Bending of an Aluminum beam

Topics: Elasticity, Second moment of area, Continuum mechanics Pages: 17 (1242 words) Published: March 10, 2014

CIVE 3202 A6E – Mechanics of Solids II (Winter 2013)

Experiment 2: Bending of an aluminum I-beam

“Beams are long straight members that are subjected to loads perpendicular to their longitudinal axis and are classified according to the way they are supported”[1]. When a beam is subjected to an external load there are unseen internal forces within the beam that one must be aware of when implementing it into any design or structure. These internal forces create stress and strain that could result in failure or deformation. This lab looked at how an aluminum cantilevered beam performed under symmetric and unsymmetrical bending as well as the stresses and strains developed as a result.

“To study the stress and strain induced in an I-beam under symmetric and unsymmetrical bending” [2].

σ – Normal stress (Mpa)
ε – Strain (mm/mm)
M – Moment (kN∙m)
I – Moment of inertia (mm^6)
E – Modulus of elasticity (Mpa)
G – Modulus of elasticity (Mpa)
v – Poisson’s ratio.
L – Length (m)
*Subscripts x, y, z indicate plane of reference.

The strain rosettes are orientated so that θb = 0, θc = -45, and θa = 45. The strain gauge equations then simplify to
εx = εb, εy= εc+ εa- εb, and γxy = εc- εa
Using Hooke’s Law:
σx= εxE, σy= -v σx, τxy=γxyG
This Experiment consisted of symmetric and unsymmetrical bending. For symmetric bending the relevant theory is as follows:

Because the moment about the z-axis here is zero the equation equates to: Where: My = PLA.

When rotated 45 degrees:
My = PLA Cos(45) and Mz = PLA Sin(45) there is compressive stress along the y-x axis The moment of inertia about the y-axis is found by determining the inertia of the shape and subtracting the imaginary parts as shown:

The max normal stress with be at the furthest distance from the neutral axis which is h/2 therefore: (σx)max =
The strains can be found by implementing Hooke’s Law:

Since σy and σz are zero in symmetric loading, the two equations simplify to:

Because the there is no shear stress in the x-y plane when the normal stress is at maximum the shear strain will also be zero. The vertical displacement of the end of the beam is determined by multiplying the area under the moment diagram and the distance between the end and the centroid of the diagram. This equates to:

For unsymmetrical bending the theory is the same however there is a moment about the y-axis and z-axis. This will affect the calculation of the normal stress and the strain in the x and y plane. Also the moment of inertia in the z-direction will need to be determined.

(a) *Mount the I-beam on to the support frame. Make sure the mounting screws are tight. (b) Measure the dimensions of the I-beam including its components. (c) Mount the magnet bases of the dial gauges at appropriate positions to permit the measurements of the deflections at the free end of the beam in the vertical and the horizontal directions. (d) *Connect properly the wires from the strain gauges to the readout unit. (e) Place weights to the hanger in increments: 4, 6, 10, 26, and 42 kg. (f) Unload the hanger in increments in the reversed order as for loading. (g) For each increment, measured the strain readings at the given locations and the vertical and horizontal deflections at the free end of the beam. (h) Repeat steps (a) to (g) by rotating the beam with the following angles: 45°. [3]

*Refer to appendix for sample calculation and calculated results. Part 1: I-beam at 0o

Loading (Kg)
Strain Gauge 1 ()
Strain Gauge 2 ()
Strain Gauge 3 ()
Displacement 1 (mm)
Displacement 2 (mm)
Load (N)

References: [1] – Mechanics of Materials Eight Edition, R.C. HIBBELER, 2011
[2], [3] - CIVE 3202 – Mechanics of Solids II (Winter 2013) Experiment 2: Bending of an aluminum I-beam. Obtained from:
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