A simply supported beam of length 18 ft. is subjected to a trapezoidally distrib

Question

A simply supported beam of length 18 ft. is subjected to a trapezoidally distributed load that has a value of 3 kip/ft at x=0 ft. and 6 kip/ft at x=18ft.. The beam is made of a low carbon steel that has a yield strength of 48 ksi. Select the lightest W-beam that will support the distributed load. Furthermore, the maximum deflection cannot be more than 0.8 in..

1. Determine the reactions at the supports due to the applied load.

2. Draw SF and BM diagrams.

3. Determine the maximum moment, shear force.

4. Use the equations for normal stress to determine the section modulus.

5. Use tables to determine the W-beam that has the lowest section modulus that is bigger than the above.

6. Calculate the I for the beam by using the allowable deflection. See if the beam from 5 has this I. If not, choose a different W-beam.

7. Take into account the weight of the beam and check to make sure that the beam can support the applied load as well as the self weight. In addition, check to make sure that the shear strength of the beam is not exceeded.

8. Provide a plot of the deflection curve of the beam.

9. Provide a plot of the slope of the deflection curve of the beam.

10. Draw Mohr's Circle for a point on top of the cross-section of the beam where the maximum bending moment occurs. Determine the principal stresses and draw an element showing the principal planes.

Must show all work for rating. If you cannot do all parts fully with work shown, do not submit a response. Please number all solutions to corresponding questions.

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Explanation / Answer

strain = dl = l2 - l1 = 0.1 mmE = stress / strain = 207GPAstress = F / A = force / areaarea = pi(r^2)207 = {F / pi(r^2)}/(0.1)solve for F (watch units)Poisson's ratio for steel is about 0.3so if the axial strain is 0.1 the lateral strain is0.3(0.1) = 0.03 cc reduction in diameterArea = pr

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