| Consider previous example with a short term transverse applied load causing a bending moment about |
| the X-X axis of 240Nm. Weak axis restraint 800mm. |
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M*x = |
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0.24 |
kNm |
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Z = b.d2 / 6 = |
60750 |
mm^3 |
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k1 = |
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1.0 |
for short term load |
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fb = |
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14 |
MPa for MSG8 timber |
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using k8 and f factors from example 1 |
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we are considering the short term loading condition here, so the k1 factor used throughout the calculation will be k1 = 1.0 |
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fMnx = f.k1.k8.fb.Z = |
0.53 |
kNm |
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fNncx = f.k1.k8.fc.A = |
23.8 |
kN |
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fNncy = f.k1.k8.fc.A = |
45.4 |
kN |
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N*c / fNncx + M*x / fMnx = |
0.96 |
< 1.0 so OK |
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N*c / fNncy + (M*x / fMnx)2 = |
0.45 |
< 1.0 so OK |
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