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Alternative Support Systems for Cantilever - National Transportation ...

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Vbfp<br />

l e<br />

tfp<br />

bfp<br />

f’c<br />

ca1<br />

= the basic concrete breakout strength in shear of one side of a flexural plate in<br />

cracked concrete (lb.)<br />

= the equivalent bearing length of the annular plate, taken conservatively as 1/8 of<br />

the circumference of the centerline of the plate (in.)<br />

= the thickness of the annular plate (in.)<br />

= the bearing width of the annular plate (in.)<br />

= specified compressive strength of concrete (psi)<br />

= the edge distance, taken from the center of the width of the plate to the nearest<br />

concrete edge (in.)<br />

Figure 3-11. The tensile and compressive <strong>for</strong>ces seen as shears acting parallel to an edge<br />

Once the basic concrete breakout strength of one plate bearing area has been determined,<br />

then the total shear breakout capacity can be determined using Equation 3-12. Equations 3-11<br />

and 3-12 are used to determine the shear strength perpendicular to an edge. To determine the<br />

shear strength parallel to an edge, the perpendicular shear strengths obtained need to be doubled.<br />

See Figure 3-12 <strong>for</strong> a visual representation of the values in Equations 3-11 and 3-12.<br />

A<br />

= V<br />

(3-12)<br />

Vcp<br />

V c ψ<br />

bfp<br />

ed,<br />

Vψ<br />

c,<br />

Vψ<br />

h,<br />

V<br />

AVco<br />

Where<br />

Vcbfp<br />

AVcfp<br />

Vt<br />

bfp<br />

= The nominal concrete breakout strength in shear of plate configuration<br />

where the plate bearing areas are not acting as a group (lb.)<br />

= The projected area of the failure surface <strong>for</strong> a single bearing location on the<br />

plate, used to determine the shear strength (in 2 )<br />

=(3.0ca1+ le)*(3.0ca1 + tfp)<br />

AVco = projected concrete failure area of a single anchor, <strong>for</strong> calculation of<br />

strength in shear, if not limited by corner influences, spacing, or member<br />

thickness (in. 2 )<br />

= 4.5(ca1) 2 , based on an ≈35° failure cone (Figure 3-4)<br />

Vc<br />

35

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