Composite
Beams
*Example: (continued) 4. Next calculate the moment of inertia of the
entire “transformed” section about its neutral axis, y. Use the parallel axis theorem. Iyy
= (1/12) b1h13
+ b1h1( c ˗
h1/2)2 + (1/12) b2h23 + (b2h2)
( h1 + h2 /2 ˗ c )2 Iyy = (1/12)(80)(0.53) +
(80)(0.5)(1.315 – 0.25)2 +
(1/12)(4)(5.53) + (4)(5.5)(3.25 – 1.315)2 So for the “transformed”
section Iyy =
184 in4 Now the maximum bending
stress can be calculated using σ
= Mc/I . So we need to calculate the maximum bending moment by constructing
the shear and bending moment diagrams as shown below. The bending stresses in
terms of the base material (wood) are as follows: Let
c1 = c
, c2 = h1
+ h2 ˗ c , c3 =
c ˗ h1 At the top of the composite beam σ = Mc2/Iyy = – (4950)(12)(6-1.315) / 184 = – 1513 psi At
the bottom of the composite beam σ
= Mc1/Iyy = (4950)(12)(1.315) / 184 = 424 psi The
maximum tensile bending stress in the wood is: σ = Mc3/Iyy = (4950)(12)(1.315 – 0.5)/184 = 263
psi The
maximum tensile bending stress in the steel is 424(Es/Ew) =
424(20) = 8480 psi
(result) |
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