. The values of the elastic modulus for steel and aluminium are 2.1 × 105 N/mm2 and 0.7 × 105 N/mm2 respectively. What is the total work done by the force P?(A) 27.5(B) 28.5Correct 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 A member formed by connecting a steel bar to an aluminium bar is shown in figure. Assuming that the bars are prevented from bucking sidewise, calculate the magnitude of the force P that will cause the total length of the member to decrease by 0.25 mm. The values of the elastic modulus for steel and aluminium are 2.1 × 105 N/mm2 and 0.7 × 105 N/mm2 respectively. What is the total work done by the force P?(A) 27.5(B) 28.5Correct answer is option ''. Can you explain this answer?.
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