SampleProblemsFinalAnswers.epdf

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    University of Wisconsin Madison; ME 306

    Problem Find the reactions , &A A BR M R due to the load P. Assume that the bar has inertia I and

    Youngs modulus E. Do you get the expected reactions in the limit 0k . You may use the

    information given on the right.

    Solution:The deflection at B is:

    ( ) ( )3 3 3 3

    3 3

    4.5 36 6

    3.5 2

    6 6

    = +

    =

    B

    B

    P RL L L L

    EI EI

    P L R L

    EI EI

    3 3

    3

    3

    3 3

    3.5 2

    6 6

    3.53.56

    2 6 21

    6

    =

    =

    = =+

    +

    BR k

    P L k L

    EI EI

    P LP LEI

    k L EI k L

    EI

    3

    3

    3.5

    6 2= =

    +B

    kP LR k

    EI k L

    From force balance:

    ( )

    3

    3

    33

    3 3

    3.5

    6 2

    1.5 63.56 2 6 2

    + = =+

    = =

    + +

    A B

    A

    kP LR P R

    EI k L

    P k L EIkP LR PEI k L EI k L

    From Moment Balance about A

    P

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    4

    3

    4

    3

    4 4

    3

    3

    3.51.5

    6 2

    3.51.5

    6 2

    3.5 1.5 (6 ) 3.5

    6 2

    1.5 (6 )

    6 2

    + = =+

    = +

    =

    +

    =

    +

    A B

    A

    A

    A

    kP LM PL R L

    EI k L

    kP LM PL

    EI k L

    kP L PL EI kPL

    M EI k L

    PL EIM

    EI k L

    In the limit 0k 0=BR ; = AR P; 1.5= AM PL , as expected.

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    Problem 4 Determine the maximum height h from which the diver can jump onto the board so

    that maximum being stress in the wood does not exceed 6 ksi. The board thickness is 1.5 inch

    and its width is 18 inches. Use6

    woodE 1.8 * 10 psi= . Assume the diver weighs 180 lbs.

    Solution:Using the energy principle, the equivalent load is defined as:0.5 ( ) = +P mg h

    We now must determine the relationship between the deflection and load. The reactions forces

    are at the left and right are:

    1

    2

    8 / 3

    11 / 3

    =

    =

    R P

    R P

    The moment diagram can be computed as follows.

    1

    2

    ( ) 8 / 3;0 3

    ( ) (11 ) / 3;3 11

    =

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