Bending
Members (Beams)
Example: (continued) The moment of inertia of
each part about their centroidal axis is: Iz1z1 =
(1/12)bh3 and I z2z2 = (1/12)hb3 Next use the parallel axis
theorem to transfer these moments of inertia to the centroidal (neutral) axis of the
composite section. For part 1: IzNA1 = inertia of part 1 about
the centroidal axis of the composite section is: Iz1z1 =
(1/12)bh3 + bh(D – h/2)2 For part 2: IzNA2 = inertia of part 2 about
the centroidal axis of the composite section is: Iz2z2
= (1/12)hb3 + bh(b/2 + h – D)2 Then the moment of inertia
of the composite section about its neutral axis is the sum. IzNA
= IzNA1 ,
IzNA2 = = (1/12)bh3 + bh(D – h/2)2 +
(1/12)hb3 + bh(b/2 + h – D)2 For the given data: b = 20 mm, h = 80 mm, D = 65 mm
IzNA
= 2.906 x 106 mm4 =
moment of inertia of composite section about neutral axis For the top fiber: σtop = MCtop / IzNA
=
10000(35) / 2.906 x 106
= 0.120 N/mm2 = σmax For the bottom fiber: σbottom
= MCtop
/ IzNA = 10000(65) / 2.906 x 106 = −
0.224 N/mm2 = σmin Top fiber has the maximum
tensile stress = 0.120 N/mm2 (result) Bottom fiber has the
minimum compressive stress = −
0.224 N/mm2 (result) |
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