Use the Java screen shown below to modify this example and see the results.
Use the Java screen shown below to modify this example and see the results.
(a) As seen in the figure above, the cross section is symmetric about the horizontal axis, therefore, the product of inertia is zero in this case. Furthermore, with the bending moment applied about the x axis, the y component of moment is zero. As a result of the previous two conditions, the NA orientation according to eqn. A13.15 will be horizontal - passing through the centroid as expected. This problem is an example of symmetric bending.
NOTE: There is no need to find the horizontal position of centroid because there is no need to calculate the moment of inertia about the y axis as y component of bending moment is zero.
(b) Because of the conditions stated in part (a) of solution, eq. A13.13 reduces to
The bending stress distribution will be linear with a zero value at the NA.
(c) In this case, the maximum stress is at the farthest point from the NA. Because of horizontal symmetry about the NA, the stress at the top and bottom of the section will have equal magnitude with the one on top being compressive. To get the maximum value of stress, the reduced equation given previously will be used.
The moment of inertia about the x axis is
This makes the maximum bending stress
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Pure Bending