# Lab1 Unsymmetrical Bending

**Topics:**Bending, Second moment of area, Beam

**Pages:**12 (2455 words)

**Published:**April 14, 2015

Department of Civil & Structural Engineering

BEng(Hons) in Civil Engineering

Structural Mechanics II

Laboratory Instruction Sheet:Unsymmetrical Bending

Objective:

To observe the two principal axes in a beam with unsymmetric cross sections; and make comparison between the theoretical and actual behavior in bending of two unsymmetrical section cantilevers:

1. An equal angle with one axis of symmetry.

2. A Z section completely unsymmetrical.

Apparatus:

Vertical cantilever system, dial gauges, standard weight, hanger, cantilever beams of L and Z section.

Procedure:

1. Mount up a beam section on the vertical cantilever system and measure the cross sectional dimensions. 2. Set the position of a geometrical axis of the beam section at zero degree. 3. Start the bending test on the cantilever beam with the angular position equals zero. 4. Add weights onto the hanger one by one. For L section, use loading 10N, 20N, 30N. For Z section, loading : 400g, 800g, 1200g. Record the dial gauge readings at each stage of load. 5. Repeat the bending test at different angular positions of the applied load with an increment of 22.5.

(Please properly show your calculations)

Training on critical thinking (20 marks out of 100):

1. If two dial gauges are not placed at a right angle to each other as shown in Figure (a), justify whether the objectives of the experiment are still achieved. (8 marks) Ans. The objectives of the experiment can be still achieved, while it is necessary to do some substitution.

Therefore this method is still work when placed two dial gauges at an arbitrary angle as long as the exact angle had been measured. However, the dial gauge may be not as sensitive as that of detecting the deflection of section of interest if the gauges are not placed at a right angle. As a consequence, the result could not be derived easily and directly and the corresponding error are expected to be larger than former one. But, overall, it still works.

2. If the load is applied to the corner of the section as shown in Figure (b), criticize whether the objectives of the experiment are achieved. (12 marks)

Ans, the objectives of this experiment cannot be achieved.

If the load is applied to corner of the section, the overall result under this force is equivalent to the force with the same magnitude act on the shear center and a moment couple. Therefore, the members will be twisted and create bending stress as well as shear stress at the same time.

Furthermore, a normal stress will also be created since the degree of deformation in every two arbitrary sections is different. Also, the change of interior normal stress will create varied shear stress.

To draw the conclusion, these stresses corresponding the additional moment couple are complicated and cannot be resolved only by equivalent equation, the relevant knowledge of statics, physical, geometry should be applied. In our experiment, this problem cannot be solved with the readings of two dial gauge.

Reports and discussion (80 marks out of 100):

Section Properties:

1. Label dimensions.

2. Find and mark the centroid of the sections.

3. Determine Second moment of area I xx, I yy

(x is the horizontal axis, y is the vertical axis) 4. Determine Cross moment of area I xy

5. Determine principal moments of inertia I u and I v

6. Determine orientation of principal axes p (show them in the figure)

1. As shown in the figure (mm).

2.x'= =10.148 mm

y'= =10.148 mm

3.I xx = ×34×4.53+34×4.5×(10.148-2.25)2+ ×4.5×29.53+29.5×4.5×(14.75+4.5-10.148)2 = 30427.084 mm4= 3.043×10-8 m4

I yy = ×34×4.53+34×4.5×(10.148-2.25)2+ ×4.5×29.53+29.5×4.5×(14.75+4.5-10.148)2

=30427.084 mm4= 3.04×10-8 m4

4.I xy =...

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