Example 6.3.1

Given: The rectangular beam, built in at the left end, having length, L, and cross-section of width, b, and height, h, is acted upon by a 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

v'(x=0) = 0 C₁ = 0.

Integrating again gives:

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

v(x=0) = 0 C₂ = 0.

Therefore, the equation of the ELASTIC CURVE is:

Deflection at the tip is then:

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

Example 6.3.2

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_f/in. The beam is made from 6061 aluminum.

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

Sol'n: Recall from Example Problem 5.2.2 that the bending moment and the moment of inertia of the beam are given by:

M(x) =

q 2

Lx - x2

and

Iz =

p r4 4

= 1.63 in4

The GOVERNING DIFFERENTIAL EQUATION for DEFLECTION is then:

v''(x) =

M EI

=

q 2EI

Lx - x2

Integrating once gives the equation defining the SLOPE of the deflection:

v'(x) =

q 2EI

Lx2 2

-

x3 3

+ C1

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

Integrating a second time gives the equation defining the ELASTIC CURVE of the beam:

v(x) =

q 2EI

Lx3 6

-

x4 12

+ C1x + C2

The constants C₁ and C₂ are determined using the boundary conditions:

• Due to the constraints at the ends,     v(0) = v(L) = 0;

• Due to symmetry in the beam,     v'(L/2) = 0.

Therefore:

C1 = -qL3 24

and

C2 = 0

The DEFLECTION of the beam is then given by:

v(x) =

q 24EI

2Lx3 - x4 - L3x

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 =

24 lb/in 24(10 Msi)(1.63 in4)

2(50 in)(25 in)3 - (25 in)4 - (50 in)3(25 in)