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I-BEAM EM327: MECHANICS OF MATERIALS LABORATORY

EXPERIMENT: I-BEAM OBJECTIVES:

L/3

(1) Investigate the theoretical beam relationships for an I-beam loaded at the third (l/3) points. (2) Investigate the stresses and deflection. (3) Determine the Modulus of Elasticity of the specimen. INTRODUCTION:

L/3

L/3 P/2

P/2

P/2 P/2

P/2 P/2

An aluminum I-Beam is to be loaded in bending. The strains, center deflection, end rotation, and modulus of elasticity are measured and compared to theoretically predicted values.

P/2 V

BACKGROUND:

-P/2

A method of solution, applicable to elastic design only, can be developed using the elastic flexure formula. The assumption that plane sections remain plane is made. It is also assumed that the intersections of materials are securely welded together to provide adequate resistance to longitudinal shearing stress. A simply supported beam loaded at the third points is shown in Figure 1. The elastic curve and shear-moment diagrams for the beam are also shown in the figure. The slope of the elastic curve at the support and the deflection at the center of the span are given by:

PL/6 M

FIGURE 1

=

PL2 18 EI 23PL3 1296 EI

(at support)

(a)

=

It is important to note that the shear force is zero and the bending moment is constant for the middle third of the beam. This implies that the shear stress as expressed by = (VQ/It) is zero and the flexural stress = -(My/I) is constant over the middle third for a specified distance from the neutral axis. Since there is a uniaxial state of stress the following equation applies: = E (c)

(center of span)

(b)

These equations can be developed using singularity equations. The procedure is outlined at the end of this experiment description.

For this loading the moment is defined as positive as shown in Figure 2. This creates

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I-BEAM EM327: MECHANICS OF MATERIALS LABORATORY

compressive deformation above the neutral axis and tensile deformation below. corresponding stress distribution can be obtained by multiplying the strain by the modulus of elasticity (Hooke's Law). The moment of inertia, I, with respect to the neutral axis is found using the standard moment of inertial equation for a rectangle along with the parallel axis theorem. Moment of inertia for a rectangle is given in equation (d).

M

M y N.A. y

c FIGURE 2

If the two vertical planes shown remain plane surfaces after rotation, the change in length per unit length at a distance y from the neutral surface is y. The deformation divided by the original length is y . This strain would vary linearly with distance from the neutral surface. A homogeneous beam with a rectangular cross section is shown in Figure 3. For this case, the neutral axis coincides with the centroidal axis located at half of the depth.

bh 3 I= 12

(d)

where b is the base of the rectangle and h is the height. Note that the base of the rectangle is always parallel to the neutral axis. If the moment of inertia is determined for each piece of the I-beam, then the total moment of inertia can be calculated using the parallel axis theorem. The parallel axis theorem is used when the centroid of the particular rectangle does not coincide with the neutral axis of the entire system. In the case of areas 1 and 2 in Figure 4 the parallel axis theorem is used by adding additional area*distance2 term (see equation (e) and Figure 4).

I total = I1 + I 2 + A2 d 22 + I 3 + A3 d 32

(e)

To determine the normal stress at a point on the beam, equation (f) will be used.

N.A.

=

FIGURE 3

The longitudinal strain distribution for this cross section is also shown in Figure 3. The

-My I

(f)

where s is the normal stress, M is the moment, I is the total moment of inertia, and y is the distance from the neutral axis to the point of interest.

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I-BEAM EM327: MECHANICS OF MATERIALS LABORATORY

1. Calculate the moment of inertia about the neutral axis for the beam. 2. Compare the calculated moment of inertia with the given moment of inertia. (use the given moment of inertia in all further calculations) TESTING PROCEDURE:

2

1.) Measure the span length, depth, and width of the beam. 2.) Determine and show on a sketch the location of the loading points, reactions, strain gages, and level scale.

+y

d2

3.) Zero strain indicators 4.) Obtain initial values indicated on center deflection scale and end rotation scale. 5.) Apply the 500 lb., 1000 lb., 1500 lb., and 2000 lb. loads by turning the crank on the back of the machine until the scale on the front of the machine zeros. 6.) Obtain readings for strains, center deflection, and end rotation at each load step. 7.) Remove the load and turn off the strain indicator box. REPORT:

NA 1 d3

3

FIGURE 4

EQUIPMENT TO BE USED: Olsen Testing Machine Strain Indicator Switch and Balance Unit Level Deflection Indicator SPECIMEN TO BE TESTED: 6061-T6 Aluminum Wide Flange Beam PROCEDURE: PRELIMINARY CALCULATIONS: REPORT REQUIREMENTS: (1) Compare the slope and center deflection determined experimentally with the theoretical values (use equations a and b). (2) For each loading condition calculate the theoretical strain at each strain gage position using equations (f) and (c). Plot theoretical strain versus beam depth. Plot the experimental strain versus beam depth on the same diagram. (3) Use equation (f) to calculate the stress at the top of the beam for each loading condition. Plot the stress versus the experimental strain at each loading

81

I-BEAM EM327: MECHANICS OF MATERIALS LABORATORY

condition to determine the modulus of elasticity for the beam. Compare the experimental modulus of elasticity to the theoretical modulus of elasticity.

82

I-BEAM EM327: MECHANICS OF MATERIALS LABORATORY

SINGULATITY FUNCTIONS: Recall that when the <x-b> quantity inside the bracket is < or = 0, the term drops out of the equation.

RL = R R = EI

P 2

1

P L 1 d2y x- = RL x - 2 dx 2 3

2

-

2

P 2 x- L 2 3

1

dy R L EI x = dx 2 R EIy = L x 6

3

P L x- - 4 3

3

P 2 x- L - 4 3

3

2

+ C1 + C1 x + C 2

P L x- - 12 3

P 2 x- L - 12 3

BOUNDARY CONDITIONS: y=0: y' = 0 : =0 x = L/2

2

therefore

2

C2 = 0

2

therefore

RL x 2

P L - x- 4 3

P 2 - x- L 4 3

gives

+ C1 = 0 C1 = - PL2 18

P Substituting = R L 2

SLOPE: x=0

= =

dy 1 PL2 = dx EI 18 - PL2 18EI

DEFLECTION: x= L/2

= y= =

1 EI

R L 6

2 3 P L L PL2 L - - + 18 x 2 12 2 3

- 23 PL3 1296EI

P/2

P/2

L/3 RL

L/3

L/3 RR

83

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