Edge shear at front face of footing for load

[rsbmusicguy]

Hi! I was recently assigned to temporarily jack a bridge girder from the toe of footing of an abutment due to constraints on other jacking alternatives.

Note the load (V) is 170 kips and the footing is unreinforced with f'c = 3000 psi. I am using ASD from the AASHTO Standard Specification 2002.

I assumed a 45 degree shear plane in order to distribute the force through the concrete using a distribution beam. There is soil that bears under the footing toe which I did not consider in my calculation to be conservative.

Per AASHTO Eq. 8-3, the design shear stress is equal to v = V / (bw*d) with a minimum vc = 0.9*(3000)^0.5 = 49 psi per AASHTO 8.15.5.2.1.

I have disagreement among my colleagues on the 'd' value. We agree that the bw = 116" considering my assumptions but disagree upon the 'd' value.

We can not agree on whether to take the d = 11", 24", or 30".

In reality I feel the toe will slip from the back and therefore I can take the full 30" as my depth (v = 170,000 / (116"*30") = 48.8 psi < 49 psi). Alternatively, when I do footing design the plane is taken from a distance 'd' from my support and therefore it will be outside of my footing limit and shear capacity will be zero which I find to be unrealistic.

Any opinions on the 'd' value considering the attached?

Thanks!

RSB

 

[BridgeSmith]

None of the above. You have to calculate the area of the failure plane from the perimeter of the distribution beam, projecting at the 45 degrees, to where it intersects either the edge or bottom of the footing. The front and back are full trapezoids (one 30"/COS 45 degrees tall and the other 11"/COS 45 degrees tall). The ends are each a trapezoid with a corner cut off. The easiest way that I can think of to do the ends is to calculate the area of the projection shown in your end view sketch (using the line dimensioned as 11" at the front) and divide by COS 45 degrees to get the area of the plane.