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Example 3
Select the
lightest W16 for the cantilever beam shown in Fig. 5.13 using A36 steel
with yield stress of 36 ksi. Load P1
is applied in the vertical plane of symmetry. Load P2 passes through the centroid of the section and
makes an angle of 30 degree with the vertical plane of symmetry. (It is
located in a plane perpendicular to the vertical plane of symmetry.) Assume
total lateral support and neglect the weight of the beam. Specify the point
where the normal stress has the largest absolute value.
Solution
This beam
is subjected to biaxial bending, that is, simultaneous bending about major
and minor axes. If we denote these two axes by x and y, the
resultant normal stress due to bending Mx
about the x-axis and My about the y-axis for a doubly symmetric
section is found from
where Sx and Sy are section moduli
with respect to x (major axis)
and y (minor axis), respectively.
Denoting the maximum bending stress due to
bending about the x-axis by fbx = Mx/Sx and the maximum bending stress due to bending
about the y-axis by fby
= My/Sy and noting that the
allowable bending stresses about the major and minor axes are different, we
must satisfy the following interaction equation for design of beams
subjected to biaxial bending:
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(5.29)
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where Fbx and Fby are the allowable
bending stresses with respect to major and minor axes, respectively. To
start the iterative design process it may initially be assumed that Fbx = 0.60Fy and Fby = 0.75Fy. Thus, the following
approximate equation for the section modulus Sx may be used for preliminary selection of the
section (depending on the span length, lateral bracing points, and the loading,
a designer may choose other initial values for the allowable bending
stresses):
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(5.30)
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where Rs = Sx/Sy.
This ratio can be estimated approximately for different rages of sections.
The most frequent values for Rs
are 7 for W27 to W36 sections, 5 to 6 for W16 to W24 sections, and 2 to 3
for W10 to W 14 sections.
For the example of Fig. 5.13, the maximum
bending moment occurs at the fixed support.
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Mx = 2(6)+(1)cos 30° (12) = 22.39 K-ft =
268.71 K-in.
My = (1)cos 60° (12) =6 K-ft = 72
K-in.
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Assume Rs = 5.
Try W16 x
26.
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Sx = 38.4 in.3
and Sy = 3.49 in.3
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The
section is compact and Fbx
= 0.66Fy = 24 ksi and Fby = 0.75Fy = 27 ksi.
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N.G.
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Try
W16 x 31.
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Sx = 47.2 in.3
and Sy = 4.49 in.3
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O.K.
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Points A and B
identified on Fig 5.13 are the points of maximum stress.
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