From Mcgregor's "Reinforced Concrete Mechanics and Design 6E"
![URL]](https://res.cloudinary.com/engtips/image/fetch/w_800,c_lfill,q_auto,f_auto,g_faces:center/[URL unfurl="true"]http://imageshack.com/a/img921/2012/wbirKv.jpg[/URL])
If you are familiar with the concept, jump to the bottom questions:
(briefing quoting Mcgregor)
"When the frame is displaced laterally through the inelastic deformations required to
develop the ductility of the structure, the reinforcement at the ends of the beam will yield
unless the moment strength is several times the moment due to seismic loads. The yielding
of the reinforcement sets an upper limit on the moments that can be developed at the ends
of the beam. The design shear forces, Ve are based on the shears due to factored dead and live loads (Fig. 19-19c) plus the shears due to hinging at the two ends of the beam for the
frame swaying to the right or to the left, as shown in Fig. 19-19a. Mpr is the probable
moment strength of the members, based on the dimensions and reinforcement at the joint
and assuming a tensile strength of 1.25 fy and alpha=1.0. For a rectangular beam without
axial loads, ACI Code Section 21.5.4.1 requires that beams be designed for the sum of
Vsway = (Mpr1 + Mpr2) / ln
Vg = wu ln / 2
Ve = Vg + Vsway"
It is stated that Vc shall be taken equal to zero if (a) the shear Vsway due to plastic hinging at the two ends of the beam exceeds half or more of the maximum shear Vu, within the span
Let say your Vsway = 88 kips (yeah Vsway is really huge)
Your Vg (1.2 LL + 1.6 DL) or shear from factored gravity load = 87kips
When Vsway is equal or more than Vg then you need to set Vc= 0
But here's the dilemma. If you add 2 kips to the Vg, then Vsway is less than Vg, and Vc=0 is no longer required.
There is a big difference in the beam to accomodate Vc=0. This means you need to design stirrups that will take on the entire Ve without any contribution from the Vc shear capacity of concrete. This would cause increase in section sizes as well as bigger diameter stirrups (can cause some congestion too). And if Vc is not zero, it simply means you can use the concrete shear capacity (the logic of having Vc=0 is "The damage to the hinging area due to repeated load reversals
greatly reduces the ability of the cross section to resist shear, requiring more transverse
reinforcements"
So in the event you encounter situation where your Vsway = 88 kips and Vg (1.2 LL + 1.6 DL) = 87kips.. you can simply add 2 kips to make Vsway less than Vg and Vc bcome no longer zero and make available the concrete shear capacity (saving a lot of member size increase or bigger diameter stirrups).
What would you do if you encounter the scenerio above. Add 2kips and make Vc not zero or make Vc=0 (but more costly and even attracting more moments from increased member sizes)?
If you are familiar with the concept, jump to the bottom questions:
(briefing quoting Mcgregor)
"When the frame is displaced laterally through the inelastic deformations required to
develop the ductility of the structure, the reinforcement at the ends of the beam will yield
unless the moment strength is several times the moment due to seismic loads. The yielding
of the reinforcement sets an upper limit on the moments that can be developed at the ends
of the beam. The design shear forces, Ve are based on the shears due to factored dead and live loads (Fig. 19-19c) plus the shears due to hinging at the two ends of the beam for the
frame swaying to the right or to the left, as shown in Fig. 19-19a. Mpr is the probable
moment strength of the members, based on the dimensions and reinforcement at the joint
and assuming a tensile strength of 1.25 fy and alpha=1.0. For a rectangular beam without
axial loads, ACI Code Section 21.5.4.1 requires that beams be designed for the sum of
Vsway = (Mpr1 + Mpr2) / ln
Vg = wu ln / 2
Ve = Vg + Vsway"
It is stated that Vc shall be taken equal to zero if (a) the shear Vsway due to plastic hinging at the two ends of the beam exceeds half or more of the maximum shear Vu, within the span
Let say your Vsway = 88 kips (yeah Vsway is really huge)
Your Vg (1.2 LL + 1.6 DL) or shear from factored gravity load = 87kips
When Vsway is equal or more than Vg then you need to set Vc= 0
But here's the dilemma. If you add 2 kips to the Vg, then Vsway is less than Vg, and Vc=0 is no longer required.
There is a big difference in the beam to accomodate Vc=0. This means you need to design stirrups that will take on the entire Ve without any contribution from the Vc shear capacity of concrete. This would cause increase in section sizes as well as bigger diameter stirrups (can cause some congestion too). And if Vc is not zero, it simply means you can use the concrete shear capacity (the logic of having Vc=0 is "The damage to the hinging area due to repeated load reversals
greatly reduces the ability of the cross section to resist shear, requiring more transverse
reinforcements"
So in the event you encounter situation where your Vsway = 88 kips and Vg (1.2 LL + 1.6 DL) = 87kips.. you can simply add 2 kips to make Vsway less than Vg and Vc bcome no longer zero and make available the concrete shear capacity (saving a lot of member size increase or bigger diameter stirrups).
What would you do if you encounter the scenerio above. Add 2kips and make Vc not zero or make Vc=0 (but more costly and even attracting more moments from increased member sizes)?
