https://communities.bentley.com/products/ram-staad/f/ram-staad-forum/112587/ram-concept--shear-reinforcementHi everyone, Why does using 5000 psi concrete strength for a beam design give you a reduced spacing (22.6&quot;) shear reinforcement as opposed to using 4000 psi concrete strength (24&quot;)? See attached model.en-USTelligent Community 12https://communities.bentley.com/thread/343806?ContentTypeID=1Thu, 22 Oct 2015 17:20:59 GMT6dad98f5-dbc9-4c4d-a9ba-e9da8dc6aa8e:10def1c1-7634-4312-b137-ee1d1feeea03Seth Guthrie<p>If you look closely at the section audit for a section from each case and compare them side by side it&#39;s a little easier to tell what is most significant. Below is a table of just a few lines from the audit to help you, ultimately is a small change in Vc that matters:</p> <table border="0"> <tbody> <tr> <td>5000 psi case</td> <td>4000 psi case</td> </tr> <tr> <td> <p>Calculating Vc using Equation 11-9:<br />(Vu dpt) / Mu = 0.9732 <br />Vc = 486.3 Kips (equation 11-9)<br />2 root-fc bw d = 95.04 Kips<br />5 root-fc bw d = 237.6 Kips<br />Using Vc = 237.6 Kips<br />Considering cross section as &quot;beam&quot; (not slab)<br />Maximum (shear) stirrup spacing = 24 inches (for shear, if stirrups required)<br />Vn Required = 161.1 Kips (from Vu/phi)<br />Vs Required = 0 Kips (from Vn - Vc)<br />Minimum stirrups (or more) required<br />Maximum phi Vn = 463.3 Kips (not considering torsion)<br />phi Vn taken by torsion = 0 Kips<br />Available Vn = 617.7 Kips (after torsion considered)<br />Designing for Vs = 0 Kips<br />Existing usable Vs = 0 Kips<br />Av/s = 0 inches (shear only)<br />Av/s = 0 inches (torsion only)<br />Av/s = 0 inches (shear + torsion)<br />Minimum additional design Av/s = 0.01768 inches<br />Final design Av/s = 0.01768 inches (shear, torsion &amp; minimum)<br />Total Av/s = 0.01768 inches (shear, torsion &amp; minimum)</p> </td> <td> <p>Calculating Vc using Equation 11-9:<br />(Vu dpt) / Mu = 0.9729 <br />Vc = 483.1 Kips (equation 11-9)<br />2 root-fc bw d = 85 Kips<br />5 root-fc bw d = 212.5 Kips<br />Using Vc = 212.5 Kips<br />Considering cross section as &quot;beam&quot; (not slab)<br />Maximum (shear) stirrup spacing = 24 inches (for shear, if stirrups required)<br />Vn Required = 161 Kips (from Vu/phi)<br />Vs Required = 0 Kips (from Vn - Vc)<br />Minimum stirrups (or more) required<br />Maximum phi Vn = 414.4 Kips (not considering torsion)<br />phi Vn taken by torsion = 0 Kips<br />Available Vn = 552.5 Kips (after torsion considered)<br />Designing for Vs = 0 Kips<br />Existing usable Vs = 0 Kips<br />Av/s = 0 inches (shear only)<br />Av/s = 0 inches (torsion only)<br />Av/s = 0 inches (shear + torsion)<br />Minimum additional design Av/s = 0.01667 inches<br />Final design Av/s = 0.01667 inches (shear, torsion &amp; minimum)<br />Total Av/s = 0.01667 inches (shear, torsion &amp; minimum)</p> </td> </tr> </tbody> </table><div style="clear:both;"></div>