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001232
2023-12-29

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, Vg = 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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