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Homework Help: Help with problem

  1. Jan 6, 2005 #1
    moments problem

    Hi everyone,

    Just wondering if you could help with this problem a formula will do cos i don't know where to start.

    A steel beam 50 feet in length is suspended with pin connections at either end and bears a load of 6000 pounds in the centre. Its elasticity is 29 000 000 pounds per square inch and its moment of inertia is 850 pounds^4.

    How far is the beam deflected?

    Thanks iin advance for any help.
     
    Last edited: Jan 6, 2005
  2. jcsd
  3. Jan 6, 2005 #2
    I too would like the answer to this question. Please note that all loads other than the initial load of 6000lbs (ie weight of the beam itself) is to be disregarded.

    I am told that there is a graph which charts information about specific modules of elasticity. Does anyone know where I could find such a graph?
     
  4. Jan 6, 2005 #3
    Check the units.
     
  5. Jan 6, 2005 #4
    Try the following:

    Draw a diagram and set the angle made by the deflacted beam be [tex] \theta [/tex]. Use the formula for the elasticity, [tex]Y = \frac{stress} {strain}[/tex]. Stress would be the tension devided by cross sectional area. [tex] Strain = \frac{length change} {original length}[/tex] which is a function of [tex] \theta [/tex].

    To find cross sectional area use the given information about moment of inertia, length of the beam and the density of steel ( which is not given) .

    gamma.

    Given moment of inertia is with respect to which axis?
     
  6. Jan 7, 2005 #5

    Astronuc

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    The units of moment of inertia should be in4 in the English system.

    Solving the equation for the deflection of a beam (of length, L) subject to a point (local) force in the center (L/2), one finds that the deflection given as a function of distance is:

    [itex]y = \frac{Fx}{48EI}(3L^2 - 4x^2)[/itex], where x is taken as positive from the left hand side, F is the point force at the center along, E is the elastic modulus, and I is the moment of inertia taken with respect to the beams major axis. By convention, F is negative (-) for downward forces and positive (+) for upward forces (assuming a horizontal beam).

    At the hinged ends, the moments are zero.

    If weight of the beam were a factor, the weight (or mass) would have to be considered as a distributed force.

    Has the instructor provided the beam equation and boundary conditions?
     
  7. Jan 7, 2005 #6
    What is x.. I don't get it. I understand everything else except for x. Help!?
     
  8. Jan 7, 2005 #7

    Astronuc

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    x is the distance from the left hinge, which increases from 0 to L/2.

    The equation is for the deflection y from the unloaded (undeformed) axis as a function of x, i.e. y=y(x), where y is the displacement at x.

    The maximum deflection is located at L/2. The deflection a the hinge is 0.

    If the force is down (-), then the deflection should be down (-).

    Also, some equations use P for load rather than F for force.
     
  9. Jan 7, 2005 #8
    So x would be where the force is on the beam? So if the force was in the center, and the beam was 50 ft(600in) long, then x = 300in!?

    F (force)= 6,000lbs
    E (Modulus of elasticity) = 29,000,000
    I (Moment of inertia) = 850
    L (length) = 600 in

    So if I understand this right, then this would be...

    y = [(6,000 x 300) / (48 x 29,000,000 x 850^4)] (3 x 600^2) - (4 x 300^2)
    y = 1.09533 ?!?!??!? ?? ??!?!
     
  10. Jan 7, 2005 #9

    Astronuc

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    In the English system, the units for this problem are:

    Load = pounds force = lbf
    Modulus of Elasticity = lbf/in2 = psi
    Moment of inertia = in4
    Length = in

    I will allow you to determine the units of deflection.

    Regarding the Modulus of Elasticity (Young's Modulus) E, this relates stress with strain in the elastic region of the stress-strain diagram. Stress has units of force per unit area (or lbf/in2, psi or ksi) just like pressure and strain is dimensionless, therefore E has units of pressure (e.g. psi or ksi).

    1 ksi = 1000 psi.
     
  11. Jan 9, 2005 #10

    Astronuc

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    I have posted the solution to this problem - https://www.physicsforums.com/showthread.php?p=423455#post423455.

    The above formula should reduce to [itex]\delta = (PL^3)/(48EI) [/itex] at L/2, point of maximum deflection.

    The term I = 850 in4, but the value 850 is not raised to the power of 4.

    The answer is then [itex] \delta [/itex] = 1.0953 in.
     
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