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2024-01-16

Shear Forces at Plastic Hinge

Shear forces at the beam plastic hinge location
Vpr+Vg = 2·MprLh + wu·Lh2
Vpr+Vg = 2·12650 kip·in299.8 in + 1.15 kipft·299.8 in2
Vpr+Vg = 84.4 kips + 14.4 kips
Vpr+Vg = 98.8 kips
Vpr Shear required to produce the maximum probable moment at the plastic hinge
Vpr = 2Mpr/Lh
Vg Shear from gravity loads at the plastic hinge location
V= wuLh/2

Mpr Probable maximum moment at the plastic hinge location
Lh Distance between plastic hinge locations
Lh = Lbeam - dc - 2Sh = 360.0 in - 15.20 in - 2*22.50 in = 299.8 in
Lh is equal to Lcf (clear length of beam) when the plastic hinge location is omitted
wu Gravity loads on the beam

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