Shear Forces at Plastic Hinge

Shear forces at the beam plastic hinge location

$V_{pr} + V_{g} = \frac{2 \cdot M_{pr}}{L_{h}} + \frac{w_{u} \cdot L_{h}}{2}$

$V_{pr} + V_{g} = \frac{2 \cdot 12650 kip \cdot in}{299.8 in} + \frac{1.15 \frac{kip}{ft} \cdot 299.8 in}{2}$

$V_{pr} + V_{g} = 84.4 kips + 14.4 kips$

$V_{pr} + V_{g} = 98.8 kips$

V_pr	Shear required to produce the maximum probable moment at the plastic hinge V_pr = 2M_pr/L_h
V_g	Shear from gravity loads at the plastic hinge location V_g= w_uL_h/2
M_pr	Probable maximum moment at the plastic hinge location
L_h	Distance between plastic hinge locations L_h = L_beam - d_c- 2S_h= 360.0 in - 15.20 in - 2*22.50 in = 299.8 in L_h is equal to L_cf (clear length of beam) when the plastic hinge location is omitted
w_u	Gravity loads on the beam

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