| 6c-ex. Deflection Examples
• Ex. 6c.1 • Ex. 6c.2
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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:
Therefore the differential equation for bending is:
Integrating with respect to x gives:
Since the slope at the built-in end is zero, then
Integrating again gives:
The deflection at the built-in end is zero, so:
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.
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.
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:
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:
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