Bending
Members (Beams)
|
Example: (continued) Next determine the
location of the centroid of the entire X-section
shown below using the principle of first moments
namely the moment of the sum equals the sum of the moments. See the figure below (NA
denotes the neutral axis passing through the centroid. Moment of sum = (bh + bh)(D) Sum of moments = (bh)(h/2) + (bh)(h + b/2) Therefore D =
h/4 + h/2 + b/4 = 3h/4 + b/4
= 3(80)/4 + 20/4 = 65
mm =
0.065 m So the distance from the
y-coordinate of the centroid to the top fiber is Ctop = ( h + b) – D = (80+20) – 65 = 35
mm and the distance from the
y-coordinate of the centroid to the bottom fiber is Cbottom
= D = 65 mm
So σtop = M Ctop / IzNA and σbottom
= M Cbottom / IzNA where IzNA is the moment of inertia of the composite
area about its neutral axis. So calculate the moment of
inertia for each part (1 and 2) about their centroidal
axes, transfer them to the centroidal axis of the composite section, and add them
together to get the total moment of
inertia IzNA . Click here to continue
this example. |
| Return to Notes on Solid Mechanics |
|---|
All rights reserved.