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Mechanics of Materials – MAE 243 (Section 002) Spring 2008

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Presentation on theme: "Mechanics of Materials – MAE 243 (Section 002) Spring 2008"— Presentation transcript:

1 Mechanics of Materials – MAE 243 (Section 002) Spring 2008
Dr. Konstantinos A. Sierros

2 Problem The simply-supported beam ABCD is loaded by a weight W = 27 kN through the arrangement shown in the figure. The cable passes over a small frictionless pulley at B and is attached at E to the end of the vertical arm. Calculate the axial force N, shear force V, and bending moment M at section C, which is just to the left of the vertical arm. (Note: Disregard the widths of the beam and vertical arm and use Center line dimensions when making calculations.)

3 Problem The cantilever beam AB shown in the figure is subjected to a uniform load acting throughout one-half of its length and a concentrated load acting at the free end. Draw the shear-force and bending-moment diagrams for this beam.

4 5.5: Normal stresses in beams (Linearly elastic materials)
Since longitudinal elements of a beam are subjected only to tension/compression, we can use the stress-strain curve of the material to determine the stresses from the strains The most common stress-strain relationship encountered in engineering is the equation for a linearly elastic material Resultant of the normal stresses A force acting in the x-direction A bending moment acting about the z-axis

5 5.5: Location of neutral axis
Consider an element of area dA in the cross-section. The element is located at distance y from the neutral axis This equation states that the z-axis must pass through the centroid of the cross-section The neutral axis (z-axis) passes through the centroid of the cross-sectional area when the material follows Hooke’s law and there is no axial force acting on the cross-section

6 5.5: Moment curvature relationship
Moment of inertia of the cross-sectional area with respect to the z-axis The moment-curvature equation shows that the curvature is directly proportional to the bending moment M and inversely proportional to the quantity EI, which is called the flexural rigidity Positive bending moment produces positive curvature and a negative bending moment produces negative curvature

7 5.5: Flexure formula We can determine the stresses in terms of the bending moment This equation is called the flexure formula and shows that the stresses are directly proportional to the bending moment M and inversely proportional to the moment of inertia I of the cross-section Stresses that are calculated from the flexure formula are called bending stresses or flexural stresses

8 5.5: Maximum stresses at a cross section
The maximum tensile and compressive bending stresses acting at any given cross-section occur at points located furthest for the neutral axis The corresponding maximum normal stresses σ1 and σ2 (from the flexure formula) section moduli

9 5.5:Doubly symmetric shapes
If the cross-section of a beam is symmetric with respect to the z-axis as well as the y-axis then we have or For a beam of rectangular cross-section with width b and height h For a circular cross-section

10 Wednesday: QUIZ on Chapter 4
Duration 20 minutes 1 question to answer


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