6c-ex. Deflection Examples
Ex. 6c.1Ex. 6c.2
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Example 6c.1

Given: A cantilever beam, is built in at the left end, has length L, and rectangular cross-section of width b, and height h. The beam is loaded by point load P at its free end.

Req'd: Determine the deflection at the end of the beam.

Sol'n: The bending moment in the beam is given by:

M(x) = –P(L – x)

Therefore the differential equation for bending is:

EIv''(x) = –P(L – x)

Integrating with respect to x gives:

Since the slope at the built-in end is zero, then

v'(x=0) = 0, so:  C1 = 0.

Integrating again gives:

The deflection at the built-in end is zero, so:

v(x=0) = 0, so:  C2 = 0.

Therefore, the equation of the elastic curve - the deflection - is:

Deflection at the tip is then:

Since the sign of v(L) is negative, the deflection is downward.

Example 6c.2

Given: A simply supported solid circular beam with radius r = 1.2 in. and length L = 50 in. is subjected to a uniform distributed load of q(x) = 24 lb/in. The beam is made from 6061 aluminum.

Req'd: Determine the maximum deflection of the beam.

Sol'n: Recall from Example Problem 6b.2 that the bending moment and the moment of inertia of the beam are given by:


The moment is constant over the entire length of the beam. The curvature of the beam's deflection is then:

Integrating once gives the equation defining the slope of beam:

The constant C1 will be found later using the boundary conditions.

Integrating a second time gives the deflection of the beam:

The constants C1 and C2 are determined using the boundary conditions (BC) . Since the beam is simply supported:

  • v(0) = 0
  • v(0) = L

Due to symmetry of the load and geometry (supports), another BC that be applied in this case is v'(L/2) = 0.

Solving for the constants:

The deflection of the beam is then::

From the symmetry of the beam we know the maximum deflection is at x = L/2 = 25 in. From the Materials Property Table, E for 6061 aluminum is 10 Msi. The maximum deflection is then:

  vmax = –0.120 in.

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Updated: 05/24/09 DJD